 Welcome to module 40 of chemical kinetics in transition state theory. In the last two modules, we looked at a theory called RRK theory to look at rate constants calculated at constant energy throughout the course we have been looking at calculating rate constants at constant temperature. Now, we have been asking the question what if we have a different ensemble where the energy is constant not temperature. In the last two modules, we looked at a simple model which we call as the RRK model. Today, we will do extend that model and fill in the deficiencies of that very simple model. The model that I am discussing today is called RRKM model which is Rise-Ramperger-Casel and Marcus. So, what I am teaching today is there in Steinfeld-Francisco and Hayes in chapter 10.7. I have also given you the original reference for those who are extra interested. So, the history of this work is as follows. In late 1920s, 27 and 28, these three authors Rise-Ramperger and Casel formulated this theory for unimolecular decay. They were trying to understand if a molecule is decaying at a constant energy. How do I calculate the rate constant and match the experimental data? And so, we studied their efforts in the last two modules, but that actually does not work well and we discussed that as well. That is a very simple model with a lot of deficiencies. Enter Marcus. Marcus is truly a pioneer in chemical kinetics. This was, he was doing a post-doc with Rise actually of this RRK, the R of RRK and he was actually before that doing some experimental work. But Marcus was a genius and before like when he joined Rise as a post-doc, he started looking at papers for the last several decades. He just started reading and reading and reading. And finally, Rise and one more person, they gave him this problem. You know this several decades ago in 20s, there was this problem of unimolecular decay that never got resolved, you know. We do not know how to deal with it. Marcus, it took him basically only a few months and he solved this problem. And he published a few papers, one I have cited here in 52. He published a few more papers in 53 and that is it. After that Marcus changed his field and he was looking at electron transfer for practically a good portion of his life. But these few papers he published are authoritative works. They are the final word on this subject on unimolecular decay. And even today, you can open any physical chemistry journal. There will be at least one paper that calculates rate constant using this RRKM theory. So, let us study what Marcus had proposed. So, we will do a quick recap of what we covered in a long time ago in module 27. We wrote an expression to calculate rate constant at a given temperature earlier. We integrated over dividing surface over momenta. This is the density, this is the transmission factor that we have discussed for some time. And this is the finally, the flux. Transition state theory assumes this rho to be the thermal density, that is the Boltzmann density. And for chi, we put chi equal to 1 if P1 is greater than 0, otherwise chi equal to 0. We put a very simple expression for chi. And if we assume that we can derive transmission state theory, this is the expression we get out of it. Excellent. We are going to essentially follow this approach now. So, this was really the genius of Marcus. This since I am presenting in this fashion now I have told you the answer. It might look trivial, but it is not. When you have a flurry of ideas, what is the right way to think about it? What is the first equation to write? That is the question and that shows the geniusness. And Marcus said, so I will write the same expression. Perhaps I have to just use different values of this rho and chi. So, this rho earlier was for at a given temperature. So, now I must use a rho at a given energy. And what he said is, this is then equal to a delta function. H of x comma p is the Hamiltonian, which represents the energy. So, I am saying that rho at that given energy, this Hamiltonian must be equal to that energy. And chi we will start with the same as transmission state theory. Transition state theory after all is very successful. So, chi is equal to 1 if p1 is greater than 0 equal to 0 if p1 less than 0. So, that is going to be our starting point. This is the expression we want to simplify. So, we have changed a density from a temperature density to a energy density. I have made a mistake. So, let me correct my mistake. They should be normalized. Remember that even in the previous slide our rho t was normalized. We had this full integral there. So, we must write a similar integral as well here. That integral will now be dq dp delta of h of x comma p minus e. So, this delta is essentially constraining those x and p points for which the Hamiltonian takes a value e. So, the energy of the system is equal to whatever I want it to be. So, just sorry I had put in this nice animation that I had forgotten. But I have written what this rho is supposed to be. This is supposed to be e. Let me give you a little bit of properties of this Dirac delta function. This function here is called the Dirac delta function. So, this can get into very deep mathematical details. But I am not going to go into that. We are going to study only a few important properties of Dirac delta function. First is what if I integrate Dirac delta function multiplied by some other function. And let us say I am sorry let me write this as x naught and integral from a to b. This is equal to f of x naught if x naught lies between a and b equal to 0 otherwise. So, the Dirac delta function can be thought of as 0 everywhere. Sorry let me write delta of x minus x naught. And it takes a very, very large value at x naught. So, it is 0 everywhere except at this x naught where it is blowing up. So, one other property I can write is if I integrate over all space then I get f of x naught because well x naught has to lie between minus infinity to plus infinity. And one final property that we will use today is that Dirac delta function is even which would be obvious if I invert it I still get the same function. So, Dirac delta function is even. So, before we move forward actually this expression here is much harder to evaluate than what we had for constant temperature believe it or not. Because of this Dirac delta function it is not easy to deal with this because this is h couples all this q a and p. So, it is not easy for me to simplify the integral over dp1. Earlier we had e to the power of minus h that I could separate out the p1 term and integrate it out separately. So, it is a slightly more complex. So, I will have to introduce a little bit more language or few more terms added to my language. So, that I can simplify this expression a little bit more. So, I define two quantities for me now. One is called g of e which represents the density of states at energy e and w which represents the total number of states with energy less than e ok. So, if I am at a given energy e how many what is the density of states? How many states per unit energy are present at that energy that is g of e and w of e is the total number of states that will be there between 0 to that energy e. So, what I can do is start writing expressions for these g of e is then given by an integral over dx integral over dp and delta of h of x comma p minus e. So, I integrate over all space all possible configurations of x and p and I find for what values of x and p is my Hamiltonian equal to the energy. So, that will give me the density that will tell me how many states per unit energy are there that is what delta is doing ok. So, that you can think of as the definition of g if this is the density of states and I want to find a total number of states between 0 and e well that task is simple. I must integrate g between 0 and e g is telling me the density at that given energy. So, if I want go from 0 to e and integrate it out I will get the total number of states I write this and as a consequence I can also write g of e equal to dw over de. So, if I differentiate it dw over de you see that I will immediately get g of e ok alright. So, let us look at this expression of w little bit de prime g of e prime well let us put in the expression of g it is dx dp delta of h of x comma p minus e let me switch around the integral sorry little bit. Now look at this function what do we get this is equal to 1 if h of x comma p sorry this should be e prime is less than e equal to 0 otherwise agreed that is what this thing says. So, if h is less than e then I get 1 otherwise I get 0 well what does that mean imagine a graph of x comma p since I have a 2d screen I can only make 2d plots. I have imagine some contour lines or at given energy this is let us say energy e 1 contour line e 2 energy contour line e will have its own contour line I am sorry if it is not e 1 it is e. So, this function this function that is there is 1 if I am inside this contour f is 1 here ok where I am calling this whole thing as a agreed outside I have f equal to 0. So, all I am doing is integrating over all x and p which is completely covered by this contour line. So, this is nothing but the volume or the area whatever word you want to use of the contour equal to e. So, I look at this contour for h equal to e in a higher dimension you will get this high dimensional surface and I look at the volume enclosed ok. So, let me just say volume enclosed. So, that is the physical interpretation of W. W is the area or volume occupied by phase space and closed by this contour line alright. So, we will do one example we will calculate G and W for a 1d harmonic oscillator ok. So, my Hamiltonian I have only 1d actually I have made a mistake let me correct my mistake I had forgotten my factors of h remember this is an integral of x and p and an integral of x and p have dimensions. If you remember our discussion on partition functions I have h to the power of 3n that is what I had forgotten to write. So, let me correct that. So, this becomes a true density I do not have any factors of h floating around. So, we divide by h to the power of 3n just like what we did in partition functions ok. So, I apologize I had forgotten that let me correct all of these things here as well. I will have this divided by h to the power of 3n here as well h to the power of 3n here as well. So, volume enclosed divided by h to the power of 3n. So, volume in units of h to the power of 3n to that minor correction. So, the Hamiltonian for a harmonic oscillator is this. So, if I have to find w I have to find the area enclosed by the contour line. So, let us say this is equal to E. So, I have to draw the contour line for h is equal to E yeah and find the area enclosed by this contour line. So, you see this expression this is actually an equation of an ellipse. So, I will rewrite this slightly in this fashion. You notice let me get my energy also in here. So, let me correct that. So, let me put my energy also in the denominator because I have done nothing I have just reorganized it my equation here and make it look like a equation of an ellipse. So, an ellipse essentially looks like this where this is A and this is B. So, for me A is equal to 2 E root 2 E over m omega square and B is equal to root 2 Me. So, I know the formula for area of an ellipse. So, the area w of E will be nothing but area divided by h I have only one dimension here. So, h to the power of 1 is equal to pi A into B divided by h happily mass cancel. I get root 2 E into root 2 E which is 2 pi E divided by h omega well I will get E over h bar omega where h bar of course is h over 2 pi. You start seeing some good expressions do not you h bar omega is somehow related to harmonic oscillators quantum mechanically. So, let us move forward let us see what we get. Now g of E is nothing but d w over d that d E. So, this is equal to d over d E of E over h bar omega which is equal to 1 over h bar omega. So, the density of states is 1 over h bar omega just think of quantum mechanics. Quantum mechanically the energy is given as n plus half h bar omega right we get equal spacing like this quantum mechanically. So, what it means is even quantum mechanically what is the density of states I have one state per h bar omega energy that is what this figure really means if I look at one quanta of h bar omega I will have one state in it which is also what I get classical. So, that was a little aside. So, now let me finish the proof of transition state theory at constant energy now that we understand this g of n E and a little bit of Dirac delta functions ok. So, we start with the expression that we wrote earlier. Let me just clarify a few things from earlier integral over dividing surface b q is nothing but it is shorthand notation for b q 2 to d q 3 n h T s of q comma p is nothing but h of q 1 equal to q 1 dagger and all other variables are varying as they wish ok. So, this is just shorthand notation just to keep my life simple nothing more ok. So, first let us look at what we had in the denominator we had this you now we have defined this g as a density. So, we quickly notice this n is nothing but g of E into h to the power of 3 n ok. So, my k becomes 1 over h to the power of 3 n g of E into all this integral d q d p 1 d p 2 to d p 3 n and delta of h minus E into p 1 over what do we do next? We are actually going to use the same set of approximations as we have used for deriving the transition state theory at constant temperature. So, the next assumption I am going to make sorry is h T s is separable. So, this is equal to I will say is equal to p 1 square over 2 m plus v of q 1 dagger plus h prime of everything else. So, again that is also what we assumed when we were doing transition state theory that the same thing we do now all right. This is nothing but E a. So, what we get I get these constants out g r of E one thing I had forgotten to mention this n that I am calculating is only over the reactant Hamiltonian just like we did in transition state theory. So, I write for that only g r ok just to signify that I am integrating only over the reactant side and not on the product side ok. I have still this giant integral that takes me 100 minutes to write d p 2 integral of d p 3 n delta. Now h I will expand h prime as a function of all this minus E have we achieved are we really achieving anything let us see. So, we get this big expression now we will do a little bit more mathematical jugglery just a little bit of focus focus we will transform our variables now. I am going to define an E dagger as E minus p 1 square over 2 m minus E a ok focus focus abracadabra. So, d E dagger is equal to E is a constant remember I am calculating everything at that energy E minus p 1 over m d p 1 E a is also a constant. So, d a dagger equal to this. So, this is the one I am looking at you see p 1 over m d p 1 here I get the same thing here that is why I did the transformation. Now if p 1 is equal to 0 E dagger equal to E minus E a I am trying to find my limits ok. What is the maximum value E dagger can take? Note that I have a delta function sitting here right. So, I claim the maximum value of E dagger is the minimum value of E dagger is only 0. So, imagine I have my reaction coordinate I have E here this is E a this here is let us say p 1 square over 2 m this is then E dagger. So, E dagger is E total energy E minus E a minus p 1 square over 2 m. So, E dagger cannot be less than 0 ok why because I have this Dirac delta function sitting here h prime is positive. So, if E prime is less than 0 this Dirac delta function will immediately become 0. So, I will anyway get 0 ok. So, what I get here is then 1 over h to the power of 3 n g r of E dividing surface d q. So, this integral becomes E minus E a to 0 I notice this thing is minus d E dagger integral of d p 2 to d p 3 n that I am not touching into delta of h prime minus E dagger. I take this integral I play around with it a little bit I take this minus and invert the limits integral over dividing surface d q integral from d p 2 to d p 3 n delta of h prime minus. So, let us just look at this expression what is this equal to you note that let me erase a little bit stuff here out of space. So, I am just erasing. So, I notice that this integral is nothing but g of E ok. So, this is then equal to 1 over h to the power of 3 n g r of E integral from 0 to E minus E a d E dagger g of E dagger of E dagger. I have put a little bit dagger here sorry this is supposed to be g E prime it was something is not ok I am sorry I am getting confused myself happens with me. So, I will correct it all of it down I am and I have put a little dagger here to denote that this integral does not involve integral of q 1 and p 1. This is an integral from q 2 to q 3 n and p 2 to p 3 n ok look at this it does not involve integral of q 1 and p 1. So, that is why I have put a dagger here to remind you of that and I look at this integral and I see this. So, this is nothing but h to the power of 3 n I apologize again this I should multiply by h to the power of 3 n minus 1 because I have this factor here ok. So, I have this integral and this integral into h to the power of 3 n minus 1 will give me g. So, I get h to the power of 3 n minus 1 into w again dagger of E minus E a. So, that is my integration limit here ok. So, at the end I get w dagger E minus E a divided by h into g r this E minus E a because of this limit. So, whatever is the limit here comes here that is why I put E minus E a w dagger again denote that the integration is only from q 2 to q 3 n and p 2 to p 3 n ok and this factor of h 3 n minus 1 immersed because I had a division by h to the power of 3 n here. So, this is really the final expression for the micro canonical transition state theory now. This expression looks like w dagger of E minus E a divided by h into g r of E and the list of assumptions is very similar to what we had made earlier transition state exists. Nuclear have been treated classically we are integrating over q and p there is no recrossing. So, the chi that we choose is the same as before. What we when we write this delta function what it assumes is that all energy states are equally possible. So, I am not discriminating between two points in a space space having the same energy they are equally possible ok. So, that is called air body city it has profound consequences to thermodynamics and statistical mechanics I have listed it here and final assumption is that Hamiltonian is separable in transition state geometry. So, following the same set of assumptions we derive the expression for the transition state theory at constant energy. Thank you.