 Welcome back. We have been discussing contributions to equilibrium constant. We have taken certain examples for discussing insights into different contributions that can go into explaining the equilibrium constant in terms of population of the various energy levels corresponding to reactants and products. In the previous lecture, we were discussing a special case where the reactants are have only one ground state available and the product P has energy states which are like uniform ladder of energy levels. And then by using an expression for equilibrium constant which is equal to the ratio of the molecular partition functions corresponding to products and reactants into exponential minus delta E naught by RT. And then we discussed that since it has only one energy level. So, q is 1 and here we used the partition function for a uniform ladder of energy levels and after substitution we got this result. And that result is that equilibrium constant is equal to Boltzmann constant into temperature divided by this energy separation into exponential minus delta E naught by RT. Note that this delta E naught is positive is a positive number here. And then we started discussing Le Chatelier principle. Recognize that this particular example is of an endothermic reaction because the energy corresponding to product is higher than the energy 0 point energy corresponding to the reactants. Now, let us discuss effect of temperature effect of delta E 0 and temperature both we will discuss. First of all case number one if delta E 0 is positive and large what I am trying to say here is that if this distance this difference is very large then this exponential term if delta E naught is very large that means what will happen if delta E naught is very large then the value of equilibrium constant is going to be small. So, if delta E naught is large that means exponential minus large quantity which is equal to 1 over exponential large quantity that is going to reduce the value of k. So, k is going to be small. If k is small what that means that the population of the various energy levels corresponding to product is also small. It makes sense because if delta E naught is large most of the molecules will be in the energy state corresponding to R that means k is going to be small. The second example or second case we will choose if delta E naught is small, but positive. Now you assume that this delta E naught is very not very large it is positive, but not very large then this is a reasonable number. Remember if we were saying if delta E naught is large then this factor is going to dominate. If delta E naught is small, but still positive this is a reasonable number, but this pre exponential factor is going to dominate. That means if we increase the temperature then more and more molecules will occupy the energy levels corresponding to the product. So, that means the equilibrium constant is also going to increase. I repeat if delta E naught is small, but still positive. In that case this is a reasonable number and the overall value of k is going to be dependent upon this temperature. So, according to this if you increase the temperature k T by E term is going to increase. So, therefore, the equilibrium constant will increase. Isn't it in accordance with the Lee Chatelier principle which says that for an endothermic reaction if the temperature is increased the value of equilibrium constant is going to increase. Therefore, based upon these considerations we took only one energy level for reactant and we took uniform ladder of energy levels for product. We derived some equation and based on that equation we have been able to explain the effect of temperature. Our explanation is consistent with Lee Chatelier principle. Remember that delta G naught is equal to minus R T log k. This particular discussion suggests that we cannot explain everything just based upon delta G naught. It is not just delta G naught, but its components delta G naught has two components. One is delta H naught and the other is T delta S naught. I would say minus T delta S naught. So, these two components are explained by this differences in zero point energy levels. This particular example belongs to endothermic reaction. So, therefore, for the reaction to be spontaneous for an endothermic reaction there has to be sufficient increase in entropy and that factor will be taken care of if more and more molecules here occupy these densely populated states of the product that will explain the increase in entropy for the process. So, it is not just the free energy its components delta H naught and minus T delta S naught become very very important in providing mechanistic insights into the process under consideration. So, these were the various contributions to equilibrium constant. It is very important to address these contributions, because delta H naught delta H in other words the enthalpy depends upon the nature of interaction. An entropy change depends upon not only the interaction, but also associated with the changes in the surroundings. So, it is not only the association between the two reactants if that association between two reactants lead to release of lot lot of solvent molecules which were earlier oriented around the reacting groups and that release can increase the entropy. And here in terms of statistical thermodynamics we are explaining them in terms of the population of various energy states. So, not only enthalpy, but entropy change is also very important in this consideration. By now, we have discussed all thermodynamic quantities starting with internal energy, we discussed entropy, we discussed Helmholtz energy, Gibbs energy, enthalpy, heat capacity, mean energy and then we have now discussed the equilibrium constant. We have connected all these thermodynamic quantities with either canonical partition function or molecular partition function. So, in a sense now we have shown that all these thermodynamic quantities can be experimentally measured with the help of spectroscopy. Now, a couple of lectures we will spend on solving numerical problems. One such numerical problem let us take it up today. The question is calculate the vibrational contribution to molar heat capacity of nitrogen gas at 1000 Kelvin. The experimental value is 3.43 joules per Kelvin per mole. The question is calculation of vibrational contribution to molar heat capacity. How do we begin answering this question? First of all, we need to first decide that how to connect this molar heat capacity with something which may depend upon the molecular partition function. Molar heat capacity or heat capacity is change in internal energy when we are in 1 degree centigrade temperature change occurs in a system at constant volume. So, what we have now is to calculate the vibrational contribution to molar heat capacity of nitrogen gas at 1000 Kelvin. The experimental value is given to us and we know that heat capacity C v is equal to del u by del t at constant volume. This we know, we also know that C v m is equal to Avogadro constant into temperature derivative of mean energy at constant volume that also we know. So, that means why do not we first get an expression for the mean energy? Mean energy is minus 1 by q into del q by del beta at constant volume let us do that. We know that we are going to now derive an expression for mean energy at constant vibrational energy which will be minus 1 by q vibrational into del q vibrational del beta at constant volume. This is what we are going to do. We also know that q vibrational is equal to 1 over 1 minus exponential minus beta e. Therefore, your mean vibrational energy is going to be minus 1 by q that means minus 1 minus exponential minus beta e. This is minus 1 by q into derivative of 1 by exponential minus beta e. This is going to be minus 1 over 1 minus exponential minus beta e square into minus exponential minus beta e into minus e. This is all going to be the expression for e v that is minus 1 by q this is minus 1 by q into derivative of this is minus 1 by 1 over 1 minus exponential minus beta e square derivative of this which is minus exponential minus beta e then derivative of minus beta e is minus e. So, what it comes to now? This is cancelling with this and we have e exponential minus beta e over 1 minus exponential minus beta e. I can write this again as e over exponential beta e minus 1 which is equal to h c nu bar over exponential h c nu bar by k t minus 1. But remember h c nu bar is equal to k times theta v. So, I can write this k times theta v divided by exponential h c nu bar by k is equal to theta v. So, theta v by t minus 1. So, therefore, my mean vibrational energy is k times theta v over exponential theta v by t minus 1. So, what I have now? Mean vibrational energy is k times theta v over exponential theta v by t minus 1. This is what I have let us double check k times theta v over exponential theta v by t minus 1. Now, we are interested in c v m which is equal to n a times derivative of delta e by with respect to t at constant volume. And since there is no volume term appearing in this I will as well may not write at constant volume, but it is ok. So, this is n a times I need to take derivative of this which is k theta v is a constant numbers into minus 1 over exponential theta v by t minus 1 square into derivative of theta v by t. This will be minus theta v by t square. What I have now? I have k times n a into there are 2 theta v's theta v square by t square minus minus cancel into 1 over exponential theta v by t minus 1 square k times n a this is r. Now, you see I have this derive this equation c v m is equal to r times theta v by t whole square into yes there has to be another term over here there has to be exponential theta v by t yes exponential theta v by t. Yes we put there. So, exponential theta v by t over exponential theta v by t minus 1 whole square. That means, now if I have the numbers for theta v and temperature I can easily get the value. Theta v can be calculated from h c new bar is equal to k theta v from the knowledge of Planck's constant speed of light wave number is already given 2344 centimeter inverse k Boltzmann constant you can get theta v. That means, your theta v is going to be h c new bar by k substitute the numbers and you get a value of theta v which is 3374 point k 3374 Kelvin now theta v by t is 1000. So, theta v by t is 3374 by 1000 which is 3.374 you have the value of theta v you have the value of theta v by t substitute what you get is constant volume molar heat capacity equal to 3.48 joules per Kelvin per mole. Continuing this further it also says calculate the fraction of molecule in v th vibrational state fraction population p i is equal to n i upon n is equal to exponential minus beta e i upon q that we know. Therefore, if I replace p i by fraction in the v th vibrational state this will be equal to exponential minus beta vibrational energy level divided by q q vibrational and then and what are these vibrational energy levels v h c new bar because we know it is v plus half h c new bar half h c new bar we ignore remember the discussion in the vibrational partition function and we also know that q v is equal to 1 over 1 minus exponential minus beta h c new bar. Once you have that substitute this you end up with this expression once you have this expression then it is a matter of just putting the numbers the fraction is given by this to calculate f 0 use v is equal to 0 to calculate f 1 is equal to 0 to 1 use v is equal to 1 that is all you are already given the wave number v value can be variable beta is 1 over k t h is Planck's constant. So, use those numbers substitute over here you will find out that the fractional population of the ground state is nearly 0 and the first excited state the fractional population is 1.31 into 10 is to the power minus 5 it is so less this comparison suggests that even at the temperature which is given temperature given to us is 1000 Kelvin even there you see the fractional population in the first excited state is only 1.31 into 10 is to the power minus 5 that means most of the molecules are in the ground state only. So, therefore, with the knowledge of the partition functions corresponding to particular mode of motion and the fractional population we can calculate the fractions of the molecules in the ground state fraction of the molecule in the first excited state provided we know we remember the formulae or we are able to derive the formulae appropriately. In the lectures ahead we are going to take up more examples and do some calculations. So, that the applications of the derived equations become easier to understand. Thank you very much.