 liquid phase equilibria liquid-liquid equilibria you said you could have this is G before mixing we have the ideal mixture we envisage the case where we pointed out that if you this is let us say x1 alpha this x1 I take a mixture of 1 and 2 with of composition x1 then the such a mixture before mixing has this gives free energy if it formed an ideal mixture of 1 and 2 it would have this free energy at B if I accepted porters model it would have hypothetically this free energy at C but if I take these two points the lowest points in the free energy curve and draw a common tangent you get this point D fact it is a geometric exercise to show that there exists a common time but if you have a mixture of composition x1 one phase of composition x1 alpha another phase of composition x1 beta you have a mixture of these the free energy would then be at D now this ideal mixing does not count because this is the only hypothetical case if they mix ideally then they are miscible in all proportions so ACD are the ones that count in D has the lowest free energy so you will have to say G at D is less than G at C is less than G at A therefore the equilibrium point is D so what you then have is alpha and beta this is x1 alpha this is x1 beta and the extent of the two phases the amount of alpha phase present and the amount of beta phase present is going to depend on mass balance what you start with so if you have these two compositions as I said for example in the case of ethyl acetate and water this is not a good model for ethyl acetate water you will have to actually use a more complicated model and the most successful liquid liquid equilibrium model is the NRTL model I said the algebra gets a little out of hand so I would not do the you might say the most to describe well only so as far as this course is concerned I will put out a table of G excess I forgot to do that I do not know if it is already there I told Krishnamurti to put up a table did you see the website is there a table of G excess you have to look at the course website and I will ask Krishnamurti to put it up there you will need to know those models specifically for each system what I would do is in the normally in the exams you specify exactly what model to use so what I will do is simply give you a problem there are suggestions on the based on past experience as to what models use for what cases for example alcohol hydrocarbon systems it would be either Wilson or Van Laar so like this there are regular suggestions for many kinds of systems these are tabulated in Wallace there is no unique answer but you will have to take the probable model and work with it so in the I think the next they are all the assignments are in but I will give I will put in some variations for it these assignments are same as last year's assignments so what I will do is simply put in some variations to emphasize only a few models about 10 models you have to really know and other than that unless you are doing research you do not need to know more than that what I want to do is I want still simply stop with this at this stage that is you got this these two at equilibrium this is what you have to first do for water ethyl acetate in the problem of other directing plus water plus ethyl acetate you will have these two you will then have a ternary system I have a water layer and ethyl acetate layer and water rich layer in the ethyl acetate layer and you are interested in this case and it is all you this is let us say this is three this is one actually this is one this is two by normal convention you put the most volatile compound as one in the less volatile as two and then three so this will be X one this is the ethyl acetate rich layer this is X one beta which is less than X one alpha actually it will be much less and then you will also have X three alpha and X three beta so what you have is a system in which you form these two phases and then you pull out the ethyl acetate phase in the water phase separately you allow it to settle you draw of the water phase and the ethyl acetate phase so if you draw a mass balance around this if this is the feed with composition X one F and X three F then all I have to do is write a mass balance around these three this will be extract layer E and this is say the water layer so you will get F is equal to E plus W then F X one F actually the layers are alpha okay does not matter E is I have marked as X one alpha plus W X one beta and then I have a balance on other direct in F X three F and then I have chemical potential of one in the alpha phase same as chemical potential of two because I am talking of small amounts of Azadiractin the equilibrium will not be affected it will be affected to some extent these numbers will change but broadly speaking you will have two layers with an ethyl acetate rich layer and a water rich layer exactly like this in the final compositions will be close to these except you have to account for Azadiractin's presence essentially you will have these equations mu one one and two are actually solvent like substances so you have gamma one alpha X one alpha equal to gamma one beta X one beta this will be again gamma two alpha X two alpha is equal to gamma two beta X two beta here we have a slightly different expression we had log of gamma three X three beta by gamma three alpha X three alpha remember we had this in terms of the heat of solution this would be the difference in the heats of solution in the alpha phase in the beta phase because mu three would be mu three star plus RT ln gamma X and mu three star by T you will different divide by T and differentiate with respect to T so you will get minus H three star where RT squared if you treat the difference between H three bar infinity in the alpha phase and the beta phase the difference between these two as constant you can integrate this so you will get essentially delta H this will produce a minus sign so this will be delta H alpha phase minus delta H three in the beta phase by RT this is a constant plus some constant C1 it is you write the chemical potential modules for the models for the solute remember these two are solvents this is solute the solute model will have mu three star so you divide by T and differentiate with respect to T you will get H three bar you recognize that when you rearrange the equation that H three bar actually refers to is independent should be independent of composition so you will go to the limit as 3 goes to 0 so you will get an infinitely dilute solution in these cases when I say alpha phase in a phase that contains X1 alpha of ethyl acetate the beta phase is water ethyl acetate mixture with composition X1 beta as long as the solutions are dilute you can take this binary equilibrium as giving you the compositions of the two solvents if you like effective solvents as far as calculation of delta H three in practice what you do is just treat this is another constant so this will be of the form A by T plus B you just treat it as some constant so what you will need is data at two points and you will have to solve these equations normally because it is a very dilute solution as X3 goes to 0 you have gamma 3 going to 1 so gamma 3 beta by gamma 3 alpha will be taken as 1 so you will have X3 beta X3 alpha so you have a total of variables are X1 alpha X3 alpha you do not have to list then X1 beta and X3 beta I do not list X2 these are the four variables that I have to solve let us first do the degrees of freedom degrees of freedom will be in this case you have water ethyl acetate and as I direct in three components two phases plus two so you have three degrees of freedom of which you fix T and P so one degree of freedom left that means if you specify X1 for example if I want the final concentration you know I do and separate analysis I take the ethyl acetate with the other direct in a net and I do analysis at what concentration will it is it economical to recover the Z actin or how much does it cost so I fix X3 alpha for example as an objective if I fix that the whole system becomes invariant everything can be calculated about the equilibrium so this is really a specific designer specification the user will say I want a Z actin in ethyl acetate at such and such a concentration I will say normally user specifies this is also user specified but user specifies X3 alpha in particular this is as far as the thermodynamics is concerned then you have the mass balances here the thermodynamics gives you X1 alpha X1 beta X3 alpha X3 beta these are known to you the feed composition is known but this is also under control because the feed comes from what you do is add this feed you also add actually this feed comes from maybe this complete the picture you should really do this problem with and in our TL calculation this is water plus as a direct in this is the ethyl acetate this is what comes from the extraction unit so what you have is neem seeds which are ground extracted with water that water contains as a direct in plus other components normally you in this analysis we are ignoring the other components if you do an actual exact calculation you have to analyze the neem seed get all the components in it that will dissolve in water then do a multi-component analysis but you have this you add a certain amount of ethyl acetate now the ratio of this this to this we will call this primary feed and this is you have already used E this is solvent addition yes so this P plus S is also equal to SF and then P X3 P is equal to F X3 F and if you like P X2 or I do not have to do the balance on 2 doing 1 and 3 all the time so S is equal to F so what you have control over is this ratio of P to S so you must use this P to S so that you get the right F here which leads to finally this X3 so look at what is specified this is specified all the rest are under your control here T and PR specified that is all but once that is specified the equilibrium gives you these values so these are calculated from these three equations you need experimental data on A and B that is you should have made the measurements at two temperatures that you get A and B these are available so once these are available you can therefore calculate out of this you can calculate all these quantities if these quantities are known I will just stick them these are known from phase equilibrium look at the totality of the problem and this is actually a simplified problem it just looks complex but the principles are trivial it looks complex because along with azadaractin you have at least 12 other compounds that come in and you have to first make sure that none of those other 11 compounds other than azadaractin you know it kills the plants because you are finally getting a pesticide you are getting a biopesticide that is friendly suppose you isolate some other component turns out to be deadly to the plant that you are growing then your biopesticide instead of killing the pest it is killing the original plant that was designed to protect so you have to first do a spectral analysis of those compounds make sure they are not lethal if they are then you will have to have another extraction unit to remove those or you must know the concentrations below which they are not lethal so you have to take those components into account when you write your models you would not write just 3 you will write balances for remaining 11 but simply more detailed calculation because all of these independently solubility of each of these are known in the form a by t plus b for many many compounds you have this data in fact there is a database called dekema I think professor Kannon has a limited version of it this is computer designed package called aspen which comes with the thermodynamics package in the thermodynamics package there is a lot of data in it dekema is very very expensive for a university it is meaningless to buy it although it is a little cheaper for the university than for the industry if you buy it you can get exact numbers but since we do not earn the money out of it there is no point buying it if you buy it you must buy it with industry support we probably will get industry support and we can buy it because cheaper for us and we can use it but if you buy it at a cheaper rate you cannot use it for consultancy that is the rule but the point is anyway such data is available so if you just put in the compounds it will give you a and b for example it will give you the it may not give you to you in that form it may give you in the form log x3 in beta alone it is in the beta phase in the presence of a certain amount of water and certain amount of ethyl acetate what is the variation of solubility with temperature similarly log x3 alpha alone will be given you are simply taking a ratio in this problem so these are known a b are known it will be in the form of two constants one for the alpha phase one for the beta phase sometimes you have to put it together you have solubility in solvent a solubility in solvent b and you have to a interpolation between these two but all that the program will do there are default options that so basically once you have all this data you simply calculate the equilibrium you go back here so given this x3 you can calculate what should be x1f and x3f and you can modify those by changing the ratio of p to s it just works backwards so you start from the other end where the user specifies what the final product he wants is then work in fact very often in these cases it will be a solid product that they want they want as a direct in which case the ethyl acetate will be evaporated and sent back to the use here so the whole process at chemical plant actually looks very very confusing so I told you the first thing is some use is usually a ball mill of some kind so these are neem seeds and even here they will be picking there will be a whole lot of labor picking off other things because you do not get neem seeds as such you get with a lot of garbage and then you have to select the neem seeds you put it in a ball mill you pass it through after its ground the powder comes out and then what you do is add some kind of a solvent in this case water and if necessary the water will pass through a heater depending on what temperature you choose to use so you have you take out the extract here this extract contains this is your water plus and then you add ethyl acetate go through a phase separator take this out this will contain the primary ethyl acetate so this will go here go to an evaporator and you will take out as a direct in evaporator then it will go through a crystallizer and in the from the crystallizer you take out as a direct in this solvent will go back to the evaporator is these lines that go back and forth this vapor this will be ethyl acetate that is removed that will go back here you have to recycle this you go to a chemical plant like this Chennai petroleum corporation the thing that strikes you first is that a chemical plant consists of pipelines occasionally connected to units okay so what I will do is I want to in order to complicate your life a little more I want to discuss a bit of reaction equilibria reaction equilibria what you do is generally the you have a general representation of a reaction I rate it as nu i ai is equal to 0 nu i are stoichiometric coefficients ai are species this is not a mathematical equations representation simply symbolic the stoichiometric coefficients by convention are positive for products and negative for reactions you know by experimental evidence that all the new eyes you also know from atomic theory now that new eyes are always rational numbers certain number of moles of one species will react with certain number of moles for another species these are integral numbers so you can present no eyes is rational if I have a set of reactions I write this as sum over i nu ij sum over i equals 1 2 etc r reactions I put down the same species I take all the species in the system and write the set of reactions in this form nu ij forms what is called the stoichiometric matrix if there are n components it is n by r matrix this is i equals 1 to n n is number of components in this representation what we do is to say nu ij equal to 0 whenever j is not a participant in reaction I one typical system we worked for for say long ago if you look at the blast furnace for example for making iron you have essentially iron ore lime and coke these are the things that are added coke is actually burned in the exothermicity of carbon going to carbon dioxide makes up for the endothermicity of iron ore being reduced to iron it is like a packed bed reactor but you have to actually write about 14 balances of 14 reactions there are many minor components which play an important role in iron making so I will have for example 14 species participating typically but nu ij will be 0 for many of them for example Fe 2 O 3 plus carbon monoxide giving you Fe O plus carbon dioxide I can balance this reaction only 4 components involved out of 14 4 10 of them will be 0 so you will essentially get a matrix stoichiometric matrix which is often described as parts that means many entries are zeros but you will get a matrix so this is the representation now question is what is the mathematical equation that describes the equilibrium of such reactions I want to know what the final equilibrium is in the case of the blast furnace I have two layers as well I have a phase equilibrium come reaction equilibrium I have molten iron and slag on top the slag has the calcium oxide essentially as a base but calcium magnesium oxide and there is the molten iron as well and both of these have a distribution of components and all the we have 14 components the care process consists of two parts I will tell you why it is also interesting because it has two this is more or less like a packed bed reactor I will tell you what it contains what you have is iron ore coming in and then you have actually you drop carbon particles from some height these are burnt and then you have oxygen this is cold what happens is cold and oxygen burn of course to give you carbon dioxide the exothermic is such that you have two layers this is the iron layer and this is the slag layer as the iron ore particles are actually fed from the top here form of pellets and as the ore falls it starts off as Fe2O3 so you go to FeO and in this reactor you go from FeO to iron the reductant is always carbon monoxide it is in gaseous form it mixes with these things it is everywhere when this happens you have CO going to CO2 you have to balance these reactions out then you have magnesium all this now this gas that goes here is also has a CO to CO2 ratio and you can control the temperature here simply by the amount of coal you burn per unit of over added you can vary this so you will get this reactor has been this is essentially iron oxide actually I should not put iron ore here iron ore will go here actually I forgot you know 15 years since we did this this is Fe2O3 sorry yeah this is Fe2O3 this is over I have got this backwards this reaction occurs here this reaction FeO goes in this goes here what you get here is FeO from the packed bed reactor Fe2O3 to FeO and then FeO gets reduced here and you get iron at the bottom and of course you have slag because the iron ore contains in addition it contains CAO, MGO, etc. If you do not have enough for the slag formation then from the linings it picks up the magnesium that is what happened they did not have enough magnesium and calcium they picked up the magnesium from the fire brick the blast furnace because of air using air the velocities of air are very high and because of that you cannot use iron ore fines I do not know if you have been to a steel plant must take a look at the blast furnace it is an absolutely optimized chemical reactor it is very hard to see anything but the size will impress you when the air goes up it will take away these particles so you have to have iron particles of a certain size and when you grind ore to particles that are suitable for a blast furnace you also get fines so if you go to a place like any place where there is a steel plant or whatever it is you will see mountains of iron dust so in order to use the iron ore fines you cannot use the blast furnace you have to use some process like this where the air velocity is much less because you are using oxygen so one-fifth velocity you will get effectively so it do not carry away the fines so these fines iron ore fines and poor quality core which is what we have in plenty we can use only in alternative process to the blast furnace they do not replace the blast furnace you have the main blast furnace and then you use additional you make an additional iron using these processes in the coal and reduction process an excellent process where you have two part control in the in red process there was one reactor in which both of these occurred and it was called in red because the heating was by induction in addition to coal burning and there was an induction coil around which did part of the heating but this is a much better process in turns out in this process they did not have the lining problem because they had made sure that there was enough magnesium in the input it was quite a simple correction but if you do this calculation all you have to do is I list out the 14 reactions and you can try doing the calculation the major problem is only doing a numerical calculation with these numbers some of these numbers are very small in doing the calculation with that is the problem but the thermodynamics itself is trivial and you have to keep all components in because finally you have to make sure that one component does not interfere with the other and for this is a lot of data because the iron ore industry is a very old industry metallurgists understand the blast furnace very well so they studied all the reactions associated with it one of the important things to remember in metallurgies the temperatures are so high the reactions are very fast so equilibrium calculations are even more valid because the rates are very fast they reach equilibrium very quickly so many many metallurgical systems can be designed exactly using thermodynamic data whereas in chemical practice you often have to take the rate of reaction into account it does not reach equilibrium so coming back to this is what I am trying to analyze so I have got some J will be I will go from 1 to 14 typically or even more so you need a good numerical analysis program but in order to analyze this what do I have to do I have to first recognize that stoichiometry plays an important role