 Welcome to module 13 of chemical kinetics and transition state theory. We have discussed the collision theory quite deeply now over the last 7 or 8 modules and we are near the end of it. We are soon going to begin transition state theory. But before we do that I want to spend little bit of time on one very interesting topic which is what is called detailed balance or the idea of thermodynamic equilibrium ok. So, it is a very interesting comment. You can find a very beautiful paper written by Bruce Mahan in 1975 titled Microscopic Reversibility and Detailed Balance. What I am covering today in the next maybe 15-20 minutes is a much simpler proof than is provided in this paper as applicable to collision theory. But for those who are interested can go to this paper and read the full proof. It is very readable and you have enough information to be able to go through this actually ok. So, the more interested readers can look at this paper. So, what is it that I want today? What I am trying to calculate is not the rate constant today but equilibrium constant ok and something interesting pops out when we do that. So, let us say I have the wrong color of the pen somehow I will change the color. I have A plus B and let us say the reaction is reversible with C plus D. I have a forward rate, I have a backward rate ok. Well what is the big deal? I can calculate both forward rate and backward rate using collision theory. I have assumed biomolecular on both sides. If that is the case my forward rate is given by this equation that we have derived and used multiple times in the last few several modules. But I can also write a backward rate as R C plus R D square. I do not know what your A, D, I are they might have changed when the collision happens some mass might transfer from A to B and the rate I can change. Route 8 k T over pi mu C D e to the power of minus A and my activation energy can also change. I am keeping everything as general as I can and k equilibrium is k s over k B ok. So, this I just divide these 2. When I do that you see the pi cancels I have R A plus R B ok. I take in the square root I have a lot of common terms 8 k T over pi cancels and I get e to the power of minus what I will call as delta e naught over k T where your delta e naught is simply eaf minus eab. So, that is eventually the potential energy difference between reactant and products delta e naught ok. I have some equation can I play around with this equation a little bit more let us see. Remember what we are assuming here is Newton's laws and Newton's laws conserve what is called the angular momentum. So, we are going to calculate the forward angular momentum and the angular momentum of the products and equate those 2 ok. So, let us look how we do that. We have moving forward colliding with B. These 2 in general have different radiac. So, this is A it has moved from here to here. This distance is R A plus R B and let us say this angle is theta I am bad at drawing straight lines apparently, but this is a straight line all right. Angular momentum is defined to be R cross P R here. So, this thing is nothing but magnitude of R magnitude of P into sin theta where theta is the angle between R and P and I am finding the angular momentum at the point of collision ok. I can do it before as well it does not matter actually you are going to get the same answer angular momentum is after all a conserved quantity. So, R is R A plus R B ok. So, that is your R that is a distance between the 2 centers. P is a mass for mass remember I should use mu the reduced mass into u because u is the relative speed and for relative speed again we derived all of this in detail you get the relative mass and at the reactant side it is mu AB and let me call this as u AB. So, this is the forward angular momentum I can do the same thing in the backward side as well. So, what I mean by the backward side of right after the collision has happened the mass transfer has happened this has ok. So, this came like this as A here this is B right now and after the collision this goes like this and it becomes D this A gets reflected like this as C. So, if this is theta the reflection will be equal according to Newton's laws ok. So, the angular momentum right after collision ok when the mass transfer has happened my radii have changed. So, again in collision theory effectively we are assuming all this is happening instantly my I go from radii R A and R B to R C and R D and masses M A and N B to M C and M D ok. So, I can write the same kind of angular momentum now as R cross P which is mod R mod B and sin theta that angle is the same theta this R is now R C plus R D the distances have changed and P has become mu C D u C D. So, my speeds have changed my masses have changed my radii have changed. So, I calculate the new angular momentum that is alright. But angular momentum however you want to transfer masses whatever you do angular momentum is nonetheless a conserved quantity. So, I equate the two angular momentum that I have calculated this should have been A B this should have been C D. So, this theta terms cancels and I get mu A B R A plus R B u A B equal to mu C D R C plus R D u C D and as is common I am going to take essentially a thermal average here instead of u A B I will write the thermal speed of A B which is root this is what we do really in collision theory pi mu A B this is equal to mu C D R C plus R D sometimes will cancel 8 kT over pi 8 kT over pi and I will get R A plus R B divided by R C plus R D this is equal to you will notice I have a full mu here and a root mu I can simplify all of that and this will become equal to mu C D over mu A B a very powerful relation the radii have to be related they cannot be arbitrary because angular momentum is conserved. So, earlier I had derived k equilibrium to be this big equation R A plus R B square all this thing just a few slides ago and now we have found the relation between the radii. So, I am going to substitute this equation here so this will become equal to mu C D over mu A B and you notice I have a square root of mu C D over mu A B here so this becomes to the power of 3 half. So, according to collision theory your equilibrium constant is given by this it is actually independent of the radii if you tell me the masses if you tell me the reactants and you tell me the products and you tell me the difference between the potential energy between the reactants and products I can calculate you the equilibrium constant rather simple proof but very very powerful it goes somewhat deep and once we discuss transition state theory and partition functions this we will revisit them in the language of partition functions. So, there is more about this that we come to later one comment that I do want to make today itself is relation with Gibbs equation. So, Gibbs much earlier had written an equation that relates the equilibrium constant with a free energy ok and Vanthoff's equation actually comes out of this. So, let us relate I have a equilibrium constant from collision theory and a equilibrium constant from Gibbs equation well let us do a little bit of our thermodynamics. So, that is how delta G is defined in our case we have A plus B going to C plus D in this case what you can and this we will examine in more detail delta H naught you can basically convert to delta E naught. You have to be a bit careful factor for RT and PVE might be there so I will leave that to you but let us just ignore constant factors right now let us say this is delta E naught. So, if I put it in the Gibbs equation you will see this is equal to and if I compare these two equations I get something interesting I get delta S naught is R into ln of 3 half mu C D over mu AB ok. So, actually collision theory is also commenting on change of entropy. Simple theory just particles colliding and the power of thermodynamics is can come in and you can use all these equations in thermodynamics and comment a lot more ok. So, I will end with this today and again this in a few more modules once we have a better idea of transition state theory we will come back to this. In summary what we have done today is looked at driving the equilibrium constant from collision theory and commented on the change of entropy of a reaction based on collision theory ok. So, this is not we will see actually that this is there is a mistake here and transition state theory is going to correct that mistake. Thank you very much.