 Welcome back, we have been discussing equilibrium constant, equilibrium constant which is a very important thermodynamic quantity because not only from academic point of view, but from industrial point of view this thermodynamic quantity derives a lot of meaning. As discussed in the previous lecture, in industry it is important to have as much yield as possible. And equilibrium constant is nothing, but the ratio of the activities of products divided by that of the reactants. And if activities are equal to concentrations, we generally write this as ratios of the concentrations weighted by their stoichiometric numbers. And then in industry we are also interested in knowing how to optimize various parameters so that you can improve upon the value of equilibrium constant. In statistical thermodynamics, we are interested in connecting equilibrium constant with partition function. And that is what we have done in the previous lecture where we connected equilibrium constant with molecular partition function by the expression which is given on this slide. Equilibrium constant is equal to product of q j m naught by n a raised to the power stoichiometric number into exponential minus delta r e naught by r t. We also discussed that suppose if we apply this to this kind of reaction a moles of a plus b moles of b in equilibrium with c moles of c, then how to write this k is the product j q j m naught by n a raised to the power nu raised to the power g into exponential minus delta r e naught by r t which is the same expression that is written about. But what I will do is now I will put positive number for the products and negative stoichiometric number for the reactants. So, how we expand it positive for the products product is c that means q c m naught by n a this will be raised to the power c into q a m naught by n a raised to the power minus a pay attention to this I am putting a negative sign because a is a reactant into q b m naught by n a raised to the power minus b again because b is a reactant then into exponential minus delta e naught by r t. This can further be written as k equal to now I will write this as q m by n a raised to the power c divided by q a m naught by n a raised to the power a continuation into q b m naught by n a raised to the power b into exponential minus delta e naught by r t. This is how we expand this product that means it is a similar way of writing equilibrium constant as you have done in discussing chemical thermodynamics. That is for the products you write in the numerator and for the reactants you write in the denominator and then there is an additional term which is an exponential term taking into account the difference in 0 point energy that has to come in this expression. Now, let us take an actual example the question is evaluate the equilibrium constant for the dissociation of n a 2 going to 2 n a this is the reaction which is going on which is given to us this is dissociation of disodium at 1000 Kelvin from the following data rotational constant is given rotational constant means it is given for n a 2 because n a cannot rotate that is given as 0.1547 centimeter inverse vibrational wave number that means this is also given for n a 2 that is 159.2 centimeter inverse and you have 0 point dissociation energy that is also for n a 2 that is 70.4 kilo joules per mole further it is given to us that the sodium atoms have doublet ground terms that means the degeneracy of electronic states for sodium atom is 2 this is the data given to us. We are supposed to find the equilibrium constant for the dissociation of disodium to sodium at 1000 Kelvin how we will approach this problem is first we will write an expression for k. As we just discussed we will write k for n a square by n a square divided by q n a 2 divided by n a right. This is for the numerator I am already squaring it this is for the denominator and then I have exponential minus delta e 0 by r t this is now equal to q square n a divided by q n a 2 into 1 over n a this n a is Avogadro constant not sodium into exponential instead of delta e 0 let me write d 0 by r t. What we have is the ratio of partition functions for sodium squared divided by partition function for disodium 1 by n a which comes from this n a square n a and then exponential term is there. Now you remember that n a is an atom sodium atom and then you have q n a 2 this is a disodium that means it is a molecule. So, therefore, we have to carefully decide what contributions come in. For atom you have only translational contribution and you may have the electronic contribution. So, therefore, what I will do is I will write here that q of sodium only that means q translational sodium square into if there is degeneracy of sodium divided by I have q translational for disodium. Now remember that this is a molecule. So, it will not only have translational degree of freedom you will also have rotational disodium you will also have vibrational disodium into 1 over n a is already there into there is 1 over g n a 2 square into is exponential minus d 0 by r t. So, since we are talking about n a 2 in equilibrium with 2 n a and we write the equilibrium constant is q translational of n a square into degeneracy of the ground state for n a divided by q translational for n a 2 into q rotational n a 2 into q vibrational n a 2 into electronic contribution degeneracy of n a 2 and there was 1 over n a term and then there was exponential minus d 0 by r t. So, we have this expression now we very much know that q translational is V m naught by lambda n a cube, but this is squared this is for the translational contribution then degeneracy of n a I will retain like this in the denominator I will write V m naught because everything we are treating as ideal gas divided by lambda cube of n a 2 this is translational contribution I will retain rotational as such I will write n a 2 into q vibrational I have n a 2 into degeneracy of the ground state n a 2 into 1 over n a into exponential minus d 0 by r t. Please note that I am writing here standard state conditions because although I am not writing naught naught naught with all the q's it is understood that these are all the standard state quantities. Now let us try to further solve it. So, now I have k is equal to V m naught square by V m naught I will just write V m naught right 1 is square 1 is 1 term and here only let me include n a I will just bring n a along with this. Now what I have is lambda n a 2 cube over lambda n a 6 this is the ratio of the thermal wavelengths into g of n a over g of n a 2 this also let me combined into 1 over q r n a 2 into q V n a 2 into exponential minus d 0 by r t this is what I have. Now we can further work upon this and see what we get P V m naught is equal to r t I am consuming n because V m m means molar. So, can I write this r is equal to k times Avogadro constant into t r is k times n a. So, therefore, V m naught by n a this is what I want is equal to k t t by P naught right this is also not because the standard state we are talking about. So, V m naught by n a this n a is Avogadro constant is k t by P naught. So, instead of this I can write k t by P naught if I write k t by P naught then I do not need the volume because we are considering the standard state conditions and applying the ideal gas equation P naught V naught is equal to r t I can get V m naught by n a in terms of k t by P naught k is a constant t is a given temperature P naught is 1 bar. That means, you need not be given the volume. Once I substitute this k t by P naught over here then I have this k t by P naught into ratio of the thermal wavelengths ratio of the degeneracies of the ground state electronic ground state. And then I need the information about the rotational partition function I need the information about vibrational partition function and I also need the dissociation energy 0 point dissociation energy and that is what is written in this expression k t by P naught you have this this is g x square naught q g x square lambda x 2 cube g x 2 q r x 2 q vibrational x 2 lambda x 6 exponential minus d 0 by r t. That means, now the next problem is to evaluate the thermal wavelengths rotational partition function vibrational partition function and then proceed for the calculation. The data given to us for lambda lambda is equal to beta h square by 2 pi m square root which is equal to h by 2 pi m k t square root beta is equal to 1 over k t h is Planck's constant. We know 6.626 into 10 raise to the power minus 34 joule per second k is Boltzmann constant 1.381 into 10 raise to the power minus 23 joules per Kelvin t the temperature given to us. So, we can calculate the thermal wavelength from the knowledge of the masses of Na 2 and Na thermal wavelength can be calculated by using these formula the thermal wavelength of Na 2 comes out to 8.14 picometer whereas that for Na itself the sodium atom it comes out to 11.5 picometer. This can be obtained by appropriately substituting the numbers over here alright. So, we know how to deal with the thermal wavelength q rotational Na 2 is a linear diatomic molecule. So, therefore the expression that you require for the molecular partition function is that for a linear rotor and we know that q r for a linear rotor is 1 over sigma h c beta p. Remember that this equation is a high temperature result that means your temperature if it is much higher than the characteristic rotational temperature then only you will use this equation that means what you will do is you will use this equation k theta r is equal to h c b calculate the value of theta r because you are given Planck's constant you know speed of light you know and the rotational constant you know the Boltzmann constant also you know substitute these numbers and get the value of rotational temperature you will see in this case where the temperature given is 1000 Kelvin is much much higher than the rotational temperature you calculate and check yourself yes it comes out to be the temperature given temperature is much much higher than the rotational temperature. So, that means I am justified now to use this expression q r is equal to 1 over sigma h c beta p we are dealing with Na 2 let us say first we will talk about Na 2. So, if I put like this Na Na a rotation by 180 degree leaves the molecule in an unidentifiable state. Therefore, in complete one rotation the same state will appear twice that means sigma in this case is equal to 2 you know the value of sigma now you know the Planck's constant speed of light beta is 1 over k t and b is given to you once you substitute this you will see that you get a q r value of 2 to 4 6 sodium atom is atom you cannot have rotation in that that means we do not need to worry about the rotational contribution to Na. Now comes vibrational contribution again sodium atom cannot vibrate, but sodium molecule can vibrate. So, when we discuss about vibrational contribution disodium linear molecule. So, number of normal number of modes of vibration is 3 n minus 5 is equal to 6 minus 5 is only one mode of one normal mode of vibration one normal mode of vibration and now if you look at the problem statement rotational constant is given we have already used that in calculating the rotational partition function vibrational wave number is the given and we are also given the sodium atoms have doublet ground terms. So, by using q v is equal to 1 over 1 minus exponential minus beta h c nu bar this is the expression to be used remember this expression which I have written over here this can be used at any temperature there is no high temperature approximation used over here. An alternate way of calculation can be just like you evaluated the value of rotational temperature similarly you evaluate the value of vibrational temperature from k theta v is equal to h c nu bar calculate theta v compare with temperature and then use the high temperature result not required because this is a very simple expression you know beta is 1 over k t you know plex constant you know speed of light and we know the vibrational wave number once you substitute these numbers what we have is 4.885 this is the number that we are going to get third one the degeneracy of the electronic ground state for sodium atom is 2. The degeneracy for the ground state electronic state for disodium molecule is 1. So, by now we have everything q t can be calculated from phi m naught by lambda q, but in fact you do not need to calculate because the expression that we are going to use k is equal to k times t this is g x square a lambda x cube 2 p naught g x 2 q r x 2 q v x 2 lambda 6 we know we know this we know this we know this this this this and dissociation energy is given to us let us look at whether it is given or not the zero point dissociation energy is 70.4 kilo joules per mole. Therefore, since we are given the value per mole you will use d 0 by r t if it were given per molecule then you would have used d 0 by k t the value of p naught is equal to 1 bar this is because we are invoking standard states why standard state because in our all calculations which are the expressions are based upon delta g naught is minus r t log k and we have connected everything with the standard state or change in Gibbs free energy under standard state condition and when we say standard state then we are saying it is 1 bar pressure. So, now we have all the information the given reaction was disodium going to 2 n a and from the knowledge of k is equal to pi j q j m standard state by n a exponential minus delta r e naught by r t by we use this expression and then that expression was converted into this particular expression and then by substituting the various numbers at 1000 Kelvin p naught you see here when you solve the numerical problems make sure that you are using the appropriate units. If you do not put the appropriate units your final answer is going to be wrong the given pressure p naught is equal to 1 bar and then you have converted this bar 1 bar into Pascal's 10 to the power 5 Pascal's why you introduce Pascal is because it will take care of the units alright. So, you have Boltzmann constant you have temperature you have this degeneracy factor you have these ratio of the thermal wavelengths and then you have this rotational and then you have vibrational contribution everything put together you see at 1000 Kelvin the value is 2.42. What we have done is that we have calculated the values of various molecular partition functions substituted those values we used the value of dissociation energy and we can easily get the value of equilibrium constant. The value of equilibrium constant obtained here is not very high only 2.42 that means this reaction does not have a high very very high yield the delta G naught of this reaction is not very highly negative. Now, it is a different matter how to improve this value improving the value of equilibrium constant depends upon various factor for that you need least at earlier principle, but in this case what we have done is we have restricted our discussion only to equilibrium constant. So, therefore, just to recap what we have done is that by starting with this equation we have identified that what different contributions go into the molecular partition function and from those contributions what we have done is evaluated this equilibrium constant. The calculations may appear little lengthy, but the procedure is very simple only you need information on translational contribution which depends upon the thermal wavelength, rotational contribution, vibrational contribution, electronic contribution and the difference in 0 point energies. Once you have this information the calculations of equilibrium constant is very easy. We will take some more examples to further clarify about what factors go into the evaluation of equilibrium constant. Remember that this equation is very simple where you have only diatomic sodium and you have sodium atom. If you have a reaction in which there are triatomic molecules and multi atomic polyatomic molecules the situation can be complex. One such example we will take and further discuss, but that we will do in the next lecture. Thank you very much.