 Welcome to module 44 of chemical kinetics and transition state theory. In the last module, we will relook at unimolecular decay in more detail. And we found that there is a little discrepancy if we build a simple model for unimolecular decay that we had looked long time ago and compare it with experimental data. So, today we are going to look in more detail with a different approach that R, R and K had come up with R, R and K again are Rice, Ramsperger and Kassel in trying to resolve the differences we get with experiment, okay. So, what is a quick recap of what we had discussed in the last module? What is again the problem? We have unimolecular decay, the overall reaction looks like this, some A going to be only 1 reactant and the mechanism of this proposed by Lindemann was as follows, 2 molecules of A collide with each other to give an excited A star and A and A star gives B. You can of course, use chemical kinetics and work out the overall rate constant for this problem. What we have this K1 is defined as rate equal to minus dA over dt equal to dB over dt and I am defining this rate to be equal to K1 into concentration of A. So, when I am defining this, I am not assuming this 8 going to be to be elementary, therefore this K of 1 is not a number, it is depends on the concentration of A, fair enough, okay. When I work out the kinetics of this and try to find K1, what you will get is and this we showed a long time ago, K1 over K minus 1 of these are all functions of temperature, you will get this quantity, I am sorry this is supposed to be A. The problem is that this result does not matches well with experiment. It has qualitative agreement, but not exactly agreement and so then this new idea came about that here we are assuming this K1, K minus 1 and K2 to be independent of each other. We are calculating them one by one as simple numbers at a given temperature, but that may not be right, suppose this is A, this is B and this is somewhere A star, K1 is this, K minus 1 is this and this is K2. So, in this model the problem is that this K1 takes me to A star and at that given energy, A star has either a choice to become a product B with rate constant K2 or come down with rate constant K minus 1. So, these are not, I cannot calculate them as functions of temperature, I have to look at each energy what is happening. So, instead of writing this, we write it as a function of energy, the same quantity and integrate over all possible energies above the barrier height, EA again is this barrier height. Below if E is less than EA, we know that no reaction will happen. So, we integrate over all energies more than this barrier height. All right. So, now the question becomes how do I calculate K1, K minus 1 and K2? So, RR and K, Rice-Ramperger and Kassel had given a recipe to calculate this and that is what we will discuss today. So, the basic underlying model for this calculation of K1, K minus 1 and K2 has what we have discussed a few modules ago, when we were calculating rate constant at a given energy. Remember, now we are thinking of one given energy calculate this K1, K minus 1 and K2 and then integrate over all energies. So, again that picture of constant energy calculation becomes important. So, the RRK model was that my molecule comprises a bunch of harmonic oscillators and all harmonic oscillators have the same frequency nu, where S oscillators with frequency nu and the rate of the reaction is proportional to the probability that mode 1 has energy E greater than EA and I have arbitrarily chosen mode 1 to be the reaction coordinate. You could have chosen mode 7 or mode 9 whatever is your favorite number, 1 is my favorite number. So, I decide that mode 1 is the reaction coordinate. So, and we have discussed this model before and today we will use this model to calculate this unimolecular decay rate constant. So, the first thing that we have already solved for is the K2 of E, once more what is K2 of E, if this is A star, K2 is that if I have come up here going from A star reactant to B in the product. So, and at that given energy to going from here to here. So, within the RRK model we had calculated this a few modules above and we had shown that this is given by this expression E is of course, the total energy EA is again the barrier height and S is the number of oscillators. So, one constant down K2 of E, the second constant K minus 1, K minus 1 the prescription given was and it is going to be a somewhat approximate prescription. We calculate K minus 1 using collision theory that is not the big approximation really. I like my 1D surfaces this is A star at that given energy E, K minus 1 is the rate of losing energy that is important it is not gaining energy but losing and the mechanism that is being thought of is that this A star is there zigzagging at very high energy and it collides with some A and loses all of its energy. So, we know how to calculate that rate constant of collisions and all that using collision theory and the collision theory of the same atom colliding with itself is given by this expression. This was discussed in great detail remember this factor of half comes because same atom is colliding with itself it is A plus A, it is A star plus A but A star still has the it is still equivalent to A. D is the diameter of A and mu is the reduced mass. There are two critical assumptions we make when we have written this. First you note that this K minus 1 or it is supposed to be K minus 1 is independent of energy. This might seem like a shocker because so far we have been arguing that we have to make this K 1, K minus 1 and all these as a function of energy and not as a function of temperature. But here I come and I say K minus 1 is a function of temperature alone it is a gross approximation agreed. But if you start thinking about it a little bit more what we are really saying is let us assume that this activation energy E A is large it is much greater than K T let us say. So, this energy is a large number compared to K T. So, what it is saying is if your energy is here or your energy is here or your energy is here it really does not matter I have to be I have to lose energy. So, this K minus 1 dependence on energy will not be very high. The idea I am saying is if you have 1000 kilojoules per mole of energy instead of 1010 kilojoules per mole of energy or 1900 kilojoules per mole of energy this K minus 1 is not going to change exponentially with it. So, let us just assume it to be a constant. Well, one question that perhaps you are asking my energy is going up to infinite. So, there should be a difference between 1000 kilojoules per mole and 5000 kilojoules per mole agreed. But also remember that if the energy is much larger than the barrier height the rate constant will anyway come down a lot and that will come out naturally in this theory. So, that is why energies that are way beyond the barrier height do not contribute anyway. So, we only end up with those energies that are close to barrier height and in that range I can assume K minus 1 to be almost a constant. Second assumption you will also notice that I have missed the collision factor here that when I was discussing the collision theory I said is very very important without that the rate constant will be completely wrong and you will get a zero mark if you do not include that. But here I come and I have a happily forgotten to write that for a good reason. The reason is that I am losing energy here not gaining energy. When you have already very high energy and if you hit another molecule the probability is fairly high that you will just lose that energy. As opposed to if you have two molecules that are moving slower and they collide with each other and you gain a lot of energy that is not as probable. In short what I am saying is you think of a ball and it is somewhat of an inelastic ball and I throw it very very hard at a ball the probability is high that it is going to lose its energy you will hear a loud thump and the ball is not going to do much. But the other way around where you have a ball that is just let us say floating around on a wall and it gains a lot of thermal energy because of collisions that is not as likely. So, that is why we assume the collision this frequency of the collision factor is 1. So, these two assumptions are applied here. The final number is K 1 but instead of finding K 1 I will find K 1 over K minus 1. K minus 1 I have already found if I find K 1 over K minus 1 I have also found K 1. Why am I insisting on K 1 over K minus 1 now? The reason is think of this A in equilibrium with A star so 2A plus A. So, K equilibrium will be nothing but A star into A divided by A square is equal to A star over A. But K equilibrium is also equal to K 1 over K minus 1. So, K 1 over K minus 1 has a neat interpretation. This is relative population of A star with respect to A. Concentration is nothing but moles by volume volume is a constant. So, I do not worry about volume. So, what I am saying is how many if I start with A how many A star are there at that energy E. So, I am I am looking at this energy E and I am finding how many A star are there. So, how do I calculate this? Now, we we have done a little bit of our stat muck and rate theories. This is a relative population is nothing but e to the power of minus beta E right. Not right. There is a little factor that I am missing here. What is the factor? This we have discussed once before. It is a little challenge to you what what mistake have I made? Of course, partition function, but other than partition function what other mistake? D-generacy the density of states. So, at that given energy G tells me the density that is already there at that energy or the degeneracy multiplied by the Boltzmann factor. So, Boltzmann factor gives me the overall ratio of energies which energy is more prominent at the different energy and G of E tells me within that energy how many states are there. So, this divided by an integral of all energies that is my partition function now. So, never forget degeneracy all right. So, the idea is let us say you have one state here at as I am moving in energy and I have more states here. The point is let us say E 4 has only very few states and E 5 has only even lesser state. So, even though E 3 is higher in energy than E 2 it still might be more probable because there are just more states there. So, you are just more likely to be found there. So, it is a competition between how high the energy is and how many states are accessible at that energy. So, you must multiply those together. Now, for our given model which is a bunch of harmonic oscillators, we already showed what this G of E is. So, what we are going to do is use this formula that we had used before and apply it to our expression here. Let me just make one simplification. This formula was calculated for a bunch of harmonic oscillators with different frequencies. But if all omega i's are equal to some constant omega then this becomes h bar omega to the power of s. So, that is our model that all oscillators have the same frequency. So, this becomes e to the power of s minus 1 divided by s minus 1 factorial h bar omega to the power of s into e to the power of minus beta e divided by an integral 0 to infinity dE e to the power of s minus 1 divided by s minus 1 factorial h bar omega to the power of s e to the power of minus beta e. s minus 1 factorial and h bar omega are independent of energy. So, I can cancel these out. This is equal to e to the power of s minus 1 into e to the power of minus beta e. Now, look at what integral I am left with and I have given you a general rule to do this kind of integral. Here my x is nothing but e, n is nothing but s minus 1 and a is nothing but beta. So, if I replace x with e n with s minus 1 and a with beta, I will get exactly this integral here. So, this integral then becomes n factorial n is s minus 1 divided by a is beta to the power of s. Beta is nothing but 1 over kT. So, this becomes. So, now we have a full prescription to calculate the RRK answer. We want to integrate this integral. In this integral, k2 we will use as this. We have not yet specified what this k dagger though is. So, we will just do it. It is some constant, but how do I calculate that constant? I will tell you in a minute. k2 I calculate this way, k1 over k minus 1 I calculated in the last slide and k minus 1 I calculate using collision theory. So, I have all the components here and I can just do this integration. If I cannot even do it analytically, then I can always use a computer to do this integration. Another important thing to note k1 over k minus 1 varies exponentially with energy. So, that is my argument that when energy is much lower larger than barrier height, the contribution will become very small. This exponential will become very very small and dominate every other term and this k1 over k minus 1 sits here. So, that is why for energies that are much higher than barrier height, the contribution is always small and that is why for those terms is k minus 1 if even if it is changing with energy did not matter. So, now let us look at this calculation of this k dagger that I have been avoiding so far. We have some expression here now. I have substituted k1 over k minus 1 right here. To calculate k dagger, I will do the following. I will take the limit concentration of A goes to infinite. A appears only here. When A goes to infinite, A is in denominator here. This whole term becomes 0. So, I am simply left with 1 in the denominator. So, this is A to the power 2 infinite dE, E to the power of S minus 1, E to the power of minus beta E divided by S minus 1 factorial kT let me make it clear kBT better put Boltzmann name there k2 is k dagger E minus EA to the power of S minus 1. So, at least one term cancels here. So, I am left with 1 over S minus 1 factorial kT to the power of S into k dagger into this integral. To do this integral, I am going to do substitution of variable. I am going to define E prime as E minus EA. DE prime will be equal to DE when E prime is equal to EA my bad when E is equal to EA E prime will be 0 when E is equal to infinite E prime will be infinite. So, with that I put 0 to infinite here DE is DE prime. I write this as E to the power of minus beta E prime into E to the power of minus beta EA E to the power of minus beta EA is a constant. So, I have pulled it out of the integral into E minus EA is E prime to the power of S minus 1. So, I have gotten this into a form which I know how to integrate it is the same integral I showed you a couple of slides ago X is E prime N is S minus 1 and A is beta. So, I get this as E to the power of minus beta EA into k dagger divided by S minus 1 factorial kT to the power of S into this integral is N factorial N is nothing but S minus 1 S minus 1 factorial divided by A to the power of N plus 1 A is nothing but beta to the power of S. So, this terms cancel no surprise this also cancels. So, this is implies limit A going to infinite of k1 is equal to. So, this is an experimental recipe to calculate k dagger. We find this k1 experimentally at high concentration of A or high pressure of A is usually a gas. So, either way even if you are in a solution you can just make the concentration of A high if you are in a gas you make the pressure of A high and then you calculate this k1 and then you can calculate EA with aid of some electronic structure calculation and then from that you calculate k dagger. So, this portion comes from typically experiment and this portion will typically come from either experiment or electronic structure calculation. And once you have these two quantities you can then calculate k dagger. So, in short what are the parameters that are required to do an RRK calculation for unimolecular decay? First is S the number of oscillators that unfortunately is ad hoc. We calculate the RRK rate for different values of S and for whichever S the fit matches best with the experiment that is the value of S that we take k minus 1 comes from collision theory, k dagger comes from high pressure limit of rate constant calculated experimentally. So, that is another unfortunate feature that you have to rely on experiment and you cannot calculate everything from calculations and E naught or EA I have been using so I should stick to my notation it is the it comes from either experimental data or electronic structure calculations that is the barrier height of course. So, let me just show you on comparison against experiment this is a very famous paper the original paper was in 1920s or 30s I cannot remember. So, this is the isomerization of cyclopropane that was the big puzzle of that day in 1910s and 20s that is the problem that everybody was trying to crack. The experimental data is given by this solid line with these squares. If you use the Lindemann answer which is the one which is you what you calculate with k 1, k 2 and k minus 1 calculated constant temperature what we were doing earlier then you see some deviation not a huge deviation agreeably but not exactly right as well. Now, when we apply RRK theory for S equal to 13 no good reason why S should be 13 by the way it just happens that S is equal to 13 you get the result which is almost matches the experimental answer over a very wide range of pressure. So, in the x axis we have essentially log of pressure which is nothing but proportional to log of concentration. So, you can find more details in this paper as well very famous paper. So, with that I summarize today that we have shown how to calculate the unimolecular decay rate constant using RRK model. Our model comprises a bunch of harmonic oscillators with the same frequency and within that model we have a few parameters we need to find S which comes rather ad hocly k minus 1 comes from collision theory k dagger typically comes from high pressure limit of rate constant and E naught or EA which is the barrier height comes from either experimental data or electronic structure calculations with these you can calculate k 2 k minus 1 and k 1 and calculate this experiment this integral using a computer usually and calculate k 1. Thank you very much.