 Welcome to module 8 of chemical kinetics in transmission state theory. This is a follow up from the previous module where we introduced the kinetic theory of collisions. So, again our aim is to calculate the rate constant and from an atomistic picture. So, that for any general reaction I will be able to calculate a number out. In the last module we had introduced the basic idea as essentially the rate is equal to the rate of collisions between hard spheres. So, we do not think of any bonds for now, we think of these reaction as between two spheres this colliding with each other and we calculate the rate of these collisions. We ended up in the last module of rate as a function of u and so today we are going to answer the question how do we calculate this u? In doing so we will introduce also the notion of a transformation to center of mass. So, again the resources for this module of or for this entire kinetic theory of collisions is chapter 4 of the Laidler's book or you can also look at this link there also you can see a sufficiently good description. So, just a quick recap how did we estimated this rate? The rate is the rate of collisions. The rate of collisions is nothing else but by definition the number of collisions occurring between a and b per unit time per unit volume this way that I have defined here and so this I rewrote as the number of a per unit volume into the number of collisions that a has with b per unit time. Now, number of a per unit volume is simply the density of a which is n a and the number of collisions per unit time we basically estimated as the collision region into density of b and the collision region last module we went into great details and showed that it is simply this volume of this cylinder that is here where this distance d is nothing, but r a plus r b ok. So, we calculate the volume of this cylinder and multiply by the density of b that is what we derived the last module. So, we put this number of collisions per unit time and the density here in this equation above to calculate the overall rate and that is where we ended in the last module. And so one very natural question that arises is what is this u? Not only this u another assumption we had made while doing this volume calculation is we assume b to be stationary. What if b is not stationary that of course, is the very ad hoc assumption b is also moving right we are at some constant temperature t and at constant temperature both a and b will be moving. So, how do we account for that? So, let us start with b being not stationary. So, let us imagine I have a moving with some velocity u a and b moving with some velocity u b both are vectors. So, what do I do now? Actually the answer is quite simple. We simply change our inertial frame and go to a frame where b is stationary. We simply change our frame to we basically sit on b. So, we have subtracted u b from every single particle. So, this now becomes some u a minus u b becomes stationary yeah. So, this u is now this speed essentially. So, how do we basically calculate this u? So, we will start with a very simple approximation. What we do is to assume this u to be the average thermal speed of this relative velocity u a minus u b. So, that of course, is an approximation and today we are going to work towards this, but in the next module we are going to improve upon that as well. But let us start with this it is at least a reasonable assumption to start with. We assume all particles are moving with this average speed and anyway this whole quantitative solutions is a very rough estimate. So, it is a bulk per quarter of calculate estimate that we are doing anyway. So, that is why we go ahead with this average thermal speed. But how do we even calculate this average thermal speed? So, to calculate this average thermal speed, we will have to do some transformations and we go to what is called the center of mass. So, what is the center of mass? So, imagine I have this particle a this is vector r a and I have b and they have their corresponding speeds and momentum. So, what we do is we define new variables u center of mass u relative to be this. Now, basically we can also define the corresponding let us define these total masses as well. So, I define capital M to be ma plus mb and we will need one more quantity that will soon enough you will see the reason for defining this. This is called the reduced mass. So, effectively we define p COM as m u COM and p relative as mu ui. So, this if you have not seen this transformation before it might seem a little bit abstract. But just bear with me soon enough this will simplify to something very beautiful. And if you have seen this before this will come for example, if you have studied quantum mechanics of hydrogen atom or even if you study planetary system the center of mass is useful its center of this transformation comes in many many contexts. The point is what I will can show is that the Hamiltonian equal to 1 over 2 ma pa dot pa plus 1 over 2 mb pb dot pb. So, I have my masses ma and mb here. So, this is my net Hamiltonian remember in kinetic theory of collisions we do not have any potential. My Hamiltonian is simply the kinetic energy that is a very important thing that you must always remember in kinetic theory of collisions no potential no bonding. So, this one can show is equal to 1 over 2 m p COM dot with p COM plus 1 over 2 mu pr dot with pr. So, if you want you can go ahead and prove this for yourself as additional slide I have shown the proof here. So, you have to be just very careful it is not a hard proof at all. So, you go by the definitions that was defined in the previous slide that I have re mentioned here and you can let us start with this Hamiltonian. This is easier to start with rather than with p a square over 2 ma plus pb square over 2 mb. I substitute p COM and pr here once you simplify you will see that sometimes are going to cancel you keep on simplifying and eventually you will get end up with this. So, this proof is not part of this syllabus but it is also not a hard proof to do you should be able to do this proof actually. Another point which is actually going to become important is that the volume element also remains the same. So, you can also show that d p a into d p b equal to d p r into d p c um. Again the proof is not part of the syllabus but I have provided you this proof as an additional slide for those who are interested and the proof is actually very simple. How do you transform? You essentially find the Jacobian of the derivative matrix some technical language if you do not know then you can look it up and once you know this language then it is very easy. All you do is to find the derivative of these variables with respect to p a and p b. You get this m a over m b m b over m and the corresponding coefficients here you find this determinant and take the magnitude of this determinant. So, these things are easy to prove although we will not take this officially as a part of this this syllabus. So, what am I doing with all of this? This all mathematics is of course good but what is the point? What is the point is that we are trying to find the average speed. So, let us define this u more formally as this is nothing but the average of this p r by mu. So, that is how we are going to estimate u. It is this relative momentum p r again is my bad u r again is u a minus u b and p r is nothing but mu u r. So, p r over mu is your relative speed and so this is what we are trying to find and we are trying to find the magnitude. We do not care about the vector a and this relative direction might be moving in any direction across space I do not care if it is moving up or it is moving this direction or towards u yeah all like here is what is overall value. So, how do we calculate this? Well we use our general strategy of how do we find averages which is equal to 1 over mu we calculate essentially dq dp rho equilibrium of q comma p that we looked a few modules ago that is also shown here into the quantity that I want to average yeah. So, I go ahead and I formally put in rho equilibrium as e to the power of minus beta h and h is nothing but half 2 m a p a square plus 1 over 2 m b p b square divided by the partition function and you can go back and revise the modules partition function is nothing but integral over let me just write otherwise I get confused dq dp integral over dq integral over dp e to the power of minus beta h I have an additional p r in the numerator. So, now you see that is separation of h in the language of p r is very useful because I do not know how to integrate this very directly. So, I transform to this center of mass Hamiltonian dq I leave aside and remember what is dp dp is nothing but integral over all momentum which is integral over dp a integral over dp b yeah. So, I could have written here dp a integral over dp b, but you note that integral over dp a into integral of dp b is nothing but integral of dp c o m into integral of dp r. So, this thing instead let me erase and I can write that as dp c o m dp r and the Hamiltonian as well I can write as all divided by the same thing these are vectors you understand that and here also I have transformed to p c o m and p r I write the same big Hamiltonian here. So, now that you what you notice that my integral here has neatly separated into center of mass component and relative component. So, that is what helps us in simplifying. So, the last integral that I have written now I can simplify that as integral over dq divided by integral over dq. So, you can go back one slide and you will note that there is no q term at all in the integral multiplied by the center of mass also will very nicely separate out and finally, we have the relative term. So, I am just simplifying what I had written in the last slide into different terms. So, this portion cancels this portion cancels and I am left with I have forgotten 1 over mu 1 over mu dp r beta p r square absolute value of p r divided by dp r p r square. But do you recall this term at all have we calculated this before does it tells you remember reminds you of anything this is nothing but a relative speed. So, remember if my we look at just the kinetic energy component before and for this we were trying to find the average speed and we defined the average speed as nothing but 1 over mass into average momentum and this was nothing but exactly the quantity that was written in the last slide beta 1 over 2 m p dot p into absolute value of p divided by the partition function. This is how exactly we calculated this if you do not remember go back a couple of modules this is module 6 and in that we had shown that this is how you calculate average speed exactly and we showed that this average speed comes out to be root 8 k t over pi m. So, I am not going to rederive it you can just look back into your module 6. The only difference in r current Hamiltonian is that r Hamiltonian looks like this p r square divided by 2 mu. So, the only thing is my here my m gets replaced with mu. So, your u in short becomes 8 k t over pi mu that is your average thermal speed where mu again is the called the relative mass k of t was pi r a plus r b whole square I am sorry this is the rate u into n a into n b and this becomes equal to pi r a plus r b whole square root 8 k t over pi mu n a where again mu is I am just writing it again. So, that you have the final answer at one place. So, that is how you calculate rate in kinetic theory of positions at least the first estimate we are going to refine upon it. So, in summary for today's module we have looked at how to calculate this u and we have identified this u as the average thermal speed of the relative motion between a and b. So, we have calculated that formally by doing that doing the center of mass transformation and in that basically my mathematics simplifies a little bit and I can calculate this relative speed u and with that we I get the final answer as this rate equal to pi r a plus r b square into this average thermal speed into n a into n b. So, in the next module we are going to continue from this point we are going to calculate rate constant and discuss a few of its properties. Thank you.