 Okay, how 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. All right, 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, all right? 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. All right, 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. So the tech says, okay, where are you from? Other words are sometimes substitute. Okay. So I would love to spend 10 minutes talking to you about these two guys. Henry Erie worked at the University of Utah for most of his career, one of the great American physical chemists of the 20th century. Michael Polanyi, Nobel Prize winning physical chemist. All right, 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 Dot structures. All right, 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, bunch of Germans. Heitler and Fritt, 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, Fritt's London. He's the guy who figured that out. All right, 1929, five years after quantum mechanics, these guys developed a quantum mechanical way to think about reaction rates. All right, these guys figured out a quantum mechanical way to think about bonding. That's pretty important. All right, but Polanian Erring figured out how to think about reaction rates from a quantum mechanical perspective. All right, so 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. All right, 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, right? 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. All right, A plus B are in equilibrium with something called AB double dagger. And that reacts in a unimolecular fashion to give us products. All right, 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. All right, 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. All right, if bonds are broken here, they're partially broken in the transition state. So 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. So 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 CH3 BR apparently, all right, OH minus, CH3 BR, that's the transition state, okay, 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, okay? And so 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, right? And that bond is partially broken. See how long it is? It's a long, weak bond also, right? That bond is going to be broken in the product state. That bond doesn't even exist in the reactant state so the transition state contains all of the bonds that are present in the reactants and in the products, all right, but the bonds that are involved in the formation of the products from the reactants are weakened, all right? See how long that is? See how long that bond is? These are two super weak bonds, all right? So 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, all right? This shows an attack in a particular geometric orientation of the OH minus to the CH3Br, all right? Thinking about this and thinking about this, we can construct what this transition state should look like, all right, using a few simple rules. Okay. Now, it turns out that transition state and activated complex should not technically be used interchangeably. There's a nuance, right? 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, all right? There's a fine point there that you should know. These are not identically or not interchangeable to say activated complex and transition state but today we're just going to use those two terms interchangeably. Okay. 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, all right? 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, okay? And so when we're done canceling these C0's, I end up with an extra factor of C0 on the numerator that ensures that the equilibrium constant is dimensionless, doesn't it? All right? 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, all right? That is the symbol for the standard concentration, one molar, for example, but I'm just going to call it C0, all right? 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. So when I write two pluses, I'm just indicating the transition state, okay? So 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 you can, if anybody knows a workaround for that, that would help me out a lot. If I, 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 and then I thought I can't do that. There's like 106 slides in this presentation. Okay. Now I said yes, yes, we could write this in terms of pressures if we want to, all right? Here's the concentration. There's 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, all right? And 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, all right? 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 a sort of an important thing that we left out. Here's what's there, all right? Here's some generic reaction, A moles of A, B moles 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, all right? So 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, M means molar, A is BC's A. What's that? That's Avogadro's number. That's not a misprint, all right? It's Avogadro's number in every single case. That is the difference in the zero-point energies between reactants and products. So if this is your generic reactant here, here are vibrational energy levels. Here are vibrational energy levels of the product, all right? That's the ground vibrational energy level. That's the ground vibrational energy level. That's delta RE0, all right? The difference between the ground, the zero-point energies of the reactants and the products. Okay, so here's our Gibbs free energy as a function of reaction coordinate, all right? The delta RE0 is closely related to this green quantity that I'm indicating here, all right? 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, all right? Because we're always talking about the zero-point energies. And we could say something about this equilibrium constant. We could calculate it using this equation right here. Now 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, all right? 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, all right? But what transition state theory says is these two guys are in equilibrium with one another. So what matters is this delta RE0 here, all right? This thing that we normally call the activation energy. It's the activation energy from the Arrhenius equation, all right? So we want to calculate that imaginary equilibrium constant, all right? Because these things may actually be in equilibrium, but this thing is not really observable, except using some exotic spectroscopic nanosecond, picosecond spectroscopy. Okay, so a normal equilibrium applied to this generic reaction right here, I could calculate the equilibrium constant using this equation. So 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. And so 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, okay? So I can calculate this equilibrium constant that applies to the formation of the transition state, all right? Now keep in mind, everything we're talking about here is kinetics, all right? We're sort of mixing thermodynamics with kinetics. They took a thermodynamic concept, the equilibrium constant, and applied that to transition state theory. Okay, so you're with me so far? We've got an equilibrium constant here. We can calculate it using statistical mechanics. Two step mechanism. Okay, that's the first thing to understand, all right? The second thing to understand is that this rate right here shown 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, right? The rate at which products are formed. That seems to be a statement of the obvious, all right? Obviously, as you cross over this barrier from reactants to products, the frequency that that happens, that's the rate of this reaction. I mean, it's totally obvious to say that. Okay, so the reaction rate, I can write as this frequency times whatever the concentration of this transition state is, all right? We're going to talk about this frequency, all right? It's the frequency with which if you think about this as being a molecule, all right, there's a vibration that has to happen here, all right? This guy moves back and forth between these two guys, all right? We get a symmetric vibration of this transition state. The frequency that characterizes that mode is the frequency that we care about. Now, your book includes something called the transmission coefficient, we're just going to assume that's one. See that kappa right there? Just forget about it, it's one. Okay, so 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, right? It's what we've been saying all along. So we can, here was our equilibrium constant expression for the transition state, and if I just solve for A, B in this expression right here, I get this and I can plug that into A, B there, all right? So I've just solved for the concentration of the transition state from here, okay, and now I'm going to plug that in to this equation right here and so I'm going to put all of this into here, and so there it is. There's the V, there's the frequency, the special frequency, here's all the rest of that, all right? That's our reaction rate. So 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 for as long as that's an elementary reaction, right? All right, and 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. All right, so 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. Okay, so here's our, so 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, okay? And so 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? 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. Okay, how do we calculate that transition state? I mean, how do we calculate that partition function? It's a transition state for goodness sakes, all right? What is it? This is the question that these guys wrestled with when they worked out transition state theory. All right, if we think about the vibrational partition function first, all right, it's the vibrational partition function that we really care about here, all right? Essentially, the transition state's undergoing a vibration. All right, how do I think about that? I mean, this bond is getting longer. This bond is getting shorter, all right? It's like an asymmetric stretch of the transition state. That's the mode that we care about. So if we could calculate the partition function for that vibrational mode, that's critical to understanding what the reaction rate is going to be. All right, so here's the generic expression for the vibrational partition function, right? For some mode that has a natural frequency or natural energy h nu and a natural frequency nu, all right? Now, 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? So tell me, these are weak bonds here, right? Do weak bonds have high frequencies or low frequencies? Low, thank you. All right, so 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, all right? Here's a picture from your chapter that I think is somewhat non-intuitive, all right? 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, all right? In other words, the frequency, you know, to go from here to here, that h nu is tiny, all right? 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, so 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 and when we do that, we just get kT over h nu, all right? 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, all right? Whatever that is for, whatever the transition state is, whatever mode the trans product bonds are getting formed, reacting 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, all right? This was the conceptual leap that Polany made. Okay, so 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, all right? Notice that there's going to be a rotational, an electronic, a translational, all right? The partition function contains many other manifolds and other vibrational modes as well for the molecule that are orthogonal, right, to this special mode, right? 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, okay? 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 there are things that are not perturbed in the molecule, right? What's perturbed in the transition state is product bonds are getting formed, reactant bonds are getting broken, all right? And we can describe that, we can roll that process all into up into one mode that has this characteristic frequency here. We're going to have to figure out what that frequency is, all right? But all the rest of this stuff, we can just roll in, right, 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, okay? 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, all right? 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 the double dagger on it, all right? And 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, all right, we don't have to measure it, all right? We get this expression right here and this is the famous erring equation, all right? We derived it going way too fast in about 20 minutes, all right? Here's, this is a very important equation. Now, why is it so important? How do we find K using K? How do we find that K using that K? Oh, that K, okay, 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. 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, okay, so here's the erroring equation. All right, I'm just substituting now for K double dagger here. All right, and here's the Arrhenius 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, all right, 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, right? How would you measure it? You'd have to have some exotic spectroscopic method to do that, okay, so the Arrhenius equation is a special case of the Erring equation, all right, and 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, all right, here are pre-exponential factors, you can calculate this one, actually it is calculated later in this lecture, all right, we can calculate these pre-exponential factors for the Errhenius equation knowing transition state theory so it's very powerful, okay, what can we do in the last 15 minutes, something very important, all right, we want to apply the same thinking to a reaction that occurs in water, okay, and 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, all right, lead chloride has some solubility in water, very low, all right, what rate would this reaction take, all right, 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, all right, if you don't think about this too hard, even if you think about it hard, you might come to that conclusion, all right, 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, all right, that turns out to be exactly wrong and it's important to understand why, all right, 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, all right, so 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, so 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, all right, 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, all right, but so this is a, 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, all right, so 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, right, 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, okay, so exactly, we can write 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, okay, so then we have to think back and remember something about activities, activities are going to be the key to understanding how this works, all right, the activity of some I in A is equal to its concentration times some activity coefficient, gamma sub A, all right, that's the activity of A, that's the activity coefficient, the activity coefficient for a neutral is one, all right, the activity coefficient only deviates from one as a consequence of Coulombic interactions with the solution, other ions in the solution, all right, if other ions are present in the solution, the activity coefficient will be less than one, the more ions, the lower the activity coefficient, okay, and there's something called the Debye Huckle 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, all right, 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, all right, 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, I and A, all right, because here's its charge, all right, so if it has a charge of one, there's a small deviation, charge of, all right, 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, all right, so activity effects have everything to do with charge, how many of you had 151, from me, okay, so this is review for all you guys, good, so when we write in 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, formic acid, acetic acid, acetic acid dissociates to give hydronium ion and acetate, all right, if I dump sodium chloride into the solution, what's going to happen to the pH? Sodium chloride's an inert salt, sodium chloride has no acidity or basicity of its own, all right, and yet the pH will change in a predictable way, what will happen, how many people think the pH is going to go down, solution'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, and you guys never had 151 from me. Now, if you add salt to this equilibrium, it will always shift to the right, it'll 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, all right, that's the most ionic state, if I add salt, the reaction will shift to the right, it's called salting in, all right, now I can prove that, all right, here's the equilibrium constant for this reaction, here are the activity coefficients, the activity coefficients for the neutrals are zero or one, rather, all right, the activity coefficient for this guy's one, the activity coefficient for this guy's one, these activity coefficients for the charged species are less than one, okay, and they will become lower as I increase the ionic strength of the solution, those guys will get smaller, okay, and so 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, all right, 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, all right, 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, all right, 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 and 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, all 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 will also alter the equilibrium in a predictable way. Okay, 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, all right? 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, so at the end of the day when we're done during 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, okay? That is the rate constant that applies for one molar or everything, that's why it's got a zero on it and that's the actual rate constant of the reaction, okay? So this is the most important slide that has to do with this second concept here. All right, 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, all right? And this is the ionic strength, all right? So the key point here is we want to understand how salt affects the reaction rate, all right, it's easy to understand how salt affects the reaction, 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've both got a plus 2 charge, the transition state has a charge of plus 4, all right? So 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's 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 of 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. Absolutely not. No. Nothing at all? Nothing to do when you add it to the voltage? 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 it, I'll say it, I'll say this one last time because I don't have any more time. There's an equilibrium between A and B, right, the reactants and the transition state, all right? 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, okay? So 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, right? 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, okay, and the reaction rate goes up. Look, it goes up here too. Look, it goes up here. Also, all right, 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's 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, all right? You're going to get it right, all 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. All right, and there's like 20 more slides that work through the equilibrium. We're not going to get to those. All right, so on Friday, we're going to work on the final. Please take the course evaluations, all right.