 Welcome to module 24 of Chemical Kinetics and Transition State Theory. Today we are going to cover a few basic properties of transition state theory. We have calculated the rate constant from transition state theory and we have even looked at numerical examples. So, today we will just play around with this rate and discover some interesting properties. So, let us start with one first property. So, this is the rate constant that we have derived from transition state theory. So, let us start with the first point which is the equilibrium constant. So, transition state theory is given here. Let us assume I have some reaction that is like this that is reversible and k equilibrium is kf over kb. So, I can calculate kf and kb both from transition state theory and take the division. So, kf will be equal to kbt over h q of transition state over qA naught qB naught e to the power of minus EA forward over kT. So, I am defining a 1D surface A plus B here, C plus D here and the transition state here. So, this is EA forward and I will also need this one. So, let me define it right away EA backward. So, this is kf divided by kbt over h. Note that a transition state is the same for both. So, I have the same qTS dagger qC naught qD naught e to the power of minus EA B over kbt. So, you note that a lot of terms will cancel kT over h cancels qTS cancels and what I am left with is what I will do is write qC into qD as a q products divided by q reactants. So, q reactants naught is qA naught into qB naught and q product naught is qC naught qD naught into e to the power of minus. What I will have is EAF minus EA back, but this is what is the exothermicity of the problem. So, EAF minus EAB is simply delta E. So, I get this for k equilibrium according to transition state theory, but this is exactly what you expect based on thermodynamics. So, this relation I derived in a few modules ago and we were discussing statistical mechanics. You can go back to it and so this relation is exactly right it matches thermodynamics perfectly. So, that is very good news we have actually made a tremendous progress then based on collision theory at least. If you go back to your collision theory we got wrong equilibrium constant in transition state theory now we get the correct equilibrium constant. The second thing I want to just discuss today rather interesting connection between transition state theory and collision theory. Let us just apply transition state theory to a special problem where the reactants are both atoms. So, I have A one atom B another atom they go to some reaction some products I do not even care about the product is. But the transition state let us assume is simply a linear structure that looks like this. It does not actually have to be, but that is a natural thing to imagine with some distance D. So, A and B are basically I am imagining coming close and the transition state is that A and B come close at some critical distance D which is the transition state. So, this is not really a bond by the way because transition state is a maximum energy structure along the reaction coordinate. So, I want to calculate the transition state rate for this problem. So, let us assume that A has some mass A this has some mass B. So, what all do I need to calculate? Well I need to calculate Q A naught Q B naught and Q transition state. So, let us do this. So, my reactants are atoms A and atom B. So, here is a question for you first. What all information do you need to gather to calculate partition function of atom A? Take a little pause think about it and then we will discuss it together. Hope that you have thought about it. It is an atom. Atom has only translational partition function and electronic partition function and atom cannot rotate and atom cannot vibrate. For rotation you need at least two atoms same for vibration. So, rotation or vibration is between nuclear never between a nuclear and the electron we do not consider electrons here. So, I just need the mass of A and the electronic degeneracy. What we are doing is a rather simple analysis today. So, we are going to assume that all the electronic degeneracy is simply 1. We are simply going to ignore it for simplicity. What we are doing is a simple minded analysis. So, my Q A naught is simply a translational partition function divided by volume which is nothing, but 2 pi m A k B T to the power of 3 half divided by h cube. I can do the same analysis for atom B. So, Q B naught correspondingly will be 2 pi m B instead of m A everything else remains the same divided by h cube. So, let us now think of the transition state. So, transition state I am assuming to be diatomic. It well it has to be diatomic it cannot be anything more. But I am assuming of this particular form where you have distance is B. So, in that case one information I can tell you from mechanics this can be calculated exactly the moment of inertia for this diatomic molecule is equal to mu D square. So, once more it is a diatomic and diatomic is by definition linear there is no way 2 atoms can be non-linear. So, the Q of the transition state will be the translational part. It will be the rotational part. What about vibrations? Will it have a vibration? Think about it once more it will not have a vibration why? I have 2 atoms 2 atoms linear. So, total number of vibrations for a linear molecule equal to 3 n minus 5 we have discussed it a few times. So, 3 into 2 minus 5 this is 1 well I have one vibration. So, what is going on? Transition state requires one less vibration than the total number of vibrations. The transition state vibrations that I have to include in my partition function is 1 minus 1 because the 1 vibration goes into the imaginary mode which is included in KT over H of transition state rate. So, go back to your proof if you do not understand this this is an important point for transition state we take one frequency less compared to the total number of frequencies. So, there is no vibrational part and the electronic part we are going to assume again to be 1. So, Q translational is Q translational divided by volume into Q rotational and this is of course for linear. So, I can quickly write this answer this is 2 pi. Now, what is the total mass of a transition state M A plus M B the total A has M A B is M B. So, total is M A plus M B KB T to the power of 3 half divided by H Q and this thing is 8 pi square I is mu D square this is rotational this is translational one thing I have just forgot to mention mu is the reduced mass that we have used before. So, mu is given by M A M B divided by M A plus M B. So, again this moment of inertia is actually very easy to derive, but I am not doing it here I am simply stating it. So, we have got all the components that I wanted I have got the translation or reactant partition functions and the transition state partition function. So, let us plug it all in here and see what happens. The transition state is 2 pi I have to be very very careful because I might make mistakes. So, you also look very carefully and see if I have making a mistake to the power of 3 half divided by H cube into 8 pi square KB T mu D square divided by H square whole thing divided by Q A naught which is sorry 2 pi M A KB T divided by H cube 2 pi M B KB T divided by H cube 3 half e to the power of minus E A over KB T. You will see that some terms will cancel for us 1st thing I have written something wrong. So, let me correct it the H is simply cube the bracket should be only here. So, yeah that you can quickly verify that what I have written was wrong. So, H cube cancels with this H cube this H into H square is H cube. So, this also cancels with this H cube. One of the 2 pi KB T cancels they all have both have power of 3 half. So, let me write what I am left with very very carefully I am left with KB T I am left with M A plus M B divided by M A M B to the power of 3 half. So, I have a M A and M B here that I have taken out. What I will do is I will write this as 1 over 2 pi KB T into 1 over square root of 2 pi KB T you will soon see why it will just help me in I will delete that arrow it is going to come in between 8 pi square KB T mu D square into the exponential. I think I have all the term I have KT here I have M A plus M B divided by M A into M B I have 2 pi KB T to the power of 3 half here I have 8 pi square KT mu D square everything here. So, let us cancel more terms KT cancels with this KT 1 of the pi is cancel 2 cancels with 8 to give you 4 excellent. Now, I note this term is nothing but 1 over mu. So, this is 1 over mu into root mu M A M B over M A plus M B is mu. So, this is 1 over mu to the power of 3 half. So, I write that as mu into root mu. Let me just extend it 2 pi KB T into 4 pi KB T mu D square 1 of the mu's will also cancel. I will just take this term and I will reorganize in a particular fashion I will note that I have a KT in the numerator and a root KT in the denominator. So, I will write that as root KT in the numerator I will take 4 and write that as square root of 16 and 16 by 2 is 8. So, I will write that as 8 over and I will take the pi as well in the numerator. So, I have 8 pi KT divided by mu into actually let me not take the pi out I am sorry I will leave the pi in there and I will write this as pi D square you will see why if you do not already see it if you have been following the modules you will see you will recognize this form very carefully. So, you can verify that this form in this bracket here is exactly the same just rearranged do you recall this expression from some time before some memory stirs. So, this expression is exactly the same as we derived for coalition theory. It was equal to pi this D was replaced by R A plus R B but otherwise this was exactly the same. So, what we have got is we have derived coalition theory result from transition state theory. So, transition state theory is a more general theory than coalition theory. So, transition state theory is basically always going to get you better results than coalition theory because it is more general transition coalition theory is a special case of transition state theory. So, essentially coalition theory is a special case for transition state theory applied to two atoms reacting together with a linear transition state that is really what coalition theory is. And we can derive this coalition theory from the way how we derived from the actual mechanics of two heart spheres colliding but you will get the same result out the the fundamentals is the same. So, you can think in the language of partition functions and get the same answer out. So, in effect coalition theory is really missing all rotations and vibrations of my transition state of my reactants and transition state I am sorry. And that is what transition state theory is doing it is looking deeper into the structure of my reactants and transition state and adding appropriate amount of vibrations and rotations as should be added to get you a better rate. So, in summary today we have discussed two points we very briefly looked at the equilibrium constant derived from transition state theory and shown that it is consistent with what you expect from thermodynamics. And secondly, we have looked at the relation between transition state theory and coalition theory. Coalition theory is a special case of transition state theory applied to two atoms colliding together to give a linear transition state. Thank you very much.