 Hello, and welcome to module 38 of Chemical Kinetics in Transition State Theory. In the remaining modules, we will switch gears a little bit and discuss the last leg of this course. What we will discuss now is calculating rate constant at constant energy. So, in a different ensemble, so far we have been looking at how to calculate rate constant at a given temperature. But let us say there is no bath, we have a constant energy simulation in state, constant energy system in state, how to calculate rate constants. We will start today with a rather simple treatment done by these three gentlemen Rice, Ramsberger and Kassel in 1920s. And today we will interestingly take a little detour of permutations and combinations. Alright, so what I am teaching today, you can find in this book by Steinfeld, Francisco and Hayes in chapter 11.5. I have also given to the more interested readers the very original papers that were given in 1920s, one by Rice and Ramsberger and the other by Kassel. Alright, so first let me be clear on what our goal is, we are trying to find rate at a given energy. So, again I like drawing 1D surfaces, so this is some kind of a potential energy surface that we have drawn many, many times. So far, we were working at a constant temperature, which means you have an ensemble of states. So, this was we were working at constant temperature. Now, we will fix one energy here. Let me just use a different color, so that this comes out well in color red. So, if I am at this given energy instead not at a temperature, how do I calculate rate constant, that will be the focus now. So, this ends is called a micro canonical ensemble, so far what we were discussing was the canonical ensemble which was at a given temperature. So, let us proceed, what I am presenting you today is somewhat instructional today and tomorrow. So, this looks at a very, very simple model, this model was developed by these 3 gentlemen Rice, Ramsperger and Kassel in 1927 and 28. So, let us go through this model and let us see what we learn from this. So, our model consists of just imagine S harmonic oscillators, so these S harmonic oscillators essentially refers to S, let us say vibrations in the molecule. So, S vibrations somewhere in the molecule and for simplicity we are going to assume all have the same frequency nu. The next point in the model is that we have one particular harmonic oscillator, let us call that one to be S1 that coordinate represents the reaction coordinate. So, therefore, if certain amount of energy gets deposited in this reaction coordinate the reaction will happen. So, I have total energy E, sorry I want the pen, so this is my mode S1, I am thinking of it as harmonic oscillator although I have drawn a little curve here. So, this is essentially representing Ea, so if I get energy in this one mode greater than Ea my reaction will happen ok. So, final thing we are going to assume is that there is fast energy relaxation among the S oscillators, the consequence of this assumption really is that we can assume everything is equally distributed ok. So, all modes have equal probability of getting the energy, it is just a matter of commutational combination. So, the rate of the reaction at a given energy let me make it absolutely clear, rate as the function of energy will be proportional to the probability that mode 1 has energy greater than Ea ok. So, this is our essential model, this is our starting point ok. This model I want to cover not because it is most accurate model, it is not very well used these days. We have a better model which we will discuss post this model, but my intention of covering this model is to give you a sense on how science works. We always start with the simplest possible model that we can get ok and this is a simple enough model and based on that simple model then we make predictions and predictions that can be experimentally verified and see how this model tests against experiments. If it tests well then our approximations are good, if it does not then we better get a better model alright. So, the rate is now proportional to probability that the mode 1 energy is greater than Ea ok. So, we are going to assume quantization of energy in harmonic oscillators. So, let us assume that the total energy E represents J quanta of energy ok. Let Ea represents M quanta of energy where M is of course less than J, so our question is what is the probability that if there are J quanta of total energy mode 1 gets M quanta or more that is what is proportional to the reaction rate. So, to calculate this we will need a little detour of permutations and combinations something that you have studied in your high school and I think I can easily say safely say that none of you would have ever thought that this permutations and combinations can possibly be used in chemistry, but here we are. So, what we learn you never know when it becomes useful. So, if your younger colleagues ask you why do they ever teach you permutations and combinations you tell them to calculate rate constants ok. So, we are going to assume a few principles from our 12th class knowledge of permutations and combinations I am not going to derive these points. If I give you n distinct objects and I ask you what is the number of ways of arranging n distinct objects that is n factorial ok. So, I am simply stating it this is not hard to prove, but we are assuming this to be true. Number of ways of arranging n objects out of which m are identical is given by n factorial over m factorial. In fact, I can generalize this that m1 and m2 are identical I have 2 separate groups of m1 objects that are identical and m2 groups that are identical. And what does what is the answer here n factorial divided by m1 factorial into m2 factorial yeah. And finally, number of ways of choosing m objects out of n objects is given by mcm which is n factorial divided by m factorial into n minus m factorial. So, we will start out by assuming these 3 points alright. So, our main question really is that we start out with some quanta of energy that I will be distributing in my modes. So, I will start with a related question that can be built up to answer that probability question we had earlier. What are the number of ways in which j identical balls can be put in s distinct containers. So, I have s distinct containers sorry and I have some number of balls with me j balls. How many ways can I put these balls in these containers? Each container can also have 0 balls 0 to as many balls as you want to put. But I want the total number of balls to put is j. So, this is a little puzzle for you if you like solving these puzzles you should pause this video. It is a very interesting way to solve these on very very creative way to solve this actually. We will solve this right now if you are interested in solving these kind of puzzles which are always fun pause the video and solve it for yourself because once you see the solution there is no turning back when you have seen the solution and that creative answer you will never be able to come on your own somebody told you that creative answer alright. So, we will start with an example sorry I over I will erase this thing. Let us start with an example and start building our intuition or how to solve it for a general problem. Let us assume I have 2 identical balls that I want to put in 3 containers. I have container 1, container 2, container 3 and I have 2 balls I want to put. So, let us see how many different ways I can put it. I can put both of the balls here and nothing here and nothing here. I can put nothing here like this my bad too many balls or I can put both the balls in the third box. Alternatively, I can put one ball here, one ball here nothing here, one ball here nothing here, one ball here nothing here. So, these are all the 6 possibilities there are only 6 ways I can do it there is no other way I can put these 2 balls in this 3 containers right and remember the 2 balls are identical. So, these are the only 6 possibilities. Now comes the creative part of the solution on how to solve this. I can write these 6 configurations as a set of balls and sticks ok, so just bear with me. So, I draw 2 balls, I have 2 balls with me, so I draw 2 balls here and I have 3 containers right. So, 3 containers means essentially I have to partition these 2 balls into 3 partitions ok. So, to partition it in 3 ways I will put 2 sticks ok, so let me draw this as these 2 sticks. When I draw a configuration like this, this is box 1, this here is box 2 and this here is box 3 ok. So, I have put 2 in box 1 nothing in box 2 in nothing in box 3 which represents configuration number 1. Second configuration is given by I draw 2 balls, first one has nothing, second one has 2 and third has nothing. So, nothing here, 2 here, nothing here ok. The third one is this configuration, nothing here, nothing here, 2 here. So, now a fourth configuration I have 1 ball, 1 stick, 1 ball, 1 stick. So, I have 1 here, 1 here, nothing here. Similarly, let me write for the others like this. So, at the end of the day I can write each configuration alternatively as an arrangement of 2 balls and 3 minus 1 sticks ok. So, if I want to generalize this, every arrangement of J balls in S containers is identical to arranging J balls and S minus 1 sticks ok. So, I have J balls with me here now and if I can put S minus 1 sticks in some fashion here, then this will correspond to some configuration and this configuration is unique. Each arrangement of J balls and S minus 1 sticks corresponds to a unique configuration and all possible configurations can be covered. So, all I have to do is to calculate the number of ways of arranging J balls, J identical balls. Let me make it absolutely clear, S minus 1 identical sticks. So, this is remember we had already told the answer for arranging N objects out of which M are identical. So, we have J plus S minus 1 objects right. I have J balls S minus 1 sticks. So, total of J plus S minus 1 divided by J are identical and S minus 1 are another group of identical objects. So, this is the number of ways in which J identical balls can be put in S distinct containers. It is a very clever trick alright. So, let us ask another puzzle to you. Given that we have found this number of ways of finding J balls in S containers. Now I asked you can you calculate my bad, can you calculate the number of ways in which a maximum of J balls can be put in S containers. So, so far we had told that you have to put all J balls into these S containers none of them have to be left out. Now, let us lift that restriction. Now I am asking you you can put 0 balls in this S containers, you are free to put only one ball among these S containers, you can put up to J balls in this S containers. So, how will you calculate that total? Again, if you enjoy solving these puzzles, this is your one chance to solve it because in a minute I am going to provide you with an answer and this is even more beautiful. This is a one line answer. If you strike the correct thought, the answer can be written in one line, very simple. You do not have to do anything complicated, there are very very complex ways of solving it, but there is one way of solving it which is one line answer. So, if you are interested, so pause the video and do it on your own, otherwise I am doing it now for you alright. So, the number of ways in which up to J balls can be put in S containers. Listen closely it is beautiful. So, I have S containers. Now, I have to put up to J balls right, so I have less than equal to J balls here. Let us consider one more container and put remaining balls here. So, I put let us say only J minus 3 balls here. So, I will put the third ball in the S plus 1th container and make it up to J. So, whatever is left over, I will put into this S plus 1th container. So, now you see the problem becomes this number of ways is the ways of putting J balls exactly J balls in S plus 1 containers. So, we will just use this formula with replacing S with S plus 1. So, this is nothing, but J plus S divided by J factorial S factorial ok, I told you it is a one line answer alright. So, we have worked a little bit of permutations and combinations today and now we want to apply it to this simple model that we have presented that the rate is proportional to the probability and we have total energy E which is corresponds to J quanta and the first mode if it has more than M quanta of energy then the reaction happens. So, we have to calculate what is the probability of distributing these J quanta among S oscillators such that one of the mode the first mode has more than or equal to M quanta ok. So, that we will do in the next module. So, today we have just presented to you the simple model that was developed by R, R and K. We usually do not call out their full names simply call their R, R, K. It has S oscillators all with the same frequency nu and the rate of reaction is assumed to be proportional to the probability that mode 1 has energy E greater than E a, I should put E 1 here and we had looked a took a little detour of permutations and combinations who nu permutations and combinations is actually useful in chemistry in calculating rate constants. But I hope to convince you that it is and what we derived today is that the number of ways of putting J balls in S containers is J plus S minus 1 factorial divided by J factorial into S minus 1 factorial and number of ways of putting up to J balls in S containers is given by this. Thank you very much.