 Welcome back, we have been discussing some applications of the connection of equilibrium constant with molecular partition function. In the previous lecture, we took the example of dissociation of disodium molecule to sodium atoms. That example was simple because on the reactant side, you had just one molecule and on the product side, you had just one type of atoms. But sometimes situation can be complex, your reaction can involve multi atomic molecules. In that case, each molecule may have different kind of modes of motion. For example, let us take the case of this gas phase exchange reaction. What we have here is water plus DCL is forming HDO plus HCl. This is a simple gas phase exchange reaction. But this is not as simple as dealing with disodium going to sodium atoms. Because number one, here the number of molecules in the reactants and products is more, you also have triatomic molecule H2O is a triatomic molecule. So, the more the number of molecules in the reactants and products, the more will be the calculations because overall partition function is going to be product of partition function for each mode of motion for each molecule. I repeat for product of each mode of motion for each molecule. For example, water molecule here, water can have translational degree of freedom, it can also have rotational degree of freedom, it can also have vibrational degree of freedom and if at all there is electronic degree. Here DCL, DCL is a linear molecule. Here you will have all translational rotational vibrational electronic is possible. HDO, translational rotational vibrational electronic similarly for HCl. So, therefore, in order to solve this question you require lot of data. The wave numbers vibrational wave numbers for H2O are 3656.7, 1594.8, 3755.8. Water, water is a non-linear molecule OHH non-linear molecule. So, normal modes of vibration are 3 and minus 6 and in this case in the case of water 3 times 3 is 9 minus 6 is equal to 3 and those wave numbers are given here 3656.7, 1594.8, 3755.8. HDO is also a non-linear molecule. HDO being non-linear molecule you have again 3 modes of vibration, normal modes of vibration. Rotational constant for water 27.88, 14.51, 9.29, non-linear rotor. Both H2O and HDO are non-linear rotor therefore, you will require the values for A, B, C, 3 rotational constants. HCl will be linear and it is a linear rotor. So, therefore, you will require only one rotational constant for each and for a diatomic linear molecule there is only one mode of normal mode of vibration. So, therefore, there is only one value given for HCl and there is only one value given for DCL. With this data now we have to proceed and evaluate the value of equilibrium constant. Equilibrium constant now let us write down the expression. I will write Q HDO by Na into Q HCl let me write in the bracket divided by Na then I have Q H2O divided by Na into Q DCL divided by Na into exponential minus delta E0 by RT. When you write like this all these Avogadro constants will cancel out. The resulting expression is Q HDO, Q HCl, Q H2O and Q DCL I have retained here the standard state conditions into delta minus delta E0 by RT alright. HDO and H2O let me write for Q HDO I am not writing not just to make it more simpler, but assume that this is standard state. This is going to be Q translational HDO it can undergo rotation also HDO it can undergo vibration also HDO and I will rule out electronic because if at all there is a contribution that is going to be nearly one. So, I am not incorporating I am not including the electronic contribution and similarly you will also have Q for HCl this also you will have Q T HCl Q it can undergo rotation also HCl HCl HCl HCl HCl HCl into Q vibrational HCl and similarly I can write for Q H2O and Q DCL. Now if you examine carefully you will require Q T Q R Q V for each molecule alright. So, you have this Q translational Q rotational Q vibrational for each molecule that you need to calculate that means it can be a very cumbersome and it can be very lengthy process. Now if you look at the problem statement calculate the equilibrium constant at 800 Kelvin for the gas phase exchange reaction with the given numbers there is no additional information given. And what we discussed is that I need translational rotational vibrational contribution for each contribution Q translational is equal to V m naught by lambda Q where lambda is equal to H over square root 2 pi m k T. See how this given problem can be simplified H is constant planks constant 2 pi k T is fixed you are given the temperature. That means this lambda is inversely proportional to square root of m at a given temperature lambda is inversely proportional to square root of m that means Q is going to be directly proportional to the 3 by 2 1 by 5 1.5 fifth power of m m is here mass and you can always express in terms of molecular weight. So, that means when you consider this ratio and since this is only the mass of let us see here it is mass of one particle and Avogadro constant everything gets cancelled. So, that means this ratio is directly you can calculate from the ratio of their molar masses you do not need to calculate anything else right. This is where we said that Q T is directly proportional to m it is with the power 3 by 2. So, molar mass of H d o molar mass of H c l molar mass of H 2 o and molar mass of H c l when you substitute all these you get a value of this ratio approximately 1. This ratio is approximately 1 this gives another information that the translational states occupied by these molecules are almost the same similar ok. So, we have found out a very easy way of dealing with the ratio of translational contributions. Now, let us look at rotational contributions for a linear rotor like your d c l like H c l the rotational contribution to partition function is given by K T by sigma H c b where we know the sigma is symmetry number and b is rotational constant. And for a non-linear rotor you have 1 by sigma K T over H c pi over A B c all right. Now the systems that we have we have H 2 o we have H d o and H c l and we have d c l sigma for H 2 o H d o H c l d c l sigma for this is 1. This is also 1 because you can distinguish when you rotate by 180 degree this is also 1 only this is 2 because in complete 360 degree rotation the H 2 o molecule will appear twice in the same state. So, when you take the ratios of the rotational contributions to partition function what will remain see K will cancel T will cancel H will cancel C will cancel only the rotational constants and the symmetry number will remain. And by using these 2 equations and these through these 2 ratios when you substitute you will see what remains is the symmetry number of water which is 2 others are all 1 1 1 1 we do not need to worry about that. And what else will remain A B c for H 2 o and A B c for H d o b for H c l and b for d c l and when appropriately substituted in that expression you have all the numbers and then you get 1.702. So, you see you do not need to evaluate full value of rotational partition function for either linear rotor or non-linear rotor when you are writing when you are expressing in the terms of the ratio many things get cancelled and things get simplified. So, this ratio is coming out to 1.702 now 1 x comes a vibrational contribution vibrational contribution q v is 1 upon 1 minus exponential minus beta H c nu bar. You are given the value of vibrational wave number you are given temperature you can easily calculate the value of vibrational partition function. Remember that q vibrational is a product of vibrational partition function for each normal mode alright. So, therefore, you will require if you look at here let us look at the problem statement. You have the vibrational wave number 3 vibrational wave number for water 3 for H d o. So, therefore, in the numerator you will have 3 for this and 1 for this 4 in the numerator and denominator 3 for this and 1 for this 4 in the denominator. And let us see that is what you have here you have 4 in the numerator and 4 in the denominator these vibrational wave numbers are chosen from that table. So, corresponding to this vibrational wave numbers you can evaluate the vibrational partition function and substitute over here this evaluation can be done by this calculate for each normal mode of vibration and substitute over there. We have discussed many times that the vibrational contribution to partition function is usually not very large. Therefore, you expect this ratio also to be close to 1 in any case calculate remember again I will write q v of k k th mode this is individual right. So, overall partition function vibrational partition function vibrational partition function for mode 1 into vibrational partition wave function vibrational partition function for mode 2 mode of vibration into vibrational partition function for normal mode 3 and continue continue that is the partition function is multiplicative. Now, what we had the expression was let us note down the equation then it will be easier for us. The equation is h 2 o plus d c l h 2 o plus d c l is equilibrium with h d o plus h c l this is the equation this is the reaction and k what we wrote was q naught of h d o into q naught for h c l over q naught for h 2 o into q naught for d c l over q naught for h 2 o into q naught for d c l into exponential minus delta E naught by R T. Our now purpose is to find out the differences in 0 point energies remember what is the value of 0 point energy you have 0 point vibrational energy oscillator has a 0 point vibrational energy which is equal to half h c nu bar half h c nu bar that means delta E naught by h c will be equal to half nu bar differences in the nu bars alright the 0 point vibrational energy is half h c nu bar and when you use half h c nu bar add up that for the product and from that subtract for the reactants you will get the value of delta E naught by h c right. So, here I will say summation j nu j nu bar j psychometric number you can take and when you substitute all these numbers you know for the products you look back for the products for h d o plus d c l this is for h d o 2 7 2 6 1 4 0 3 7 0 7 and h c l wave number is 2991. So, these 3 plus 1 4 take the addition of that and then for the reactants you subtract the corresponding number when you do that this difference comes out to minus 162 centimeter inverse this is delta E naught by h c ok and when you multiply by h c you can get the value of delta E naught difference in the 0 point energies alright. Now, we know the value of delta E naught this all we have expressed as the ratio of translational rotational vibrational partition functions. We calculated for 1.041 we calculated for 1.0707 I a left for you to calculate for vibrational deliberately I left for you to calculate vibrational by, but I will tell you since vibrational energy levels are separated far separated you expect this ratio also to be close to 1. And then exponential minus delta E naught by r t or delta E naught by k t if you are expressing per mole or not. So, substituting all those numbers what you have is a final answer of 2.41 the question was complex, but we discussed how to simplify a complex problem because the ratio of these partition functions they take a shape where many factors cancel out and you have a simplified version either in terms of molecular masses molar masses or in terms of their rotational constants and vibrational wave numbers. We also discussed how to get this 0 point difference in 0 point energies from the vibrational wave number data. So, I hope that by now you have understood how to calculate equilibrium constant for a reaction from the knowledge of molecular partition functions. When I say from the knowledge of molecular partition function I mean that from the knowledge of corresponding spectroscopic data which give information about rotational constants or vibrational wave numbers. Electronic contribution we deliberately did not include because we always find that the electronic contribution to molecular partition function is usually close to 1 or close to degeneracy of the ground state. In any case we will further solve similar questions and try to clarify if there are any problems in dealing with such questions, but that we will be doing in the coming up lectures. Thank you very much.