 We will discuss reaction equilibrium, I have brought some data just illustrative examples I will give you 2, 3 these are part of your one is the classical problem is ammonia synthesis second problem isomerization fact I should probably bring you some data from a consultancy I did long ago is chloromethane, chemplast the sun mark chemplast group has a huge chloromethane we will discuss it it is chlorination of methane you get all the 4 compounds up to CCl4 very funny actually it was long ago the problem is still I mean the same kind of problems arise again even now I will discuss it later let me here you have n butane going to isobutane is never so clean in actual industry this for example some data 93 input gases 93% 2% HCl 5% isobutane is there is a little isobutane in the input itself question is what you get here isobutane is the major thing that you are looking at here and normal butane you want to find out what the mix you will get is the temperature is given 300 degrees F and ?g0 is given okay then take another 3 problems I will put down all of these are fairly straightforward this is benzene plus hydrogen giving you C6 H12 this is cyclohexane it is a hydrocarbon mix but look at the process incidentally the reason this is so expensive is because hydrogen is not a raw material that is available right it is always available in the form of hydrocarbons which you have to split you have to produce hydrogen in various ways in the cost of producing hydrogen turns out to be very very high now if you have this you usually have a catalyst usually the ratio is in this ratio 3 of hydrogen to 1 of nitrogen that is the ratio here so you have ammonia nitrogen and hydrogen incidentally this is also coupled to a phase equilibrium problem so I may as well discuss it in a coupled fashion if you actually look at the process you have nitrogen and hydrogen coming in maybe I should draw a mixer here only a symbol in the ratio 1 is appropriate ratio this is a mixing unit you send it in this is the reactor this is a catalyst you get this typically if you we are going to do the calculation you will get about 10% ammonia production and only 10% of the raw metal of nitrogen and hydrogen will be converted to ammonia this ammonia will have to be condensed typically the temperature here I think I have the temperature for the data here since 500 degrees k original process was 450 degree centigrade improved the catalyst therefore they can get the same rates at lower temperature remember that catalyst is not going to give you a different equilibrium okay the equilibrium conversions will be the same at a given temperature but you will get the rates will be different so you have 500 degrees k typically the pressure is 70 atmospheres that means these gases I should have put in a compressor here do that sorry so you put in work the feed to this comes from a mixture this is nitrogen and hydrogen they are compressed and sent in here and you have a throttle valve you have a separator normally you have a cooler or a coolant cooler come separator what this does is condense out the ammonia this is liquid ammonia and then the gases the part of it is condensed the remaining gases must remember there is already a lot of nitrogen and hydrogen here these will have to be recycled this is liquid ammonia simply because at room temperature ammonia happens to be a liquid and it is also very convenient to separate it out this way because this is liquid the rest are gases you can separate them out so what I need is the following in the actual process I have to calculate reaction equilibrium here I have to calculate phase equilibrium since you already done the phase equilibrium let us look at the phase equilibrium the solubility of hydrogen and oxygen hydrogen and nitrogen in liquid ammonia is negligible so we will treat ammonia as pure phase so we look at phase equilibrium in separator I am looking at ammonia alone let us say nitrogen is component 1 hydrogen is component 2 ammonia 3 so I have p y 3 times p 3 this p is the pressure in the separator will market pl because it is a low pressure there is a high pressure part of it this will or some other symbol p s for separator p subscript s okay p separator is equal to we will get p 3 saturation p 3 saturation pointing correction for component 3 gamma 3 there is no gamma 3 that is all because the liquid is pure so gamma is 1 pure liquid phase so this should be equal to this so what is y 3 equal to what we will do is for the purposes of this calculation we will take p 3 by p 3 saturation by p 3 is approximately 1 you can do this calculation but anyway let us just take this right now so if you look at y 3 y 3 is simply p 3 saturation by p in the separator thing I will forget this subscript I say it is a damn nuisance to carry this just keep it as p this by this into exponential of v 3 or I will write I will also write an expression for this sorry let me write this in terms of virial equations we will use virial equation use Lewis and Randall rule Lewis and Randall rule is a good assumption that means p 3 in the mixture is the same as p 3 in the pure state so in the gas phase I have hydrogen nitrogen and ammonia p 3 for ammonia is the same as if I use the virial equation this is simply exponential of B B is the second virial coefficient for ammonia because it represents you put two subscripts on the second virial coefficient we will write B 3 3 p saturation will be p 3 saturation by RT by p 3 so minus p by RT this is p 3 saturation by p 3 the pointing correction is v 3 liquid into p minus p 3 saturation by RT y 3 is p 3 saturation by p this is the pointing correction then I must add this this is B 3 3 into p 3 saturation minus p so p minus p 3 saturation so it will simply be v 3 liquid minus B 3 incidentally this phase equilibrium problem arises in many many situations I will tell you see here I have taken I will give you an example let me plot this first and then describe the physics of it at a given temperature incidentally there is some common sense conditions the coolant you can use in large amounts is only water in water at room temperature because this water will be put on top of a cooling tower it will come down a cooling tower it will cool by evaporation so you will get water at what is called the wet bulb temperature little lower than the room temperature so typically 80 degrees F you can take in say 30 degrees C so if you take about 30 degrees C the maximum temperature to which you can cool this tower will be about 35 you have to give a difference so 35 degrees will be the practical temperature here so this will be at this whole graph will be at say 35 degrees C this is fixed I fixed the temperature and I ask how does the composition vary with pressure as the remember some things v 3 liquid is positive number B 3 3 is the second variable coefficient which is a function of temperature this can be plotted from experimental data if you plot B versus T you get a curve that looks like this typically up to what is called the boil temperature at which the variable coefficient is 0 second variable coefficient exactly 0 this is called T boil after the famous scientist boil below T boil B is always negative and T boil is fairly high for most substitutes it is well above the normal boiling point so this is it is usually below the critical point but about the boiling point but this is very high so B 3 3 is generally negative the idea being this is the first correction to ideal gas behavior ideal gas behavior tells you PV is equal to RT the next correction is P is equal to RT by V I am sorry I should write it this way P is equal to you writing it as RT by V plus B by V B by V into this is okay PV by RT is equal to 1 plus B by V right so if I do this approximation you just want B by RT is it into V yeah right thanks this is this is V by RT getting it all backwards I will write it out you know I could make such mistakes V by V so the pressure itself or take this is RT by V which is the ideal gas pressure plus B into RT by V squared this is the pressure that you would have if it was an ideal gas but if there are attractive forces the pressure would be less because the number of molecules that pull the molecules that collide against the wall is proportional to 1 by V the number of molecules that are pulled away from the wall is also proportional to 1 by V so there is a correction due to attractive forces proportional to 1 by V squared because it is a correction that reduces the pressure B must be negative so unless you are talking of very very high temperatures at the very high temperatures this becomes positive because you are getting into let me explain why that happens also if you look at the potential energy between two molecules molecules do not have a size this is your potential energy the derivative of this with a minus sign by convention is the force so here this region is repulsive this region is attractive so the thumb rule is if you have a molecule at the center and you have another molecule moving towards it the force is attractive up to this point and then it gets repulsive this distance sigma is taken as a measure of size it beyond that the force becomes so steep so repulsive and unlikely the molecule but actually what is meant by size if I take a second molecule to a particular kinetic energy it will keep coming towards this till the kinetic energy gets converted to potential energy when these two are equal it will stop and then it will go back so where it stops depends on the temperature depends on its initial kinetic energy so the higher the temperature the smaller the size of the molecule appears to be so effectively these sizes will get smaller and smaller so what you are talking about is if you are if you are talking about collisions against the wall if the number of molecules if these if the temperature gets very high then and the density gets high you get a number of molecules within the repulsive range so the pressure will actually begin to increase in a real gas compared to an ideal gas when the repulsive forces dominate you get an increase in pressure when the attractive forces dominate you get a decrease in pressure so B will generally be negative and then it will turn positive as you go to very high temperatures so I am telling you this because in bulk of the applications of such nature that is you are talking about the solubility of or in this case you are talking about the mole fraction in the gas phase at which you get condensation in this particular case so the mole this is the mole fraction at which condensation occurs if you want liquid ammonia you must have a minimum of Y3 of this beyond this excess over this will condense as a liquid that is how you take out a product now if you look at this mole fraction V-B33 is positive so V-B33 by RT will call it a then the form of that expression is essentially Y3 is equal to P3S by P is let us say P greater than P saturation at room temperature the saturation pressure of ammonia will be below 1 atmosphere and you are looking at pressures here that are quite large so you get an exponential increase in pressure and a fall in pressure due to this 1 by P factor clearly this is going to have initially when P is equal to P saturation there is no this is P-P saturation at very low pressures this is effectively Y3 is 1 so if I plot it will be at 1 here and then it will fall off because of this P then it will increase because of the Z power alpha P finally it will go to infinity it will go very high alpha is generally small RT is very large V3-B33 by RT is small in this point you can verify the minimum will occur when this times the differential of this is equal to this times the differential of this so P is equal to 1 by alpha Vdu is equal to Udv denominator that is how we mugged it because U by V you differentiate U by V you get V squared Vdu-Udv think you mug it differently is my daughter mugged it as numerator denominator and all that I do not know how you mugged it okay this goes to actually this is sort of you point it up here this pressure already is much later than saturation pressure and then you have a gap it will be 1 at P is equal to P saturation so it comes down here it comes to so you have this curve now when will I get condensation in that this is the equilibrium curve actually this curve is not quite correct because as you go to very high pressures I have to take the third virial coefficient into account and all these corrections are the second virial correction overcorrects for attractive forces so then it will decrease so the actual curve if you plot the actual curve it look like this it will turn around sort of flatten out at some point here the third virial coefficient is important you have to add third virial coefficient of it up to this somewhere here second virial coefficient will do so you will get a linear portion e power a p then you will get another p squared term e power third virial coefficient times p squared so the actual curve will turn like this but the principle is like this supposing you want condensation here you want to take out ammonia here if this is actual y3 at equilibrium this is the equilibrium curve you are getting ammonia out of the reactor supposing the mole fraction of ammonia from the reactor is here supposing the mole fraction y3 reactor is here then you can get no condensation at all right supposing y3 reactor is here this is case one then this is y3 reactor case two then if the mole fraction in the reactor is greater than y3 equilibrium then the liquid will condense right unless I have more ammonia than the equilibrium demands I will not get condensation so if y3 actual is this line anywhere in this region I will get condensation so if I want to remove liquid ammonia as a product y3 from the reactor should be higher than the equilibrium value then at equilibrium you will reach the equilibrium value the balance of ammonia will condense out in the liquid phase this clear so for example if I am operating at some pressure here for example 5p is equal to 1 by a this much of ammonia will have to condense out till the composition actually reaches y3 equilibrium same problem arises here in our liquefaction unit it is a slightly different problem if you are liquefying air you have carbon dioxide as an impurity and the way you do liquefaction of air is you bring it to a lower temperature and do a joule Thomson right when you do a joule Thomson expansion you get cooling but you know the joule Thomson coefficient can be positive or negative it goes into the cooling region only below a certain temperature we can work that out so what they do is say at 130 degrees k we will pass it through a needle valve throttling from high pressure to low pressure this causes cooling when you throttling through a needle valve in the needle valve you are worried about I have a needle valve this is air liquefaction problem the problem is the same thermodynamically I have here air plus carbon dioxide and I throttling through this valve this needle valve I do not know if you have seen a needle valve really what it says it is just got a pinhole through which the whole thing goes out now in this case carbon dioxide goes directly from gas phase to solid if a solid carbon dioxide forms it will plug this needle valve so what is the thermodynamic problem solid carbon dioxide vapor phase air plus carbon dioxide right I have the same identical thermodynamic problem this is CO2 forget that it is only a spec this is air plus CO2 I am asking what is the mole fraction of carbon dioxide at which solid carbon dioxide will form identical formula I will get the same curve this will be Y carbon dioxide versus pressure alpha will be different now alpha will be B carbon dioxide minus B carbon dioxide carbon dioxide by RT I can do the same calculation I will get the same result here my worry is that this carbon dioxide should not solidify if it solidifies it will block my needle valve we actually had an accident here 30 years ago in our liquefaction plant they had a small explosion I mean this is known they were careless what you normally do is the here this instead of reactor I will have Y carbon dioxide in the ambient which will be typically 10 to the power minus 2 mole fraction or 10 to the power minus 3 mole fraction this air is bubbled through mononethanolamine solution so here before it gets here they will bubble it through actually it is not this way this is MEA this is air plus CO2 of course it is bubbled through so the CO2 composition here will be Y is equal to 10 to the power minus 4 so it will be below this curve the same curve you will get below the curve you will get ambient carbon dioxide there will be no solidification you have to make sure that the composition of a carbon dioxide in air is below this lowest point at any given temperature then you are safe at a higher point you will have solidification here you want solidification you want liquefaction so you are looking at this range here if the mole fraction is here for carbon dioxide you will operate either below this pressure or above this pressure on either side the phase equilibrium curve lies above the ambient composition so you would not have any liquefaction condensation the same problem arises in another context you have if you look at oxygen this is oxygen in cylinders this oxygen or asiatic oxygen all of them make oxygen by finally bubbling it through a liquid solution to remove other impurities so this oxygen is said to be packed when it is wet that is it is bubbled through water in the final stage so it will have some water vapor so typically in the cylinder you will have H2O liquid at the bottom you do not see it the cylinders come at very high pressure there will be a small layer of water can even be few drops it can be a 1 millimeter thick layer now when you pump this oxygen out this oxygen will carry water with it water vapor with it the amount of water vapor that the oxygen carries is given by the same formula liquid phase vapor phase contains oxygen plus water vapor how much water vapor Y3 is equal to same equation now in some experiments water is very detrimental to the catalyst so you have some organic chemistry experiments where they are using this oxygen in oxidation they want this oxygen to contain water less than 10 to the power minus 5 they will give you a pre-specification they will say tolerance for water is 10 to the power minus 5 mole fraction if that is true what you need to do is operate at a pressure where this Y3 is below that value of course this has this lowest value has to be 10 to the power minus otherwise you cannot reach that value so you will choose a pressure here such that the phase equilibrium value is lower than the required value that is here the design specification is the other way around I tell you this limit I draw the limit on Y3 that I can tolerate for the water to be below that value I have to operate again between these pressures so same problem will just appear in different ways in all these cases the thermodynamic diagram is simply a beta two phases in the new equate chemical potential so let me get back here in this phase equilibrium problem I have already plotted I want Y reactor to be certainly above this minimum if it is below the minimum I cannot get any product out of the plant so what I need this is my specification this is design specification that is Y3 greater than Y3 minimum Y3 minimum is here in order to be sure you do not get ammonia in spurts say much greater than Y3 minimum so Y3 minimum is 0.03 you will probably ask for 0.1 so you need 10% so that is a specification then you have to ask what are the conditions in the reactor here that will give me that concentration in the outlet now let me tell you another thing suppose I reduce the pressure too much here or even if I reduce the pressure when I send this back I have to compress it again so I will need a compressor here so you will need more work input normally this is avoided by operating the separator at the same pressure as the reactor you have to work out the economics of it whether it is worth putting a compressor here or not but if it is at the same pressure so this pressure is fixed from this consideration the temperature is fixed so simply solving a phase equilibrium problem asking what is the ammonia you can get what is Y3 equilibrium so essentially the pressure may be fixed at some point here this is P reactor you can choose the P reactor but it is also equal to P separator so you choose this pressure so that Y reactor is greater than Y in the separator and how do I calculate Y reactor I go to this equation I have KF is equal to FnH3 by Fn2 power half F hydrogen to the power 3 by 2 this is equal to exponential of – ?g0 by RT be careful about taking ?g0 data because sometimes the data is given for 2 moles of ammonia then this will be twice this will be half of what the data is in the tables you must write it for 1 mole of ammonia otherwise you write the reaction with for 2 moles and write this as FnH3 squared times the trouble about reaction equilibria is that if you make mistakes even silly mistakes your numbers will be so different everything is an exponential so instead of exponential of – 2 you may get exponential of – 4 the huge difference between e power – 2 and e power – so this is the same as the P to the power the change in number of moles actually you recall that this P to the power was sum over ? ij for that particular reaction that sum over for reaction I sum over ? ij just sum over the species is called ? no the change in the number of moles ? no for ith reaction so you will get P to the power ? no times Kp times Ky so in this case ? no is 1 – this thing so this is – 1 Kp is a function of T and P normally for gas phase reactions we replace P by P pure use Lewis and Randall rule so you take P for the pure substances for you won't take P ammonia in the mixture you will take P ammonia in the pure in this Kp has been studied very extensively in old paper by Newton and dodge dodge was a chemical engineer so this is equal to of course this is known exponential this is F of T this is known at a given temperature so you solve this equation Ky for Y you solve it for Y ammonia let us look at the degrees of freedom in this case or let us write out that expression to see what it looked like Ky is Y ammonia by Y hydrogen to the power 3 by 2 nitrogen to the power half this is definition this is equal to some function of T which is known times pressure by Kp Kp is a function of T and P because you are using Lewis and Randall rule use Lewis and Randall rule P becomes P pure so it is no longer a function of composition these are known functions this is known actually I start off maybe I should write this in terms of N so we will start off with 1 is to 3 of nitrogen to hydrogen then I have nitrogen hydrogen and ammonia so if X moles are formed ammonia will be X hydrogen will be 1-X by 2 this is 3 by 2-X by 2 or 3-X by 2 I will write this is half-X by 2 this is you start with half and 3 by 2 moles so your total moles is 2-just 2 no 3 by 3X by 2 so this is 1-X by 2 this is 3 times 1-X by 2 this is 2-X so you can solve in terms of X so you will get essentially 1-2-X into ammonia which is X by 1-X by 2 to the power half into 3 into 1-X by 2 to the power 3 by 2 this is equal to f of T into P by Kp so you will solve for X and therefore solve for mole fraction ammonia X by 2-X in case you can see the way the reactions are written down although this looks like a complex algebraic expression X will increase if P increases and this side Kp is a weak function of pressure basically increase if P if the right hand side will increase if P increases and if P increases X will increase because it is X by something into 1-X more ammonia is produced so it will increase so if you look at the graph here I just erase this graph this is phase equilibrium the reaction equilibrium curve will go something like this as the pressure increase so this range this is reaction equilibrium this is phase equilibrium so only when the reaction equilibrium produces ammonia more than is required for condensation you can get a product ammonia out of the react so this is the range of feasible operation range of thermodynamically feasible operation I think dodge uses the notation ? for ? but you can get this K ? data Kp data from this book by dodge chemical engineering thermodynamics original references I and EC sometime in the 1940s it is available in our library it is called industrial engineering chemistry it is still one of the good journals in chemical engineering it is by Newton and dodge