 Hello and welcome to module 30 of chemical kinetics and transition state theory. So, this is going to be kind of the last module directly related to transition state theory. We have covered as much theory as we want to actually and today we are just going to solve problems alright. So, let us start with that alright. So, let us start with our very first problem. So, we are going to look at the reaction of H 2 plus F 2 going to 2 H F and let us assume that the transition state is non-linear. So, the first part is that assuming translational partition function is given to be 10 to the power of 32 meter minus 3 per molecule, rotational partition function is 10 per degree of freedom and all other vibrational and electronic partition functions are 1. Use TST theory to estimate the pre-exponential factor at 300 Kelvin in specific units of later mole inverse and inverse. So, I really want you to pause this video right now, solve this problem on your own. This is not a hard problem actually. You should be able to solve this, pause the video and solve this problem ok. Hopefully, you have solved it on your own and you have a number with you. Now, we are going to do it together ok, not then alright. So, well as always whenever formulas are needed you will be provided formulas. You do not have to memorize formulas, I have provided all the formulas here. It will be up to you generally to figure out which formula will be useful and which formula will not be useful. So, we have to essentially figure out the partition functions right. So, K TST will be equal to KB T over H into the partition functions. So, let us simplify that partition function of transition state. So, that will be Q translational of transition state. I have been using small Q. So, let me not break my own notation. So, let me use small Q only Q not of transition state translational into Q not rotational. And since vibrational and electronic I am going to assume to be 1 anyway, I will not even bother writing ok. This divided by Q of H2 translational Q of H2 rotational into Q of F2 I am sorry translational Q of F2 rotational. So, this is your pre exponential that is what we want to calculate ok. So, KB is 1.38 into 10 to the power of minus 23 kilograms meter square per second square Kelvin into 300 Kelvin that is the temperature that is given to us divided by 6.6 into 10 to the power of minus 34 kilograms meter square per second ok that is H ok. Q translational per molecule is given to be 10 to the power of 32. So, transition state is 1 molecule. So, its partition function is 10 to the power of 32 rotational. So, the transition state is non-linear. How many rotational degrees of freedom are there for a non-linear molecule? 3. So, the rotational will be 10 into 10 into 10. Remember it is 10 per degree of freedom and I have 3 rotational degrees of freedom. So, I get 10 cube ok. H2 translational will be 10 to the power of 32. Now, how many rotations will H2 have? H2 is definitely linear and linear molecule has 2 rotational degrees of freedom. So, I get 10 square F2 will be exactly the same 32 meter minus 3 into 10 square. So, let me just pull out my notes first we have to make sure that our units are all right. So, kilogram meter square per Kelvin second kilogram meter square second. So, I will be left here with meter cube per second ok. So, let me just write this thing 1.38 I have not yet plugged it on a calculator divided by 6.6 into 10 to the power of minus 34 into 10 to the power of 32 cancels with 10 to the power of 32. I have 10 cube in the numerator 10 to the power of 4 in the denominator 3 of them I can cancel easy math minus 33 meter cube second inverse second inverse because of this meter cube because of this ok. So, let us write this once more I have 1.38 into 10 to the power of minus 23 I have forgotten into 300 sorry into 300 divided by 6.6 I am sorry 6.6 into 10 to the power of minus 34 into 10 to the power of 33 minus 33 my bad sorry for the interruptions meter cube second inverse we can write second like this, but remember that the unit I want is this liter mole inverse second inverse. So, we have done this a few times I am sorry 1000 liters in 1 meter cube in 1 meter cube into 6.02 into 10 to the power of 23 moles divided by mole yeah. So, that is our conversion from per molecule to per mole we multiply by the Avocandro number. So, meter cube cancels and I get units of liter mole inverse second inverse. So, now it is easy we just plug it all of it in a calculator and what I get is what I have written here 3.77 into 10 to the power of 6 liter mole inverse second inverse. So, this is easy you have to be just careful of units you have to remember to multiply by this do not if the units ask this liter mole inverse second inverse that is the unit you have to provide the answer in and look the answer is also somewhat in what we expect it is a bit lesser typically you get into the power of 8 or 9, but not too far off. So, we have little bit of confidence that we are not completely off the charts. So, now we will look at part B of this problem let us recalculate this, but from collision theory now the collision theory if you have forgotten by now the answer rate is provided here. So, again take a pause solve this problem get the answer in liter mole inverse second inverse and then come back. So, pause the video now. Hopefully you are back hopefully you have punched the numbers on a calculator and got a number out those are very important skills. So, let us calculate it on our own. So, k will be equal to pi r a plus r b r a is 0.12 nanometer plus 0.15 nanometer, but we will write everything in si units which is meter and 1 nanometer is 10 to the power of minus 9 meter into root 8 kb is 1.38 into 10 to the power of minus 23 kilogram meter square second square Kelvin into 300 Kelvin divided by pi into mu. Mu is the reduced mass is m a m b over m a plus m b. You can calculate it exactly, but I know that since hydrogen is so much lighter than it will flow in we will just say this is the mass of hydrogen, but you can work it out. So, the mass of hydrogen will be 2 into 1 gram per mole, but again we want it in si units 1 kilogram per 1000 gram. I am doing all of these problems so explicitly just so that you get used to this notion of units. They are very important and unless you work them out on your own and keep on listening to me you will not be able to solve them. You have to practice. So, if I plug the all of this in I get 1.6 into 10 to the power of minus 27 kilograms moles cancels with mole gram cancels with gram. So, I will use 1.6 into 10 to the power of minus 27 kilograms here. If you want to improve you can use the mass of fluorine as well and calculate what you get exactly, but it would not be very different. So, kilogram cancels with kilogram Kelvin cancels with Kelvin. So, if I plug all of this in I got 4.16 into 10 to the power of minus 16 units. Units are I have meter square here and I get another meter here. So, meter square into this meter square I take the square root I get meter. So, this becomes meter cube second inverse because I have a second square here. So, if I take the square root of that I get 1 over second. But we have to do the same thing as for transition state. We multiply it by 1000 liters over 1 meter cube into 6.02 into 10 to the power of 23 over moles. So, meter cube cancels here I get liter mole inverse second inverse I multiply all of this and I get 2.6 into 10 to the power of 11 liter mole inverse second inverse. So, these are numerics, but one point I want you to now notice notice what we got out of transition state theory roughly 4 into 10 to the power of 6. Now, notice what I have got out of collision theory 10 to the power of 11 5 orders of magnitude more. So, does that make sense this is not really written as a problem, but now you know transition state theory you now know collision theory can you explain this. So, the explanation really is the collision factor in collision theory. In collision theory we have assumed all collisions are equally reactive. So, when I write this formula here what I am assuming is that h2 and f2 can come in any orientation and they hit each other and a reaction is going to happen. But you of course know that is wrong h2 and f2 have to approach in a very special way for the reaction to happen right. So, h2 looks like this f2 looks like this and they have to come like this if it comes like this the reaction would not happen right. If it comes like this the reaction would not happen there are only a few orientations where the reaction will be happening. And so, as is usual collision theory will grossly overestimate the rate constant that is the usual case with collision theory. And now transition state theory has actually corrected for that you see the transition state theory naturally gave you a much lower rate constant. Final problem and this is a hard problem now, but I still want you to think about it at least. If h2 is replaced with d2 can you estimate the pre-exponential factor using transition state theory and collision theory. You can assume vibrational partition functions are independent of mass and you can assume geometries are independent of mass and you can assume that the mass of Lorentz is much greater than mass of h. This is m underscore f my bad m h, but hopefully you understand that. So, this is slightly hard here you will have to make certain assumptions. I also emphasize that this is slightly open ended without providing you full information and such a problem I will never give you in an examination setting. Well, but all teaching is not about examinations anyway right. So, this is you think about it and then we can discuss it together and I will provide you my thoughts on it. So, again take a pause and first attempt the problem on your own. Hopefully you have taken a pause and you should think on this problem because this is a slightly deeper problem alright. We will first start with collision theory that one is actually easy that one is not very easy. So, collision theory rate is provided here. So, if I have to work with k collision d 2 over k collision h 2 well you notice that first of all all geometry factors we have assumed to be same. So, things like r a or r b are same ok. So, they will cancel out k t will cancel out, pi will cancel out. The only thing I will be left with is mu of h 2 divided by mu of d 2. Also please note that it is inversely proportional to mu the rate constant. So, k of d 2 over k of h 2 is mu of h 2 over mu of d 2 square root ok. So, this is easy I know that mass of deuterium is double of mass of hydrogen. So, the reduced mass will also be double. So, I get this is square root of 1 over 2 ok plain and simple. Now, let us work with transition state theory that is one is slightly more complex, but it is doable. So, k t s t of d 2 divided by k t s t of h 2 k t over h now I am looking at this formula. So, k t over h is the same for d 2 over h 2 the temperature is the same. What I have is q transition state of d 2 divided by q transition state of h 2 divided by q of d 2 divided by q of h 2 naught and fluorine is the same ok. So, this is one happily enough and the exponential also cancels. We are assuming all energetics remains the same ok. So, we have to now find these ratio of partition functions. We will start with q d 2 naught over q h 2 naught ok. So, the partition functions will comprise of rotation, vibration, translation and so on and so forth alright. Vibration we are ignoring and the reason actually why we are ignoring vibration is because frequency is large. So, either of d 2 or h 2 they are different, but they are much larger than room temperature nonetheless. And therefore, this factor does not matter this still comes out close to 1, beta h bar omega is still much less than 1 and therefore, this comes out close to 1 anyway for both d 2 and h 2. Therefore, I am ignoring vibrational partition function fine, but I still have translational and rotational. So, the translational partition function I will have m of d 2 to the power of 3 half here all other factors are going to cancel. So, I am not looking at this 2 pi k t and all that into the rotational. So, this is translational now I have to write the rotational. So, the rotational this is linear yeah d 2 has to be linear h 2 has to be linear and for linear the moment of inertia will go as mu r square. So, again this will be proportional to mass some proportionality constant let us say m of d 2 multiplied by some distance, but distance again we are assuming to be same. So, even if I write d 2 square I do not care similarly I will write this, but we are assuming all distances are equal. So, r h 2 equal to r d 2 that term cancels. So, I get this is equal to m d 2 over m h 2 to the power of 5 over 2 alright which is nothing, but 2 to the power of 5 over 2. Now somewhat harder thing to understand is for the transition state q d 2 of the transition state divided by q h 2 of the transition state. Again I will have translation and rotation sorry not d 2 and h 2 this will be h 2 f 2 let me clean this up ok d 2 f 2 and h 2 f 2 the mass I have the translational part translational part again is here which is m of d 2 f 2 to the power of 3 half divided by m of h 2 f 2 to the power of 3 half, but note that the mass of fluorine is much more than mass of hydrogen or deuterium. So, to a very good approximation I can say that mass of d 2 f 2 is almost the same as mass of h 2 f 2 this is remember the total mass. So, I am adding the masses. So, when I add the mass if I write h 2 which will give you a mass of 2 or if I write d 2 which will give you a mass of 4 the factor difference will be only 2. So, if I take this ratio and I add fluorine as well remember that and fluorine is very large. So, this is almost equal the hard part is the rotational part. So, this will be I a, I b, I c of d 2 f 2 divided by I a, I b, I c, h 2 f 2 again we have not discussed much of mechanics in this course. So, we have not went into this moment of inertia in great detail, but what do these moment of inertia physically represent they represent 3 kind of motions. So, let us just think about what this transition state is you know a little bit of your chemistry some structure like this. Now, now this structure can rotate in 3 possible ways. So, let us imagine this is my transition this is my center of mass I can draw a line like this and rotate the whole thing like this. I can draw a line like this and rotate about this axis and I have a line going through the plane of my board through the plane of the screen like this. So, those are my 3 axis and along all 3 axis I am rotating the molecule. The question is how will moment of inertia change with mass? What is your intuition? So, that is slightly open ended you do not you can calculate it if you want, but let us try to make educated guesses without doing proper math. The point is any motion that will involve moving fluorine a lot will require moment of inertia of fluorine atom, but the kind of motion where fluorine is not moving much assume fluorine is infinitely massive. So, in that limit one axis which is this axis which is almost passing through the 2 fluorines. If I rotate around this axis you see fluorines are not moving the 2 hydrogens are moving. So, the moment of inertia corresponding to this will be proportional to mass of hydrogen or deuterium depending whether this is H2 or D2. But if I rotate let us say about this axis you see this will be dominated by mass of fluorine because here fluorine is moving. So, this is a slightly more involved concept, but hope it has some sense also you can think of the third axis which is passing through the screen that also involves rotation of fluorine. So, that will also involve rotation of mass of fluorine. So, I have only one I which will depend on the mass of hydrogen or deuterium. So, this thing will become M D over M H2 to the power of only one half I B will cancel and I C will cancel. So, what have we got in total? This will be in the numerator I have the ratio of transition state. So, let me just write this as 2 to the power of half. So, this becomes equal to 2 to the power of half divided by 2 to the power of 5 over 2 which is what we had shown for the reactants. So, this will be equal to 2 to the power of minus 2 or 1 over 4. So, you can compare it against the collision theory which was 1 over root 2. So, again collision theory and transition state theory will give you qualitatively different results the ratio is very very different. So, we will stop here and in the next module we will solve a few more problems. Thank you very much.