Lecture 26. Transition State Theory
Penner, Reginald
 In Chemistry 131C, students will study how to calculate macroscopic chemical properties of systems. This course will build on the microscopic understanding (Chemical Physics) to reinforce and expand your understanding of the basic thermo-chemistry concepts from General Chemistry (Physical Chemistry.) We then go on to study how chemical reaction rates are measured and calculated from molecular properties. Topics covered include: Energy, entropy, and the thermodynamic potentials; Chemical equilibrium; and Chemical kinetics. 
 Okay, are you guys doing? Two more lectures. We're going to review the final for the final on Friday. So that's all we'll really be doing on Friday. I'll tell you in some detail what's going to be on it. Did I mention that the electronic course evaluations are available? I didn't look to see if any of you guys did these in the last two days. Monday we talked about enzyme kinetics, but we're not going to say anything more about that today, but as you'll see on Friday, that's definitely going to be on the final. So I jammed a lot of information into that lecture on Monday. All of enzyme kinetics, all of enzyme inhibition, it's all in there. All right, some of that stuff is not in your book, like the enzyme inhibition stuff. Okay, so you can either find another source to study that from or you can study it directly from the lecture. I think everything that I put in there is approximately correct. Now, today we're going to talk about transition state theory. At the beginning of the quarter, this is what we hope the course would look like. We had this beautiful blue rectangle of reaction dynamics here at the end of the course, and we were optimistic that we'd be able to talk a lot about this, maybe for more than one lecture. But what actually happened was it fell off the end of the course, and all we've got left is a single lecture that you're going to hear about today. That's all that's left of reaction dynamics. Could we have just left it out? We could have, but enough damage has been done to you guys. In 131A, all you heard about was quantum mechanics from Professor Renzepis. In 131B, all you heard about was spectroscopy from Professor Martin. And I know you know a lot about quantum mechanics and spectroscopy, and that's good. But everything else is important too, and so this subject is actually very, very important to us. And so I've tried to pick out the one thing that I cannot allow you to leave this class without knowing. I've tried to pull the one thing out of reaction dynamics that you've got to know about. This is the one thing, this lecture. I'm going to tell it to you. Basically what we want to understand is where does the Arrhenius equation come from when we measure the temperature dependence of reactions? This is the reaction rate. Most of the time we find out that it conforms to this equation. We make a plot of log of the reaction rate versus 1 over T. We get a straight line. We get the activation energy of the reaction from that straight line, and we report that, and we think about that. That guides a lot of what we do as physical chemists, but we have never explained where this equation comes from. In some ways this is the most important equation in chemical kinetics, but we haven't talked about what its origin is. Fundamentally, where does this equation come from? So we're going to do that today. I can do that in one lecture. It reminds me, of course this statement is grammatically incorrect. It's a dangling preposition. It reminds me of one of my favorite jokes. The Texas student goes to Harvard, and he asks a Harvard student, where are you from? The Harvard graduate says, well, I come from a place where we do not end our sentences of propositions. The Texas says, okay, where are you from? Other words are sometimes substitute. I would love to spend 10 minutes talking to you about these two guys. Henry Ahring worked at the University of Utah for most of his career, one of the great American physical chemists of the 20th century. Michael Polany, Nobel Prize winning physical chemist. Both of these guys are responsible for what we're going to be talking about today, transition state theory. They worked it out in the 1930s. So here's the short version of the history that matters here. Way back in 1916, G.N. Lewis, the first great American physical chemist, worked out Lewis-Dott structures. There was no quantum mechanics in 1916. It wasn't invented until 1924. He came up with a very compelling model for chemical bonding involving pairs of electrons that turned out to be correct. And much of what it predicted way before the fundamental underpinnings of quantum mechanics were described. He figured all that out. He was a genius. Heisenberg, Dirac, and Schrodinger discovered quantum mechanics in 1924. Much of Germans, Heitler and London, rather, took quantum mechanics and applied it to bonding for the first time. They did the first quantitative description of chemical bonding. Their names are usually lost. We hardly even mention it. Sometimes we talk about London's dispersion forces. You guys remember that? Vanderwall's forces? This is the guy. Fritz London. He's the guy who figured that out. 1929, five years after quantum mechanics, these guys developed a quantum mechanical way to think about reaction rates. These guys figured out a quantum mechanical way to think about bonding. That's pretty important. But Pellanian Erring figured out how to think about reaction rates from a quantum mechanical perspective. We're going to zoom through what they taught us today. Here's the basic idea. A plus B goes to products. We're just going to talk about this generic reaction today. We've stripped down this lecture so that we've taken everything else out. It should be three weeks' worth of lectures. We know the rate law is this. The rate constant times the concentration of A times the concentration of B. These can be pressures or concentrations. Transition state theory says that this reaction actually occurs through this mechanism. A plus B are in equilibrium with something called AB double dagger. That reacts in a unimolecular fashion to give us products. Transition state theory basically takes all of the reactants, puts them on one side, all of the products puts them on the other side, and in between it constructs something called an activated complex or a transition state. That's why it's called transition state theory. This thing here is the transition state. It's an entity that is intermediate between the reactants and the products. If bonds exist in the products that don't form in the reactants, they exist partially in a weak form in the activated complex or in the transition state. If bonds are broken here, they're partially broken in the transition state. The transition state has structural attributes, bond lengths, and so on that are intermediate between the reactants and the products. One of the key points about transition state theory is it postulates that an equilibrium exists between this transition state and these reactants. Here's the picture in terms of energy that you've seen a million times, but what we're presenting here now, the notation applies specifically to transition state theory. You've got reactants that's represented here by these two molecules. This looks like it could be an OH, that's CH3, that's CH3Br apparently, OH minus, CH3Br. That's the transition state and here are the products. Notice that in this reaction, a new bond is formed between this oxygen and this carbon and this bond here between this carbon and this bromine is broken. In the transition state, notice that the transition state, what we're depicting here is this bond is partially formed. You see how long it is? It's a long, weak bond and that bond is partially broken. See how long it is? It's a long, weak bond also. That bond is going to be broken in the product state. That bond doesn't even exist in the reactant state. The transition state contains all of the bonds that are present in the reactants and in the products, but the bonds that are involved in the formation of the products from the reactants are weakened. See how long that is? See how long that bond is? These are two super weak bonds. We can construct a transition state by thinking about what the products look like, what the reactants look like, products, reactants and then thinking about how the products are formed from the reactants geometrically. How does that happen? This shows an attack in a particular geometric orientation of the OH minus to the CH3Br. Thinking about this and thinking about this, we can construct what this transition state should look like using a few simple rules. It turns out that transition state and activated complex should not technically be used interchangeably. It's a nuance. The activated complex actually exists over a range of this reaction coordinate, whereas the transition state in principle exists only at a particular point in time. We're not going to bother ourselves with that distinction today. There's a fine point there that you should know. These are not identically or interchangeable to say activated complex and transition state, but today we're just going to use those two terms interchangeably. Now, with that as a premise, let's see if we can work out what the reaction rate is. Here's our mechanism, our transition state theory mechanism for this reaction. Yes, so we can write an expression for this equilibrium constant. What is it? Product over reactants. Notice that I'm normalizing and being very careful here to write activities, the activity of the transition state, the activity of A, the activity of B, because I'm dividing by the standard concentration here. When we're done canceling these C0s, I end up with an extra factor of C0 on the numerator that ensures that the equilibrium constant is dimensionless, doesn't it? Now, unfortunately, I used the Microsoft word equation editor to write a lot of these equations. It does not allow me to write a zero with a line through it. That is the symbol for the standard concentration, one molar, for example, but I'm just going to call it C0. There's another issue. See this double X here, this double plus? Microsoft equation editor doesn't contain a diocese. What's a diocese? It's that thing, the double dagger. The double dagger is the same thing as the diocese. When I write two pluses, I'm just indicating the transition state. That refers to the transition state. That's the equilibrium constant that involves the transition state. That's the unimolecular reaction rate constant that involves the transition state. You see how? I'm going to write that as that. If you see the double plus, it's just the diocese. If anybody knows, I'll work around for that. That would help me out a lot. How do I write a diocese in Microsoft equation editor? Here I cheated. I put a white square here and I pasted this on top. Then I thought, I can't do that. There's like 106 slides in this presentation. Now I said yes, yes. We could write this in terms of pressures if we want to. Here's the concentration. Pressures, no difference. Now, if we flash back, we're talking right now about chapter 20. If we flash back to chapter 17, it turns out we can calculate equilibrium constants from partition functions. Recall, partition functions are very important to us because they allow us to make a connection between statistical mechanics and thermodynamics. Partition functions contain information about the actual molecule. We can look at a molecule. If we know something about its state distribution, we can write its partition function. What we learned in chapter 17, what we didn't have time to talk about in the class this quarter is the fact that you can also use these partition functions to calculate equilibrium constants. If you want to read more about that, it's on page 670. Turns out to be an important thing that we left out. Here's what's there. Here's some generic reaction. Amoles of A, binomoles of B, c-moles of C, d-moles of D. Here's what the equilibrium constant expression looks like written out in terms of the standard molar partition functions for A, B, C and D. If I can calculate these partition functions for all of these guys, and I know that, so what is this? This is the standard molar partition function. M means molar, A is species A. What's that? That's Avogadro's number. That's not a misprint. Avogadro's number in every single case. That is the difference in the zero point energies between reactants and products. If this is your generic reactant here, here are vibrational energy levels. Here are vibrational energy levels of the product. That's the ground vibrational energy level. That's the ground vibrational energy level. That's delta Re0. The difference between the ground and the zero point energies of the reactants and the products. Here's our Gibbs free energy as a function of reaction coordinate. The delta Re0 is closely related to this green quantity that I'm indicating here. If this thing was in its ground vibrational energy level, and this thing was in its ground vibrational energy level, both of these guys, and both of these guys, then this would be delta Re0. We're always talking about the zero point energies. We could say something about this equilibrium constant. We could calculate it using this equation right here. In transition state theory, that's not the equilibrium constant we care about. What we say in transition state theory is this is an equilibrium with this. I don't care about the equilibrium of this with this. That's going to give me the normal equilibrium constant that I can learn about in chapter 17. What transition state theory says is these two guys are in equilibrium with one another. What matters is this delta Re0 here. This thing that we normally call the activation energy. It's the activation energy from the Erranius equation. We want to calculate that imaginary equilibrium constant. These things may actually be in equilibrium, but this thing is not really observable except using some exotic spectroscopic nanosecond picosecond spectroscopy. A normal equilibrium applied to this generic reaction right here, I could calculate the equilibrium constant using this equation. Now I'm going to apply that same thinking to the transition state theory. Here's the transition state equilibrium we care about. A reacts with B to give this transition state. Now what do I want? I want to put the partition function for that guy in the numerator for these guys in the denominator and then that's Avogadro's number. That's just left over from, that's an extra factor of Avogadro's number because I've got two reactants in one product. I can calculate this equilibrium constant that applies to the formation of the transition state. Keep in mind, everything we're talking about here is kinetics. We're sort of mixing thermodynamics with kinetics. They took a thermodynamic concept, equilibrium constant, and applied that to transition state theory. So you're with me so far? We've got an equilibrium constant here. We can calculate it using statistical mechanics. Two-step mechanism. That's the first thing to understand. The second thing to understand is that this rate right here showing in gray, that's going to be approximately equal to the frequency with which the transition state crosses over the top of the barrier. We've got a barrier here. We've got a transition state. We are moving along this reaction coordinate from reactants to products. The frequency with which this thing moves across the top of this barrier, that's going to closely approximate this rate. The rate at which products are formed, that seems to be a statement of the obvious. Obviously as you cross over this barrier from reactants to products, the frequency that that happens, that's the rate of this reaction. It's totally obvious to say that. The reaction rate I can write as this frequency times whatever the concentration of this transition state is. We're going to talk about this frequency. It's the frequency with which if you think about this as being a molecule, there's a vibration that has to happen here. This guy moves back and forth between these two guys. We get a symmetric vibration of this transition state. The frequency that characterizes that mode is the frequency that we care about. Your book includes something called the transmission coefficient. We're just going to assume that's one. See that cap right there? Just forget about it, it's one. The rate of the reaction is given by this special frequency that applies to the reaction coordinate times the concentration of the transition state, but we know we can also write the rate in the normal way with a rate constant times A times B. That's what we've been saying all along. Here was our equilibrium constant expression for the transition state. If I just solve for AB in this expression right here, I get this and I can plug that in to AB there. I've just solved for the concentration of the transition state from here. Now I'm going to plug that in to this equation right here. I'm going to put all of this into here and so there it is. There's the frequency, the special frequency, here's all the rest of that. That's our reaction rate. In essence, this is the phenomenological rate expression that we would normally write for this reaction. A plus B goes to products. We know that the reaction rate is K times A times B as long as that's an elementary reaction. What we've said is look, that rate constant is given by this expression right here. That frequency times that equilibrium constant divided by this concentration term just to keep the units right. Transition state theory has already, and the important thing is these two parameters here relate directly to physical parameters of the transition state that we can think about calculating. In other words, we can calculate this rate constant from fundamental properties of the transition state because we know enough physical chemistry to do that already. We have the key point is we have expressions for this equilibrium constant and for this rate constant right here. We said that's just equal to V, this guy's equal to C0 times K double dagger divided by V. Let's say that we actually do want to calculate now what the rate constant is. Let's say that we actually want to calculate that K right there. We have to be able to calculate big K double dagger and we've got to be able to calculate little K double dagger which is just equal to V. How are we going to do that? Well, here's the expression for big K double dagger. We know what the partition function of A and B are. We know how to calculate that already. We've done that. How do we calculate the partition function of the transition state? How do we calculate that partition function? I'm just confused on your K double dagger expression. That's the same as that. That's the same as that. Well that's a little confusing, isn't it? What the heck did I? That's supposed to be K. Sorry, that shouldn't be K double dagger. That should just be K. Sorry, you're right. Those double daggers shouldn't be there. Thank you. How do we calculate that transition state? How do we calculate that partition function? It's a transition state for goodness sakes. What is it? This is the question that these guys wrestled with when they worked out transition state. If we think about the vibrational partition function first, it's the vibrational partition function that we really care about here. Essentially the transition state is undergoing a vibration. How do I think about that? This bond is getting longer. This bond is getting shorter. Just like an asymmetric stretch of the transition state, that's the mode that we care about. If we could calculate the partition function for that vibrational mode, that's critical to understanding what the reaction rate is going to be. Here's the generic expression for the vibrational partition function. For some mode that has a natural frequency or natural energy h nu and a natural frequency nu. What can we say about this magic mode that we care about, this asymmetric vibration along this axis? Is that going to be a high frequency mode, a low frequency mode? What do you guys think? High frequency? Hello? So tell me, these are weak bonds here, right? Do weak bonds have high frequencies or low frequencies? Low, thank you. We're going to expect this to be a weak, we're going to expect this to be a very low frequency mode. We call these soft modes. Here's a picture from your chapter that I think is somewhat non-intuitive. Here's the transition state up here. What this picture is trying to convey is that the transition state has a very shallow vibrational energy well with vibrational energy levels that are very, very close together in energy. In other words, the frequency to go from here to here, that h nu is tiny. Yes, but the key point is that the frequency that characterizes these transitions in the transition state along this direction right here is very small. What that allows us to do is simplify this equation. We can write this exponential as a series and we can truncate it at the first term. When we do that, we just get kT over h nu. We take this normal expression for the partition function, we truncate it, we write it as a series expansion, the exponential and we truncate it at the first term and we just get kT over h nu. This is a special nu. This is the frequency that describes motion over the barrier. Whatever that is for, whatever the transition state is, whatever mode the product bonds are getting formed, reactant bonds are getting consumed, we can think about that process as involving a single vibration that has a very low frequency in general. That's what they realize. Conceptually this is not obvious. I think everybody in this room would agree. None of this stuff that I'm talking about in the last five minutes is obvious. This was the conceptual leap that Polani made. I can write the whole partition function for the transition state as the partition function for this low frequency mode and then the whole rest of the partition function. Notice that there's going to be a rotational, an electronic, a translational. The partition function contains many other manifolds and other vibrational modes as well for the molecule that are orthogonal to this special mode. We're going to roll that all into this guy over here. This is just the partition function for the magic mode that corresponds to the reaction. This is the rest of Q, all the rest of the partition function for all the other modes, manifolds, and so on. So that's the partition function. I can just plug that in then. Here's the partition function that we were wondering about. Here's the expression we now have for it. Notice that the rest of this partition function involves things that we can already calculate because they're things that are not perturbed in the molecule. What's perturbed in the transition state is product bonds are getting formed, reactant bonds are getting broken. We can describe that. We can roll that process all into one mode that has this characteristic frequency here. We're going to have to figure out what that frequency is. But all the rest of this stuff, we can just roll in. Bonds that are orthogonal to the reaction rate, we can just calculate their vibrational partition function, their rotational partition function, all of that stuff using the conventional methods that we already know about. So now I rewrite this in terms of this guy right here. Look at this KT over H nu. Now that's right there. Here's the rest of the partition function, this Q with a line over it here. And I'm getting close to being able to calculate this guy. This is the contribution along the reaction coordinate only. Yes, yes, this is all the rest of it. Yes. Okay, so our expression for the rate constant becomes this and we're just plugging now this expression in for this equilibrium constant with a double dagger on it. So we can actually, when we do that, we can cancel this frequency. Hey, turns out we don't need to know what it is. It cancels for gosh sakes. We don't have to measure it. We get this expression right here and this is the famous Erring equation. We derived it going way too fast in about 20 minutes. This is a very important equation. Now why is it so important? Yeah? How do we find K using K? How do we find K using K? How do we find that K using that K? No, the little K within the equation. Oh, that K. This could be confusing. That's Boltzmann's constant. Sorry. This is the rate constant, phenomenological rate constant for the reaction. That is the equilibrium constant that we calculate using this expression right here. Right there, that whole thing. Now you might ask, are you going to be called upon on the final exam to calculate all of this stuff? No. But I have to be able to sleep at night and so I am going to disclose all of this information to you even though I don't think it would be fair of me to write a question where you have to calculate all of it. If you look at the end of this lecture, when I post this lecture after class, it does in fact have like 120 slides and the last 20 slides are a calculation. It allows you to calculate the rate constant for H plus H2 goes to H2 plus H. The very first reaction that was studied using this equation, we can work out exactly what the rate constant is and if we had enough time we would do it but we don't. So if you're interested in this rate theory stuff, this transition state theory stuff, you might want to just look down at those slides because we're never going to get to them today. It doesn't matter. It does matter but. So here's the Erring equation. I'm just substituting now for K double dagger here. Here's the Erring equation. Do you see the parallels? The activation energy is this delta E0 that we were talking about. The difference in the zero point energies for reactants and products. This pre-exponential factor, A that we've never said anything about, that's given by this collection of variables right here. RT over H times this guy, what's left over from the partition function, we strip out from the normal partition function for the transition state, we strip out the part of it that pertains to the vibration along the reaction coordinate and that's what's left over because remember that frequency just canceled for us. We don't need to know it. That's a good thing because who knows what it is. How would you measure it? You'd have to have some exotic spectroscopic method to do that. Well the Erranius equation is a special case of the Erring equation. We can calculate these things and if you don't believe me go to slide 101 and we go through and we do it laboriously for a particular reaction. We're not going to say more about it sadly. Here are pre-exponential factors. You can calculate this one, actually it is calculated later in this lecture. We can calculate these pre-exponential factors for the Erranius equation knowing transition state theory so it's very powerful. What can we do in the last 15 minutes? Something very important. We want to apply the same thinking to a reaction that occurs in water. The key point is that we want to think about what the influence of charge is on the reaction rate. It's not inconceivable that there could be a question like this on the final. For example two chlorides react with lead to give lead chloride. All of these things can be in solution. Lead chloride has some solubility in water, very low. What rate would this reaction take? What is the influence of the charge on the reaction rate? You might think naively if the reactants and products are oppositely charged they're going to be clumpically attracted to one another and the reaction rate is going to be accelerated. If you don't think about this too hard, even if you think about it hard you might come to that conclusion. Negatively charged reactants and positively charged reactants are going to be clumpically attracted and boom they're going to react really fast. That's going to accelerate the reaction rate. That turns out to be exactly wrong and it's important to understand why. It's great once in a while in chemistry when you encounter a concept that is completely counterintuitive at first and later on hopefully intuitive. I've got to make this happen in like 12 minutes. Here's the idea. A reacts with B to give products. That's the charge on A and B, Z, A and ZB. That's the charge. The key point here is reactants are charged. There's a version of transition state theory but it's not strictly speaking transition state theory. It's a thermodynamic version of it. You can't use transition state theory in its normal form to describe reactions in solution because it doesn't account for reactions with solvent. It doesn't transition state theory in its normal form does not account for the complexities imparted by having the solvent present in the reaction. This looks like transition state theory but strictly speaking it's something that's related to it, not exactly the same thing. We're going to simplify this notation. We're going to call this mess. Notice something. If we think about this in terms of transition state theory, when A reacts with B, if we form a transition state, the transition state will have a total charge equal to A plus B. The total charge of the transition state will be the sum of the charges on A and B because charge has to be conserved here. That's a key point. So exactly, this is the reaction rate. Now in terms of, this is just that rate constant right there. This is the normal phenomenological rate that we would write for this reaction. Then we have to think back and remember something about activities. Activities are going to be the key, the understanding how this works. The activity of some I in A is equal to its concentration times some activity coefficient, gamma sub A. That's the activity of A. That's the activity coefficient. The activity coefficient for a neutral is one. The activity coefficient only deviates from one as a consequence of Coulombic interactions with the solution, other ions in the solution. If other ions are present in the solution, the activity coefficient will be less than one. The more ions, the lower the activity coefficient. And there's something called the Debye-Hockel limiting lot. That's the activity coefficient. That's the ionic strength and the ionic strength is just given by this expression where this is the concentration of the ion, sorry, this is the concentration of the ion and this is the charge on the ion for every ion in the solution. I add up all the ions in the solution, multiply by the square of their charge, take that times one half and I've got the ionic strength. The higher the ionic strength, it's never obvious to me when the equation has a log in it like this. The higher the ionic strength, the lower gamma becomes. If I is zero, gamma is one. This is the ionic strength on this axis. This is gamma on this axis. If ionic strength is zero, gamma is one and it deviates from one as the ionic strength goes up. And that size of this deviation here depends on what the charge is on the ion of interest, ion A. Because here's its charge. So if it has a charge of one, there's a small deviation, charge of, that's the size. If it's small, you get a little bigger deviation. If the charge goes up to three, you get a huge deviation. So activity effects have everything to do with charge. How many of you had 151? From me. So this is review for all you guys. Good. So when we write an equal, here's an equilibrium constant. We write in terms of activities. Every activity is an activity coefficient times a concentration. You guys all know that. And what's more important is you have intuition about what the influence of ions are on equilibria. Here's an important piece of intuition that everybody should have in the room, especially those of you who took my class. If you look at some equilibrium like this, all right? Sodium chloride is an acid. Acetic acid dissociates to give hydronium ion and acetate. If I dump sodium chloride into the solution, what's going to happen to the pH? Sodium chloride is an inert salt. Sodium chloride has no acidity or basicity of its own. And yet the pH will change in a predictable way. What will happen? How many people think the pH is going to go down? It's going to get more acidic. How many people think the pH is going to go up? It's going to get more basic when I add salt. Get those hands way up there because I want to see. You guys never had 151 from me. Now if you add salt to this equilibrium, it will always shift to the right. It will always shift to the right. The addition of inert salt to any equilibrium always shifts it in the direction of the most ionic state of the equilibrium. See how there's ions here and there are no ions here? That's the most ionic state. If I add salt, the reaction will shift to the right. It's called salting in. Now I can prove that. Here's the equilibrium constant for this reaction. Here are the activity coefficients. The activity coefficients for the neutrals are zero or one rather. The activity coefficient for this guy's one. These activity coefficients for the charged species are less than one. They will become lower as I increase the ionic strength of the solution. Those guys will get smaller. What that means is that the equilibrium constant that applies if we think about these activity effects is going to get bigger. If these guys get smaller, that gets bigger. That's the activity coefficient that applies when we add sodium chloride to the solution or any other inert salt. It's going to get more acidic. What about this guy? What happens to the solubility of lead sulfide if I add sodium chloride to the solution? Sodium chloride doesn't appear in this equilibrium. Why would it affect this equilibrium? And yet it does in a predictable way. What's going to happen? Is the lead sulfide more soluble or less soluble when I dump sodium chloride in? More. Why? Because, look, this side of the equilibrium has got a lot of ions. This side's got no ions. If I add more ions to the solution, equilibrium is favored by the addition of more ions. I can prove that's true by working out what the new equilibrium constant is. These are the two activity coefficients. They both get smaller when I add sodium chloride. So the equilibrium constant shifts to the right. Every time you see an equilibrium, you can affect its position by adding an inert salt or removing an inert salt from the solution. That'll also alter the equilibrium in a predictable way. What does all this have to do with transition state theory? Very simple to make a long story short. We can work out the math, but we're just going to skip over it here because we're almost out of time. Here's the bottom line. The transition state has the total charge of the reactants in it. If the transition state has a high charge, then the rate of the reaction is going to be influenced by the presence of other ions in the solution. At the end of the day, when we're done doing the derivation, we get this equation right here, which is the equation for the kinetic salt effect. What is this big K right here? It's just the collection of the activity coefficients for the transition state and the reactants. That is the rate constant that applies for one molar, everything. That's why it's got a zero on it. That's the actual rate constant of the reaction. This is the most important slide that has to do with this second concept here. What am I plotting here? This is the reaction rate for reaction that involves A with some charge plus B with some charge going on to products. This is the ionic strength. The key point here is we want to understand how salt affects the reaction rate. It's easy to understand how salt affects the equilibrium constant. How does it affect the reaction rate? The way to think about that is if A reacts with B and they both have a plus 2 charge, they're both positively charged, they both got a plus 2 charge, the transition state has a charge of plus 4. If there's an equilibrium between the transition state and the reactants, it's going to be favored by the addition of salt. The reaction is going to get accelerated. Isn't that counterintuitive? The reaction rate goes up as I add salt to the solution. Even though the reactants and products have the same charge, they have to overcome a coulombic barrier to react because they're both positively charged, the reaction rate goes up, not down. Check this out. If a 2 plus ion reacts with a 2 minus ion, you'd expect there to be a big coulombic attraction, right? Reactions should go faster. It goes slower. Does this have to do with the orthogonal components of the transition state's partition function? Absolutely not. No? No. Nothing at all? No. Nothing to do with the ionic development? Zero. Okay. It's not a bad question, but the answer's no. Does everybody see why this happens? There's an equilibrium. I'll say this one last time because I don't have any more time. There's an equilibrium between A and B, the reactants, and the transition state. What we just agreed on is that we can decide which side of the equilibrium will be favored when we add salt to an equilibrium. The most ionic state of the system is favored. If the charge on the transition state is higher than the charge on either one of the ions because the only way that can be true is if the ions have the charge of the same sign like plus 1 and plus 2, plus 2 and plus 2, plus 1 and plus 1, then the charge on the transition state will be like 3, 4, and so on. The transition state gets favored by the addition of salt and the reaction rate goes up. Look, it goes up here too. Look, it goes up here. So if there's no charge, nothing happens. And if the charges on the reactants are opposite to one another, then the charge on the transition state is lower than the charge on the ions. The transition state has a lower ionic, it's the least ionic state of the system. The reactants are more ionic than the transition state and the reaction is slowed down by the addition of salt. That's totally counterintuitive. If you understand that, you understand something that most chemists even are going to get wrong. You're going to get it right. You say those charges, that reaction is going to go slower. You can think about, okay. So on Friday, yeah, there's more here. Yes, oh my goodness. There's like 20 more slides that work through the equilibrium, we're not going to give to those. So on Friday, we're going to work on the final. Please take the course evaluations. All right. Thank you.
 Okay, are you guys doing?  Two more lectures.  We're going to review the final for the final on Friday.  So that's all we'll really be doing on Friday.  I'll tell you in some detail what's going to be on it.  Did I mention that the electronic course evaluations are available?  I didn't look to see if any of you guys did these in the last two days.  Monday we talked about enzyme kinetics, but we're not going to say anything more  about that today, but as you'll see on Friday, that's definitely going to be on the final.  So I jammed a lot of information into that lecture on Monday.  All of enzyme kinetics, all of enzyme inhibition, it's all in there.  All right, some of that stuff is not in your book,  like the enzyme inhibition stuff.  Okay, so you can either find another source to study that from  or you can study it directly from the lecture.  I think everything that I put in there is approximately correct.  Now, today we're going to talk about transition state theory.  At the beginning of the quarter, this is what we hope the course would look like.  We had this beautiful blue rectangle of reaction dynamics here at the end of the course,  and we were optimistic that we'd be able to talk a lot about this,  maybe for more than one lecture.  But what actually happened was it fell off the end of the course,  and all we've got left is a single lecture that you're going to hear about today.  That's all that's left of reaction dynamics.  Could we have just left it out?  We could have, but enough damage has been done to you guys.  In 131A, all you heard about was quantum mechanics from Professor Renzepis.  In 131B, all you heard about was spectroscopy from Professor Martin.  And I know you know a lot about quantum mechanics and spectroscopy, and that's good.  But everything else is important too, and so this subject is actually very,  very important to us.  And so I've tried to pick out the one thing that I cannot allow you to leave this class  without knowing.  I've tried to pull the one thing out of reaction dynamics that you've got to know about.  This is the one thing, this lecture.  I'm going to tell it to you.  Basically what we want to understand is where does the Arrhenius equation come from  when we measure the temperature dependence of reactions?  This is the reaction rate.  Most of the time we find out that it conforms to this equation.  We make a plot of log of the reaction rate versus 1 over T. We get a straight line.  We get the activation energy of the reaction from that straight line, and we report that,  and we think about that.  That guides a lot of what we do as physical chemists, but we have never explained where  this equation comes from.  In some ways this is the most important equation in chemical kinetics, but we haven't talked  about what its origin is.  Fundamentally, where does this equation come from?  So we're going to do that today.  I can do that in one lecture.  It reminds me, of course this statement is grammatically incorrect.  It's a dangling preposition.  It reminds me of one of my favorite jokes.  The Texas student goes to Harvard, and he asks a Harvard student, where are you from?  The Harvard graduate says, well, I come from a place where we do not end our sentences  of propositions.  The Texas says, okay, where are you from?  Other words are sometimes substitute.  I would love to spend 10 minutes talking to you about these two guys.  Henry Ahring worked at the University of Utah for most of his career, one of the great American  physical chemists of the 20th century.  Michael Polany, Nobel Prize winning physical chemist.  Both of these guys are responsible for what we're going to be talking about today, transition  state theory.  They worked it out in the 1930s.  So here's the short version of the history that matters here.  Way back in 1916, G.N. Lewis, the first great American physical chemist, worked out Lewis-Dott  structures.  There was no quantum mechanics in 1916.  It wasn't invented until 1924.  He came up with a very compelling model for chemical bonding involving pairs of electrons  that turned out to be correct.  And much of what it predicted way before the fundamental underpinnings of quantum mechanics  were described.  He figured all that out.  He was a genius.  Heisenberg, Dirac, and Schrodinger discovered quantum mechanics in 1924.  Much of Germans, Heitler and London, rather, took quantum mechanics and applied it to bonding  for the first time.  They did the first quantitative description of chemical bonding.  Their names are usually lost.  We hardly even mention it.  Sometimes we talk about London's dispersion forces.  You guys remember that?  Vanderwall's forces?  This is the guy.  Fritz London.  He's the guy who figured that out.  1929, five years after quantum mechanics, these guys developed a quantum mechanical way to  think about reaction rates.  These guys figured out a quantum mechanical way to think about bonding.  That's pretty important.  But Pellanian Erring figured out how to think about reaction rates from a quantum mechanical  perspective.  We're going to zoom through what they taught us today.  Here's the basic idea.  A plus B goes to products.  We're just going to talk about this generic reaction today.  We've stripped down this lecture so that we've taken everything else out.  It should be three weeks' worth of lectures.  We know the rate law is this.  The rate constant times the concentration of A times the concentration of B. These can  be pressures or concentrations.  Transition state theory says that this reaction actually occurs through this mechanism.  A plus B are in equilibrium with something called AB double dagger.  That reacts in a unimolecular fashion to give us products.  Transition state theory basically takes all of the reactants, puts them on one side, all  of the products puts them on the other side, and in between it constructs something called  an activated complex or a transition state.  That's why it's called transition state theory.  This thing here is the transition state.  It's an entity that is intermediate between the reactants and the products.  If bonds exist in the products that don't form in the reactants, they exist partially  in a weak form in the activated complex or in the transition state.  If bonds are broken here, they're partially broken in the transition state.  The transition state has structural attributes, bond lengths, and so on that are intermediate  between the reactants and the products.  One of the key points about transition state theory is it postulates that an equilibrium  exists between this transition state and these reactants.  Here's the picture in terms of energy that you've seen a million times, but what we're  presenting here now, the notation applies specifically to transition state theory.  You've got reactants that's represented here by these two molecules.  This looks like it could be an OH, that's CH3, that's CH3Br apparently, OH minus, CH3Br.  That's the transition state and here are the products.  Notice that in this reaction, a new bond is formed between this oxygen and this carbon  and this bond here between this carbon and this bromine is broken.  In the transition state, notice that the transition state, what we're depicting here is this  bond is partially formed.  You see how long it is?  It's a long, weak bond and that bond is partially broken.  See how long it is?  It's a long, weak bond also.  That bond is going to be broken in the product state.  That bond doesn't even exist in the reactant state.  The transition state contains all of the bonds that are present in the reactants and in the  products, but the bonds that are involved in the formation of the products from the  reactants are weakened.  See how long that is?  See how long that bond is?  These are two super weak bonds.  We can construct a transition state by thinking about what the products look like, what the  reactants look like, products, reactants and then thinking about how the products are  formed from the reactants geometrically.  How does that happen?  This shows an attack in a particular geometric orientation of the OH minus to the CH3Br.  Thinking about this and thinking about this, we can construct what this transition state  should look like using a few simple rules.  It turns out that transition state and activated complex should not technically be used interchangeably.  It's a nuance.  The activated complex actually exists over a range of this reaction coordinate, whereas  the transition state in principle exists only at a particular point in time.  We're not going to bother ourselves with that distinction today.  There's a fine point there that you should know.  These are not identically or interchangeable to say activated complex and transition state,  but today we're just going to use those two terms interchangeably.  Now, with that as a premise, let's see if we can work out what the reaction rate is.  Here's our mechanism, our transition state theory mechanism for this reaction.  Yes, so we can write an expression for this equilibrium constant.  What is it?  Product over reactants.  Notice that I'm normalizing and being very careful here to write activities, the activity  of the transition state, the activity of A, the activity of B, because I'm dividing by  the standard concentration here.  When we're done canceling these C0s, I end up with an extra factor of C0 on the numerator  that ensures that the equilibrium constant is dimensionless, doesn't it?  Now, unfortunately, I used the Microsoft word equation editor to write a lot of these equations.  It does not allow me to write a zero with a line through it.  That is the symbol for the standard concentration, one molar, for example, but I'm just going  to call it C0.  There's another issue.  See this double X here, this double plus?  Microsoft equation editor doesn't contain a diocese.  What's a diocese?  It's that thing, the double dagger.  The double dagger is the same thing as the diocese.  When I write two pluses, I'm just indicating the transition state.  That refers to the transition state.  That's the equilibrium constant that involves the transition state.  That's the unimolecular reaction rate constant that involves the transition state.  You see how?  I'm going to write that as that.  If you see the double plus, it's just the diocese.  If anybody knows, I'll work around for that.  That would help me out a lot.  How do I write a diocese in Microsoft equation editor?  Here I cheated.  I put a white square here and I pasted this on top.  Then I thought, I can't do that.  There's like 106 slides in this presentation.  Now I said yes, yes.  We could write this in terms of pressures if we want to.  Here's the concentration.  Pressures, no difference.  Now, if we flash back, we're talking right now about chapter 20.  If we flash back to chapter 17, it turns out we can calculate equilibrium constants from  partition functions.  Recall, partition functions are very important to us because they allow us to make a connection  between statistical mechanics and thermodynamics.  Partition functions contain information about the actual molecule.  We can look at a molecule.  If we know something about its state distribution, we can write its partition function.  What we learned in chapter 17, what we didn't have time to talk about in the class this  quarter is the fact that you can also use these partition functions to calculate equilibrium  constants.  If you want to read more about that, it's on page 670.  Turns out to be an important thing that we left out.  Here's what's there.  Here's some generic reaction.  Amoles of A, binomoles of B, c-moles of C, d-moles of D. Here's what the equilibrium constant  expression looks like written out in terms of the standard molar partition functions  for A, B, C and D. If I can calculate these partition functions for all of these guys,  and I know that, so what is this?  This is the standard molar partition function.  M means molar, A is species A. What's that?  That's Avogadro's number.  That's not a misprint.  Avogadro's number in every single case.  That is the difference in the zero point energies between reactants and products.  If this is your generic reactant here, here are vibrational energy levels.  Here are vibrational energy levels of the product.  That's the ground vibrational energy level.  That's the ground vibrational energy level.  That's delta Re0.  The difference between the ground and the zero point energies of the reactants and the products.  Here's our Gibbs free energy as a function of reaction coordinate.  The delta Re0 is closely related to this green quantity that I'm indicating here.  If this thing was in its ground vibrational energy level, and this thing was in its ground  vibrational energy level, both of these guys, and both of these guys, then this would be  delta Re0.  We're always talking about the zero point energies.  We could say something about this equilibrium constant.  We could calculate it using this equation right here.  In transition state theory, that's not the equilibrium constant we care about.  What we say in transition state theory is this is an equilibrium with this.  I don't care about the equilibrium of this with this.  That's going to give me the normal equilibrium constant that I can learn about in chapter  17.  What transition state theory says is these two guys are in equilibrium with one another.  What matters is this delta Re0 here.  This thing that we normally call the activation energy.  It's the activation energy from the Erranius equation.  We want to calculate that imaginary equilibrium constant.  These things may actually be in equilibrium, but this thing is not really observable except  using some exotic spectroscopic nanosecond picosecond spectroscopy.  A normal equilibrium applied to this generic reaction right here, I could calculate the  equilibrium constant using this equation.  Now I'm going to apply that same thinking to the transition state theory.  Here's the transition state equilibrium we care about.  A reacts with B to give this transition state.  Now what do I want?  I want to put the partition function for that guy in the numerator for these guys in the  denominator and then that's Avogadro's number.  That's just left over from, that's an extra factor of Avogadro's number because I've  got two reactants in one product.  I can calculate this equilibrium constant that applies to the formation of the transition  state.  Keep in mind, everything we're talking about here is kinetics.  We're sort of mixing thermodynamics with kinetics.  They took a thermodynamic concept, equilibrium constant, and applied that to transition state  theory.  So you're with me so far?  We've got an equilibrium constant here.  We can calculate it using statistical mechanics.  Two-step mechanism.  That's the first thing to understand.  The second thing to understand is that this rate right here showing in gray, that's going  to be approximately equal to the frequency with which the transition state crosses over  the top of the barrier.  We've got a barrier here.  We've got a transition state.  We are moving along this reaction coordinate from reactants to products.  The frequency with which this thing moves across the top of this barrier, that's going  to closely approximate this rate.  The rate at which products are formed, that seems to be a statement of the obvious.  Obviously as you cross over this barrier from reactants to products, the frequency that  that happens, that's the rate of this reaction.  It's totally obvious to say that.  The reaction rate I can write as this frequency times whatever the concentration of this transition  state is.  We're going to talk about this frequency.  It's the frequency with which if you think about this as being a molecule, there's a  vibration that has to happen here.  This guy moves back and forth between these two guys.  We get a symmetric vibration of this transition state.  The frequency that characterizes that mode is the frequency that we care about.  Your book includes something called the transmission coefficient.  We're just going to assume that's one.  See that cap right there?  Just forget about it, it's one.  The rate of the reaction is given by this special frequency that applies to the reaction  coordinate times the concentration of the transition state, but we know we can also  write the rate in the normal way with a rate constant times A times B. That's what we've  been saying all along.  Here was our equilibrium constant expression for the transition state.  If I just solve for AB in this expression right here, I get this and I can plug that  in to AB there.  I've just solved for the concentration of the transition state from here.  Now I'm going to plug that in to this equation right here.  I'm going to put all of this into here and so there it is.  There's the frequency, the special frequency, here's all the rest of that.  That's our reaction rate.  In essence, this is the phenomenological rate expression that we would normally write for  this reaction.  A plus B goes to products.  We know that the reaction rate is K times A times B as long as that's an elementary  reaction.  What we've said is look, that rate constant is given by this expression right here.  That frequency times that equilibrium constant divided by this concentration term just to  keep the units right.  Transition state theory has already, and the important thing is these two parameters here  relate directly to physical parameters of the transition state that we can think about  calculating.  In other words, we can calculate this rate constant from fundamental properties of the  transition state because we know enough physical chemistry to do that already.  We have the key point is we have expressions for this equilibrium constant and for this  rate constant right here.  We said that's just equal to V, this guy's equal to C0 times K double dagger divided by  V. Let's say that we actually do want to calculate now what the rate constant is.  Let's say that we actually want to calculate that K right there.  We have to be able to calculate big K double dagger and we've got to be able to calculate  little K double dagger which is just equal to V. How are we going to do that?  Well, here's the expression for big K double dagger.  We know what the partition function of A and B are.  We know how to calculate that already.  We've done that.  How do we calculate the partition function of the transition state?  How do we calculate that partition function?  I'm just confused on your K double dagger expression.  That's the same as that.  That's the same as that.  Well that's a little confusing, isn't it?  What the heck did I?  That's supposed to be K.  Sorry, that shouldn't be K double dagger.  That should just be K. Sorry, you're right.  Those double daggers shouldn't be there.  Thank you.  How do we calculate that transition state?  How do we calculate that partition function?  It's a transition state for goodness sakes.  What is it?  This is the question that these guys wrestled with when they worked out transition state.  If we think about the vibrational partition function first, it's the vibrational partition  function that we really care about here.  Essentially the transition state is undergoing a vibration.  How do I think about that?  This bond is getting longer.  This bond is getting shorter.  Just like an asymmetric stretch of the transition state, that's the mode that we care about.  If we could calculate the partition function for that vibrational mode, that's critical  to understanding what the reaction rate is going to be.  Here's the generic expression for the vibrational partition function.  For some mode that has a natural frequency or natural energy h nu and a natural frequency  nu.  What can we say about this magic mode that we care about, this asymmetric vibration along  this axis?  Is that going to be a high frequency mode, a low frequency mode?  What do you guys think?  High frequency?  Hello?  So tell me, these are weak bonds here, right?  Do weak bonds have high frequencies or low frequencies?  Low, thank you.  We're going to expect this to be a weak, we're going to expect this to be a very low frequency  mode.  We call these soft modes.  Here's a picture from your chapter that I think is somewhat non-intuitive.  Here's the transition state up here.  What this picture is trying to convey is that the transition state has a very shallow vibrational  energy well with vibrational energy levels that are very, very close together in energy.  In other words, the frequency to go from here to here, that h nu is tiny.  Yes, but the key point is that the frequency that characterizes these transitions in the  transition state along this direction right here is very small.  What that allows us to do is simplify this equation.  We can write this exponential as a series and we can truncate it at the first term.  When we do that, we just get kT over h nu.  We take this normal expression for the partition function, we truncate it, we write it as a  series expansion, the exponential and we truncate it at the first term and we just get kT over  h nu.  This is a special nu.  This is the frequency that describes motion over the barrier.  Whatever that is for, whatever the transition state is, whatever mode the product bonds  are getting formed, reactant bonds are getting consumed, we can think about that process  as involving a single vibration that has a very low frequency in general.  That's what they realize.  Conceptually this is not obvious.  I think everybody in this room would agree.  None of this stuff that I'm talking about in the last five minutes is obvious.  This was the conceptual leap that Polani made.  I can write the whole partition function for the transition state as the partition function  for this low frequency mode and then the whole rest of the partition function.  Notice that there's going to be a rotational, an electronic, a translational.  The partition function contains many other manifolds and other vibrational modes as well  for the molecule that are orthogonal to this special mode.  We're going to roll that all into this guy over here.  This is just the partition function for the magic mode that corresponds to the reaction.  This is the rest of Q, all the rest of the partition function for all the other modes,  manifolds, and so on.  So that's the partition function.  I can just plug that in then.  Here's the partition function that we were wondering about.  Here's the expression we now have for it.  Notice that the rest of this partition function involves things that we can already calculate  because they're things that are not perturbed in the molecule.  What's perturbed in the transition state is product bonds are getting formed, reactant  bonds are getting broken.  We can describe that.  We can roll that process all into one mode that has this characteristic frequency here.  We're going to have to figure out what that frequency is.  But all the rest of this stuff, we can just roll in.  Bonds that are orthogonal to the reaction rate, we can just calculate their vibrational  partition function, their rotational partition function, all of that stuff using the conventional  methods that we already know about.  So now I rewrite this in terms of this guy right here.  Look at this KT over H nu.  Now that's right there.  Here's the rest of the partition function, this Q with a line over it here.  And I'm getting close to being able to calculate this guy.  This is the contribution along the reaction coordinate only.  Yes, yes, this is all the rest of it.  Yes.  Okay, so our expression for the rate constant becomes this and we're just plugging now this  expression in for this equilibrium constant with a double dagger on it.  So we can actually, when we do that, we can cancel this frequency.  Hey, turns out we don't need to know what it is.  It cancels for gosh sakes.  We don't have to measure it.  We get this expression right here and this is the famous Erring equation.  We derived it going way too fast in about 20 minutes.  This is a very important equation.  Now why is it so important?  Yeah?  How do we find K using K?  How do we find K using K?  How do we find that K using that K?  No, the little K within the equation.  Oh, that K.  This could be confusing.  That's Boltzmann's constant.  Sorry.  This is the rate constant, phenomenological rate constant for the reaction.  That is the equilibrium constant that we calculate using this expression right here.  Right there, that whole thing.  Now you might ask, are you going to be called upon on the final exam to calculate all of  this stuff?  No.  But I have to be able to sleep at night and so I am going to disclose all of this information  to you even though I don't think it would be fair of me to write a question where you  have to calculate all of it.  If you look at the end of this lecture, when I post this lecture after class, it does in  fact have like 120 slides and the last 20 slides are a calculation.  It allows you to calculate the rate constant for H plus H2 goes to H2 plus H. The very  first reaction that was studied using this equation, we can work out exactly what the  rate constant is and if we had enough time we would do it but we don't.  So if you're interested in this rate theory stuff, this transition state theory stuff,  you might want to just look down at those slides because we're never going to get to  them today.  It doesn't matter.  It does matter but.  So here's the Erring equation.  I'm just substituting now for K double dagger here.  Here's the Erring equation.  Do you see the parallels?  The activation energy is this delta E0 that we were talking about.  The difference in the zero point energies for reactants and products.  This pre-exponential factor, A that we've never said anything about, that's given by  this collection of variables right here.  RT over H times this guy, what's left over from the partition function, we strip out  from the normal partition function for the transition state, we strip out the part of  it that pertains to the vibration along the reaction coordinate and that's what's left  over because remember that frequency just canceled for us.  We don't need to know it.  That's a good thing because who knows what it is.  How would you measure it?  You'd have to have some exotic spectroscopic method to do that.  Well the Erranius equation is a special case of the Erring equation.  We can calculate these things and if you don't believe me go to slide 101 and we go through  and we do it laboriously for a particular reaction.  We're not going to say more about it sadly.  Here are pre-exponential factors.  You can calculate this one, actually it is calculated later in this lecture.  We can calculate these pre-exponential factors for the Erranius equation knowing transition  state theory so it's very powerful.  What can we do in the last 15 minutes?  Something very important.  We want to apply the same thinking to a reaction that occurs in water.  The key point is that we want to think about what the influence of charge is on the reaction  rate.  It's not inconceivable that there could be a question like this on the final.  For example two chlorides react with lead to give lead chloride.  All of these things can be in solution.  Lead chloride has some solubility in water, very low.  What rate would this reaction take?  What is the influence of the charge on the reaction rate?  You might think naively if the reactants and products are oppositely charged they're going  to be clumpically attracted to one another and the reaction rate is going to be accelerated.  If you don't think about this too hard, even if you think about it hard you might come  to that conclusion.  Negatively charged reactants and positively charged reactants are going to be clumpically  attracted and boom they're going to react really fast.  That's going to accelerate the reaction rate.  That turns out to be exactly wrong and it's important to understand why.  It's great once in a while in chemistry when you encounter a concept that is completely  counterintuitive at first and later on hopefully intuitive.  I've got to make this happen in like 12 minutes.  Here's the idea.  A reacts with B to give products.  That's the charge on A and B, Z, A and ZB.  That's the charge.  The key point here is reactants are charged.  There's a version of transition state theory but it's not strictly speaking transition  state theory.  It's a thermodynamic version of it.  You can't use transition state theory in its normal form to describe reactions in solution  because it doesn't account for reactions with solvent.  It doesn't transition state theory in its normal form does not account for the complexities  imparted by having the solvent present in the reaction.  This looks like transition state theory but strictly speaking it's something that's related  to it, not exactly the same thing.  We're going to simplify this notation.  We're going to call this mess.  Notice something.  If we think about this in terms of transition state theory, when A reacts with B, if we form  a transition state, the transition state will have a total charge equal to A plus B. The  total charge of the transition state will be the sum of the charges on A and B because  charge has to be conserved here.  That's a key point.  So exactly, this is the reaction rate.  Now in terms of, this is just that rate constant right there.  This is the normal phenomenological rate that we would write for this reaction.  Then we have to think back and remember something about activities.  Activities are going to be the key, the understanding how this works.  The activity of some I in A is equal to its concentration times some activity coefficient,  gamma sub A. That's the activity of A. That's the activity coefficient.  The activity coefficient for a neutral is one.  The activity coefficient only deviates from one as a consequence of Coulombic interactions  with the solution, other ions in the solution.  If other ions are present in the solution, the activity coefficient will be less than  one.  The more ions, the lower the activity coefficient.  And there's something called the Debye-Hockel limiting lot.  That's the activity coefficient.  That's the ionic strength and the ionic strength is just given by this expression where this  is the concentration of the ion, sorry, this is the concentration of the ion and this is  the charge on the ion for every ion in the solution.  I add up all the ions in the solution, multiply by the square of their charge, take that times  one half and I've got the ionic strength.  The higher the ionic strength, it's never obvious to me when the equation has a log  in it like this.  The higher the ionic strength, the lower gamma becomes.  If I is zero, gamma is one.  This is the ionic strength on this axis.  This is gamma on this axis.  If ionic strength is zero, gamma is one and it deviates from one as the ionic strength  goes up.  And that size of this deviation here depends on what the charge is on the ion of interest,  ion A. Because here's its charge.  So if it has a charge of one, there's a small deviation, charge of, that's the size.  If it's small, you get a little bigger deviation.  If the charge goes up to three, you get a huge deviation.  So activity effects have everything to do with charge.  How many of you had 151?  From me.  So this is review for all you guys.  Good.  So when we write an equal, here's an equilibrium constant.  We write in terms of activities.  Every activity is an activity coefficient times a concentration.  You guys all know that.  And what's more important is you have intuition about what the influence of ions are on equilibria.  Here's an important piece of intuition that everybody should have in the room, especially  those of you who took my class.  If you look at some equilibrium like this, all right?  Sodium chloride is an acid.  Acetic acid dissociates to give hydronium ion and acetate.  If I dump sodium chloride into the solution, what's going to happen to the pH?  Sodium chloride is an inert salt.  Sodium chloride has no acidity or basicity of its own.  And yet the pH will change in a predictable way.  What will happen?  How many people think the pH is going to go down?  It's going to get more acidic.  How many people think the pH is going to go up?  It's going to get more basic when I add salt.  Get those hands way up there because I want to see.  You guys never had 151 from me.  Now if you add salt to this equilibrium, it will always shift to the right.  It will always shift to the right.  The addition of inert salt to any equilibrium always shifts it in the direction of the most  ionic state of the equilibrium.  See how there's ions here and there are no ions here?  That's the most ionic state.  If I add salt, the reaction will shift to the right.  It's called salting in.  Now I can prove that.  Here's the equilibrium constant for this reaction.  Here are the activity coefficients.  The activity coefficients for the neutrals are zero or one rather.  The activity coefficient for this guy's one.  These activity coefficients for the charged species are less than one.  They will become lower as I increase the ionic strength of the solution.  Those guys will get smaller.  What that means is that the equilibrium constant that applies if we think about these activity  effects is going to get bigger.  If these guys get smaller, that gets bigger.  That's the activity coefficient that applies when we add sodium chloride to the solution  or any other inert salt.  It's going to get more acidic.  What about this guy?  What happens to the solubility of lead sulfide if I add sodium chloride to the solution?  Sodium chloride doesn't appear in this equilibrium.  Why would it affect this equilibrium?  And yet it does in a predictable way.  What's going to happen?  Is the lead sulfide more soluble or less soluble when I dump sodium chloride in?  More.  Why?  Because, look, this side of the equilibrium has got a lot of ions.  This side's got no ions.  If I add more ions to the solution, equilibrium is favored by the addition of more ions.  I can prove that's true by working out what the new equilibrium constant is.  These are the two activity coefficients.  They both get smaller when I add sodium chloride.  So the equilibrium constant shifts to the right.  Every time you see an equilibrium, you can affect its position by adding an inert salt  or removing an inert salt from the solution.  That'll also alter the equilibrium in a predictable way.  What does all this have to do with transition state theory?  Very simple to make a long story short.  We can work out the math, but we're just going to skip over it here because we're almost  out of time.  Here's the bottom line.  The transition state has the total charge of the reactants in it.  If the transition state has a high charge, then the rate of the reaction is going to  be influenced by the presence of other ions in the solution.  At the end of the day, when we're done doing the derivation, we get this equation right  here, which is the equation for the kinetic salt effect.  What is this big K right here?  It's just the collection of the activity coefficients for the transition state and  the reactants.  That is the rate constant that applies for one molar, everything.  That's why it's got a zero on it.  That's the actual rate constant of the reaction.  This is the most important slide that has to do with this second concept here.  What am I plotting here?  This is the reaction rate for reaction that involves A with some charge plus B with some  charge going on to products.  This is the ionic strength.  The key point here is we want to understand how salt affects the reaction rate.  It's easy to understand how salt affects the equilibrium constant.  How does it affect the reaction rate?  The way to think about that is if A reacts with B and they both have a plus 2 charge,  they're both positively charged, they both got a plus 2 charge, the transition state  has a charge of plus 4.  If there's an equilibrium between the transition state and the reactants, it's going to be  favored by the addition of salt.  The reaction is going to get accelerated.  Isn't that counterintuitive?  The reaction rate goes up as I add salt to the solution.  Even though the reactants and products have the same charge, they have to overcome a coulombic  barrier to react because they're both positively charged, the reaction rate goes up, not down.  Check this out.  If a 2 plus ion reacts with a 2 minus ion, you'd expect there to be a big coulombic attraction,  right?  Reactions should go faster.  It goes slower.  Does this have to do with the orthogonal components of the transition state's partition function?  Absolutely not.  No?  No.  Nothing at all?  No.  Nothing to do with the ionic development?  Zero.  Okay.  It's not a bad question, but the answer's no.  Does everybody see why this happens?  There's an equilibrium.  I'll say this one last time because I don't have any more time.  There's an equilibrium between A and B, the reactants, and the transition state.  What we just agreed on is that we can decide which side of the equilibrium will be favored  when we add salt to an equilibrium.  The most ionic state of the system is favored.  If the charge on the transition state is higher than the charge on either one of the ions  because the only way that can be true is if the ions have the charge of the same sign  like plus 1 and plus 2, plus 2 and plus 2, plus 1 and plus 1, then the charge on the  transition state will be like 3, 4, and so on.  The transition state gets favored by the addition of salt and the reaction rate goes  up.  Look, it goes up here too.  Look, it goes up here.  So if there's no charge, nothing happens.  And if the charges on the reactants are opposite to one another, then the charge on the transition  state is lower than the charge on the ions.  The transition state has a lower ionic, it's the least ionic state of the system.  The reactants are more ionic than the transition state and the reaction is slowed down by the  addition of salt.  That's totally counterintuitive.  If you understand that, you understand something that most chemists even are going to get wrong.  You're going to get it right.  You say those charges, that reaction is going to go slower.  You can think about, okay.  So on Friday, yeah, there's more here.  Yes, oh my goodness.  There's like 20 more slides that work through the equilibrium, we're not going to give to  those.  So on Friday, we're going to work on the final.  Please take the course evaluations.  All right.  Thank you.
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Okay, are you guys doing? Two more lectures. We're going to review the final for the final on Friday. So that's all we'll really be doing on Friday. I'll tell you in some detail what's going to be on it. Did I mention that the electronic course evaluations are available? I didn't look to see if any of you guys did these in the last two days. Monday we talked about enzyme kinetics, but we're not going to say anything more about that today, but as you'll see on Friday, that's definitely going to be on the final. So I jammed a lot of information into that lecture on Monday. All of enzyme kinetics, all of enzyme inhibition, it's all in there. All right, some of that stuff is not in your book, like the enzyme inhibition stuff. Okay, so you can either find another source to study that from or you can study it directly from the lecture. I think everything that I put in there is approximately correct. Now, today we're going to talk about transition state theory. At the beginning of the quarter, this is what we hope the course would look like. We had this beautiful blue rectangle of reaction dynamics here at the end of the course, and we were optimistic that we'd be able to talk a lot about this, maybe for more than one lecture. But what actually happened was it fell off the end of the course, and all we've got left is a single lecture that you're going to hear about today. That's all that's left of reaction dynamics. Could we have just left it out? We could have, but enough damage has been done to you guys. In 131A, all you heard about was quantum mechanics from Professor Renzepis. In 131B, all you heard about was spectroscopy from Professor Martin. And I know you know a lot about quantum mechanics and spectroscopy, and that's good. But everything else is important too, and so this subject is actually very, very important to us. And so I've tried to pick out the one thing that I cannot allow you to leave this class without knowing. I've tried to pull the one thing out of reaction dynamics that you've got to know about. This is the one thing, this lecture. I'm going to tell it to you. Basically what we want to understand is where does the Arrhenius equation come from when we measure the temperature dependence of reactions? This is the reaction rate. Most of the time we find out that it conforms to this equation. We make a plot of log of the reaction rate versus 1 over T. We get a straight line. We get the activation energy of the reaction from that straight line, and we report that, and we think about that. That guides a lot of what we do as physical chemists, but we have never explained where this equation comes from. In some ways this is the most important equation in chemical kinetics, but we haven't talked about what its origin is. Fundamentally, where does this equation come from? So we're going to do that today. I can do that in one lecture. It reminds me, of course this statement is grammatically incorrect. It's a dangling preposition. It reminds me of one of my favorite jokes. The Texas student goes to Harvard, and he asks a Harvard student, where are you from? The Harvard graduate says, well, I come from a place where we do not end our sentences of propositions. The Texas says, okay, where are you from? Other words are sometimes substitute. I would love to spend 10 minutes talking to you about these two guys. Henry Ahring worked at the University of Utah for most of his career, one of the great American physical chemists of the 20th century. Michael Polany, Nobel Prize winning physical chemist. Both of these guys are responsible for what we're going to be talking about today, transition state theory. They worked it out in the 1930s. So here's the short version of the history that matters here. Way back in 1916, G.N. Lewis, the first great American physical chemist, worked out Lewis-Dott structures. There was no quantum mechanics in 1916. It wasn't invented until 1924. He came up with a very compelling model for chemical bonding involving pairs of electrons that turned out to be correct. And much of what it predicted way before the fundamental underpinnings of quantum mechanics were described. He figured all that out. He was a genius. Heisenberg, Dirac, and Schrodinger discovered quantum mechanics in 1924. Much of Germans, Heitler and London, rather, took quantum mechanics and applied it to bonding for the first time. They did the first quantitative description of chemical bonding. Their names are usually lost. We hardly even mention it. Sometimes we talk about London's dispersion forces. You guys remember that? Vanderwall's forces? This is the guy. Fritz London. He's the guy who figured that out. 1929, five years after quantum mechanics, these guys developed a quantum mechanical way to think about reaction rates. These guys figured out a quantum mechanical way to think about bonding. That's pretty important. But Pellanian Erring figured out how to think about reaction rates from a quantum mechanical perspective. We're going to zoom through what they taught us today. Here's the basic idea. A plus B goes to products. We're just going to talk about this generic reaction today. We've stripped down this lecture so that we've taken everything else out. It should be three weeks' worth of lectures. We know the rate law is this. The rate constant times the concentration of A times the concentration of B. These can be pressures or concentrations. Transition state theory says that this reaction actually occurs through this mechanism. A plus B are in equilibrium with something called AB double dagger. That reacts in a unimolecular fashion to give us products. Transition state theory basically takes all of the reactants, puts them on one side, all of the products puts them on the other side, and in between it constructs something called an activated complex or a transition state. That's why it's called transition state theory. This thing here is the transition state. It's an entity that is intermediate between the reactants and the products. If bonds exist in the products that don't form in the reactants, they exist partially in a weak form in the activated complex or in the transition state. If bonds are broken here, they're partially broken in the transition state. The transition state has structural attributes, bond lengths, and so on that are intermediate between the reactants and the products. One of the key points about transition state theory is it postulates that an equilibrium exists between this transition state and these reactants. Here's the picture in terms of energy that you've seen a million times, but what we're presenting here now, the notation applies specifically to transition state theory. You've got reactants that's represented here by these two molecules. This looks like it could be an OH, that's CH3, that's CH3Br apparently, OH minus, CH3Br. That's the transition state and here are the products. Notice that in this reaction, a new bond is formed between this oxygen and this carbon and this bond here between this carbon and this bromine is broken. In the transition state, notice that the transition state, what we're depicting here is this bond is partially formed. You see how long it is? It's a long, weak bond and that bond is partially broken. See how long it is? It's a long, weak bond also. That bond is going to be broken in the product state. That bond doesn't even exist in the reactant state. The transition state contains all of the bonds that are present in the reactants and in the products, but the bonds that are involved in the formation of the products from the reactants are weakened. See how long that is? See how long that bond is? These are two super weak bonds. We can construct a transition state by thinking about what the products look like, what the reactants look like, products, reactants and then thinking about how the products are formed from the reactants geometrically. How does that happen? This shows an attack in a particular geometric orientation of the OH minus to the CH3Br. Thinking about this and thinking about this, we can construct what this transition state should look like using a few simple rules. It turns out that transition state and activated complex should not technically be used interchangeably. It's a nuance. The activated complex actually exists over a range of this reaction coordinate, whereas the transition state in principle exists only at a particular point in time. We're not going to bother ourselves with that distinction today. There's a fine point there that you should know. These are not identically or interchangeable to say activated complex and transition state, but today we're just going to use those two terms interchangeably. Now, with that as a premise, let's see if we can work out what the reaction rate is. Here's our mechanism, our transition state theory mechanism for this reaction. Yes, so we can write an expression for this equilibrium constant. What is it? Product over reactants. Notice that I'm normalizing and being very careful here to write activities, the activity of the transition state, the activity of A, the activity of B, because I'm dividing by the standard concentration here. When we're done canceling these C0s, I end up with an extra factor of C0 on the numerator that ensures that the equilibrium constant is dimensionless, doesn't it? Now, unfortunately, I used the Microsoft word equation editor to write a lot of these equations. It does not allow me to write a zero with a line through it. That is the symbol for the standard concentration, one molar, for example, but I'm just going to call it C0. There's another issue. See this double X here, this double plus? Microsoft equation editor doesn't contain a diocese. What's a diocese? It's that thing, the double dagger. The double dagger is the same thing as the diocese. When I write two pluses, I'm just indicating the transition state. That refers to the transition state. That's the equilibrium constant that involves the transition state. That's the unimolecular reaction rate constant that involves the transition state. You see how? I'm going to write that as that. If you see the double plus, it's just the diocese. If anybody knows, I'll work around for that. That would help me out a lot. How do I write a diocese in Microsoft equation editor? Here I cheated. I put a white square here and I pasted this on top. Then I thought, I can't do that. There's like 106 slides in this presentation. Now I said yes, yes. We could write this in terms of pressures if we want to. Here's the concentration. Pressures, no difference. Now, if we flash back, we're talking right now about chapter 20. If we flash back to chapter 17, it turns out we can calculate equilibrium constants from partition functions. Recall, partition functions are very important to us because they allow us to make a connection between statistical mechanics and thermodynamics. Partition functions contain information about the actual molecule. We can look at a molecule. If we know something about its state distribution, we can write its partition function. What we learned in chapter 17, what we didn't have time to talk about in the class this quarter is the fact that you can also use these partition functions to calculate equilibrium constants. If you want to read more about that, it's on page 670. Turns out to be an important thing that we left out. Here's what's there. Here's some generic reaction. Amoles of A, binomoles of B, c-moles of C, d-moles of D. Here's what the equilibrium constant expression looks like written out in terms of the standard molar partition functions for A, B, C and D. If I can calculate these partition functions for all of these guys, and I know that, so what is this? This is the standard molar partition function. M means molar, A is species A. What's that? That's Avogadro's number. That's not a misprint. Avogadro's number in every single case. That is the difference in the zero point energies between reactants and products. If this is your generic reactant here, here are vibrational energy levels. Here are vibrational energy levels of the product. That's the ground vibrational energy level. That's the ground vibrational energy level. That's delta Re0. The difference between the ground and the zero point energies of the reactants and the products. Here's our Gibbs free energy as a function of reaction coordinate. The delta Re0 is closely related to this green quantity that I'm indicating here. If this thing was in its ground vibrational energy level, and this thing was in its ground vibrational energy level, both of these guys, and both of these guys, then this would be delta Re0. We're always talking about the zero point energies. We could say something about this equilibrium constant. We could calculate it using this equation right here. In transition state theory, that's not the equilibrium constant we care about. What we say in transition state theory is this is an equilibrium with this. I don't care about the equilibrium of this with this. That's going to give me the normal equilibrium constant that I can learn about in chapter 17. What transition state theory says is these two guys are in equilibrium with one another. What matters is this delta Re0 here. This thing that we normally call the activation energy. It's the activation energy from the Erranius equation. We want to calculate that imaginary equilibrium constant. These things may actually be in equilibrium, but this thing is not really observable except using some exotic spectroscopic nanosecond picosecond spectroscopy. A normal equilibrium applied to this generic reaction right here, I could calculate the equilibrium constant using this equation. Now I'm going to apply that same thinking to the transition state theory. Here's the transition state equilibrium we care about. A reacts with B to give this transition state. Now what do I want? I want to put the partition function for that guy in the numerator for these guys in the denominator and then that's Avogadro's number. That's just left over from, that's an extra factor of Avogadro's number because I've got two reactants in one product. I can calculate this equilibrium constant that applies to the formation of the transition state. Keep in mind, everything we're talking about here is kinetics. We're sort of mixing thermodynamics with kinetics. They took a thermodynamic concept, equilibrium constant, and applied that to transition state theory. So you're with me so far? We've got an equilibrium constant here. We can calculate it using statistical mechanics. Two-step mechanism. That's the first thing to understand. The second thing to understand is that this rate right here showing in gray, that's going to be approximately equal to the frequency with which the transition state crosses over the top of the barrier. We've got a barrier here. We've got a transition state. We are moving along this reaction coordinate from reactants to products. The frequency with which this thing moves across the top of this barrier, that's going to closely approximate this rate. The rate at which products are formed, that seems to be a statement of the obvious. Obviously as you cross over this barrier from reactants to products, the frequency that that happens, that's the rate of this reaction. It's totally obvious to say that. The reaction rate I can write as this frequency times whatever the concentration of this transition state is. We're going to talk about this frequency. It's the frequency with which if you think about this as being a molecule, there's a vibration that has to happen here. This guy moves back and forth between these two guys. We get a symmetric vibration of this transition state. The frequency that characterizes that mode is the frequency that we care about. Your book includes something called the transmission coefficient. We're just going to assume that's one. See that cap right there? Just forget about it, it's one. The rate of the reaction is given by this special frequency that applies to the reaction coordinate times the concentration of the transition state, but we know we can also write the rate in the normal way with a rate constant times A times B. That's what we've been saying all along. Here was our equilibrium constant expression for the transition state. If I just solve for AB in this expression right here, I get this and I can plug that in to AB there. I've just solved for the concentration of the transition state from here. Now I'm going to plug that in to this equation right here. I'm going to put all of this into here and so there it is. There's the frequency, the special frequency, here's all the rest of that. That's our reaction rate. In essence, this is the phenomenological rate expression that we would normally write for this reaction. A plus B goes to products. We know that the reaction rate is K times A times B as long as that's an elementary reaction. What we've said is look, that rate constant is given by this expression right here. That frequency times that equilibrium constant divided by this concentration term just to keep the units right. Transition state theory has already, and the important thing is these two parameters here relate directly to physical parameters of the transition state that we can think about calculating. In other words, we can calculate this rate constant from fundamental properties of the transition state because we know enough physical chemistry to do that already. We have the key point is we have expressions for this equilibrium constant and for this rate constant right here. We said that's just equal to V, this guy's equal to C0 times K double dagger divided by V. Let's say that we actually do want to calculate now what the rate constant is. Let's say that we actually want to calculate that K right there. We have to be able to calculate big K double dagger and we've got to be able to calculate little K double dagger which is just equal to V. How are we going to do that? Well, here's the expression for big K double dagger. We know what the partition function of A and B are. We know how to calculate that already. We've done that. How do we calculate the partition function of the transition state? How do we calculate that partition function? I'm just confused on your K double dagger expression. That's the same as that. That's the same as that. Well that's a little confusing, isn't it? What the heck did I? That's supposed to be K. Sorry, that shouldn't be K double dagger. That should just be K. Sorry, you're right. Those double daggers shouldn't be there. Thank you. How do we calculate that transition state? How do we calculate that partition function? It's a transition state for goodness sakes. What is it? This is the question that these guys wrestled with when they worked out transition state. If we think about the vibrational partition function first, it's the vibrational partition function that we really care about here. Essentially the transition state is undergoing a vibration. How do I think about that? This bond is getting longer. This bond is getting shorter. Just like an asymmetric stretch of the transition state, that's the mode that we care about. If we could calculate the partition function for that vibrational mode, that's critical to understanding what the reaction rate is going to be. Here's the generic expression for the vibrational partition function. For some mode that has a natural frequency or natural energy h nu and a natural frequency nu. What can we say about this magic mode that we care about, this asymmetric vibration along this axis? Is that going to be a high frequency mode, a low frequency mode? What do you guys think? High frequency? Hello? So tell me, these are weak bonds here, right? Do weak bonds have high frequencies or low frequencies? Low, thank you. We're going to expect this to be a weak, we're going to expect this to be a very low frequency mode. We call these soft modes. Here's a picture from your chapter that I think is somewhat non-intuitive. Here's the transition state up here. What this picture is trying to convey is that the transition state has a very shallow vibrational energy well with vibrational energy levels that are very, very close together in energy. In other words, the frequency to go from here to here, that h nu is tiny. Yes, but the key point is that the frequency that characterizes these transitions in the transition state along this direction right here is very small. What that allows us to do is simplify this equation. We can write this exponential as a series and we can truncate it at the first term. When we do that, we just get kT over h nu. We take this normal expression for the partition function, we truncate it, we write it as a series expansion, the exponential and we truncate it at the first term and we just get kT over h nu. This is a special nu. This is the frequency that describes motion over the barrier. Whatever that is for, whatever the transition state is, whatever mode the product bonds are getting formed, reactant bonds are getting consumed, we can think about that process as involving a single vibration that has a very low frequency in general. That's what they realize. Conceptually this is not obvious. I think everybody in this room would agree. None of this stuff that I'm talking about in the last five minutes is obvious. This was the conceptual leap that Polani made. I can write the whole partition function for the transition state as the partition function for this low frequency mode and then the whole rest of the partition function. Notice that there's going to be a rotational, an electronic, a translational. The partition function contains many other manifolds and other vibrational modes as well for the molecule that are orthogonal to this special mode. We're going to roll that all into this guy over here. This is just the partition function for the magic mode that corresponds to the reaction. This is the rest of Q, all the rest of the partition function for all the other modes, manifolds, and so on. So that's the partition function. I can just plug that in then. Here's the partition function that we were wondering about. Here's the expression we now have for it. Notice that the rest of this partition function involves things that we can already calculate because they're things that are not perturbed in the molecule. What's perturbed in the transition state is product bonds are getting formed, reactant bonds are getting broken. We can describe that. We can roll that process all into one mode that has this characteristic frequency here. We're going to have to figure out what that frequency is. But all the rest of this stuff, we can just roll in. Bonds that are orthogonal to the reaction rate, we can just calculate their vibrational partition function, their rotational partition function, all of that stuff using the conventional methods that we already know about. So now I rewrite this in terms of this guy right here. Look at this KT over H nu. Now that's right there. Here's the rest of the partition function, this Q with a line over it here. And I'm getting close to being able to calculate this guy. This is the contribution along the reaction coordinate only. Yes, yes, this is all the rest of it. Yes. Okay, so our expression for the rate constant becomes this and we're just plugging now this expression in for this equilibrium constant with a double dagger on it. So we can actually, when we do that, we can cancel this frequency. Hey, turns out we don't need to know what it is. It cancels for gosh sakes. We don't have to measure it. We get this expression right here and this is the famous Erring equation. We derived it going way too fast in about 20 minutes. This is a very important equation. Now why is it so important? Yeah? How do we find K using K? How do we find K using K? How do we find that K using that K? No, the little K within the equation. Oh, that K. This could be confusing. That's Boltzmann's constant. Sorry. This is the rate constant, phenomenological rate constant for the reaction. That is the equilibrium constant that we calculate using this expression right here. Right there, that whole thing. Now you might ask, are you going to be called upon on the final exam to calculate all of this stuff? No. But I have to be able to sleep at night and so I am going to disclose all of this information to you even though I don't think it would be fair of me to write a question where you have to calculate all of it. If you look at the end of this lecture, when I post this lecture after class, it does in fact have like 120 slides and the last 20 slides are a calculation. It allows you to calculate the rate constant for H plus H2 goes to H2 plus H. The very first reaction that was studied using this equation, we can work out exactly what the rate constant is and if we had enough time we would do it but we don't. So if you're interested in this rate theory stuff, this transition state theory stuff, you might want to just look down at those slides because we're never going to get to them today. It doesn't matter. It does matter but. So here's the Erring equation. I'm just substituting now for K double dagger here. Here's the Erring equation. Do you see the parallels? The activation energy is this delta E0 that we were talking about. The difference in the zero point energies for reactants and products. This pre-exponential factor, A that we've never said anything about, that's given by this collection of variables right here. RT over H times this guy, what's left over from the partition function, we strip out from the normal partition function for the transition state, we strip out the part of it that pertains to the vibration along the reaction coordinate and that's what's left over because remember that frequency just canceled for us. We don't need to know it. That's a good thing because who knows what it is. How would you measure it? You'd have to have some exotic spectroscopic method to do that. Well the Erranius equation is a special case of the Erring equation. We can calculate these things and if you don't believe me go to slide 101 and we go through and we do it laboriously for a particular reaction. We're not going to say more about it sadly. Here are pre-exponential factors. You can calculate this one, actually it is calculated later in this lecture. We can calculate these pre-exponential factors for the Erranius equation knowing transition state theory so it's very powerful. What can we do in the last 15 minutes? Something very important. We want to apply the same thinking to a reaction that occurs in water. The key point is that we want to think about what the influence of charge is on the reaction rate. It's not inconceivable that there could be a question like this on the final. For example two chlorides react with lead to give lead chloride. All of these things can be in solution. Lead chloride has some solubility in water, very low. What rate would this reaction take? What is the influence of the charge on the reaction rate? You might think naively if the reactants and products are oppositely charged they're going to be clumpically attracted to one another and the reaction rate is going to be accelerated. If you don't think about this too hard, even if you think about it hard you might come to that conclusion. Negatively charged reactants and positively charged reactants are going to be clumpically attracted and boom they're going to react really fast. That's going to accelerate the reaction rate. That turns out to be exactly wrong and it's important to understand why. It's great once in a while in chemistry when you encounter a concept that is completely counterintuitive at first and later on hopefully intuitive. I've got to make this happen in like 12 minutes. Here's the idea. A reacts with B to give products. That's the charge on A and B, Z, A and ZB. That's the charge. The key point here is reactants are charged. There's a version of transition state theory but it's not strictly speaking transition state theory. It's a thermodynamic version of it. You can't use transition state theory in its normal form to describe reactions in solution because it doesn't account for reactions with solvent. It doesn't transition state theory in its normal form does not account for the complexities imparted by having the solvent present in the reaction. This looks like transition state theory but strictly speaking it's something that's related to it, not exactly the same thing. We're going to simplify this notation. We're going to call this mess. Notice something. If we think about this in terms of transition state theory, when A reacts with B, if we form a transition state, the transition state will have a total charge equal to A plus B. The total charge of the transition state will be the sum of the charges on A and B because charge has to be conserved here. That's a key point. So exactly, this is the reaction rate. Now in terms of, this is just that rate constant right there. This is the normal phenomenological rate that we would write for this reaction. Then we have to think back and remember something about activities. Activities are going to be the key, the understanding how this works. The activity of some I in A is equal to its concentration times some activity coefficient, gamma sub A. That's the activity of A. That's the activity coefficient. The activity coefficient for a neutral is one. The activity coefficient only deviates from one as a consequence of Coulombic interactions with the solution, other ions in the solution. If other ions are present in the solution, the activity coefficient will be less than one. The more ions, the lower the activity coefficient. And there's something called the Debye-Hockel limiting lot. That's the activity coefficient. That's the ionic strength and the ionic strength is just given by this expression where this is the concentration of the ion, sorry, this is the concentration of the ion and this is the charge on the ion for every ion in the solution. I add up all the ions in the solution, multiply by the square of their charge, take that times one half and I've got the ionic strength. The higher the ionic strength, it's never obvious to me when the equation has a log in it like this. The higher the ionic strength, the lower gamma becomes. If I is zero, gamma is one. This is the ionic strength on this axis. This is gamma on this axis. If ionic strength is zero, gamma is one and it deviates from one as the ionic strength goes up. And that size of this deviation here depends on what the charge is on the ion of interest, ion A. Because here's its charge. So if it has a charge of one, there's a small deviation, charge of, that's the size. If it's small, you get a little bigger deviation. If the charge goes up to three, you get a huge deviation. So activity effects have everything to do with charge. How many of you had 151? From me. So this is review for all you guys. Good. So when we write an equal, here's an equilibrium constant. We write in terms of activities. Every activity is an activity coefficient times a concentration. You guys all know that. And what's more important is you have intuition about what the influence of ions are on equilibria. Here's an important piece of intuition that everybody should have in the room, especially those of you who took my class. If you look at some equilibrium like this, all right? Sodium chloride is an acid. Acetic acid dissociates to give hydronium ion and acetate. If I dump sodium chloride into the solution, what's going to happen to the pH? Sodium chloride is an inert salt. Sodium chloride has no acidity or basicity of its own. And yet the pH will change in a predictable way. What will happen? How many people think the pH is going to go down? It's going to get more acidic. How many people think the pH is going to go up? It's going to get more basic when I add salt. Get those hands way up there because I want to see. You guys never had 151 from me. Now if you add salt to this equilibrium, it will always shift to the right. It will always shift to the right. The addition of inert salt to any equilibrium always shifts it in the direction of the most ionic state of the equilibrium. See how there's ions here and there are no ions here? That's the most ionic state. If I add salt, the reaction will shift to the right. It's called salting in. Now I can prove that. Here's the equilibrium constant for this reaction. Here are the activity coefficients. The activity coefficients for the neutrals are zero or one rather. The activity coefficient for this guy's one. These activity coefficients for the charged species are less than one. They will become lower as I increase the ionic strength of the solution. Those guys will get smaller. What that means is that the equilibrium constant that applies if we think about these activity effects is going to get bigger. If these guys get smaller, that gets bigger. That's the activity coefficient that applies when we add sodium chloride to the solution or any other inert salt. It's going to get more acidic. What about this guy? What happens to the solubility of lead sulfide if I add sodium chloride to the solution? Sodium chloride doesn't appear in this equilibrium. Why would it affect this equilibrium? And yet it does in a predictable way. What's going to happen? Is the lead sulfide more soluble or less soluble when I dump sodium chloride in? More. Why? Because, look, this side of the equilibrium has got a lot of ions. This side's got no ions. If I add more ions to the solution, equilibrium is favored by the addition of more ions. I can prove that's true by working out what the new equilibrium constant is. These are the two activity coefficients. They both get smaller when I add sodium chloride. So the equilibrium constant shifts to the right. Every time you see an equilibrium, you can affect its position by adding an inert salt or removing an inert salt from the solution. That'll also alter the equilibrium in a predictable way. What does all this have to do with transition state theory? Very simple to make a long story short. We can work out the math, but we're just going to skip over it here because we're almost out of time. Here's the bottom line. The transition state has the total charge of the reactants in it. If the transition state has a high charge, then the rate of the reaction is going to be influenced by the presence of other ions in the solution. At the end of the day, when we're done doing the derivation, we get this equation right here, which is the equation for the kinetic salt effect. What is this big K right here? It's just the collection of the activity coefficients for the transition state and the reactants. That is the rate constant that applies for one molar, everything. That's why it's got a zero on it. That's the actual rate constant of the reaction. This is the most important slide that has to do with this second concept here. What am I plotting here? This is the reaction rate for reaction that involves A with some charge plus B with some charge going on to products. This is the ionic strength. The key point here is we want to understand how salt affects the reaction rate. It's easy to understand how salt affects the equilibrium constant. How does it affect the reaction rate? The way to think about that is if A reacts with B and they both have a plus 2 charge, they're both positively charged, they both got a plus 2 charge, the transition state has a charge of plus 4. If there's an equilibrium between the transition state and the reactants, it's going to be favored by the addition of salt. The reaction is going to get accelerated. Isn't that counterintuitive? The reaction rate goes up as I add salt to the solution. Even though the reactants and products have the same charge, they have to overcome a coulombic barrier to react because they're both positively charged, the reaction rate goes up, not down. Check this out. If a 2 plus ion reacts with a 2 minus ion, you'd expect there to be a big coulombic attraction, right? Reactions should go faster. It goes slower. Does this have to do with the orthogonal components of the transition state's partition function? Absolutely not. No? No. Nothing at all? No. Nothing to do with the ionic development? Zero. Okay. It's not a bad question, but the answer's no. Does everybody see why this happens? There's an equilibrium. I'll say this one last time because I don't have any more time. There's an equilibrium between A and B, the reactants, and the transition state. What we just agreed on is that we can decide which side of the equilibrium will be favored when we add salt to an equilibrium. The most ionic state of the system is favored. If the charge on the transition state is higher than the charge on either one of the ions because the only way that can be true is if the ions have the charge of the same sign like plus 1 and plus 2, plus 2 and plus 2, plus 1 and plus 1, then the charge on the transition state will be like 3, 4, and so on. The transition state gets favored by the addition of salt and the reaction rate goes up. Look, it goes up here too. Look, it goes up here. So if there's no charge, nothing happens. And if the charges on the reactants are opposite to one another, then the charge on the transition state is lower than the charge on the ions. The transition state has a lower ionic, it's the least ionic state of the system. The reactants are more ionic than the transition state and the reaction is slowed down by the addition of salt. That's totally counterintuitive. If you understand that, you understand something that most chemists even are going to get wrong. You're going to get it right. You say those charges, that reaction is going to go slower. You can think about, okay. So on Friday, yeah, there's more here. Yes, oh my goodness. There's like 20 more slides that work through the equilibrium, we're not going to give to those. So on Friday, we're going to work on the final. Please take the course evaluations. All right. Thank you.
source to study that from or you can study it think about reaction rates. These guys figured with kinetics. They took a thermodynamic concept, properties of the transition state because we and B are. We know how to calculate that already. understanding what the reaction rate is going to energy. In other words, the frequency to go from equilibrium, it will always shift to the right. charge going on to products. This is the ionic between the transition state and the reactants, another, then the charge on the transition state