 వాదిన్ణదిందిందాలెర్న్ అషిస్రంత్మండాప్నెదిందా. మికంత్సుపిలె, వాట్టిత్నండిందిందిందిందిందాంద్లె బిలాపిలెపిసికిక్లు. the concentration of unoccupied sides is S, these are capital is S and that occupied by various adsorbed species is S A, S B, S C, etc. Before that we have S naught and S, okay. Let me explain this, so I think you know happily would have written without even thinking you know what is the meaning but the meaning of that is given catalyst will have S naught number of active sides, right. So out of that S A, S B, S C are the active sides that are occupied by the species if you have A B C, if you have A B only two S A, S B like that then the remaining sides are S, okay. So now based on this how many sides have been occupied by the species, how many getting adsorbed, how many getting desorbed, all that will come in this and then we will assume that we have a reaction A gas, B gas going to C gas plus D gas. We have the catalyst, it is a reversible reaction, okay. This is the kind of reaction that is going on now on the catalyst. So now we have to develop a rate expression for this stoichiometric equation, okay. So now our imagination for catalytic reaction is that molecule A has to go get adsorbed, molecule B has to also get adsorbed. If they are side by side then molecule A will form some intermediate, molecule B will form some intermediate, both will react and then that is surface reaction step and then desorption, desorption of A and now there is no A, desorption of product. If there is only one product only one will come otherwise if I have C and D and C has to desorbed, D has to desorbed and then reach, come through the pores and through the film and reach the bulk that is what is the overall picture what we imagine for all catalytic reactions, okay. What is that? Once more the assumption is that molecules of A will get adsorbed on one side, molecules of B also get adsorbed on other side and both of them form intermediates and then these intermediate will react and they form the products C and D and C and D also they have to be desorbed off after the formation because they are the products where the concentration is high on the surface low in the bulk so that concentration difference will make it move through the pores, through the film and all that. So then out of these that means adsorption of A, adsorption of B, desorption of D, desorption of C and also surface reaction we have now five steps. Out of that any one may be rate controlling, okay and may be adsorption of A alone may be rate controlling or desorption of D alone may be rate controlling or five may be rate controlling. So how do we know this? We will know only when you first develop a model the general procedure is writing the main equation afterwards imagining the steps adsorption of A, adsorption of B and all that then developing a kinetic model, right. What you do is in general if it is a new process we do not know what is the rate controlling step. We write all the models that means for example here you will have lot of combinations simplest ones if you take adsorption of A, adsorption of B, desorption, desorption two steps and surface reaction so we will say that okay surface reaction is controlling one model, adsorption of A is controlling another model, adsorption of B is controlling another model. Then all these models before me now mathematically I have these models but how do you know which one is really rate controlling? I have to conduct an experiment, right and then I have to measure the concentrations of A, B if I know A, B, C, D I can calculate, okay or whatever is possible if I measure C, D then A, B I can calculate at any time. So once I have that information concentration versus time data like your normal kinetics then I have to fit this data to the overall kinetic models. So out of that five models one may fit if all of them are not fitting because each individual one which you have thought you know that is adsorption alone, desorption alone, surface reaction alone then you have to take the combinations so that is why it is hell to develop this LHHW kinetic models in catalytic reactions that is why we happily ask chemists to do that, okay but I think as chemical engineers also we should know how to do that and we should also develop because these are the macroscopic models still we are not worried about how the molecules are actually getting dissociated, what are the activation energies all those things we are not worried here, we need a macroscopic rate which is a function of concentration of A, concentration of B and if it is reversible reaction then you have concentration of C and concentration of D that is all what we require as an engineer but if you go deeper and deeper on the surface what is actually happening, what are the surface energies, where are these active sites, how do you make these active sites much more active, how do you generate more active sites in a catalyst all these things again is a big science, catalysis science and it is interface between chemical engineers and chemists where if both of us work together we can do much more because the idea is at the end with minimum amount of catalyst how do you produce maximum I mean as engineers we understand that, okay but at present what we are trying to do is that let me develop or let us develop a macroscopic model where it gives me a rate expression because this rate expression is necessary for me in my performance equation, in performance equation I have a term called kinetics and I have another term called input and another term called contacting, so contacting is given by you know the type of reactor I am choosing and kinetic models from this only I have to get and of course input you know how to get it, so that is why now let us write the steps, step 1 is A, A is the concentration and S is the active site and all the steps are reversible steps and we can also say this is K 1, K 1 dashed reverse yes giving me some complex A S, okay then step 2 B plus S giving me this is K 2, K 2 dashed this is B S then the reaction between these two step 3 A S plus B S this is K 3, K 3 dashed this will be C S plus D S those are the complexes then step 4 is desorption of C S K 4, K 4 dashed now this will be C plus S and step 5 I hope I will write here is D S this is K 5, K 5 dashed reverse D plus S, so if I call these numbers may be before that also I have to write one equation this is S naught equal to S plus S A plus S B plus S C plus that is equation 1 this is equation 2 and equation 3, 4, 5, 6, 7 definitely you know the meaning of equation 1 right in the notes which I have given you that we have S vacant sites S A occupied by A B C all together must be your S naught which we have taken in the beginning as a constant for a given catalyst, okay so out of this again equation 3 is adsorption step correct and next one also is adsorption step yes this is reaction step surface reaction and next one is desorption next one also is desorption good so this is what I have been telling any one of these steps may be rate controlling or all may be rate controlling okay two extremes okay good so now let us take that surface reaction is the rate controlling step this step surface reaction is the rate controlling step surface reaction is rate controlling so we will write a rate expression here or equal to this one we have to treat it as normal stoichiometric equation okay it is reversible reaction so the rate expression can be written as K 3 concentration of this concentration of this and minus K 3 dash concentration of this and concentration of this okay that is what we write there then we have K 3 S A S B S minus K 3 dashed C S D S so this is equation number 8 good so as usual I think you know I have been telling you this one for all heterogeneous systems this rate should be in terms of measurable quantities what are the measurable quantities here in this equations A because this is the concentration of A in the bulk which you have to we will see that and then B and C and D and all other things I cannot intermediate concentrations I cannot measure right so that is why we have to now write that equation in terms of measurable quantities the procedure is universal you know for all heterogeneous systems finally we should have then if I am able to express this rate in terms of all measurable quantities then that is called the overall rate observed rate measured rate global rate all these various names are there I am just repeating again good so now the general procedure here is that when you take reaction as the rate controlling step all other steps are so fast always you can see equilibrium of those steps in fact these what you have done even for homogenous systems let me recall your memory that you know you have non elementary reactions non elementary reactions if A plus B going to C is not actually A plus B going to C in between there are many many steps that means A may be first forming an intermediate B may be forming an intermediate then those may be reacting so all these again models you will imagine in non elementary reactions procedure is same there also you write all that imaginary steps okay how A is reacting forming the intermediates or B is forming the intermediates and then try to write for each and every equation what you write there as a elementary equation elementary means what is the definition of elementary reaction stoichiometry and then order will be the stoichiometry coefficients will be the same okay so similarly here when I write an equation this is what what you have written this is first order with respect to this first order with respect to this first order like this then overall second order whole second order because coefficient one here one here one here one here so this is the general procedure except that rate controlling step you have all other steps as I mean in equilibrium okay so that means thermodynamic equilibrium so how do you write those equations thermodynamic equilibrium okay I think you may forget later so I think better let me tell this also when step three and not step three here when step five is rate controlling okay step three is fine equation five or step three when step three is rate controlling all other steps are assumed to be under equilibrium then from step one yeah the thermodynamic equilibrium is A S A S equal to K 1 which is nothing but your K 1 by K 1 dash correct no right yeah so this is equation nine then from step two B S this is concentration of B separated concentration of A separately this is K 2 which is again K 2 by K 2 dashed 10 then we have step 4 step 4 C S by this is K 4 which is slightly change here that is K 4 dashed by K 4 that this constant I have just reversed for easy writing okay that you have to notice that okay I hope you notice it that normally it will be reactants by products by products by reactants but here I have reversed and then this is defined like that this is only for easy manipulation of algebra that is all nothing else okay then step five is B S D dot S this is K 5 which is equal to K 5 by K 5 dashed by yeah this is the one good. So now I can write A S B S C S D S here in terms of these equations correct no right so that means A dot S will come here B dot S come here and C dot S here D dot S there now let us write that substituting equations 9 to 12 in 8 substituting equations 9 to 12 in 8 first let us write that and then we can rearrange we will write here K 3 capital K 1 A dot S all this we can have a bracket like this then K 2 B dot S minus K 3 dashed K 4 C S K 5 D S yeah this equation also we can call as 13 yeah so this after rearranging a little bit then you will get R equal to K 3 K 1 K 2 now I have A B minus K 4 K 5 by K 2 K 1 K equilibrium or capital K also we can write C D bracket close I have now here S square this is okay where K equilibrium or K in general capital K I can write as K 3 by K 3 dashed good still can I use this equation immediately or do you have to do something else yeah we have to also eliminate the number of you know vacant sites okay that can be done because still this one these constants anyway I have to evaluate only from the experiments okay but this S square I can also eliminate by taking that equation this equation 1 yeah equation 1 gives me S naught equal to S plus S A S B S C here S D S D and I can also have here some units okay units also occupy some active sites okay good. So now I have I have to substitute for S A S A is nothing but this concentration A S okay B S C S like that that is the concentration on the active sites so that is why if I write this S plus A S B S C S D S and I can also have I S I is the inert component like for example nitrogen which is may not be participating in the reaction okay. So this equation now this is S naught this also of course these things I do not know but I have to write that in terms of again known measurable quantities what are the measurable quantities again these equations A S I know A S is nothing but K 1 A S right or K 2 B S is nothing but K 2 B dot S okay so you have to substitute again equation 9 to 11 here then you get S plus you have K 1 A S K 2 B S and K 4 C S K 5 D S K I K I okay I S this is the equation now you know I have S here S here S here S here S here S here S here all that S I can take out common like okay so this equation if I take as this is 15 this is 16 then equation 16 can be written as S equal to S naught divided by 1 plus K 1 A K 2 B K 4 C K 5 D plus K I I that is the equation this is equation 17 excellent good. So now equation 17 has to be substituted in 14 now you will see the format which normally you see for L H H W kinetics okay so what is that please take that substituting equation 17 in 14 what you get is R equal to K 3 capital K 1 capital K 2 S 0 square now I can write A B minus C D by I write capital K naught this is the bracket so now all this term 1 plus K 1 A K 2 B K 4 C K 5 D plus K I I whole square yeah so this is the equation actually this is equation number 18 that is equation number 18 and in the normal form when you write this I can write this as R equal to all this as some K and concentration of A concentration of B minus concentration of C concentration of D K naught whole thing divided by 1 plus K 1 C A K 2 C B K 4 C C K 5 C D plus K I C I whole square this is one format there is another format also which you will see of course this K is all that together okay let me also write that where small K equal to K 3 K 1 K 2 S 0 square good so I can also write this one this is another format normally we will see as P A P B minus P C P D by K naught I am retaining as the same constants but constants will be actually different so 1 plus K K 1 P A K 2 P B K 4 P C K 5 P D plus K I P I whole square this is equation 20 21 these are the two formats normally you will see in all the books okay many things can be understood from these equations right many things like this denominator whenever you have a rate expression you know this is the actual rate expression or this one C A minus C B minus this is normal rate equation you know when you have A plus B going to C plus D this is the kind of rate equation you will have right for reversible yeah but in catalytic reactions this is the denominator which comes and that is actually inhibiting the reaction that is why in biochemical reactions when inhibition is taken into account some terms are added in the denominator those terms are not just like that added but you know through mechanism 1 1 plus no but the constants here are not same from the previous that is what I mentioned yeah that is what I told you even though I am writing this in partial things I am retaining the same constants that is what I have told you okay yeah so you know you can the entire derivation you can also do in terms of partial pressures from the beginning you will have one only you will not have some other value here okay so that is why yeah these constants make a note I told that whenever I am writing that these constants even though I am maintaining the same they are not same because of partial pressures okay yeah good but what is that we can understand here is that you have the square here normally that square comes when two active sides involving the reaction two active sides if there are three like A plus B and also plus C going to the products then three active sides when they are involved then you will get here Q okay for surface reaction when you have surface that is why when someone gives you model if they give you in this format and also you see in the step 3 what you have that reaction step that will not be there in the denominator any constants K 3 will not be there in the denominator so in the denominator you will have only those steps which are not the rate controlling steps so here okay step 1 2 4 5 are not rate controlling steps they are the fastest steps right so that is the reason why you will have in the denominator why I am telling all this is when you read a when you see a book or when you see an equation even in the even now people are publishing papers on this when you look at that you can easily find out what could be the rate controlling step from that what could be the rate controlling step with experience even just without just by seeing you will know what is the mechanism of the you know based on which that particular equation is derived okay good so this is one thing and then what we have to do is now evaluate K 1 K 1 K 2 K 4 K 5 K i that means you have to now conduct experiment and experiment you conduct to find out what is the partial pressure of A partial pressure of B or concentration of A concentration of B then concentration of C and D you can definitely get from stoichiometric equation 1 mole 1 mole 1 mole that is no problem then with time this rate is nothing but minus depending on your base right most of the time for catalytic reactions we base our rate based on weight of the catalyst okay so how do I write that it is minus 1 by W D N A by D T so the number of moles of A changing with respect to time I should know right when you know the concentrations you can find out how many moles are changing various times you are finding out the concentrations then you know what is the change in moles of A right so you can calculate easily rate once you measure the change in concentration or the change in moles of a particular component good so that means rate I will know and then corresponding concentrations normally it is time and concentrations then you can convert that into rate and concentrations and once you have that information then you have to go to this particular model and then try to find out k1 k2 k3 there are many many methods available for finding out very accurately k1 k2 k3 k4 k5 and also other other constants right the many methods that is why again you know here it becomes purely experimental I mean mathematical techniques and whole type engineers where the complicated mathematics when they have not used what they did was that they conduct the reaction forward reaction and then avoid as far as possible this term that is why you know there is a method called initial rates for finding out kinetics I do not know whether you still remember that or not initial rates and these initial rates mainly came from for forward reaction first and then you take the products because you are trying to find out the kinetics okay so then you have to take C and D and then have the backward reaction and those initial rates so that means separately without this term you should have all this and without this term you should have all this and then conduct one experiment for the entire normal A plus B going to C plus D and then check that data with the initial rate data that is given that you have checked with so that is the overall procedure and my God to actually find out K1 K2 K3 it takes lots of time but there is a pleasure in doing that there is a real pleasure in doing that you know evaluating these constants and accurately measuring the rate of reaction because all your design now depends on this rate I can tell you very quickly here if you want to find out weight of the catalyst in a packed bed what is that you know the equation what you use the equation what you use this F A not equal to D X A by minus R A and this is minus R A for me this is the equation which you have to substitute here and many times it may not be possible to have an analytical expression you cannot integrate just like that so that is why you go for either graphical integration or numerical integration or some other technique finally to get what is W by F A not and what is this equation for which reactor P of R which is packed bed plug flow reactor all packed beds are treated as plug flow reactors so that is why this rate expression finding out is the first step and then I have been telling you in homogeneous in heterogeneous reactions most difficult steps are only finding out the rates okay finding out the rates so this is one thing I think very quickly I can also tell you another rate controlling step that is let us take adsorption of A rate controlling procedure is same again exactly same that means first writing an equation for this alone minus R A equal to this one K 1 into A S minus K 2 into sorry K 1 dashed into A S so now you have to try to eliminate this A S using all other steps that is all the procedure is same so let us do that very quickly I think we have time or no time I think because already we have some steps there okay so let us take another very simple one okay this is adsorption control I am not taking the earlier one because it takes time so we have a reaction A going to R A going to B also one can write because I wrote there A plus B do not want confuse there so now here this is the reaction that is going on the catalyst what is step one the way we imagine here I have catalyst step one is A plus S giving me A S that is adsorption of A step two is A S giving me R S that is the surface reaction step okay then step three is R S giving us R plus S you see this active site again is ready regenerated after this step so now this one as usual I will write K 1 K 1 dashed K 2 K 2 dashed K 3 K 3 dashed now we take this one is the rate controlling step so if I take this one as one two three so when this is rate controlling step so we have the equation R equal to K 1 A S minus K 1 dashed A S okay good that is the actual rate expiration for now we have to find out from the other two steps because they are under equilibrium the other two steps so I can write from step okay from step two what we have is K 2 equal to R S by A S from step three it is K 3 equal to R S by R dot S so this is equation number five equation number six now we have to substitute this in this okay equation four so if we do that we will have R equal to K 1 A S minus K 2 sorry K 1 dashed R S by from this R S by this also I have bracket that is very important step two R S by K 2 this is equation seven right but still this is unknown to me R S right so now we have to substitute equation six in seven because R S R dot S right confusing okay good so R equal to K 1 A S minus K 1 dashed for this R S I have to substitute K 3 capital K 3 capital K 3 R dot S by K 2 this is equation eight so now I have to write this equation as K 1 S into A minus K 3 by capital K 1 K 2 capital K 1 K 2 into R okay where I will write here where capital K 1 equal to K 1 by K 1 dashed so this is yeah and S is there already so this is nine this is ten what is the other equation now I have to eliminate S procedure is same I have to eliminate S okay so when I eliminate S to eliminate S I have S 0 equal to S plus S A S R yeah plus S I in it so this is equation eleven good so now I have to substitute S plus for S A it is A S plus R S plus in it also if I want to have that there it is I S so the equation twelve now I have to substitute for A S R S I S okay we have the equations here first one if I want to substitute for A S this will be R S by yeah R S by K 2 right A S what is the K 1 K 1 capital K 1 A S can we use that capital K 1 how can you use that okay capital K 1 cannot come there no capital K 1 is only for this reaction thermodynamic equilibrium capital K 1 okay so here when I substitute this I will have here S plus R S K 2 yeah plus R S I am writing that slowly now I S this is equation thirteen now these two together S plus for R S I have to use this equation K 3 R S right so this one is what K 2 yeah okay so for R S I have to use K 3 actually I can take common also there so this will be 1 by K 2 plus 1 multiplied by K 3 R S R dot S in fact okay plus I S correct no K I S okay so now S I can take common so this is S not equal to 1 plus K 3 by K 2 plus K 3 R plus K I okay whole thing multiplied by S yeah that is why I am also running fast yeah right here also here also I am is there no okay dot S yeah S not by 1 plus K 3 K 2 plus K 3 into R plus K I so that is the one yeah good so now I have to substitute there and then you will get the substitute in which equation yeah equation nine substituting in equation nine may be I think I have to write here I will give only one expression substituting here also I have to write thirteen fourteen this is fifteen yeah substituting equation fifteen in nine what you get is R equal to K 1 S not A minus K 3 K 1 K 2 R whole thing divided by 1 plus K 3 by K 2 plus so this is the expression so this is of course one can write also this one as some constant K C A minus another constant for example R by C R by K not whole thing divided by 1 plus K 3 K 2 K 3 C R plus K I of course you can also write this one in terms of partial pressures okay where K not equal to all this K 1 yeah you can go okay tomorrow we will discuss