 welcome to module 9 of chemical kinetics and transition state theory. So, this is the third installment of kinetic theory of collisions. In the last two installments we had looked at the basic idea which once again is assuming our reactants looks like hard spheres and they are colliding with each other and we calculate the rate of the reaction is nothing but the rate of these collisions ok. So, last module we tried to figure out this thermal speed u as essentially the relative speed u and finding a thermal average and we did that by doing the center of mass transformation and all. Once we have a rate constant, we are going to analyze it a little bit, we are going to calculate the rate constant out of the rate, we are going to look at a spatial case when a is equal to b when both reactants are identical and and we will end with a comparison with the famous Arrhenius equation. So, again the resources are the chapter 4 of the Laidler's book or you can look at this link by chemlibor.dex.org. Again a quick recap, we note that this kinetic theory of collisions is valid only and only for bimolecular reactions. We identified the thermal speed as this u 8 kT over pi mu and this mu entered as a reduced mass ma mb over ma plus mb and we had gotten this final rate in the last module pi r a plus r b square root 8 kT over pi mu into na into nb. So, first let us simplify and calculate the rate constant. So, what is the difference between rate and rate constant? So, again let us write the reaction clearly a plus b going to products and remember we are assuming this is elementary, this is one step nothing more is happening, a is coming and b is coming and they are colliding and giving these products. So, my rate I can write essentially as k na nb, you might be wondering why I am not using concentration of n to it is really the same thing, it is a matter of change of units. Remember what is na? Na is nothing but the particle density which is the number of a molecules or atoms divided by the volume. Let me write volume clearly and similarly b is defined as the number density of b. Well that is effectively the same thing as concentration. So, as long as we are talking in the language of na nb, rate will become equal to k into na into nb. So, you can go back to the first module we discussed and go back to the definition of rate which is d of extent of reaction over dt and you can identify that to be the same as this. It is really the same thing it is just a transformation. So, then you notice that the rate I have calculated is this. So, if I compare these two, I can easily get if I have got this rate constant and of course, the rate is k na nb. I want to spend just a few minutes and just think of dimensions or units. Just make sure that our final result makes sense. There is no catastrophe that has happened in my whole proof. Everything is still logical, everything is still sensible. That is always a good check. Anytime you have derived something make sure that the end result is something that you can verify. I mean something that makes sense and one of the most important thing that we often do is dimensional analysis. So, what is the first of all the unit of rate? So, again my reaction is a plus b going to some products and rate is minus d na by dt. Again you could have written concentration of a by dt that does not matter. It effectively is the same thing, it is just a transformation. If you write concentration of a then in our definitions here and here as well this will be transformed to concentration. So, the unit of this, what is the unit of na then the units? The units is nothing but number of particles divided by volume. So, the unit of rate is nothing but d na over dt. So, number of particles divided by volume into time, time comes from dt. So, na is number of particles by volume and te is time. So far so good. So, k is rate divided by na nb. So, this will be number of particles divided by volume into time divided by number of particles square divided by volume square. So, if you simplify this, this will be equal to volume divided by number of particles into time. So, let us see the rate constant that I have got here. How does that look like? So, our k has the dimensions of pi does not have any dimension. So, I am looking at this equation now. Ra plus rb square that has dimensions of length square and this thing is nothing but speed and speed is length divided by time. So, I have length q which is nothing but volume by time. So, the point is that this rate constant should be understood as eventually divided by the number of particles. So, we compare this with this. The final thing I the next thing I want to compare is a somewhat special case if a is equal to b. So, I have 2 a going to some products and so a is reacting with itself. You might be a bit surprised perhaps, but actually we have seen such an example when we were discussing the isomerization of cyclohexane. Sorry, isomerization of cyclopropane 2 propene. Then we discuss this 2 a. The question is can I write k as the same pi of ra plus ra square into root 8 kT over pi mu. Actually the answer is no, there is a little bit of a trick here. So, let us go back to the basics. What are we calculating? We are calculating the average number of collisions happening per unit time per unit volume. That is a definition of rate number collisions per unit time per unit volume. So, imagine you have this big box, you have a here. Let us say you have a lot of a's and you have a lot of b's. So, I am again going over back to the case when a and b are distinct. So, what I am really doing is I look at one particular a. I sit on a1 and I wait for time dT, delta T whatever and I find the number of collisions a1 had. So, this essentially is an average quantity. Remember all statistical mechanics is average. So, this we are calculating as collisions of a1 in time delta T. Well, plus collisions of a2 in time delta T plus dot dot dot till the collisions of all an divided by total number of a. So, that is why we used na as a probability density. So, yeah this is again I am going back to the very basics. Now, let us do the same thing. If a is equal to b, you will notice something very interesting will happen now. I had a1, a2, a3 and b is the same. I have a4, a5, a6, yeah and I am looking at the collisions between a and a. So, when I do this averaging number of collisions of a1 in time dT plus number of collisions of a2 in time delta T plus so on divided by total number of a. What you notice is that I am counting the collisions twice because a1 might have collided with a2. That collision will appear here and the same collision will appear here. So, every collision I will count twice if I am considering collisions between same entities yeah. So, this rate will be half of total number of collisions per unit time per unit volume because why did this half appear each collision has been. So, spend some time on it think about it a little bit and you will immediately see that this each collision is counted twice. So, in effect my rate will be half of pi or 2 rA square it was rA plus rB and rB equal to rA into root 8 kT over pi mu and mu also becomes simplified mu is ma into mB. So, B is again A divided by ma plus mB but B is nothing but A. So, this becomes ma over for A with A. So, always remember this factor of half when you are looking at bimolecular reaction where A is colliding with itself. So, we have got a rate constant k is equal to this pi rA plus rB square 8 kT over pi mu maybe a factor of half if you are more particular about it A being equal to B or not ok. But let us take a broader picture. Arrhenius and Vanthoff had earlier written a more general equation k is equal to A into e to the power of minus eA over RT yeah and this was found to be experimentally true. So, we know this equation is more or less true the do these two equations now compare that is the question. So, a few important points to note first we are basically relating this A with this quantity. So, the first thing to note is that A has this temperature dependence here. So, A is proportional to root T. So, that is the point number 1. Second which is the more important point the collision theory so far is missing a critical factor this exponential this exponential is the critical part that is what Arrhenius looked at that part and postulated that a transition state must exist this transient species between the reactant and product must exist because of this exponential form and that is simply missing here ok. So, in the next module we will look how to include that and the assumption we have made several assumptions so far and the assumption that really goes wrong is that. So, we have assumed that all collisions are reactive so far and that of course cannot be true just think from the perspective of how Arrhenius was thinking Arrhenius basically was thinking of a transition state that looks like this and so Arrhenius said that you need a minimum activation energy for reaction to happen. But what have we said in this collision theory every collision is reactive at any energy. So, these two molecules might be approaching very very slowly this is coming out very very slowly like this they will still react. So, there is no sense of any activation energy here what we have to build into this theory is that molecules that are moving slower should not react molecules should react only if they have sufficient energy. So, that is the thing we are going to build in next module. So, in this module in summary we have discussed the distinction between rate constant and rate. We have looked at a special case when a biomolecular reaction is happening of a reactant with itself and finally, we have compared and analyzed with Arrhenius equation and two important points emerge one the pre-exponential factor in Arrhenius equation depends on temperature a square root of t and the second is that currently the exponential itself is missing. So, we have to work somehow to include that which we will do in module number 10. Thank you very much.