 to module 27 of chemical kinetics and transition state theory. This really is a continuation of the last module. What we discussed in the last module is a very general formula to calculate rate constants with minimal assumptions. Today we are going to take forward that formula and make a few more assumptions to get to the transition state rate. So, again what we are doing is a different derivation of transition state theory from a dynamical perspective. Earlier we had derived it from statistical mechanics. Now we are deriving it from direct dynamics as flux across a dividing surface or the number of trajectories passing per second across a surface that divides reactants and products that is what our rate is. Again the work I am following you can either find in Steinfeld-Francisco and Hayes chapter 10 or this excellent paper by Bruce Mahan in general chemical of education you can also find the same arguments there. So, in the last module we ended with this particular formula for rate constant. We had integrated over all coordinates that is not the reaction coordinate. So, we have one coordinate that we call the reaction coordinate. So, this is an integration over dividing. So, I am sitting at a transition state one coordinate is my reaction coordinate every other direction is perpendicular to this and constitutes my dividing surface. So, I am integrating over all this dividing surface this is the here is reactant here is product and this is the surface in between. I am integrating over all momenta d here tells me density transition state. So, I have used a little bit of a shorthand notation, but this is exactly q 1 equal to q 1 dagger that is my q 1 is set at the transition state and all other coordinates. So, this is my density per unit length along q 1. So, I have my q 1 here this is my dividing surface. So, I am looking at the density in this direction how many particles will fall into this little region is what we defined as transmission that is if I am at a transition state at some point with some speeds or momentum what is the probability that that point originated from the reactant and will end up in the product. So, we are only interested in the forward reaction of going from reactant to product and this chi tells me that this coordinate point q comma p is it coming from reactant in the past and will it end up in the product in the future. And finally, this u 1 equal to p 1 over m is nothing but the flux that is the technical language we use, but that basically tells me how many trajectories per second will go through. So, now let us think of calculating this d tilde. So, this again is the number of particles per unit length along q 1. So, now is when we will start making the little bit more assumptions here we are going to assume we have thermal equilibrium. So, at thermal equilibrium we have seen the equilibrium density. Equilibrium density is given by at any given point this big integral over 6 n dimensions. Now, I have to find the number of particles per unit length along q 1 only. So, let us be more clear what I mean I should have d q 1 to d q 3 n into d p 1 into d p 3 n that is really the probability density I should always have these infinite symbols with me. So, the number of particles per unit length along q 1 is nothing but rho e q over d q 1. So, this is what d tilde really is. So, I am looking at the direction of q 1 and finding the number of particles per unit length. So, I have to divide by d q 1. So, I eventually get and I set q 1 equal to q 1 dagger. So, again I am calculating this d tilde at the transition state and the transition state is specified when q 1 is equal to q 1 dagger when the reaction coordinate corresponds to the transition state geometry q 3 n p 1 to p 3 n or d q 1 disappears and I get d q 2 to d q 3 n d p 1 to d p 3 n divided by this big integral e to the power of minus beta h q comma p. We have made some progress. We have a sense of what d tilde is. Let us just look at this denominator for a moment. One important thing that I want to bring about this denominator is that this does not integrate over all coordinates. This is a hard concept to understand. Remember what we are doing? We are looking only for transitions that occur from reactant to product. To do that, we basically get rid of all population on the product side. We only have population in a reactants and we are looking at how many trajectories are becoming products. So, this integral that we have, we integrate over only coordinates corresponding to reactants. That is very important to understand that this integration here is not over all possible coordinates. All possible coordinates will include both reactant and product. So, we integrate over reactant configurations only because we are interested only in one sided flux. We are looking only from reactant to product forward rate only. To calculate forward rate, I do not want anything to be there in the product side. I want the entire thing to be normalized in the reactant side only. What I am going to do is write this thing as e to the power of minus beta h r, just a notation. So, this thing is exactly the same as this thing, just a notation, just a shorthand way of writing things. So, the next question I have to answer is how do I estimate this chi? Actually, that is a very hard question. We are going to make a very simple assumption and this hopefully will help you clarify what transition state theory really is. We assume chi is a big factor which depends on q 2 to q 3 n, p 1 to p 3 n. This is 1, if p 1 is greater than 0, this is equal to 0, if p 1 is less than 0. Let us make it equal to as well to be more general. So, this is the transition state theory approximation, one of the central approximations that is made in transition state theory. That is if p 1 is positive, p 1 again is the momentum along the reaction coordinate. So, this is my q 1, this is my dividing surface. If I have a momentum in the positive direction, then I am reactive, then I must have come from reactant in the past and I will end up in the product in the in the future, independent of whatever this q 2, q 3, q 3 n is or whatever this p 2, p 3, p 3 n is. I do not even care about what their values are. I only look at p 1, I see if p 1 is positive, you are travelling forward, then you will end up in the product. If p 1 is not less than 0, you are coming in the opposite direction, you are not allowed. So, that is the essence of transition state theory. So, we have our formulas for D and chi and all that is needed is to put into this big integral. So, let us see where that goes. So, this chi here will be 1 only if p 1 is greater than 0. So, otherwise it will be 0. So, I change the integral over p 1 from 0 to infinity because if p 1 is less than 0, chi will be equal to 0 and that will not contribute at all to the integral and then I can get rid of chi to be 1. If p 1 is greater than 0, then chi is 1. Now, what I will do is I will look at this edge and simplify. I will assume the separability of the momentum of the reaction coordinate. I am separating out just the momentum of the reaction coordinate only. I will also take out the constant energy EA. I will just specify what EA is, please give me a moment of q 1 dagger q 2 to q 3 n, p 2 to p 3 n. So, I have taken the momentum 1 out only, ok. EA is essentially the minimum of h of this. So, that is your activation energy really because transition state structure is the minimum structure along all directions perpendicular to the reaction coordinate. So, if I look at this edge where q 1 is fixed at a transition state geometry and I am varying all other geometries, then transition state is the minimum not maximum. This is a very common confusion. This is the minimum energy structure on this edge. So, EA is the energy of this transition state and that is your activation energy. EA is the energy of the transition state. So, if I do that this k, what I will do is just separate out the p 1 term e to the power of minus beta p 1 square over 2 m into p 1. So, I have taken this integral from e to the power of minus beta h. I have taken this term out and this p 1 over m over m I have separated out and I leave everything else as it was. So, that will be integral of d q 2 to d q 3 n d p 2 to d p 3 n e to the power of minus beta h T s with all the coordinates which I am not writing for simplicity divided by the n and n is integral over all coordinates corresponding to reactants only d p e to the power of minus beta h r. So, this integral basically means integrating over all coordinates corresponding to reactants and I should not forget e to the power of minus beta e a. So, e a is a number it is not dependent on any coordinates and so, I take that outside the integral. So, I arrive with this big formula I have written it out explicitly and once more in the denominator here in this integration this corresponds to reactant coordinates only. So, this is an integral that I can actually do this formula I have provided you here. I will look at this formula I will think of a as beta over 2 m ok. So, this will be 1 over m that I will take constant out into 1 over 2 a 1 over 2 into a a is beta over 2 m 2 2 cancels m m cancels. So, this integral is equal to 1 over beta which is nothing but k b t. So, that is an easy integral for you to do given that this formula is provided. So, I substitute that integration as k b t ok. Now, we realize that the partition functions really look like these integrals. So, the partition function in 3 n dimension is really defined as h to the power of 3 n 6 n sorry d q 1 to d q 3 n d p 1 to d p 3 n into e to the power of minus beta h ok. So, if I look at the denominator e to the power of minus beta h r this is nothing but h to the power of 3 n into q reactants q reactant really by is the partition function of reactants only ok. So, I am integrating only over the coordinate of the reactants and this h to the power of 3 n I have taken here. Similarly, the numerator h of the transition state h into 2 3 n minus 1 remember note that 1 integration is less here I have 3 n minus 1 coordinates only. So, once more transition state partition function is 1 dimension less compared to the reactants into q of transition state ok. So, if I put it here I will get h to the power of 3 n minus 1 q transition state divided by h to the power of 3 n q reactants, but you I cancel lot of h. So, I get k t over h q transition state over q reactants one final thing we have to be careful about this formula really looks very close to what we have derived earlier from statistical mechanics. One tiny detail that is missing what we have derived is per unit second. So, I am sitting at the dividing surface and simply counting the number of particles that went through in one second, but that is not what exactly rate constant is remember rate constant is the number of particle passing through per unit volume. So, the rate constant is frequency per unit volume, but that actually has a easy remedy all we do is find the partition functions per unit volume where my q naught is q over v. So, I find the reactant partition functions per unit volume I find the transition state partition functions per unit volume that is all that I have to do then to get to the rate constant per unit volume. So, we end up getting the very famous formula that we had derived earlier we get k t over h into q transition state divided by q r naught we have been writing it for bimolecular where it becomes q a naught into q b naught, but it to be honest it always has been q r naught we just has been looking reactants as a a plus b into the Arrhenius exponential, but now we have this derivation from a very different perspectives. Now, we have a much more intuitive way of deriving this which is that I have a dividing surface this dividing surface separates the reactant from the products. I look at all possible positions on this dividing surface and look at the positive flux across this dividing surface that is really what all rate is the number of particles passing per unit second per unit volume across this dividing surface. Thank you very much.