 welcome to module 25 of Chemical Kinetics and Transition State Theory. So, we have been playing around with transition state theory for some time we looked at its derivation from statistical mechanics, we solved numerical problems and we analyzed its relation with collision theory in the last module. What I want to do today and in the next module is look at the derivation from a very different perspective which is a dynamical proof in phase space. This is not usually covered in many courses or in many books. So, this proof is not there in Laidler's book I will provide you with appropriate reference in the next module. But today I just want to build towards this dynamical proof I want to provide you an intuition. So, the statistical mechanics proof is perfectly correct, but it does not provide you intuition particularly it does not have the intuition of dynamics in it. It is also very static, it is they come from the language of partition functions and equilibrium constant and it gets you the final answer. But you know what we are studying is dynamics we are studying kinetics. So, the motion of atoms and molecules have to be present somewhere and we will show that we can derive the transition state theory result from looking at this perspective as well. So, today let us build this intuition we will have today we will not stick to very mathematical rigor we will be doing hand wave proofs today in one dimension just to understand just to build an intuition ok. So, let us start simple ok. So, let us first let me build a model for you what I want to think about today. So, let us imagine some reaction I have let us imagine I have some A H H is hydrogen and A is some hydrogen or some reactant and I have a proton acceptor a hydrogen acceptor B. So, I have a reaction that is happening here where H let us say transfers from A to B this happens actually inside our body all the time the hydrogen transfer proton transference whole. And let us say I have some kind of a transition state in between where H is in between neither connected to A neither connected to B, but somewhere in between ok. So, this is my reactant this is transition state and this is product. Well of course, it is a complex system A and B can be some large molecules you can have solvent surrounding all of this today I just want to keep it simple. I want to look at one dimension only which is let us say this distance. So, this distance increases here and distance is much larger for product and I want to think of this distance as my reaction coordinate and this is the only distance that exists or the coordinates of within A or AB distance just for simplicity I will not consider today. In the next module we will think about how to include those as well. So, if I let us say want to draw a potential energy with respect to this Q well when Q is small corresponding to this one when Q corresponds to this reactant distance you know I will have some kind of a minima which basically says that A H is forming a bond. So, whenever you have a bond you have a minima in energy surface ok. The transition state distance well transition state by definition should have a maxima ok along the reaction coordinate I have only one coordinate which is the reaction coordinate and when Q is very large you will have another minima. So, this is my reactants this is the transition state this is the product. I simply want to understand transition state theory for this simple one dimensional model today ok. So, nothing complex ok. Let me before I do transition state theory let us think of a simple minded back of the envelope estimate based on our intuition of dynamics ok. So, let me draw this energy surface once more this is reactants transition state product ok. I do think that this thing at the minima reactants has some frequency omega the spring constant omega is related to the spring constant I am sorry ok. So, if I want to calculate the rate constant of this problem at some temperature how do I calculate it? Well the simple thing is let us say I have my trajectory here that is moving on this what I am going to do is think of different trajectories at different energies. So, my trajectory number 1 is at energy E 1 I calculate the rate for this trajectory at E 1 I think of another trajectory at energy E 2 and I calculate that rate and I basically find a average. More mathematically what I am doing is I am integrating over all possible energies I have to find the probability density of being at that energy multiplied by the rate at that energy. So, I will call this k of E just so that it is easier for me to write. So, what I have is rho of E d E is the probability being energy E and this k of E is my rate constant if trajectory is ok. So, I have to calculate this rho and I have to calculate this k and I have to integrate. What should be the integration limits? It should be 0 to infinite energy is always positive because you always have kinetic energy which is positive and I am assuming this is E equal to 0 and this is E activation energy. So, I am integrating for all positive energies here. So, let us calculate these quantities. So, we will start with this rho of E d E. This is nothing but the Boltzmann distribution. So, rho of E d E is the my Boltzmann distribution into some normalization constant because the probability that I am at energy E is simply given by this. The only trick remains in calculating this N. So, I know that my probability should be normalized. So, if I integrate the probability density from 0 to infinity I should get 1. So, N into 0 to infinity E to the power of minus beta E d E must be 1. So, N I integrate this from 0 to infinity should be 1. I solve for this I get N equal to beta. You can quickly verify this I have skipped 1 or 2 steps here. So, I get rho of E d E equal to beta E to the power of minus beta E d E. I do not know why I have put this thing here. So, let me remove that. Now, let us calculate this k of E. So, how do I calculate k of E? That is a harder question. So, we are going to again everything is simple minded. We are going to make simple minded assumptions here. So, remember a transition or a surface looks like this and the frequency here is omega. What I am going to do is as follows. I will imagine this to be a harmonic oscillator like this. So, this I have converted into a harmonic oscillator with frequency omega and what I am doing is I am sitting at the transition state and I am asking the question at energy E what is the rate of moving forward. So, if you are at energy E what is the rate constant at which you will hit this energy surface with a positive velocity. That is what is transition state theory. It only looks at positive velocity. So, this thing is rate. So, let me be more precise number of crossings transition state at energy E with positive velocity. We do not include negative velocities in transition state theory. That is an extremely important fact to realize and of course, it is an approximation. So, how do I estimate this? So, the first thing I realize is if E is less than E a if my energy is below this well then I will not hit the reaction coordinate. I will not hit the transition state I am sorry. So, if my energy is here I will oscillate between this point and this point. I will have no possibility of reaching transition state. So, k of E will be 0. The number of crossings at transition state is equal to 0. If E is greater than E a that is when I will invoke this harmonic approximation. I will think of this problem as completely this parabola. This is my transition state. So, per second what is the frequency of I will hit this transition state given that I have high enough energy now. So, k of E then will be omega over 2 pi. So, omega over 2 pi is the frequency of a harmonic oscillator. So, each point in one time period is covered twice. Remember that I go from here in one time period I do this motion and I return back, but I am looking only the crossing with positive velocity. Therefore, I ignore the negative portion and so, k of E is simply omega over 2 pi. If I included the negative one as well then I would have to double it, but those are ignored in transition state. So, I have a k of E now with me. So, k of T we argued is given by this integral here. We argued p of E to be the thermal equilibrium p of E and k of E we have argued to be this again a simple minded model. So, I get k of T I just substitute things here D E, p of E is beta into e to the power of minus beta E into k of E. So, k of E will be 0. So, I will take this integral and take it into two parts and note that the k of E here is 0. So, this is 0, this portion from 0 to E m. And so, I will get this one where k of E is simply omega over 2 pi. So, this is an integral with omega over 2 pi that I take outside the integral E A to infinity D E beta e to the power of minus beta E. So, this is an easy integral to do again. So, let us do it. This I integrate as from E A to infinity. So, this will be equal to omega over 2 pi into beta will cancel with beta here. And you will get is e to the power of minus beta E A after you put in the limits this you can easily verify. So, this is what I have got for k of T a simple minded result. I have a surface that looks like this I have a frequency omega here I have activation energy. So, actually this result is very physical omega over 2 pi is my flux. So, that is the rate constant that I will traverse the reaction this transition state. This thing is probability that energy is greater than E A for 1 D only. So, I multiply by the flux with the probability that my energy is greater than E A. It is as simple as that and that is my rate constant. Let us get back to transition state theory which gives you a much more sophisticated rate formula. This looks much more involved is it at all somehow related to the formula that we have just written. So, we are going to think about it. Here I have only one reactant this A I have my reactant is simply this one structure A H B. So, it is governed only by this Q and my transition state is once more some structure that looks like this. So, we are going to make very rough approximations for that partition functions. We have to calculate this Q A naught and Q T S. So, let us make a rough estimate of this Q A naught and Q T S. Q A naught well I will have a very simple minded model. So, I will not include any only I will include only one vibration. I have some translational I have some rotational of the entire molecule and I will include only one vibrational only 1 D. All other vibrations I am for simplicity I will ignore just to be able to correlate with my previous simple minded answer. Q T S I will actually the translational part I will assume is equal. I will leave that is a very good approximation because the mass will not change in this simple minded structure. More importantly even the rotational part we will assume is equal and the vibrational part we will assume is 1. So, that 1 vibrational coordinate that was there is my reaction coordinate and so there is no vibrational component to transition state. Remember that transition state always has 1 vibration less compared to the total number of vibrations. I am considering only here 1 vibration. So, the total number of vibrations becomes 0. Now, Q vibration in 1 D classically is a quantum mechanically is given by 1 minus beta h bar omega. This we can go back and see this we derive this and so we are going to make an approximation now. We are going to assume high temperature. So, this will become 1 minus 1 minus beta h bar omega in high temperature limit. So, this is equal to K T over h bar omega which is the same as the classical answer. So, we are going to replace this by K T over h bar omega. So, I have got Q A naught as some Q translational into Q rotation into K T over h bar omega. Q transition state is the same as Q T R naught into Q R. So, K T S T if I divide the two first of all I have K T over h and then if I divide the two this will cancel and this will cancel if I am dividing them and I will be left with h bar omega over K B T. So, Q A naught is in the denominator. So, I invert it into e to the power of minus E A over K T. Now, you notice something beautiful happens and I get h bar divided by h which is nothing but 1 over 2 pi notice something beautiful you get the same simple minded answer. So, the idea is this thing we derived remember how we calculated this. We assume that the rotational component is the same we also assume one dimensional simple answer and that answer we are treating harmonically. So, the vibration is treated harmonically that is a simple harmonic oscillator and classically high temperature limit. So, in that limit our simple 1 D estimate back of the envelope estimate basically works and this is actually a very powerful estimate you know this is your somebody who throws a reaction at you and of the top of your head you have to come up with a number on what the rate constant is going to be. This is what you do actually you think on what is the reaction coordinate at the transition at the reactants you think of what that frequency is you estimate the activation energy somehow and you calculate this and you will get a rate constant. It is a very simple minded, but nonetheless like it is a starting point you can say calculating this involves more calculations and it will be of course, more accurate. So, I want to stop here and today what we have done really is we have calculated a 1 D rate as a flux across this dividing surface. What we have done is we have we have put a marker on my transition state and I was looking at the flux of trajectories moving in the positive direction at the transition state and from that we calculated a rate constant as omega over 2 pi to the power of minus beta ea and that we saw actually matches in some limit with the transition state answer we have already derived. So, we will take this idea forward in the next module we will make it much more precise. Today was hazy, today was I waved my hands next time I am going to be more accurate and put everything in correct perspective in the dynamics in phase space and derive the transition state rate again from this idea of a flux at the transition state surface. Thank you very much.