 Now we have been talking about kinetic models for non catalytic reactions and all the reactions we have also divided into 5 or 6 categories A, B, C, D, F. Now we will discuss kinetic models for reactions of type A and B that we will write here. Kinetic models for reactions of type A and B. I think I have to tell you few words for connection may be you can take this one as notes, okay. Type A reactions are type A reactions are the most common and are exemplified and are exemplified by reduction, oxidation and roasting of ores. The general stoichiometric scheme is A g plus B B S solid going to R G plus S S G that is the general stoichiometric scheme and you have to notice this one very carefully because A does not have a coefficient. So if it has the stoichiometric coefficient you have to divide with that then make it as 1 and automatically B or S will be not exactly their own stoichiometry but divided by A small a. This is S and this is gas, good yeah please continue. This reaction scheme belongs to type A as we know we have discussed it already. This reaction scheme belongs to type A as you know but can easily be applied to type B, type B in which no gaseous product is involved in the bracket S equal to 0, full stop there. Similarly two types of models have been considered, one in which the solid is non porous and one in which it is porous. Since most of the ores are non porous shrinking core model is widely used, shrinking core model is widely used for developing the kinetic model, for developing the kinetic model. Now we are talking about shrinking core model, you can also write this. This model can also be called as sharp interface model, I think let me also write that interface model and also shell progressive model. If you suddenly some other books other than Levenspiel or some other books then you will find sometime sharp interface model and shell progressive model. You should not think that it is a different model. You know broadly in chemical reaction engineering the way the subject is explained can be divided into two groups, one is Levenspiel group, the other one is Smith group, right. So all Levenspiel group people will follow his notation, that is why you know A plus B going to R plus S. In many other books like Smith and Carberry, they write A plus B going to C plus D, okay. And of course you would know or definitely you should have studied about the epsilon as the volume increase in factor, okay. Yes, that epsilon was introduced by Levenspiel whereas if you see Smith group like Carberry and other books it is not there. So they write every time for gas phase reactions, the number of moles and then find out mole fraction and find out partial pressure and then try to develop the kinetic model. This difference also you should know because there are two approaches. Epsilon is blind approach like our coaching centre, you do not have to explain anything calculate epsilon, substitute there and then I mean integrate if it is a design expression, okay. Otherwise the other group every time you have to write the equation how many moles of A converted, how many moles of B converted, the others formed and then total moles mole fraction, total pressure, you will get partial pressure that you introduce and then that also can be used. At the end both will be same. In my reactor theory course I have I normally tell these two approaches and also tell that both will give exactly the same result but Levenspiel approach is direct substitution without thinking much, right. But whereas there every time you have to use your brain but normally we do not like to use our brain every time. So that is the reason why that coaching, epsilon, sticking and all that is easy for us, okay. So these are the other names for the sinking core model. Then as we know definitely this is used for non-porous, right and what is our general procedure? Our general procedure is first to identify a single particle and then draw the profiles and after that write the equations for each profile mass transfer steps and reaction steps, equate all of them and then finally you will get a rate which is called the global rate of reaction because every intermediate step where you cannot measure the concentrations are eliminated and finally it is expressed in terms of only measurable quantities. So many times I am repeating this you know you should remember if someone asks what is global rate of reaction, what are the other names for this? Overall rate, observed rate, okay, sorry, yeah measurable rate, all these different books use different things but that is what is you know automatically comes for heterogeneous systems. So homogenous systems does not matter both are same, okay good. So now this model before drawing before drawing the figure picture we also should have some assumptions, right like any model. So the first assumption what we have is assumption 1, yeah we have isothermal conditions, second one easy to understand, constant particle size and third assumption you know equimolol, counter diffusion of gaseous reactants and products and fourth one pseudo steady state approximation, pseudo steady state approximation and the fifth one to make our mathematics simpler is first order reaction, first order okay not only first order reversible reaction, irreversible reaction with respect to gas, okay good. So these are the assumptions and of course this is easily understandable constant particle size you know during the reaction I told you one example where calcium oxide when it is reacting with calcium sorry SO2 in the presence of oxygen calcium sulphate where the particle slightly grows bigger because the molar volume. So that is the reason why in many reactions this increase in particle size many times happens but right now it is mathematically slightly complicated that is why we are not considering that particular aspect but we assume that we have constant particle size, right. Then this equimolar counter diffusion because we have the gas phase going into the pores for the reaction to take place on the surface of the core, right and the products have to come out so if you take multi component diffusion again mathematically it is more complicated so to start with we can assume that equimolar counter diffusion for the products and gas so that assumption is clear, pseudo steady state assumption is a very beautiful assumption for gas solid reactions and I will explain that when actually we are deriving an equation for one of the steps that I will explain later and of course first order with respect to gas irreversible first order and again to simplify the mathematics, okay. If I do not have that again mathematics will be really complicated and we lose track of the physics what is happening in the problem that is why to start with we always start with simple assumptions so that we will understand what is going on, yes, Shekhar? Then we say equimolar diffusion because A has only one mole and the product S has S moles so then how can we say equimolar? Yes that is why because it is not equimolar we are assuming here equimolar and then trying to do that, right if you are actually taking the actual number of moles we do not have to assume that, right. So that is why if it is one mole and then this is two moles the effect may not be that much but if it is one mole that is seven moles definitely there is effect, right that comes in the form of you know your diffusion, product diffusion that also I will tell you in one of the steps. So this is the reason why to make the mathematics simple so that we will understand the physics that is the reason why we are going for these assumptions, okay otherwise in actual practice if you want to do who cannot I mean for example here I have not told implicitly one all particles are spherical for us but you can never see that kind of beautiful spherical particles except when you are playing your marbles, okay marbles are beautiful spherical particles other than that in chemical industry no one makes that kind of spherical particle this is ore please remember you are now digging out the ore and then you are using those particles you can never expect spherical particles but in spite of that we use you know mathematical equations only in terms of spherical geometry because it is very easy for us to imagine how the symmetric diffusion is going through the this particle from all sides you know equal diffusion symmetrically so that is why all these assumptions are to make our mathematics simpler, right so our next step is now to plot the particle this is spherical particle and we are talking about after some time it has already reacted so this is the actual particle and this is the product ash layer I have been telling let me write that later yeah, okay this is film and this is ash ash is nothing but product, okay let me also write once product then we have here, okay b reacted b and in fact what we have is here gas is moving so a, a so now we have to plot the profiles so that is the center this is r c this is capital R and this is the film so this is film again of course film and here we have the center as 0 r is increasing in this direction I have here what is called r c this is the core and here I have the capital R which is the size of the particle, right and this side also I will try to yeah so this is r c capital R and this is the film good so anyway I do not have to ask you many times but I can simply draw this this is c a g and through this we may have a profile like this this is c a s and now after some reaction as I told you we should have the this ash layer or product layer must be porous, right otherwise there is no reaction if it is non porous so through the pores a again diffuses because this is a product which is inert as far as reaction is concerned because it is irreversible if it is reversible then you have the problem again this product also will react with the the product gas will again react with the product layer and then you will have again some more reaction you know backward reaction for that is not there and most of the time for overreaction that will not happen, okay so it is only irreversible reaction which is a good assumption, right so that is why it would not bother us then inside you have the diffusion as if it is a porous particle simply reactant is diffusing so this on this surface of this core we have c a c because it is non porous it cannot go inside reaction is taking place only like this and then slowly because of the reaction this core shrinks and then finally everything is converted and our aim here is in the kinetic model to find out what is the time required for 100 percent conversion of this particular particle, right so our kinetic model should give us that information then afterwards we will try to use that information in the actual reactor design, good so this is one R C and C A C C A S and C A G now we have to also plot a profile for the solid this is very easy we have also discussed in the last class so if I start with C B or C B not for example then I will have here it is 0 then only C B not here which is C B not of course here I have to put dash dash this is C B not, right okay so with the time only thing is this is becoming smaller and smaller but there is no decrease here that we have to remember good okay so that is all I think here if I say maybe this one will be C B not equal to 0 somewhere so outside of this it is 0 here here everywhere it is 0 except on this point good so now what are the steps we have to take to develop the model this is universal procedure for all heterogeneous systems we have to first draw this diagram I tell you this diagram is not that easy for you just to copy now you can copy easily but if you are developing on your own a new model you have to imagine what is the process that is going on, right and you have to identify what are the steps, right and for each step you have to write an equation then only eliminate all the intermediate concentrations that is universal procedure but the physics of this that phenomena in your mind that is more important to draw this picture as simple as possible that is why even though I am telling you many times it is not that easy for you the moment I say that okay I have some other particle with two gases for example A plus B also going to A plus C and B B is the solid and I have A and another gas also C and if I ask you to draw that picture you cannot immediately draw unless you imagine the process unless you have the clear understanding of the process, right so this is why most of the time in all these processes one should have the physics behind your mind that physics has gone long time back from our mind because our training for you is always writing examinations so equations are more important for you but at least now in the PG courses and also in PhD level that imagination is very very important for us for anything, okay so the moment any process is told to you then the entire process should run in your brain as a movie and then you should be able to imagine that okay this is step one this is step two this is what happens inside and this what happens outside all that once you have that understanding the next one to prove is mathematics you have to write and then prove that your idea is right or wrong that is what is all research about, right so if you are doing PhD the guide will give you the problem and you have to sit down that is what I was telling you know first six months if you give total thinking period for PhD scholars probably we will get very good you know quality work but of course they should also use their brain properly not for seeing DC plus plus and all that but if they are able to really think and after thinking you should be able to propose a hypothesis this is what is going to happen in my problem and just note down that on a notebook and at the end when you are submitting just think how far you are right and if you are intuition at that time and also you if you are mathematics or if it is experiment if both are matching you have done wonderfully well you became excellent research scholar or researcher later for life so that is the training actually what we are supposed to give, right but I think it is not happening in reality most of the time but individual students who have got that passion for research they will understand that so they will become an excellent people but most of the time education is by force unfortunately in this country or it is necessity I want this degree that is all whatever means you use you should finally get that that is unfortunate really, okay so now let me write these steps, step one step one is mass transfer of reactant reactant through film cross the film step two diffusion of reactant reactant gas only that is what I mean diffusion through ash layer through ash layer and step three reaction between gas and solid I am writing gas there but I think it can be also fluid what swami you are asking something step three is not everything is combined as reaction step if it is a catalytic particle yes, okay what you are thinking is catalytic particle where molecule gets absorbed and another molecule will come and sit side by side and then decomposition of you know surface reaction and then dissociation then desorption all that that is for catalytic steps this is for non catalytic reaction even though things like adsorption first should happen and then reaction to take place with the solid whereas in catalytic in catalytic reactions the solid is not a reactant so that is why step four is diffusion of diffusion of product gases through ash through ash layer to the surface that is the one and step five is mass transfer of product gas through film to the bulk to the bulk these are the five steps and we know very well that because it is irreversible reaction these two steps can be knocked off because they are not affecting our rate of reaction that is why simply so we have only these three steps, okay for irreversible reaction yes of course if I have type B again same model right but only thing is there is no gaseous diffusion anyway these two steps will not be there the reason is there is no yes the product gas coming out so that is why the same model can be used good so now at steady state all these three steps must be equal right so at steady state diffusion mass transfer of reactant A equal to diffusion of A through ash which is also equal to rate of reaction these three steps now we have to write in terms of an equation all these three right so the mass transfer step minus dna by dt equal to 4 pi r square k g that is why this helps cag minus c as this is equation 1 yeah this is also equal to minus dna by dt second step second step is yeah okay by the way what are the units of minus dna by minus dna by dt moles per time so if I if I want to express that one as a flux I have to take one of those areas okay so now I am multiplying by this area 4 pi r square so that is why it simply becomes rate of mass transfer that is one so now the other one is diffusion through the ash layer that we can express anywhere inside you know I can express here I can express also here I can express anywhere but only thing what you need is a concentration profile and a derivative of that with respect to r okay so if I know that means it is nothing but fix fix law na equal to you have rja equal to minus d into dca by dr or dca by dz that concentration gradient I want so that is why we will write that equation and then try to think later how do you get that 4 pi r c square d e a few sentences about d I have to tell this is dca by dr now there is a condition at r equal to rc otherwise I cannot take this area if it is r equal to capital R still I can use no problem right but this will be 4 pi r square right okay good so the next one is rate of reaction equal to we have now 4 pi rc square that is where the reaction is taking place okay and k ks surface reaction and ca ca c here surface is the outer surface okay but actual reaction is going on the core this is the core okay I think may be at also right here core that is the core so that is why please remember that this is where most of the time we will write we will make the mistake even though it is simple things I say I am only trying to find out point out very very simple things because most of the students make simple mistakes and think that in the examination they have done correctly and then scold the teacher that he is not giving the marks okay I mean this is the normal thing okay good these three equations what we have and what is the procedure now I have to ca c I do not know right and ca s I do not know this contains ca s and ca g I know so ca c ca s I have to eliminate and then finally get the equation in terms of ca g right but of course this is dna by dt but you know this is extensive rate and intrinsic rate I have to have later this minus dna by dt divided by 1 by 4 pi r square if I take then that becomes the intrinsic rate but based on outer surface of the particle right so that is also important because in heterogeneous system most of the confusion is how do you express the rate by the by what is the rate here I think what should be the rate units here in the step chemical reaction rate units chemical reaction rate units dimensions normally what are those rate of reaction moles per per time per I have to make intrinsic right but in this case I have already made into yeah this this is the rate in fact k s e ac that is minus r a right but now we have to multiply that one by area because my rate is based on unit area so here it is not our unit weight of the particle or unit volume of the particle it is unit surface area of the particle okay so that is why one has to remember that and you know the moment we say that we have rate based on moles per based on meter square surface area the units of k s will be different they are no more it is first order reaction in general the moment I say I ask you what are the units for first order rate constant what do you say all of us will say time inverse okay it is like l k g alpha betas you know always we remember that only right because always first time we remember in the class only rate of reaction first order homogeneous so that is why it becomes time inverse but later in heterogeneous systems it can be any you cannot if the anyone ask you what are the rate units for first order rate in heterogeneous system you have to ask okay you tell me what is the rate based on whether it is volume or weight or surface area or interfacial area whatever that is why you have to be more careful about that so then here the most difficult thing is I cannot eliminate so easily this one c ac and I do not have explicitly here what is c ac or c as I need an equation for that this is the flux this is fix law and this d e is the diffusivity coefficient d e effective what do you mean by this effective diffusivity coefficient it is not simple binary diffusion you calculate d a b but this is happening d a b is between molecules coming a coming this way b coming this way but here this diffusion is through the pores and in that pore you also have the r r s so what is that yes gas coming this way and a going in the opposite direction but all this happening within the pores but these pores also having different dimensions right the pore size so if I have very large pores then I can say maybe 100 times 200 times the molecule size of the the gas then it is fine I have bulk diffusion I can happily consider that one as equivalent to normal bulk diffusion but if they are very very small pores you should have heard of what are called what is called Knudsen diffusion Knudsen diffusion is when the mean free path is almost equivalent to your pore size mean flow path of the gas so then we have the Knudsen diffusion we have another diffusion called configurational diffusion so there are some pores where the size of the pores is almost same as the molecule size there it is like q one mole will push one molecule will push the other molecule that will go inside other molecule will push it is like exactly plug flow one behind the other so all this I don't know what is the fraction why because I cannot exactly find out the pore fraction of each and every I mean the pores in the particle so all together I measure and that is called effective diffusivity there are methods how to find out effective diffusivity and this is very widely used even catalyst particles it is same even catalyst particles you have the same problem because the the pore sizes are not uniform right yeah so that is why if you have one pore that is why zeolite catalysts are very easy to characterize because the pores are almost uniform for the for a given catalyst whereas here when you are making the catalyst it is not that easy so that is why this effective diffusivity is a fitted parameter what we call that means you take the data and fit that DE value so that your experimental data and model will be fitting satisfactorily okay so but still we call this one as fixla right fixla normally used only for gas gas most of the time or or liquid diffusing to the other liquid okay I mean inside the liquid phase that is what what we say even here effective diffusivity will come because of the various contributions from various diffusions what you have to remember is bulk diffusion motion diffusion and configurational diffusion these three diffusions we will stop here