 and welcome to module 28 of Chemical Kinetics and Transition State Theory. Over the last several modules, we have now learnt the basics of transition state theory. We have looked at a transition state theory from 2 different perspectives and we have looked at numerical problems. And we are getting close to the end of transition state theory. Today, what I want to do is a more qualitative discussion. And the discussion I want to have today is to look very closely at all the assumptions that go into transition state theory, when do they hold and when do they do not hold. So, before doing that, let me just tell you what are the 5 assumptions we have made. We assume that the transition state exists. Second, we assume a classical treatment of nuclear specifically the reaction coordinate. We assume no recrossing that is the particles that are going from the reactants to the products. Once they cross the transition state, we will never return back. Transition state is in thermal equilibrium with the reactants. And finally, at a transition state geometry, we separate the Hamiltonian along the reaction coordinate and along all other coordinates ok. So, under these 5 assumptions, we derive this equation. So, we have provided you 2 proofs and today I just want to give you a quick recap of what those 2 proofs were. So, in the first proof, we start with the rate as integral over speeds along the reaction coordinate d into rho into u. This d was the density per unit length that is how many particles per length will be present at the transition state. And along the reaction coordinate that per length is along reaction coordinate, rho of u is the probability density of speed u. And finally, u is the speed which is nothing but the flux. So, we start with this formula for rate and we have already made 2 assumptions. One it is classical that is speed itself can is well defined concept. In quantum mechanics, speed it is differently defined. And second is no recrossing assumption that is we integrate over only positive speeds. So, we basically estimate this d and rho after that to calculate d. We first assume that this transition state exists and this transition state is in equilibrium with the reactants ok. So, that is the fourth assumption. And we can use then essentially some ideas from statistical mechanics to calculate this k equilibrium in terms of partition functions. And then we have made the final assumption that this transition state partition function is separable along the reaction coordinate and along all other coordinates. So, we write QTS not as QR into QTS dagger. And we calculate QR as this translational partition function. We then calculate d as this concentration of Ts over Lx. To get this quantity for d for rho, we make the same assumption as this one that we have thermal equilibrium. If we have thermal equilibrium, we can use Maxwell Boltzmann distribution for rho. And we basically put this value of rho here and this value of d here. And from that we integrate it formally and calculate the rate constant and show it comes out equal to this. Another way of deriving the same thing with the same set of assumptions. What we do is instead we do a integration over all phase space not only along the speed. So, well we again start with the same set of assumptions. So, we start making a classical assumption. So, we can write positions and momentum. And we have assumed here that the transition state exists or a dividing surface exists. So, we integrate over all coordinates along the dividing surface all coordinates and momentum. And so, one coordinate does not appear here dQ1. It is not here because we are integrating along the dividing surface and not along the reaction coordinate. d here is the probability density per unit length. And this is also a assumption of x comma p. Chi here is in general the transmission coefficient. So, transmission coefficient again tells you that if I am at that position Q comma p on the dividing surface, what is the probability I am reactive that is I start from the reactant and end in the product. And finally, this P1 over m is the plus. So, this formula by the way is extremely general. We have made only two assumptions in writing this rate formula classical assumption and transition state this dividing surface exists. Now, to get transition state rate or be able to simplify this integral, we first assume the thermal approximation. So, this one is thermal equilibrium and this one is the no re crossing assumption that is chi is 1 only P1 is greater than 0 and chi is 0 otherwise. And P1 again is the momentum along the reaction coordinate. So, if I have positive momentum, I will definitely be reactive. And finally, we assume the separability of Hamiltonian. So, we take this Hamiltonian here and this Hamiltonian we separate as h along Q1 which is nothing but P1 square over 2 m this P1 square over 2 m plus h transition state. So, with that we substitute all of this in here, we look at every integral very carefully and we can derive this equation again. So, no wonder we get the same equation because we have made the same set of approximations in both the derivations. But they provide you different perspectives. So, let us get back to these assumptions and think about what are the criterias when they hold and when they do not hold. So, the first assumption itself transition state well the first assumption is actually almost always true with a few exceptions. Few exceptions I can provide you are photo excited dynamics for example. So, these are problems where you have a potential energy surface you excited to an excited state. So, this is S naught this is S 1 you might have if you have ever studied spectroscopy you would have seen these kind of surfaces and then the trajectories come from to the top to the bottom like this. Here there is no barrier here this is not really have a transition state. And so, application of a regular transition state theory is very hard here a lot of assumptions we are going to break down. Another problem I can tell you are barrier less problems a simple example I can give you is H plus H going to H 2. So, if I draw the energy surface for this problem this is how it looks likes and your trajectories are starting here and it is falling here well it is barrier less again where is a transition state here it does not have one. So, these kind of problems application of transition state theory is very hard. But these problems are also not very common in general condensed phase chemical dynamics. The second one I want to discuss is separability of reaction coordinate that is H is equal to H along the reaction coordinate plus H along everything else. This is also an excel inter approximation almost always. So, again if I can draw a 2 dimensional energy surface this is something we have drawn earlier this is your reaction coordinate and this is your dimension perpendicular to the dividing surface. So, this separability is in classical mechanics at least it is almost always true, but you might have some very exotic quantum mechanical phenomena then this might not hold. So, we will not discuss get into those details right away, but fair enough to say that this is one of those assumptions that is almost always true. The assumption that are more problematic are the following one is quantum effects of bath. So, there are there can be many many quantum effects or by bath I used a wrong word here I should not use bath, but nuclear coordinates. There are many many ways this quantum effects appear. So, we are going to think of this let me just draw one dimensional surface like this this is my transition state first one is what is called tunneling. So, let us say my energy is this much only this is my energy of the system. In that case if classical mechanics holds then I am bound between these 2 points I cannot access the transition state ever not accessible if energy is less than E a where E a is equal to this much. So, classical mechanics will say that the rate constant at this energy is equal to exactly 0 you cannot go to the product side. However, quantum mechanics is just weird quantum mechanics tells you do have some probability of transmitting even at these energies that are less than E a. So, what it implies is that rate constant with tunneling included is always going to be greater than k t s t at least the one we have derived because that one treats it only classically. So, all the transmission that is happening at E less than E a are ignored second one is the 0 point energy effects. So, what this means is let me just redraw the surface this is your activation energy that we have been discussing. But as it turns out quantum mechanics says that the 0 energy is not equal to this but is equal to somewhere here. So, your true activation energy will be E a minus E a E not. So, your true activation energy can be different from the transition state activation energy. So, one has to be careful about that as well when applying transition state theory that is another effect that shows up. And the final thing where quantum effects can play a role is the Born-Oppenheimer approximation that is a technical word let me just write it Born-Oppenheimer. So, remember that what we are assuming is that our Hamiltonian is kinetic energy plus potential energy and potential energy is a function of all coordinates. That is what we have been working on throughout and these queues again are nuclear coordinates. Quantum mechanics again is just strange. Quantum mechanics tells that under some conditions you do not have only one potential, but multiple potential energy surfaces which are essentially the excited states. So, think of a molecule some reaction is happening you also have an excited state of the molecule. So, this is electronic ground state and this is the excited state and so far we have been discussing of dynamics only on one energy surface. But there can be effects of the excited states as well that is also not included in the transition state and the effects can be quite complex when excited state effects are also included. It is not straightforward to say what will be the effect and the model theories do try to look into including it, but we will not cover those within this half semester course. The fourth one is the most critical assumption, which is the no recrossing assumption. This is the one that gives you the most trouble. Imagine a potential energy surface just let us imagine our again 1D surface that we have been looking at that is my transition state. What we are assuming is that I start from here I go here and I will definitely go to the product, but there is a chance that I might turn back here. So, think of some kind of Brownian dynamics you are going through the transition state. Now, some other particle might collide with your reaction coordinate and might make it reverse and go back. So, in that case your rate constant is certainly going to decrease because these trajectories which I had assumed to be reactive might end up being not reactive. So, a trajectory that is doing this kind of dynamics is not a reactive trajectory at all and should not be counted towards rate. So, rate constant recrossings included will always be less than KTST. KTST is in upper bound. So, remember we wrote this formula essentially a more general formula which had this chi in there which is the transmission factor. So, if you want to correct for this recrossing assumption in your calculation this chi cannot be assumed to be 1. So, this assumption is what does not hold it breaks down. So, in practice what people often do is to rate at k equal to kappa into KTST kappa is called transmission factor and kappa is really an integral over this dividing surface dq dp rho equilibrium chi p1 over m divided by an integral over dividing surface dq dp rho equilibrium p1 over m. You can quickly verify that this is what kappa is the denominator is KTST and numerator is K and there are strategies on calculating this and we will look at a few strategies later on in this course after a few modules after we have looked at molecular dynamics. So, no recrossing assumption is always will increase the rate or if you include recrossings it will decrease the rate. So, kappa is a factor that is between 0 and by the way just an additional fact for you another way to that many people look at this transition state theory is to think about where to locate this dividing surface. So, the point is if recrossings is the main problematic assumption then what we do is we write this formula again and we look at this formula and say let us choose a different dividing surface and integrate over that and so the dividing surface at which k will be maximum is my best dividing surface because remember KTST is an upper limit to the actual rate. So, we try to find the dividing surface we move the dividing surface such that rate is maximized. So, that basically recrossings are minimal that is what it means and that is called variational transition state theory. It is very popular nowadays. So, just an additional fact and the final assumption that transition state is at thermal equilibrium. This also sometimes can run into trouble. So, let me once more draw my 1D energy surface like this. Let us imagine these you have these energies here what we are really assuming is that you have a thermal equilibrium of these energies at all times that is what it really means we just assume that the density is rho equilibrium there is no time dependence of the density right. So, what it basically is saying is that the thermal relaxation rate is much much faster than this vibrational frequency effectively vibrational frequency is basically telling you in some sense what is the frequency at which you are hitting the dividing surface. So, within that time let us say a particle is here it comes down like this it comes up like this and it let us say it goes to product within that time scale you will equilibrate. So, that the trajectories that are going away from the reactants get equilibrated. So, in other words this might be a bit confusing you have to think about it a little bit more. See what happens is that the trajectories at higher energy are the ones that are reactive. So, imagine I have some populace number of trajectories that are at this energy some number of trajectories here some number of trajectories here at excited state I have some number of trajectories. These trajectories are not reactive they cannot unless tunneling is included and let us say tunneling is not included for a moment let us say tunneling is not important then only these trajectories are the one that are reacting which means that trajectories at higher energies are being depleted. So, this trajectory goes to the product side now and I have a vacancy here vacancy so I lose thermal equilibrium by that I should have had one trajectory there but that went ahead and became a product. So, I am out of equilibrium now and if I am out of equilibrium essentially your rate will keep on going down because my trajectories at excited state are being depleted continuously. So, if not at equilibrium usually the rate will again be less than k TST why again because excited state energy is population is depleting and excited state populations is what contributes towards rate. So, what we are really saying in a transition state theory is this excited state as soon as a deplete immediately a thermal equilibrium is re-established the trajectories that are at the bottom will immediately jump that thermal equilibrium is so fast. So, that is the essence of thermal equilibrium because it is a very subtle point very fine point. So, I just want to show you a very beautiful plot shown by Kramer called Kramer's turnover. I have given you a reference here that you can look up it is a review. So, imagine I have 1D surface once more this has some frequency omega and I have some friction eta. This friction eta is essentially telling me how frequently my molecule is 1D is interacting with the vat. So, this will decide how quick my thermal relaxation is. So, let us imagine this eta I am sorry this is gamma. So, my x axis is gamma if gamma is low what will happen is that my equilibrium would not be established. Why because my excited state energies are depleting but I do not get enough energy for the ground lower state energies to replace these excited state population. So, your rate will be less than k TST in the limit when gamma is low that is what you see you get a line like this also when gamma is very large you again get rate that is very low. In this case the opposite effect happens at this point because you are colliding so frequently what happens is that a lot of recrossings happen and when gamma is low loss of thermal equilibrium. So, in both these two limits when thermal equilibrium is not there and when you have too many recrossings your rate will be always less than k TST and somewhere in between at some sweet spot you get a rate constant that is approximately k TST close to. So, in summary today we have look at the assumptions very very carefully. The quantum effects of the nuclei generally will have multiple effects it is complex a quantum tunneling will always give a rate that is more than k TST, 0 point energy will change the activation energy and Born-Oppenheimer approximation if that is failing then you can have very complex dynamics including recrossing will always lead to a rate constant less than the k TST. And finally, if non equilibrium if your equilibrium assumption is not valid at transition state then also you can run into trouble typically your rate constant will be less than k TST. Thank you very