 We are discussing reaction equilibria in particular we are discussing ammonia synthesis here we are discussing both reaction and phase equilibria. So we said you have these gases coming in you mix them you compress them goes through a reactor goes to the separator let us say this is the work this is my reactor I have more Q removed from here this is a condenser there is a WS dot so if you are doing an overall analysis of the process first of all let me discuss these individually as far as the reactor is concerned we said KY is equal to not KY KN I think through to the mole numbers it is we said at equilibrium we had half into 1-x this is 3 by 2 into 1-x and this is x we had essentially KN okay KN is suppose I want KY the total moles this is 2-x so you had 2-x into x divided by this is 1-x to the power half this is 1-x to the power 3 by 2 times there is a half 3 by 2 to the power 3 by 2 half of 1-x by 2-x then 3 by 2 into 1-x by 2-x to the power 3 by 2 this is to the power half this is half sorry so this whole thing is 4 by 3 root 3 can we raise this and write this as 4 by 3 root 3 this is equal to this is just from the definition equal to P times is it P or 1 by P it is P to the power – 1 it will be here so it is P times KF by KF I told you when you plot this is if you plot Y ammonia which is 1-x by 2-x this is Y in the reactor versus pressure you get a curve that looks like this as far as a separator is concerned you have Y separator is equal to effectively with P saturation for ammonia by P times exponential of B liquid – B into P – P saturation by RT so at P saturation it becomes 1 then it comes down of course like this curve is Y separator actually the Y separator would go exponentially up but it falls down if you take into a third variable coefficient so you have to do this correct I will put it in the assignment I will put in a special assignment with additional problems this region of feasible operation that is there is a P minimum and a P maximum between which the mole fraction coming out of the reactor is higher than the mole fraction required in the condenser for you to produce liquid ammonia that is the range of feasible operation this is typical of coupling in the chemical industry where this has only reaction requirements so you can alter its condition so that you get some optimal output but you have a subsequent equipment that does phase separation and it has its own requirements and you have to meet both requirements in order to meet the technological output needs you also notice that if I did this this is always typical the problem is always simpler if you can use a larger system if I use the entire thing as a system this is my plant I get hydrogen and nitrogen and I supply ammonia that is all I care about and finally this ammonia is supplied at room temperature these enter at room temperature if any heating is required I do it in between if I do that then you know that this is an isothermal process effectively 25 degrees so ws dot is equal to m dot times delta g the minus sign so all I need is g for delta g for this process that is I have to take the chemical potential of ammonia at 25 degrees in one atmosphere subtract from it the chemical potential of these in the ratio 1 is to 3 and I take x I take one mole there I take one 3 by 2 moles and one half if I subtract the chemical potentials then I get exactly delta g you have to verify what you need to do is this is the energy requirement actually I am only interested in ws dot by m dot you can take a look at the cost of energy and present T and Eb rates and ask if the price of ammonia in the market corresponds to this there will be a factor which is greater than one but you will find this is strict proportionality that is what I was talking to you about at the beginning separation controls most of the cost but it is a very important to process and this ammonia goes on into a urea reactor you get ammonia plus carbon dioxide what is urea CO NH 2 twice is it forgotten NH 2 no CO NH 2 that is what I thought CO NH 2 twice do the balance is one oxygen extra no two what else is extra C is okay O is okay it is direct synthesis is it NH 2 twice is there a H 2 left so plus water if you do I do not know if you know that in the madras refined is this ammonia comes in in one place and they have this reactor carbon dioxide is supplied to the reactor these two are mixed as always and sent to the reactor out of this urea comes out of course H2O there is a reactor bend and this is also done at very high pressure and turned out that the valve for this was located here in this is long ago what 20 years ago we have a madras fertilizers here we make this plant and they had a control room here and control rooms are not so recent I mean they have been around for many years so this is what 80s thing early 80s or so we had somebody come and talked to us there was an accident in madras fertilizer when this hot area urea see the urea was coming out here and taking a bend and at this point by erosion the pipe had completely eroded then this broke urea came in it broke the glass here and about three people three scientists were killed here in the inside the control room before they could escape and nothing afterwards everything was brought under control but it is a very elementary thing I just mentioned this after that after a great study with huge committee and all that they said the mistake lies in placing the valve here this valve should be here because valves are always inspected the inside of pipelines are not inspected regularly so when the valve corrodes it will start leaking as soon as it leaks you can fix it in this case you do not notice that this is corroded at all till suddenly the whole thing blows because this valve does not leak okay let me discuss a couple of other problems is one other I wanted to discuss I brought some data and the second is isomerization this is actually more typical it is a very simple problem this is butane to n butane I mean n butane to isobutane then 2% HCl this must be coming from another stream in isobutane here they have the question is the temperature is given 300 degrees it is given Fahrenheit but does not matter and delta G0 is known for this reaction at this condition 97.5 calories per gram more the reaction is n butane to isobutane this is just calories per gram more not kilo calories because it is just an isomerization reaction so delta G delta H are very small next problem I have done to turn these things wrong then the data wrong so I have to go back and look at okay we will just look at this problem again in this case you have all I have to do is write down here butane is the n butane then isobutane and HCl so let us assume that I have 1 mole here and 0 actually it is not 1 mole and this thing 0.93 moles and 0.05 moles initially at equilibrium I have 0.93 minus x this is 0.05 plus x and you can put down HCl is to or 0.02 0.02 total moles remains unchanged you have KF is equal to exponential of minus 97.5 by RT delta G is positive or negative you return a positive number does not matter it is a small number so again your KF would be simply P P by NT to the power 0 no change in number of moles so pressure and total moles cannot affect the reaction so you cannot alter this reaction by altering inerts inerts is one form of control that you have then there is no point changing the pressure so it will be carried out at atmospheric pressure times KN KF if you like since pressure does not matter nobody will bother to use high pressure so KF will be 1 and then you have only KN is 0.05 this is the plus x by 0.93 minus x that is it is no nothing else this is equal to KF KF course is constant this is known this is 1 this is 1 so all you have to do is calculate x just add these two you get 0.05 plus x by 0.98 is equal to KF by 1 plus KF then you have a KX calculation you have to use R and T in the right units do not make make a mess of that the turbulent reaction equilibrium is if you use somebody puts in 0.08 in litter atmosphere since of calories in your number will be so different that I cannot give you any credit for it because ultimately in engineering the order of magnitude of the number is important and in these cases since you have never worked in the industry of no feel for those numbers so you will have to be familiar with the numbers and make sure you do not make mistakes in the units normally for number numerical mistakes I do not give I do not punish you but is the which order of magnitude of you will get a minus 1 so be careful as far as that is concerned so these are the KF is exponential of – ? G by RT KP is 1 because pressure has no effect you would be foolish to use any high pressure right essentially P has no effect industry never uses therefore P would be approximately 1 atmosphere whatever is convenient usually low pressure will be used and the invariably chemical engineers and chemists are blamed for all pollution thing to remember is that you are not polluting there for your own benefit you are actually minimizing the pollution that anyway would occur because if finally a plastic bucket is made only because the consumer wants it the question is making the plastic bucket what is the minimum pollution you will create I mean you cannot not pollute the environment if you want a plastic bucket okay let me discuss this other problem is both phase and reaction equilibrium problem told you about pluromethane I forgot to bring the data with me this is actually a consultancy that I did many years ago it is quite long time since I did any consultants what you have essentially is a reactor in which you have methane and chlorine these are not mixed premixed usually because chlorine is a dangerous chemical so you have one reactor which has all the provisions for safety so you send them in otherwise normally the ingredients are premixed because in chemical reaction engineering the important thing is mixing the product that comes out is several things you have four reactions I mean it is essentially corresponding to the number of chlorine atoms you can add CH3Cl plus HCl then you can add you can add more chlorine you get CH2Cl2 plus HCl CHCl3 plus HCl and then CCl4 plus HCl essentially in the reaction you will just get I will write sake of HCl plus HCl comment all so you get all these you get all possible forms of chlorinated methane now this whole thing is then sent to a distillation column distillation column has many trays what you do is pull out at different points a component this will be essentially CCl4 because it is the least volatile of the compounds when I say CCl4 it would not be pure CCl4 it will be 99.9% and so on the other components will be present and on top of course you will get HCl and then CH3Cl was I need one more thing CH2Cl2 and CH3Cl3 in general in all distillation columns you have certain amount of the top this thing is fed back what you do is actually not take out at CL like this to pass this through a condenser and part of it is taken out as product part of it is recycled this is called reflux you reflux the product back and by controlling the reflux you can control the composition of the products you will do that anyway in mass sense similarly this bottom product is also not taken out as is normally what happens is you have to supply energy somewhere you take out energy here this is a condenser then you have this is called reboiler it reboils the bottom product if you like and part of it is taken out and the remaining goes back as vapor to the distillation column normally because the latent heats molar latent heats of compounds are somewhat equal you will find the amount of heat taken out in the condenser in the component of heat are comparable but this is at a higher temperature you are supplying the heat here you are getting it at lower temperatures so there is a loss of availability that is there is a loss of energy in terms of what is available for you to do work at higher temperature the quality of heat is better than right this is actually the this part of it is fairly straight forward you do these reactions you have to assume that x1 x2 x3 let us assume x1 moles are formed x2 moles x3 and x4 and we will call this one two three four call this five if you like this is six this is seven so what you have in the product reactor product is what counts at equilibrium assume you are doing a calculation at equilibrium so you have CH4 you will get essentially CH4 will be let us start with one mole of this and a moles of chlorine I want to know if I can choose this a suitably so you will get one minus x1 minus x2 minus x3 minus x4 that is the methane part the chlorine will be a one mole of chlorine will be used up here you will have two moles used up right CH2 Cl2 plus 2 HCl or 3 by 2 right so you will get minus x1 minus 3 by 2 x2 maybe we should write this out for every mole here you have got two moles of chlorine used up now here it is half sorry a minus x1 by 2 minus x2 then CHClC this takes 3 by 2 moles minus 3 by 2 x3 minus 2 x4 pardon no CH3 Cl then here also I must so all of them must be corrected this one is x1 this one is what Cl2 2 HCl so Cl2 plus Cl2 2 then next one next one will be CHCl3 so 3 by 2 here and 3 HCl right so 3 and 4 this one I think that is all right yeah it is always best to be careful with reaction equilibrium do not carelessly write what I wrote but I mean if your life depended only you will spend time writing it correctly so it is not then HCl is x1 plus 2 x2 plus 3 x3 plus 4 x4 right 234 correct and then you have to write these down of course CH3 Cl is x1 CH2 Cl2 is x2 CHCl3 equal to x3 CCl4 all are in the gas phase so you can do a summation you will get 1 plus a minus okay all of it is balanced okay fine so essentially if there is no is there is no change in moles in any of the reactions so you have to write down for each of them Kf1 is equal to let us assume that the pressure is low it is because none of them are affected by pressure you would be foolish using any pressure it will be you will necessarily use a low pressure so your Kf1 will be essentially P by NT will be equal to KN right Kf is 1 KN times P by NT to the power change in number of moles because change in number of moles is 0 this is simply KN KN1 is you have the product it is x1 times HCl this is x1 plus 2 x2 plus 3 x3 plus 4 x4 divided by you have CH4 which is 1 minus x1 minus x2 minus x3 minus x4 and Cl2 was yeah this is equal to of course exponential of minus delta G1 0 by RT so you will simply similarly write KN2 KN3 and KN4 you would not write these out in each case this will be exponential of minus delta G2 by I will say and so on so if I give you the temperature you can essentially solve for x1 x2 x3 etc there is no effect of the total moles but you still have an effect of A on the tell me what will determine in what considerations will you base your choice of A on this common sense that is your desirable product but from here you are using two raw materials simply because of mass action you will use the cheaper material in excess so what you will do here is really use the cost of the raw material and also chlorine in excess is dangerous for the environment so in this case you would go with methane so you would go with the minimum chlorine requirement so you probably use a stoichiometric amount of chlorine required but it is a useful computation to do to see how sensitive the calculations are to the ratio A is to 1. Coming back what you do is solve this for the reaction equilibrium and this again is coupled to this tell you another interesting anecdote here this is chemical industry practice is full of this only thing is you have to do careful calculation the calculations are trivial in thermodynamics here you can design this distillation column you will do this next semester and this is not a problem at all because this mixture happens to be an ideal mixture because these are very similar compounds these are chlorinated methane's and they all mix ideally so gamma in this case in this case fee all your fugacity coefficients everything will come out to be one so you are only treating the system that is you have liquid phase because you are using a distillation column you are taking out a liquid product this product is CCl4 and they were making CCl4 we are making CCl4 of 99.99% purity Gujarat fertilizers started a new unit and they were supplying 99.9999 purity and this last two nines are important for the pharmaceutical industry their purity is of great importance and you will find that if you want to go from 99.99 to 99.9999 you will the number of plates will increase tremendously that last bit of purification is what requires large number of plates you need a large number of equilibrium stages because your incremental purification is very small manager use a general manager there and he called me and said we have this problem will you look at it so I look at looked at this problem calculate the distillation column design calculate the purity that you should get they had some 74 plates in the column for 99.9 I calculated from theory Treybal you have a Mars transfer book practically a worked example you can do the calculation straight away so I did these calculations and I said the 43 plates will do you got 31 plates extra he said it can't be then I also told him if you do have 74 plates as you tell me your purity should be 0.49 it can't be what you are getting I told him 30 of your plates must not be working the way these works I told you what you have is every plate looks like this you have liquid and this gas vapor comes from here it has to bubble through this liquid cannot go out this way it will go out I mean the vapor will go out this way so it bubbles through the liquid and escapes to the next each of these is called a bubble cap you have a cap here so when the when it there is a hole in the plate and the vapor comes through there is a cap preventing it so it will have to go out and come through the liquid but the caps are not in place this vapor will shoot right through and you won't have contact to the liquid two days later they opened in exactly 30 plates they had forgotten to put this cap this was assembled immediately after the war and they had the workers had forgotten to put that cap there but because the market requirement was only 99.99% and it was working well so they never bothered I mean it was so there are many practical problems in operation but here what you will do is you calculate the reactor contribution this in calculation you have to first maximize your CCL4 production because in this particular case the other products were not so saleable depends on the market conditions I mean the very little demand for chloroform and most of the demand was for CCL4 these two were not in demand at all so you will try to reduce this as much as possible do the chlorination so that you take this more to X3 and X4 and X4 will be the maximum so you can play with this and ask what will alter X4 most of the times you can't control it because your a is the only variable that you had in your equations by changing a you can't alter much so you will be stuck with a certain fraction of CCL4 which you have to extract.