 This may be part of your assignment problem also first one is this is a decon process in which you have two HCl plus half O2 use you this is for chlorine manufacture data is available delta GF 298 and delta HF 298 is available and CP average is available over the range of temperatures this is we call this one two three and four one this is minus two is oxygen okay three is water I have got this mixed up then this is one two four HCl it is minus 22 778 this is minus 57 798 minus I think I leave the CP average you do not need this skip the data the data is not given you just assume delta CP is 0 for the reaction finally is another curious information this is typical of industrial optimization by T squared you have to do this operation at some temperature and pressure the question is what is the optimum value of temperature at which you should run this reactor and this cost has been this is based on 80% conversion of oxygen to flow to this is data the first thing is you have to get delta G as a function of temperature for the reaction so you are looking at basically the reaction formulation is simply at equilibrium you are looking at 2 minus 2x this is HCl plus half oxygen giving you H2O plus Cl2 2 minus 2x you start with two moles and half a mole half into 1 minus x and this is x moles your kf is simply sorry kf times P by total moles to the power delta no then kn this of course is equal to exponential of minus delta G0 by RT this cost that they have worked out is purely based on operational considerations and they have a compressor they have a heating element in note the cost of energy for heating and so on usually given empirically for a range so I have this no I at the moment we will set kf equal to 1 this pressure to the power delta no is 2 here and 2 and a half so delta no is minus half it is just the sum over no I so it is 1 and 1 for water in chlorine in 2 and have minus 2 and minus half so you get essentially total moles a minus 2x minus half x I have so 2 and half minus half x or 5 minus x by 2 so I have 5 minus x by 2p to the power half kn is of course x squared by 2 squared is 4 this to the power half by root 2 or 2 into 1 minus x this is to the power 2 it is to the power half so this is to the power 5 by 2 so I have essentially exponential of minus delta G0 by RT oxygen conversion is simply x in this case it is the desirable oxygen conversion is given as 0.8 thermodynamics gives you some equations in every reaction engineering problem in industry you will have some special constraints they will decide how much of conversion they need what is the minimum conversion they will accept and this has been worked out at 80% conversion it is possible that downstream use of chlorine in this particular application may have had in a particular plant may require some oxygen to be present with it depends on the application to is the final product also they may have found that it is we can also do that calculation we can found that it is uneconomical the cost increases tremendously when you try to increase the conversion beyond that but we can double check that but this is so first you have to find delta G0 by RT you have to know this is a function of temperature is equal to some function of pressure and conversion so if x is known P becomes a function of temperature so the cost becomes a function of temperature alone that function of temperature differentiated and set equal to 0 will give you the optimal temperature if you look at delta G0 D of delta H0 by DT is delta CP I am going to take this as 0 then D delta G0 by T by DT is equal to minus delta H0 by T squared so if you integrate this you get delta G0 by T minus delta G0 298 by 298 I am integrating this over T you give you minus delta H0 which is now a constant or plus delta H0 into 1 by T is 1 by T minus 1 by 298 my recommendation in reaction engineering is in reaction equilibrium calculations do not simplify these numbers carry them as it is and do the calculation only in the last step and ways you make silly mistakes here and there and sometimes they cancel out you are unnecessarily calculating this now at this stage you have delta at 298 first of all delta H0 you just take this difference you have to take H2O minus 2 HCL this is H2O that is 2 HCL so it is 57 this is constant so you get minus 57 798 plus 44 128 this is at 290 this is the same at all temperatures delta G0 298 it is because delta CP is 0 otherwise you would have carried it this is minus 54 635 26 22 so 45 556 so I need delta G0 by T by RT will give you delta G0 by R into 298 I think this is given in all of this is calories per gram I have not noted the units must be calories per gram you have a factor of 2 again do not calculate delta G0 and then calculate delta G0 by RT again you need only delta G0 by RT so delta G0 by R into 298 this is 9079 by 2 into 298 minus delta H0 is minus or I will make this plus 13670 by 2 this is T minus 298 by 298 into T so delta G0 by RT is equal to have this 9079 and this is something wrong now this is okay 2 into 298 so this is 13670 minus 9079 91 134591 by 2 into 298 this is a plus in minus 13670 by 2 into T what is this equal to 7 okay 7.70 minus this is again 1635 6835 by T so exponential of this exponential of minus of this is equal to 5 minus x power half x is equal to 0.8 so it is 4.2 by 2 or 2.1 square root of 2.1 by square root of P and then there is a 2 root 2 root 2 in the denominator I put 2 root 2 here in x squared by 1 minus x so 0.8 squared or 0.64 divided by 0.2 to the power 5 by 2 all those are numbers into just 1 by square root of P so finally your cost P to the point 1 by T squared easier to solve for so D of cost by DT is equal to 0 implies P to the power 0.1 into 2 T will be equal to T squared 0.1 P to the power minus 0.9 this T will cancel this is when P if you bring this to the side this is just P P is equal to 0.05 T it should be sorry this is DP by DT this must be multiple this must be multiplied by DP by DT P prime so P is equal to P by P prime so you really have to because if you are putting this equal to 0 you are saying essentially P prime by P should be equal to 20 by T so that you have to substitute in this you have to solve these two equations to get T or P whichever you tell me what T comes out to be you do this by trial and error what you are doing is you are saying P is equal to what is this equal to that sound right I have 0.2 to the power okay so you have essentially P is equal to 18.3 squared by exponential of minus 14 15.4 plus e power minus so minus 15.4 plus 6835 so P prime easier to keep the exponential in numerator and keep to the numerator the exponential P is equal to this change the sign 68 minus 6835 by T yeah into 2 you are right 70 136 70 by T plus 15.4 so P prime it is going to give me back this P will get exponential of minus 13 670 no I have to just differentiate this that that is included in T and if I differentiate this with respect to T I get into 13 670 by T squared P prime is DP by DT so P by P prime is T squared by 13 670 you are saying this should be equal to 0.05 or T is equal to 13 670 into 0.05 advantage of not simplifying anything as far as possible and you would have computed all those exponentials and gone down to details how much has come to 683 yeah incidentally in all these empirical correlations where they give you cost they will say this is valid only within a range in this case it says 300 to 800 degrees so this comes within that you have to verify otherwise these things can go away where because in cost you never know suddenly there will be a step change it be a temperature up to which mild steel is allowed in heat exchangers then they will say it is not strong enough so change to copper or something or change to something else and the cost difference will be huge so in all empirical cost correlations you have to be very careful it simply depends on the local market prices so what is the pressure come out to be three point actually I the cost correlation was actually much more complicated than this I think it is given in an old issue of industrial engineering chemistry 1947 or 58 I do not know if that issue is even available here because it is only to illustrate a principle I simplified it to this form it is actually given as a long polynomial in pressure and so on some come but the principle is trivial simply and eventually you will probably solve it on the computer anyway so this is typical it is just cumbersome calculations but otherwise the principle of it is fairly straightforward because it is three your K phi calculate you have to verify one more step once you know P you can calculate your K phi and check that it is in fact approximately one in any case you do not tell an industry to operate at 3.36 atm you will probably give them some three and a half some rounded off number but the fellow has a pressure gauge that fluctuates widely so he would not know whether he is actually reading 3.5 or 3.5 you know probably within point plus or minus 0.5 I will take another example this is simultaneous reactions I have got some more data so every time what happens is you have the equilibrium equations plus special constraints could be in the form of a volume of a reactor it could be in the form of total volume being constant could be in the form before I take simultaneous reactions let me give you another example this is hf hf is known to polymerize in the gas phase it does not exist as hf it exists as hnfn the question is what is n usually there is some guess about it n is equal to 4 or 6 we are asking you to find out from experimental data on the reaction whether n is 4 or 6 the data given are in the following form at temperature and degrees came the pressure is given in atmospheres and then the vapor density is given in gram moles per liter if it says vapor density of hf it means the density has been measured assuming that the gases hf it may actually be h4 f4 h6 f6 etc so the data given is 393 I am sorry 298 the other way 298 and n2 measurements at 323 in the pressure is given this is two different pressures and then 373 also in 5.6 and vapor density 190 273 0.070 0.199 what you are asked to do is first what is n and then find ?g0 and ?h0 this is given at different temperatures these are asked at 298 one of the silly thing but you have to be careful in reaction equilibrium because you are taking these exponentials so in you end up with the order of magnitude errors and because you have no industrial exposure as such you do not have a feel for numbers you do not know as a conversion suppose we 0.1 or 0.9 we will end up so you have to be careful numbers as far as possible do not do that calculation till the end then if you are fortunate you do not have to do the calculation at all you do not have to do an exponential except for a constant how will you proceed to this how will you find out how will you test this hypothesis start off as always initial you have say n moles and 0 moles at equilibrium unit using hf let us assume there are x moles here then this is n into 1-x so total moles is n-n-1 into x so compared to the initial moles total moles by initial moles in the factor of 1- when I am just dividing by n if you had measured the density as hf this is what you would have got but the actual number of moles is 1-1-1 by n into x it depending on the fraction converted to right so the density if you want to you have p the pressures are quite small we will assume ideal gas you will get a sensory number of moles which is 1-1-1 by n into x into drt this is the vapor density is the assuming ideal gas they give you an equation of state you can use the equation of state if it had remained as hf then you can verify that p by drt if you put n equal to 1 for example in that term will disappear you can verify if p is equal to drt you can see it is not how you proceed further what is your kf for the reaction kf is kf times p by total number of moles to the power delta nu into kn we will take kf as 1 the pressure is of course there you know total number of moles n-n-1x to the power of the change in number of moles is 1-n and this is n moles giving you 1 mole so it is 1-n then you have kn kn is simply x by n into 1-x to the power n now I have got two readings at 323 the same temperature put at different pressure kf has to be the same for those two so we will do the kf calculation here for n equal to 4 and kf for n equal to 6 this 4 and 6 are guessed on the basis of the structure of hf the way the valence bonds are stuck and so on so you have no bit of chemistry some feel in this case you have told that it is either 4 or 6 that they do not know or else you can try out n equals 1 2 3 4 5 6 so in this case you get two possibilities kf for n equal to 4 is simply 1 by p cube is 4-3x x by 4 into 1-x to the power 4 so you have p by drt that you can calculate here you strike this off before you do this you have to do this p by drt d is given p is given cc atmosphere so make sure r is in the right units you calculate your p by drt from p by drt you can get n in terms of you get you can get n and x and in terms of x or x in terms of n and x for n equals 4 n equals 6 then you do the kf calculation here from here x is equal to p by drt it is 1-p by drt divided by 1-1 by n once if k then you can calculate kf for n equals 4 and n equals 6 so can you run through this with the calculator what is p by drt here rt you have to take the temperature given I think it says 300 degrees oh no temperatures here sorry so r of course you can take r as 82 cc atmospheres per gram so you have to do this this 1.4 by d into this into this if this is litters sorry so this is 0.082 litters you have to take it in litters so do it in litter atmospheres so tell me what p by drt is in these 4 cases this one 0.30 yes 38 okay next one 920 okay then x for x is simply 1- x for n equals 4 1-p by drt this is 1 by 4 so this is 3 by 4 into 4 by 3 and x for n equals 6 6 by 5 into 1-p by drt 1.9 you have to do the numbers carefully first one is 817 is it is 931 okay second one point this is all for n equals 4 okay for n equals 6 it is actually only 80% of this you can just calculate 0.8 times because you are doing the same 1- if you do 0.8 times much as your original calculations are good to 3 decimal places I would not take more this is 4 pardon this is 0.8 times this right because it is 6 by 5 by 4 by 3 into this this is how much you should get 0.8 times this because this is simply 6 by 5 this divided by this is simply 0.8 right this by this because you get 6 by 5 by 4 by 3 2 and 3 or 9.9 times sorry you are right okay till 0.83809 okay now finally for the truth kf this is the test fact you need to do these two first because if kf is not constant at a given temperature then that model is not right you need to do the middle ones in the expressions are different remember for kf for n equals for n equals 6 this is 6-5x to the power 5 by p to the power 5 x by 6 into 1-x 6 equal to 4 what is the value 0.34 okay not 6 to 4 so this sort of rules it out but what about n equals 6 since your last option as well be right not 698 not 698 and not 693 saved so once you have got that you take an average value for this then you calculate this and you calculate this value then you calculate your delta g0 and your delta h0 by differentiation that means you should really plot in these cases you got a lot of scatters so you have got to plot delta g0 by 298 I am sorry by t it is d by dt of this against t d by dt is delta h0 so you will get a straight line we will get to got only three points so you draw a straight line and take the slope of that it should give you delta h by t square you are actually asked to calculate it only 298 if you do this once you get some idea of the kinds of errors you can expect because these vapor this particular vapor density measurement is a very accurate measurement it was taken directly from journal of chemical engineering data and it has the this is about as accurate as you can get with vapor density measurements this itself is pretty unambiguous you got only 1% error essentially but in many cases you have to make a judgment I mean you have to decide that this is good enough or within 5% this is all right so the actual crucial part here was this relating the vapor density but it is always given in terms of the description of the molecule if it says hf then means vapor density is reported assuming that all molecules are hf.