 Hello and welcome to module 10 of Chemical Kinetics and Transition State Theory. So, what we are discussing in several last modules is the collision theory and we are trying to calculate a rate constant. In the last module what we had covered is the comparison of the rate constant calculated from collision theory versus Arrhenius equation. And we noticed one very important thing that the collision theory is missing the exponential factor itself that is the Holy Grail that is what we really really want that is what made Arrhenius that famous. So, today we are going to see how do we get that exponential factor in collision theory. This concept is basically called reactive cross section that we will be looking at today ok. So, quick recap the Arrhenius equation is k equal to e to the power of minus a over r t with collision theory what we have derived so far is this equation here. What you notice that this thing although is temperature dependent, but it does not have this exponential ok. So, this thing is be essentially your pre exponential this is actually the A of Arrhenius equation. So, what we notice that A is temperature dependent A depends as the square root of t and we are missing this e to the power of minus a over r t. What we will show today the reason we miss this exponential factor is because so far we have assumed that all collisions are reactive every time A and B will collide you will get a reaction. So, today we are going to go beyond that approximation. So, how do we account for non-reactive collisions that is the question. So, let us start ok. So, the way we do it is as follows. First so, think of this A moving forward with some velocity u colliding with B that is how we derived this whole collision theory right. And we had approximated u as the average thermal speed. Today we are going to do something more accurate. So, we will start with our basics we are going to write k of u equal to pi r A plus r B square into u. So, this is the equation we had derived 3 modules earlier that if you are moving at speed u what is your rate of collisions great. We are not going to write away right u equal to root 8 k t over pi m alright we are going to do something better. Before doing that though we are going to add an additional factor which is going to improve which is going to get our Arrhenius factor which is P r of u ok. So, this factor was rate of collisions at speed u and this thing is the probability of reaction if collision occurs at speed u. So, what we are saying is we are having these many collisions per second at this speed u good, but not all these collisions are reactive some of the collisions just A and B will collide and it will remain A and B nothing happens ok. So, we are attaching a probability we are saying that at any given speed u you have a probability of reacting. At the end though what we are interested in is rate at a given temperature. So, how do we calculate that this essentially is then a thermal average of k of u. So, what we do is ok. So, well the idea is you have a rho equilibrium of u this is the probability of the probability density to be more accurate of being at speed u and this is the reactive speed u ok. So, I am at a given speed there are two questions what is the probability that I am at given that given speed u multiplied by what is the probability that I will have a reaction what is the rate of having the reaction at that speed u. So, if I multiply that and integrate over all speeds I will get the net rate at that temperature t and again rho equilibrium of u we derived a few modules earlier is I am just reminding you we derived this is equal to this. So, we substitute this big equation here and we substitute this equation this is your k of u and this is your rho equilibrium of u. I have just written it down multiply them together and have to integrate from 0 to infinity. Always remember the limits of speed is 0 to infinity and not minus infinity to infinity that is a very very common mistake speeds are always positive there is no such thing as a negative speed in the language of mathematics it is always a absolute value ok. Now, to make progress I will just make a variable transformation this will be very useful. So, I will transform into energy units rather than speed and the reason is this probability that is here it is just much more natural to calculate in energy units hopefully that will become clear in a few more slides. So, just allow me a few more slides and hopefully this the reason for doing this variable transformation will become clear. So, I am going to make this transformation of E t equal to half mu u square I can of course find a differential and I get this. So, I am going to what I am it is a big integral. So, I will be very very careful check carefully if I am making mistakes or not ok your responsibility. So, I will take all the constants and put them outside pi r a plus r b whole square that is a simply a constant outside the integral this big giant mu over 2 pi k B t to the power 3 half I have a 4 pi as well let me put all these constants let me pulled outside pull outside the integral 0 to infinity ok. First I note d u equal to d E t let me actually note that I have a u here as well and let me combine u into d u I note I have a u into d u here. So, I note u into d u is d epsilon t over mu all the constants I have taken outside the next factor is p r and instead of u I will just write epsilon t ok. So, this is kind of a new function p r of epsilon t it is a mathematically not the same you understand I am what I am doing I am converting from u to epsilon t or the constants have been taken care of I have u square and u square is 2 epsilon t over mu into e to the power of minus beta and I have half mu u square which is nothing but epsilon t. So, I think I have got all the factors right let me just go over this I have all these constants here pi r a plus r b square comes here this mu over 2 pi k B t comes here and 4 pi comes here all the constants are out integral 0 to infinity u into d u becomes d E over mu I get p r of epsilon u square is 2 epsilon over mu that is here and e to the power of minus beta half mu u square is epsilon ok. So, I have got this equation I will just simplify this little bit more. So, what I will do is I will note that I have a few extra constants I will take these outside. So, I will get pi r a plus r b square let me just rewrite this slightly differently just to simplify into 4 pi. Now, I have a 2 outside divided by mu square 1 mu cancels here I have a 2 pi that cancels with this 2 pi. So, I am left with pi r a plus r b square 1 over k B t. Now, you notice I have a square root of mu here and the mu here that is 1 over square root of mu. So, that becomes a square root of and what I will do is take 4 inside. So, I will get 16 divided by 2 is 8 multiplied by this big integral. So, you can just double check with yourself whether the all the manipulations are right or not. So, after simplifying I get this. It is common actually to take this term here and multiply by this term and this is a very important quantity this is called the reactive cross section. So, I can actually rewrite the same equation here and write it in the language of reactive cross sections. And what exactly is this reactive cross section? Reactive cross section is the probability it is not the probability I am sorry it is the area that the particle B has to be in such that a reaction happens. So, I have a moving with some velocity u what is the area in which B has to be in. So, such that reaction happens. So, this reactive cross section area in which B has to be in such that reaction happens. So, far we have been assuming that area to be pi r a plus r b square that is what we derived earlier. So, we have been assuming P r is 1. So, now we are going to go ahead and make more sophisticated choices, but first let me convince you that if I choose P r to be simply 1 what happens I should get the old result back what I derived in the last module. So, let us prove that. So, let us assume P r of epsilon is always 1 for all epsilon. So, I have to solve this integral now P r is simply 1. So, this integral I have provided you here in this integral note that a is equal to 0. So, I just copy from this integral here I a is 0. So, e to the power of minus beta a is 1. So, I have 1 over beta into a is 0 plus 1 over beta which is 1 over beta square which is nothing, but KBT square. So, I take this KBT square and I substitute it in the above equation this integral. So, I have taken this and substitute KBT square and then I simplify a little bit and you notice that one of the KBT's will cancel and you notice that, but I have a square root KT here and a full KT here. So, I can write this as 8 KT over pi nu. So, you can go back a couple of last module and you will see this is exactly what we had derived. So, this convinces us that the choice P r equal to 1 is our old choice. So, let us make better choice now that choice really does not work because we do not get the exponential. Well, maybe one reasonable choice might be that this reactive probability is 0 up to certain energies that is the reason we went to this units of energy the transformation of variables. So, I am saying let us choose P r to be 0 if energy is low enough it is below some given energy epsilon or not which is my activation energy and 1 above it. So, if your energy is below this threshold it will not react if it is above this threshold it will react that is a natural choice. So, let us assume that is true. So, let us do this integral very quickly. So, this integral what I will do is to break into components epsilon t e to the power of minus beta e t plus epsilon not to infinity d e t P r e t e t and what you notice this integral is 0 because P r is 0 between 0 to epsilon not by my choice. So, I have to integrate this portion, but again I have this integral given here and here a becomes equal to epsilon not. So, this is simple this is e to the power of minus beta epsilon not over beta epsilon not plus 1 over beta. So, I take this and substitute here and so I get basically this integral now this equation. What you notice that this is now a much more complex equation we actually do not end up getting what we expected as a simple Arrhenius equation, but it is still progress because I do get Arrhenius factor here. But we I get something that is a bit awkward it does not looks right this is actually an experimentally also this is not right. But we have the reason is that we have made one mistake we had said that the probability is 1 if your epsilon is above some epsilon not. But this is not very physical because I have these collisions happening sometimes this collision might be a very grazing collision here. It is r is just it is coming and just touching here like this and some collisions might be a very head on collision. We have not actually accounted for that correctly imagine if you are at some energy energy basically translational energy basically specify speed. But at a given speed it is not so obvious what it is not 1 or 0 right because if this particle was coming head on then the probability will be much higher. And if it was just grazing through like this then the probability is very small. So, I have to account for that and so how do we do that we are going to do trigonometry. So, let us say I have to get my pen I keep on forgetting this is a this is b and I am moving with some speed u and it collides here at some angle theta. Let me say this distance is d and let me say this distance is a where d is nothing but r a plus r b and a is some parameter that we will soon find out it should vary from 0 to d 0 is head on d is grazing. So, the point is I want to find out u relative, u relative is the speed along this component that is the speed I care about that is the kinetic energy that will be used for reaction. The one that has the perpendicular component plays no role in reaction. So, that is that energy will not be used for reaction. So, u relative well you guys know a little bit of your trigonometry that is u into cosine theta. So, it is less. So, well what we have here is a triangle with this being d this being a and this being theta. So, this length is d square minus a square. So, cosine theta is nothing but root of d square minus a square over d. So, I get my energy that is useful is actually half mu u relative square which is nothing but half mu u square into d square minus a square by d square. So, this is my incident energy epsilon t into 1 minus a square by d square. So, I have this epsilon relative equal to epsilon t 1 minus d square over a square 1 minus a square over d square my apologies. And we want this relative velocity to be greater than epsilon naught. So, this relative velocity that I am putting in must suffice some activation energy for the reaction to happen that is a much more natural choice to make. So, this let me simplify I will have 1 minus should be greater than epsilon naught over epsilon t. I will just juggle around a little bit and I will get sorry a square to be less than d square into 1 minus epsilon naught over epsilon t. So, this is pretty easy to just manipulate around I hope you can do that. Now, reactive cross section is really pi a square pi a max square. So, the maximum a for which the reaction will happen, but that will be equal to then pi d square into 1 minus epsilon naught over epsilon t. But remember what was d was this and d is equal to r a plus r b. So, I have pi r a plus r b square. So, essentially what I get is that p r of epsilon t is 1 minus epsilon naught over epsilon t. Remember that sigma equal to pi r a plus r b square into p r of epsilon t that is what we proved two or three slides ago. So, we get now this equation this choice that p r is 1 minus epsilon naught over epsilon t if epsilon is greater than epsilon naught. And if it is less than epsilon naught then no reaction will happen. So, now we will do this integral 0 to infinity I better start with epsilon naught to infinity because 0 to e naught will be 0 just like we did in the last case last choice d epsilon t 1 minus epsilon naught over epsilon t epsilon t e to the power of minus beta epsilon t this becomes equal to two integrals integral over 1 I am just opening the bracket minus in the second one you will notice that this will cancel with this. So, I will have epsilon naught d epsilon t this integral we have already done and this is equal to basically I will again use this with a equal to epsilon naught I will have e to the power of minus beta epsilon naught over beta epsilon naught plus 1 over beta minus epsilon naught and this integral you can do comes out to be e to the power of minus beta epsilon naught divided by beta. So, go ahead and do this integral on your own it is not very hard. So, what I notice is I take this to be a constant and you notice this cancels. So, this becomes equal to e to the power of minus beta epsilon naught over beta square. So, I have just replaced that equation here and put it back into this integral to get this and so now you see what happens something very beautiful which is what our linear equation looks like. So, this is your Arrhenius factor and this is your pre exponent. So, in summary today we have looked at different reactive cross sections and defined a reactive cross section. Reactive cross section again is the area in which B has to be in such that the reaction will happen and if I do my math and calculate the reactive cross section and put a condition that the reaction will happen only if the energy is above e naught then I can derive the Arrhenius equation out given by this term here. In the next module we will see how we can use this to solve for problems actually. Thank you very much.