 Hello and welcome to module 29 of Chemical Kinetics and Transition State Theory ah. So, this is formally going to be the last module on transition state theory ah. We might discuss more problems later on, but the theory biases will be the last. And the final thing I want to discuss in transition state theory are two important questions. First one which is a very common question ah the activation energy that we talk about is it a difference of potential energy, enthalpy or free energy. Secondly, we are also going to look at today ah is there an alternate to calculating partition functions, is there some other way we can write the transition state theory ok. So, ah let us start with this question ah Arrhenius had originally written this formula well went off head and Arrhenius had interpreted this as ah some activated state. But what exactly is E a? Exactly ah. So, to answer that let us analyze our theory a little bit, but to do that I will need a few thermodynamic relations today. One that we had already derived that this is supposed to be k equilibrium I am sorry ah k equilibrium ah is ratio of partition functions into exponential of delta E over k t, where delta E is difference of potential energies, k equilibrium can be written as minus delta G naught over k t ok that is another relation from thermodynamics that we are not deriving we are just stating. So, if I just take derivative of ah l n k equilibrium with temperature I will get delta U naught over RT square ok. Then delta U naught is your internal energy ah sorry this is your internal energy. And finally, ah you perhaps already know ah the definition of delta G itself it is delta U naught the internal energy plus pressure into delta V minus T into delta S. So, these relations I will be needing today. So, let us look at our ah k TST relation that we have derived. Now, what I am going to say is this is equal to now this thing is nothing, but k equilibrium of the transition state, where this is nothing, but A plus B in equilibrium with transition state. Remember this is how we actually started. So, this has to be true ok ok, but k equilibrium is nothing, but e to the power of minus delta G naught over k T ah. So, this relation this activation energy is really a difference of free energy your entropy is important remember that you have a free of your enthalpy as well as entropy ok. So, that is the first question the free energy the energy difference is that of free energies not enthalpy not potential ok. But let us analyze this a little bit more. So, in the last slide we had written this k TST as k T over h into k equilibrium ok. Let us just play around with it let us take the ln of this this will be ln of this constant plus ln of T plus ln of k equilibrium. If I take a derivative of this with temperature well this is a constant. So, I do have the 0 here ln of T the derivative of that is 1 over temperature and the derivative of k equilibrium is delta U naught this again is dagger over k T square as in some slide you might see is R or in some slide you might see k B they are the same thing it is a just a matter of units ok. I try to use k B at all places, but if you see R somewhere do not get worried it is not a mistake really it is just a different dimension ok. So, let me just rewrite this as k T plus delta U naught over k T square, but if I compare this with the Arrhenius equation. So, if I do the same thing here and d ln k over d T the posthum will vanish and I will get is E a over k T square and if I compare these two what I get is E a is delta U naught dagger plus k T or delta U naught is E a minus k T. So, remember this fact actually internal energy is not exactly the same as potential energy internal energy also remember as average kinetic energy included in it and because of that average kinetic energy you get this factor of k T. So, E a and delta U naught are not same that is number 1 point ok, but well so what. So, k we had shown is equal to k T over h e to the power of minus delta G naught over k T, but delta G is given by delta U naught plus p delta v minus t delta s, but delta U naught in the last slide we showed is related to activation energy ok. So, delta U naught we showed is equal to E a minus k T. So, we are going to put that here E a minus k T I will minus k T plus p delta v naught minus t delta s naught over k T. So, I will just simplify this a little bit I will write this term first. So, you see I have k T over k T that gives me 1 and minus into minus gives me plus 1 then I will write e to the power of minus p delta v naught over k T and then I will write e to the power of minus plus delta s naught over k V and finally, e to the power of minus E a over k B T ok. So, let me just write this once more and simplify this a little bit. So, my k is k T over h e e to the power of minus p delta v over k T e to the power of plus delta s naught over k T into e to the power of minus E a over k T. Now, for gases even for liquid it can be partially true p delta v naught we will approximate as by ideal gas. So, p v equal to n k T ideal gas law. So, that is what we are going to do it is just a proved approximation really and. So, I will get is p delta v I will write as delta n naught k T and k T will cancel here. So, I will get is e to the power of minus delta naught dagger plus 1 into e to the power of delta s naught dagger over k T into e to the power of minus E a over k T. So, this is another common relation people often use to calculate rate constants. Here you do not have to calculate partition functions. Here you have to calculate a free energy this enthalpy I am sorry here you have to calculate this entropy difference activation energy and delta n. So, if you can calculate that somehow experimentally then you can calculate the k T s t. While partition functions require a bit more rigorous numerical calculation you have to be much much more careful with partition functions. It is just an alternate way I am not saying this is the better way, but a different way of calculating rate constant. And delta n naught again let me just define very clearly delta n naught is the change in moles between reactants sorry between transition state and reactants. So, for diatomic A plus B going to transition state your n here is 1, your n here is 2. So, delta n naught is 1 minus 2 equal to minus 1 ok. For mono atomic similarly delta n naught will be 0 because you have one reactant and one transition state one mole of each right. So, depending on what your how much reactant number of reactants are you can calculate this delta n naught. So, this is usually trivial delta n naught. The harder part is delta s naught and E a alright. So, today we have just briefly looked at thermal formulation of transition state theory. And we have looked at two important points one that transition state rate is really a free energy difference ok, the activation energy. And second we can write transition state rate in the language of thermal quantities delta s E a and delta n ok. So, next time we will solve a few problems and then move on to molecular dynamics. Thank you very much.