 Good afternoon. So, for the last few lectures, we have been discussing the chemistry part of rocket proportion that is the combustion. We have talked about how to estimate the adiabatic flame temperature, we have defined the heat of reaction, heat of formation etcetera. When we discussed the adiabatic flame temperature, we stated that the final composition of the product is very important in order to estimate the final temperature. At that time, we have said that for the time being let us assume that the composition is known and we estimated the final temperature based on that. However, we pointed out that when we have to do the actual calculations then that will also be unknown. So, we have to follow an iterative technique in order to converge on both the concentration as well as the final temperature. So, today what we are going to discuss is the how do we estimate the chemical composition of after a reaction, the chemical composition of the products. So, the topic is determination of chemical composition that is our topic of discussion today. Let us consider that we have a chemical reaction at say T naught which is 298.15 degree Kelvin. Let us consider a very simple reaction hydrocarbon, so methane burning in air. So, this is the reaction. We say that we have after the reaction we have n CO 2 moles of CO 2 being produced n H 2 O moles of water vapor is produced. Apart from that we have nitrogen which remains unburned and some oxygen may also remain unburned because this is if I look at the stoichiometric reaction then according to that this reaction is fuel lean. So, therefore, some oxidizer will remain unburned. So, let us say that this is our chemical reaction. Here in the product side we have carbon dioxide, water vapor, nitrogen and oxygen, but we do not know the composition. We do not know how many moles of each of the species are present in the product. We want to estimate that. So, first let us use the conservation of atomic species like we have discussed earlier. So, what are the atomic species that we have? We have carbon, we have hydrogen, oxygen and nitrogen. So, first let us look at the carbon. In the reactant side we have one mole of or one atom of carbon. So, therefore, this is equal to n CO 2 because in the product side also we should have exactly same number of atoms of carbon. Similarly, if I look at hydrogen we have 4 hydrogen here. So, that will be equal to every molecule of water vapor contains 2 atoms of hydrogen. So, 2 times n H 2 O. That is going to be the number of atoms of hydrogen. Similarly, for oxygen in the reactant side we have 15 times 0.21 into 2 because every molecule of oxygen contains 2 atoms of oxygen. So, this is equal to our now in the product side oxygen appears in the carbon dioxide. So, for every carbon dioxide molecule we have 2 atoms of oxygen. So, we write 2 n CO 2. Similarly, every water vapor molecule has one atom of oxygen plus we have the unburned oxygen available. Similarly, for nitrogen we have in the products a reactant side we have 15 times 0.79 times 2 and in the product side we have 2 n 2. So, now if I look at this I have 4 equations 1 2 3 4 and 4 and I have 4 unknowns n CO 2 n H 2 O n n 2 n n O 2. So, 4 equations and 4 unknowns can be easily solved. So, this is quite straight forward we can solve for all the unknowns and I can get the composition final composition. So, this is something which the number of atoms and then or the primary atoms and the number of unknowns are same, but that may not be the case always. Let us look at another reaction. So, here we take a heavier hydrocarbon octane burning oxygen. For this I have all kind of products present and once again it depends on the final temperature also how much of each product will be present and what are the products present that we will discuss little later. So, let us say here I have all these products present. Now what are the primary atoms we have? Atomic species we have carbon, hydrogen, oxygen. So, essentially we have 3 atomic species, carbon we have 8, hydrogen we have 18, oxygen we have 30 atoms. Now in the product side carbon is appearing here and here and here A plus E plus F. Hydrogen is appearing here in water vapor plus in the molecular hydrogen plus as the atomic hydrogen plus as part of the OH. Similarly, oxygen is appearing in carbon dioxide. So, 2 A is appearing in water vapor, it is appearing in molecular oxygen, it is appearing in CO then as atomic oxygen and in OH. So, from the number of atomic species we can form this 3 equations. However, total number of unknowns are 1, 2, 3, 4, 5, 6, 7, 8, 9. So, we have 3 equations and 9 unknowns. So, we cannot solve this and get all the unknowns. We need to have some additional equations to complete this problem. So, therefore, we need the 6 additional equations which can be obtained from chemical equilibrium theory. So, that is the justification of what we are going to discuss today. So, today we are going to discuss the chemical equilibrium theory which gives us the additional equations. So, first let us go back and start discussing chemical equilibrium theory. Now, I like to point out here one thing before we go there that this reaction is not a fundamental reaction because typically the fundamental reactions will be collision of 2 molecules giving at the most 2 or 3 products, but here in the products side we see 9. So, this cannot be a fundamental single step reaction. Actually, this involves a reaction sequence containing various reaction steps. So, the theory we are going to discuss now will also give us some indication of what will be the reaction steps. So, with that let us now look at the chemical equilibrium theory. If we consider a reacting system without any velocity, let us say it is a stationary system. So, we can neglect the potential energy and kinetic energy term. One of the energy terms which are very important for reacting systems is called Gibbs free energy. So, we define is a defined property Gibbs free energy, which is defined as G equal to enthalpy H minus T s. This is a in thermodynamics you must have seen that this Gibbs free energy is defined like this. Now, here H is the enthalpy by definition enthalpy is internal energy plus P V. So, this can be written as U plus P V minus T s, where U is the internal energy P is pressure, V is the volume of the system, T is temperature, S is entropy. So, this is Gibbs free energy. Now, the change in internal energy D U can be written as then because my U is equal to H minus P V. So, therefore D U is equal to D H minus D P V which is equal to D H minus P D V minus V D P. And in thermodynamics you have seen that T D s is equal to D H minus P D V. This is called Gibbs equation and you have seen this before in thermodynamics. So, now if I put this back into this equation D H minus P D V becomes equal to T D s minus sorry D H minus V D P minus P D V. So, T D s equal to D H minus V D P. Therefore, this is equal to D U equal to T D s minus P D V. Now, let me take this and put it back into this equation. For that first let us differentiate this expression for Gibbs free energy G. So, this will be equal to G is equal to U plus P V minus T s. So, if I expand this, this will be D U plus P D V plus V D P minus T D s minus S D T. Now, T D s minus P D V, T D s minus P D V is my D U. So, this will cancel off. So, I get D G equal to nothing but V D P minus S D T. This is the first simple expression representing the change in Gibbs free energy for a particular process. Now, how do we define our chemical reaction? We define the chemical reaction occurring in isothermal condition. The reaction was occurring in isothermal temperature was not changing. What we said is that after the reaction whatever heat of reaction is released that goes on to increase the temperature, but the reaction is isothermal. So, for isothermal reaction temperature changes 0. Therefore, this term here is 0. Therefore, now my D G is equal to V D P where V is the volume P is the pressure. Now, since we have been saying that our working fluid is a puffed gas. So, for a puffed gas we use puffed gas equation of state, which says that P V equal to N, N is the number of moles, universal gas constant R times T. So, therefore, the volume can be represented as a function of temperature and pressure like this V equal to N R T by P. Now, let us take this and put it back into this expression for Gibbs free energy change. So, my D G will be equal to then N R T D P by P. Now, this is nothing but equal to N R T D ln P log natural of P. Now, let us say that if you integrate this from some reference pressure P naught to some actual pressure P. So, now, we are bringing in the effect of pressure also. So, if we integrate it from some reference pressure P naught to some actual pressure P, then we get delta G is the change in Gibbs free energy in this process, which is equal to G at this pressure minus G at P naught standard pressure. This is equal to then N R T ln P minus ln P naught. Now, this G P naught is my standard pressure. So, instead of writing it every time as P, what I would say is that I will write for any other pressure G is G P is represented as G only, only for standard state I represent as G naught. So, then I can write it as G minus G naught, where G naught is the Gibbs free energy at the standard pressure. Previously, we had defined everything for standard temperature. Now, it is standard pressure equal to N R T ln P by P naught. Now, what is our standard pressure at S T P is 1 atmosphere. So, if I consider this as 1, then this becomes nothing but equal to ln P. So, therefore, we get G is equal to N R T ln P by P naught. So, Gibbs free energy at any pressure G, a P equal to G naught that is Gibbs free energy at the standard pressure plus N R T ln P. So, this is our expression for Gibbs free energy for any species. Now, if I go to the system, chemically reacting system, where I have i species. So, for every species i, I can define its Gibbs free energy. So, that is equal to then I write G i equal to G i naught plus N i R T ln P i, where P i is the partial pressure of that particular species. N i I can take it inside. So, I can write this as plus R T ln P i to the power N i. So, this is for species i in the system. This is my expression for Gibbs free energy at any given pressure P. Now, let us consider a chemical reaction. Let us say this is my chemical reaction A plus B giving C plus D. Some N A moles of A reacting with B moles of B giving C moles of C and D moles of D. So, for this delta G, which is the change in Gibbs free energy is nothing but Gibbs free energy of the products minus Gibbs free energy of the reactants. Now, this is equal to G C plus G D minus G A plus G B, where G C is the Gibbs free energy of the species C. Similarly, this is for D, this is for A, this is for B. Now, what I do is I use this expression here. What I will get is G C naught plus G D naught minus G A naught plus G B naught plus R T and then ln of P C to the power C, P D to the power D by P A to the power A, P B to the power B. I just combine this two, I get this. Now, what is this? Is the Gibbs free energy of C at standard pressure plus that of D minus. So, this is the total Gibbs free energy of the products at the standard state minus Gibbs free energy of the reactants at the standard pressure. Therefore, this difference is nothing but total change in the Gibbs free energy at standard pressure. So, I can write this as then that total change in Gibbs free energy is equal to change in Gibbs free energy at the standard pressure plus universal gas constant times D, then ln of P C to the power C, P D to the power D by P A to the power A, P B to the power B. Now, at the beginning what we said is that, we are interested in chemical equilibrium. We are talking about equilibrium thermodynamics. So, how much will be the change in Gibbs free energy at equilibrium? It will not change. So, at equilibrium total change in Gibbs free energy is 0. So, delta G is 0, which essentially means that minus delta G naught is equal to RT ln P C to the power C, P D to the power D by P A to the power A, P B to the power B. Now, by definition this ratio, what is this ratio? This is defined as K P, this is equilibrium constant K P. So, this is defined as P C to the power C, P D to the power D by P A to the power A, P B to the power B. Here, all these things P A, P B, P C, P D are partial spatial of different species. So, then this is a defined property. Therefore, we get minus delta G naught is equal to RT ln K P. So, now, what does this relationship give us? This relationship gives us the relation between the standard free energy change and the equilibrium constant at any arbitrary pressure and temperature. So, as you can see here, temperature is present here and K P will include pressure, because the partial pressure includes the partial pressure is mole fraction of C C i times the total pressure. So, that is how partial pressure is defined. Therefore, this total pressure information of the system is already included in the definition of K P and temperature is there. So, now, we have both pressure and temperature. That is the deviation from what we have talked about in adiabatic flame temperature, where we said it was in standard temperature. We were completely silent about pressure, but now we are bringing in the pressure term as well. So, here now that is how pressure comes in into the reaction. So, coming back to this, then let us look at it little more closely. K P here, first of all, is the equilibrium constant, is known to be a function of temperature. So, K P is a function of temperature. That is a known thermodynamic property. So, now, if we know the Gibbs free energy at standard condition minus delta G naught, then we can estimate K P for some other temperature or pressure and based on that, then we can estimate the mole fraction also as you can see here. Otherwise, if K P is known, we can get this. So, both of them can be estimated. Now, I would like to point out few properties of this, K P and delta G naught. If I look at delta G naught, what is this? This is the standard Gibbs free energy. So, it is at standard pressure, at standard temperature as well. Temperature we have not talked about, but standard temperature as well. So, this can be actually related to the standard free energy of formation. Now, in general, if I look at this term, delta G naught, more negative the value of delta G naught is. So, delta G naught, if delta G naught is negative, then this becomes positive. So, K P is positive. More negative the value of delta G naught is, K P is more positive. So, we have larger value of K P. So, I can say that, more negative delta G naught, delta G naught, delta G we get larger K P. Now, what does a larger K P represent? K P is higher, K P is more. What does that mean? Either this is large or this is small or both. So, therefore, what it says essentially is that, P c by P d, this product is more than this product, which essentially means that, the partial pressure typically for the products is more than that of the reactors. So, essentially what it means is that, we have more products in the final equilibrium mixture than the reactors, which essentially means that, this is the spontaneous direction of reaction. Because, if you recall in chemical thermodynamics, we have said that, all the reactions will have the opposing reaction as well. In equilibrium actually, the way we had defined in the previous courses, at equilibrium what we have is that, we have a forward reaction and we have a backward reaction. There is certain rate of this forward reaction given by say K f or say naught K f omega f and there is certain rate of this backward reaction given by omega b. At equilibrium, we say that, the rate of forward reaction is equal to the rate of backward reaction. That is what chemical equilibrium is, but now what we are saying is that, if K P is positive and more positive, then the extent at which we get this equilibrium, it will have more products than the reactors. So, the spontaneous direction is this, not this. So, this is favored, this is not favored if you have higher value of K P. So, that is what is essentially the definition of chemical equilibrium is and now, we show that, if you have higher value of K P, then we get more of the products and the rest of the reactants will remain at equilibrium. And therefore, if I plot the variation, let us say of concentration of C, initially it will be 0, it goes like this whereas, say concentration of A, it goes like this. So, this is the equilibrium. So, at equilibrium, we see that, we have more of the products and less of the reactors, just because that will be possible if delta G is negative and more negative. So, this is something that comes out from this discussion. Now, one more thing I like to point out. Here, this discussion we have is primarily based on delta G naught, which is the standard Gibbs free energy. Now, in general, we have this term delta G, which is equal to delta G naught plus R T, delta G naught plus R L N K P. That is why change in Gibbs free energy for a particular system at any arbitrary pressure and temperature. Now, when this term is 0, when delta G is 0, which means that, the Gibbs free energy for the product, total Gibbs free energy for the product is equal to the total Gibbs free energy for the reactors. In that case, the reaction will not have any tendency to proceed in any direction. So, reaction actually takes an equilibrium position. That is why we say that, at equilibrium, this is because of the fact that, this Gibbs free energy is the measure of the tendency of a reaction to proceed in particular direction. Therefore, when this is 0, then it has reached equilibrium. It will not have any spontaneous tendency to proceed in any preferred direction. Now, let us consider the general chemical reaction we have been talking about so far in this course. This is how we have represented the general chemical reaction, where we have N species and represent the general chemical represented by M i. This is the reactance. So, this is the stoichiometric coefficient for the reactance, stoichiometric coefficient for the products. So, this is our general chemical reaction that we have been describing so far. So, for this general chemical reaction, now how do we represent K P? So, K P will be equal to, we have represented it for a particular reaction A plus A A plus B B equal to is going to C C plus D D. Now, if I look at this reaction, then this will be equal to the product pi i equal to 1 to N partial pressure at equilibrium of species i given by P i e and then the difference in stoichiometric coefficients. So, the difference in stoichiometric coefficients given like this, this is the partial pressure at equilibrium of species i. The product of that gives us the value of K P. Once again you will see that when we are in the product side, mu i double prime is greater than mu i prime in the product. So, therefore, that comes in the numerator, whereas in the reactant mu i double prime is less than mu i prime. So, that is negative. Therefore, that comes in the denominator. So, it is exactly same as the definition that we have given here. So, this is how we define K P for the generic chemical reaction we have been discussing so far. So, now, this is our definition of K P. By the way, we have taken one approach to prove these things using some other thermodynamic approach also we can prove that. So, I am not going to the details of that. Various things we can use, we can use even the entropy balance and all. Now, let me point out here one more thing. How do we get this value minus delta G naught for a chemical reaction? We want to estimate the change in standard Gibbs free energy for a given chemical reaction delta G naught. How do we get this? Actually, the molar standard free energy of formation, molar standard free energy of formation, which is designated by G F naught is tabulated typically in thermo-chemical tables. Now, first of all, what is our standard free energy of formation? This is similar to our standard heat of formation. So, when we say standard, it means that the formation reaction is from its elements in their standard state at standard temperature and pressure. So, standard free energy of formation is the free energy change for a given chemical reaction when a particular molecule is formed from its elements at their standard state at standard temperature and pressure. Now, we have already incorporated the standard pressure here, P naught is already there and P naught is the standard temperature. We can assume that the reactions occurs at standard temperature. That is how we have been working out these details. So, if you consider the reaction, isothermal reaction is at standard temperature, then this definition is applicable to this case as well. Therefore, we can get now for a given reaction delta G naught. Now, this reaction actually involves i species. So, every species has its own standard free energy of formation. So, we can estimate the total standard free energy of formation for the given chemical reaction by looking at the standard free energy formation for this reaction. So, then that will be equal to sigma i equal to 1 to n. The total standard free energy of formation for the reactants, sorry products minus the total standard free energy of formation for the reactants. So, once again I have to reiterate the point that these values are typically tabulated and we can take a step forward and say that G is equal to H minus T s. This is how it is defined. This is the standard energy of formation can be tabulated. This is the standard temperature. This is the entropy of formation. So, entropy change from the absolute zero temperature is also tabulated. So, we can use this also to estimate that or directly we can look at the tables and get this. So, this is the way we get the standard free energy of formation for any, sorry the standard free energy change in any chemical reaction. Let us take an example to explain this little more. Let us consider a reaction where oxygen, which is of course, in gaseous state converts to ozone in gaseous state and this is the equilibrium reaction. That is why I have given it this kind equilibrium sign. So, for this reaction we have to find delta G naught is what. So, we can get delta G naught by using this equation. Our product is ozone. We can look up at the tables and get the molar free energy of formation of ozone. This is 39.06 kilo calorie per mole and that of the ozone oxygen. Now, what will be the free energy of formation for oxygen? Zero. So, this is zero. So, this comes out to be equal to 39.06 kilo calorie per mole. So, that is the value of delta G, which we can get by looking at the thermo chemical tables. We can take a step further and estimate the K p for this also. So, we can say what is the K p for this reaction? For that let us first define K p. K p is equal to the partial pressure of oxygen divided by partial pressure of sorry partial pressure of ozone divided by partial pressure of oxygen and from that definition of K p from here, since we are talking at equilibrium. So, this term is zero. So, we can get minus delta G naught equal to R T L N K p. Therefore, K p is equal to e to the power minus delta G naught by R T. So, from here delta G naught by R T. So, from here delta G naught I have estimated R is the universal gas constant. So, which is known for this unit is 1.982. T is the temperature in Kelvin. So, the temperature we are talking about is standard temperature. So, it is 298 Kelvin. So, I put all these values. So, I will write it here K p is equal to e to the power minus delta G naught which is equal to 39.06 kilo calorie per mole divided by 31.06 kilo calorie is 10 to the power 3 divided by 1.982 sorry 987 that is the value of universal gas constant in V P S unit F P S unit times 298. So, this gives me the value of K p to be equal to 2.25 10 to the power minus 29 that is the value of K p. So, if the value of K p were given we can estimate what is delta G naught also. So, this is the example we wanted to talk about. Once again let us go back to our discussion on large values of delta G naught. As we have said that if we have large negative values of delta G naught then K p is large. Now, what does large K p means or large K p represents large K p which is when large negative delta G naught. In that case for these conditions when the value of K p is large once the reaction starts the conversion of the reactants to products in their standard state of course will be quite complete at equilibrium. So, essentially we have mostly products in the final state and very little reactants. In general for any values of delta G naught K p is typically greater than 1. So, for this cases when delta G naught is like this very large values K p is typically greater than 1. Now, let us now continue with this discussion and let us look at some examples. Look at so far we have defined K p, but what we wanted to do was to get the additional equations that we are looking for. So, now let us continue with this discussion and try and see that how do we get those additional equations because that was our initial objective. So, for that once again let us look at general chemical reaction first of all. Let us say that what we have is a chemical reaction after if I look at the product final product. Let us say that the final product is given as this is the total change in the Gibbs free energy at standard state where these are any species i. So, we are looking for the looking at the general equation. So, this is the total change in Gibbs free energy. Now, this is equal to the essentially change in the Gibbs free energy of every species. So, this is nothing but equal to let us consider first of all the reaction that we are considering. Let us say that we are considering this reaction. So, this will be S naught B and S. So, this is the chemical reaction we are considering. So, total change in Gibbs free energy at the standard standard Gibbs free energy is nothing but this. Now, in this case for the particular case that we are looking at the change in number of moles of A and B are essentially negative. So, d n A is equal to minus A d n B is equal to minus B, but for R and S these are positive because these are the products. So, therefore, in general if I generalize it total change in the Gibbs free energy is equal to the change in Gibbs free energy for the products minus the change in Gibbs free energy for the reactants that is in general. Now, we have defined all of this. Now, for every reaction for every reaction at equilibrium we can define K p as P R by P naught to the power R by P S P B by P naught B. We define K p for this reaction like this. Here P naught is the pressure. We are considering it at standard state. So, that is why it is atmospheric pressure. So, therefore, this reduces to the same expression that we had earlier seen. Now, this is applicable to any reaction at equilibrium. So, let us look at some common reactions and see what happens. In general, let K p sub i is the equilibrium constant for for for for for for for for for formation of species i. That is how we are defining. So, now, if we have few reactions occurring together let us say one of them is this, other is this and other is this. These are all equilibrium reactions occurring together. Then we get K p from this reaction. K p for for for this is the formation reaction of H 2 O. So, K p for H 2 O is equal to nothing but P H 2 O divided by P H 2 P O 2 to the power half. So, K p for H 2 O is equal to nothing but P H 2 O. Similarly, this is the formation reaction for H. So, I can write K p H is equal to P H upon P H 2 to the power half and this is the formation reaction for O H. So, I can write this as. Now, let us consider some other reaction for which I do not know the K p. Let us consider another reaction. Let us say I consider this reaction H 2 O is equal to H plus O H. Now, what is the value of K p for this? For this reaction it is P H P O H by P H 2 O. Let us write this as P H P O H upon P H 2 O times P H 2 O divided by P H 2 P O 2 to the power half upon P H 2 P O 2 to the power half. I can write it like this right. This can be then further simplified as equal to P H upon P H 2 to the power half into P O H upon P H 2 to the power half P O 2 to the power half P O 2 to the power half divided by P H 2 O upon P H 2 P O 2 to the power half. Now, this term here is this. So, this is my K p H. This term here is this K p O H. This term here is this K p H 2 O H. Now, this is no longer a formation reaction. This is where formation reactions. What we have seen now is that from the formation reaction we can get these values for different species. Now, when we go to a complex reaction like this or different reaction like this, we can get the K p as equal to for this example. It is K p H times K p O H divided by P K p H 2 O right. So, just in this expression for of K p, the partial pressures appearing here are just replaced by K p right. Therefore, for this reaction we can get the expression for K p. So, that is an important tool which we can use to get the K p. And some values. So, I will stop here today. In the next class, first what we will do is we have defined here K p the equilibrium constant. There are some other definitions of equilibrium constant like for at based on concentration etcetera. These are interrelated. So, first we will give some more definitions of equilibrium constant and then continue with this discussion to show how do we get finally, the product composition. And we will also solve some problems in the next class to show how to estimate the product composition.