 Hello and welcome to module 31 of chemical kinetics in transition state theory. In the last module we solved a numerical problem. Today it is a continuation of that we will continue solving more problems. Today our problems will although be more not numerical but more focused on concepts ok. So, let us start the problems. So, the first problem is a relatively simpler problem, but a conceptually very important problem. So, I really want you to take a pause draw the potential energy for along the reaction coordinate and a coordinate that is perpendicular to the reaction coordinate sitting at transition state. So, this please do take a pause and you should be able to draw do these plots do take a pause now ok. Hopefully you have taken a pause. So, now let us solve it together. So, for the first one we have to draw the potential energy along the reaction coordinate. So, this is the figure that you have commonly seen probably in every single textbook in kinetics right. So, you get a figure like this where this is reactants, this is products and this is the transition state the maximum struck point ok. So, this is your very familiar figure. However, what I want you to also understand from this course is that if I choose a coordinate perpendicular to reaction coordinate at the transition state. I will get a energy surface that is going to look like this and this is my transition state. This is a very important concept transition state is a minima for any coordinate that passes perpendicular to the reaction coordinate alright. So, once more a contour plot if I draw these kind of contour plots that I have drawn a few times now. This is reaction coordinate. Let me just change the color in color red. This is a coordinate that is perpendicular to reaction coordinate. So, this is reaction, this is perpendicular to reaction. So, along the reaction coordinate you see a figure like this along a direction perpendicular to reaction coordinate which is if you are walking along this you will get something like this alright get back to black color alright. Second problem is slightly harder, but it is an interesting problem and it shows you how to include quantum mechanical effects in transition state theory. So, let us imagine I have only a one-dimensional well and somehow a temperature is maintained at some temperature T. Let us say the frequency is given to you omega here and the barrier height is much more than kT. This ea is much greater than kT and we are going to include tunneling in this question now. So, in a module some modules ago 5, 6 modules ago we solved this problem without any tunneling. So, you can go back to that module see how we solved this problem and now the question is can you also include tunneling effects in it. So, what I have provided you here is that the probability of tunneling at an energy e less than ea is given by this formula where alpha is some constant and for energy greater than ea you can assume that the transmission is 1. So, now can you estimate the rate constant? You can make any approximations or assumptions that you want to make pause the video and attempt this problem on your own alright. Hopefully you have paused the video and you have attempted the problem at least this is not an easy problem, but we will you have to make at least attempts. So, let us try to attack this problem well we have to go back to our fundamentals. How we solved this problem earlier? Well idea is the same this is what we are going to do let us say my and I am set some energy e at that energy e what is the rate constant and what is the probability of being at that energy e. If I integrate this together I will get the total rate and energy cannot be negative because I cannot be lower than this minima. So, this is what I have to calculate fine rho I am going to approximate as we have always done in transition state theory as the equilibrium. So, that is given by e to the power of minus beta e in one dimension. So, we will find this n as always as 0 to infinity this should be 1 so now this is easy to integrate this if I integrate I get something like this. So, you can quickly show this easily enough. So, the first part I am good I know rho of e second part how do I calculate this k of e start by saying that this is equal to at least for e less than e a and e greater than e a we will calculate them separately. So, this is equal to nothing but omega over 2 pi for e greater than e a. So, if my energy is above this point then my rate is omega over 2 pi and once back we argued this in some more detail in our module it is omega over 2 pi because omega over 2 pi is the frequency of hitting the transition state. So, this is my transition state I am sitting here the probability that I will hit this point it is one time period in every one time period I will hit this exactly once in the forward direction. So, again this is approximate I am making the transition state approximations here. So, transition state approximation is that I only looked at rate in the forward direction and in the forward direction I will hit this transition state once in one time period. So, what about e less than e a. So, we are going to approximate that as omega over 2 pi that is. So, my energy is let us say here I am sorry this was e for e greater than e a and this is e less than e a well in one time period I am going to assume my tunnel once. So, I come here and then I have some probability of leaking through. So, again this is somewhat open ended problem it is doing estimates and so this I am going to say is e to the power of minus alpha e a minus. So, that is the way to estimate things. So, k of t then becomes 0 to infinity d e rho of e which we showed is beta e to the power of minus beta e into k of e. Now, k of e is divided into 0 to e a omega over 2 pi e to the power of minus alpha e a minus e plus e a to infinity d e beta. So, what we had discussed several modules ago was that we ignored this part the first part because we were assuming no tunneling and we had only this second part, but now we have the tunneling contribution as well. So, let us solve these two integrals the second integral is easier. So, that I will start with beta e to the power of minus beta e omega over 2 pi. So, this is beta omega over 2 pi which are constants I integrate e to the power of minus beta e. So, this will be equal to omega over 2 pi. So, this you can quickly verify this is not hard at all e to the power of minus beta infinite is 0. So, you are left with only e a part. So, let us also do the first part now alpha e a minus e. Well, this is also not hard really beta omega over 2 pi you have to just be careful 0 to e a d e e to the power of minus alpha e a into e to the power of alpha minus beta e. So, I have just taken the exponential separated this term out because it is a constant and taken the parts that had e in common and taken them together. So, this is nothing, but beta omega over 2 pi e to the power of minus alpha e a into e to the power of alpha minus beta e over alpha minus beta from 0 to e a. So, this is nothing, but beta omega over 2 pi e to the power of minus alpha e a into e to the power of alpha minus beta e a minus 1. Well, so the total rate is nothing, but the sum of these two beta over alpha minus beta and I just take this thing inside. So, I get e to the power of minus beta e a minus e to the power of minus alpha e a. So, I have just simplified the second part a little bit here, but that is not very important like the important is the concept how to set it up. And so, the setup was we calculate the rate constant as integral over all energies and for e less than e a we estimated it to be omega over 2 pi multiplied by tunneling probability for e greater than e a we simply took omega over 2 pi. And once you have that the rest is just numerical. So, now, the next problem I have for you is that I have a particular reaction which is of importance for atmospheric chemistry C H plus n 2 going to some transition state to H C n plus n. And people experimentally found that the rate constant fits this formula. So, you use transition state theory and estimate what alpha must be. We have given you a few things to assume, you can assume transition state to be linear and vibrational and electronic partition functions to be independent of temperature. So, take a pause and calculate alpha. Hopefully, you are back you have a value of alpha with you, we will calculate it together and see if we are results match. So, we are going to use this formula and we are going to look only at the temperature part. So, I am going to use proportional to first I have this t coming from this. Then I have Q transition state. So, that Q transition state will be Q sorry Q translational into Q rotation divided by Q CH translational into Q CH rotational into Q n 2 translational into Q n 2 rotation. I have gotten rid of vibrational part or electronic part. So, this is proportional to temperature and again I am not going to look at any other factor other than temperature. Translational part that is dependent on t to the power of 3 half. Rotational part. Rotational part we are assuming the transition state to be linear. For linear, I use this formula. So, I get t divided by CH translational. Well, all translations are 3 half rotational of CH. CH is also linear 2 translational n 2 rotational or linear molecules proportional to t. So, this cancels with this, this cancels with this, this cancels with this and I am left with t to the power of minus 3 half. This implies alpha is remember I still have that exponential factor that I have simply chosen not to write so far. Other common factors there that I have been not writing all the mass and 2 pi and m and all that. So, if I compare this formula with this, you can easily see n 5 minus 3 half. This is a conceptual question. For the following one dimensional surface, I really like one dimensional surfaces in solution at some temperature t, where should I draw the transition state which is the best line representing transition state? Is it A if I draw it here and do my transition state theory with this dividing surface? Should I draw it at B and do my transition state theory with this dividing surface or should I do my transition state theory with the C dividing surface? So, take a pause, think about it and come up with your own answer. So, hopefully you have thought about it and let us so, let us discuss it together now. So, the correct answer is B, this is the best choice why? So, you have to look into the assumptions made in the transition state theory, this is what this question is about. A, this is my reactant, this is my product. If you draw the dividing surface at A, you can have trajectories that are going up and they just have a more chance of reverting back compared to B. You can have that happening at B as well, but at B, a trajectory that crosses here sees a force that looks like this. So, this is more inclined to keep on moving forward. So, at A, a trajectory that is going up might not have enough energy and it just turns back. So, A more recrossings, which is not good. In transition state theory, remember our assumption is no recrossing. So, the best choice is where the recrossings are minimal and A has more recrossing compared to B. What about C though? In C, I have this kind of energy surface, I have my trajectories coming here and they are simply falling down. Isn't C better than? No, here the other assumption comes in picture. The equilibrium assumption will be bad. Remember that we also assume that transition state is at equilibrium with the reactants. So, transition state will not be at equilibrium. Well, again you can think of it intuitively, you have these trajectories that are coming up and they are just forward crashing into the products like this. So, there is no reason that at C, you can expect a good Boltzmann ratio, good Boltzmann distribution, e to the power of minus beta h, that the velocities are much more faster because trajectories are coming up and then crashing like this. So, the transition state will not be at equilibrium. So, the other assumption does not hold very well. So, B is the optimal point where you will get minimum recrossings as well as the equilibrium assumption will hold better. The final question for today, you think of a reaction like O plus HCl going to OH plus Cl and this is again a conceptual question. Calculate the total number of frequencies that are needed, assuming transition state to be linear versus nonlinear. Again, take a pause, solve this problem on your own. This is an important problem as far as this course is concerned and get an answer. Pause this video now and solve this problem on your own. Hopefully, you are back. So, let us solve this problem together. Remember that for transition state, I need frequencies of reactants transition state. I do not worry about products, reactants here O, well that is an atom, atom does not have frequencies, that is there is no vibration there, no frequency. HCl, one frequency, the stretch, right. So, HCl can vibrate one frequency only. Linear transition state, how many frequencies are required for a linear transition state is the question. So, I have OHCl, 3 atoms. For 3 atoms, I have 9 total degrees of freedom, right, x, y, z corresponding to each atom, out of which I have 3 translational rotations. Remember that a linear molecule only has 2 rotation. So, whatever remains are the number of vibrations. Now, there is still a little bit of trick. At transition state, we need one frequency less compared to the total number of frequencies. Why? Because in transition state theory, the frequency of the reaction coordinate is not considered. So, minus 1 reaction coordinate frequency which is an imaginary frequency. So, 3 frequencies. So, if the transition state is linear, I need to calculate one frequency for HCl and 3 frequencies for OHCl. Now, for B, nonlinear, I will do the same math. I will have 9 total minus 3 translational minus 3 rotations. So, a nonlinear molecule has 3 rotations instead of 2. So, I get 3 frequencies. I subtract 1 reaction coordinate frequency. So, I get 2 frequencies, okay. So, for A, I will need 4 total number of frequencies, 3 for transition state, 1 for reactant and for B, I will need 3 total number of frequencies, 2 for transition state and 1 for reactants, okay. So, we will stop with this problem session now. Hope you got to learn and hope you tried these problems on your own. Thank you very much.