 Good morning. In the last lecture, we looked at how to derive or synthesize a rate law for heterogeneous catalyst, when the reaction is taking place on the surface. There are different steps occurring, while the reaction is taking place on the surface of the catalyst. We already looked at these steps. First is of course, the external mass transfer, then the internal mass transfer or diffusion through the pores and actual catalytic site is present on the surface, which is nothing but the wall of the pore and on this side, the reaction would take place. So, after the diffusion that is the second step, we have adsorption of the reactant on the catalytic site, then the adsorption species or the adsorption component, we will undergo the reaction. We looked at isomerization, dehydrogenation reactions, there are many possible reactions and this transformation will occur on the surface that is nothing but a surface chemical reaction. Then the product, which is possibly in the adsorbed state will get desorbed the fourth step and then again the back diffusion through the pores of the product and then the external mass transfer of the product. So, these are the different steps, we have already looked at them, but while deriving the rate law, we looked at only three steps that is adsorption, surface reaction and desorption. Only three steps we looked at and we synthesize the rate law and then I told you how to determine or how to validate this rate law on the sense how to know whether the rate law that we have synthesized is correct or not. That is through the experimental data, performing experiments in laboratory under different conditions and seeing the effect of partial pressures of the reactant, sometimes the products and see whether the rate equation that we have derived fits well in the experimental data that we have got. So, I gave you an example last time of cyclohexane dehydrogenation and I told you how it fits well for the rate equation derived based on the assumption that the chemical reaction controls, but sometimes it is quite possible that adsorption or desorption would control. So, it all we have to check that through the experiments. Now, today we are going to look at the other steps, which we had ignored before and these are external mass transfer and internal mass transfer that is diffusion. So, we are going to spend some time on diffusion first and then we will look at external mass transfer. So, on a catalyst particle we have a reactant going first to the external surface then inside and somewhere you have a catalytic side where the adsorption reaction and desorption would occur and then the diffusion back diffusion and then back mass transfer to the bulk of the product. So, these are the steps we are now going to concentrate more next 3-4 lectures on this particular step that is pore diffusion. A very important step I told you that the catalysts are porous in nature and there are different or there are many or several pores inside the catalyst. In fact, that is the main reason the catalysts would have very large surface area. So, even if the main active catalytic components say for example, palladium or platinum or Nikkei or Rudenium these metals may not have very large surface area what we do is we put them on supports. Supports have very large surface area we disperse this catalyst on the supports so that we get very large surface area and nice dispersion maximum number of atoms of the catalytic component will be available for the reaction and I need very less catalyst loading that is the idea. So, we are going to look at the porous catalysts now as I told you before the first lecture on catalysis like there are different types of catalysts sometimes you may have to deal with non porous catalysts where the reaction takes place only on the external surface like for example, gauze say platinum gauze, silver gauze and all that. In that case we do not need to worry about this pore diffusion effect, but there are many catalysts which are porous in nature for them we need to consider the effect of pore diffusion. The rate equation that we have derived will further get modified because of the pore diffusion effects and we are going to look at the effect of pore diffusion in this particular lecture. Now, I have told you that we have a catalyst I will redraw the picture of the catalyst again. So, we have this catalyst in which you have different pores remember I had done this before as well and then I have this particular space which may be considered as solid continuous solid actually there are micro pores inside, but then for our analysis purpose let us say that I have these spaces where only solid is present and these are the empty spaces which are nothing but pores which are nothing but pores. And the actual catalytic site will be present somewhere on these surfaces or the walls of the pores and in order to access these sites a molecule has to diffuse. So, it will enter the pore from various locations there will be several such locations through which it will go inside and will get diffuse. So, if the catalyst particle has a radius capital R overall movement of the reactant molecule will be inside a catalyst particle towards the center though it will not directly go to the center because there is hurdle here, constriction here it will go this way. Why I am explaining all this in detail because later on you are going to introduce a concept called effective diffusivity and I am going to relate it to the normal diffusivity then with the molecule is moving without any constriction hurdles and so on. Now, imagine a situation where I have a molecule which is sitting at a center of this classroom and for some reasons there is a concentration gradient and this molecule has to diffuse. Diffusion takes place when you have a concentration gradient or other chemical potential gradient and this has to diffuse now this molecule does not know where the walls are walls are far away from the molecule and this molecule will flow assuming or other will get transferred rather there will be flux because of its movement assuming that there are no walls at all because they are far away from the molecule. Now, what is this particular motion molecule like how do we characterize it is a fixed law and then there is a constant called as diffusivity. This diffusivity has nothing to do with the walls it is a free molecule present in the bulk and it is moving because of the concentration gradient. Now, this is not going to experience any resistance otherwise from the walls, but there will be some resistance and because of its molecular weight because of the medium and all that and that is why we have called we have a parameter called diffusivity and this diffusivity will decide how fast the molecule will move now compare this situation as against a molecule which is present inside the pores inside the pores here now if molecule is moving here it is going to experience some resistance from the walls why because look at the pores what is the typical size of the pore the pore size is very small and that is the reason we get such a large surface area. So, this size is of the order of few nanometers or it can be angstroms the different types of pores micropores, macropores, mesopores the very small and the order of magnitude of this diameter of the pore is comparable with the size of the molecule what is the typical size of a molecule say benzene it is close to 6 angstroms. So, 6 angstroms molecule size and suppose you have a pore of say 15 or 20 angstroms then it is going to definitely face some resistance look at a mean free path of the molecule definitely it will experience the existence of the wall around it and the diffusivity will go down. So, this wall effect this wall effect what else there is for example, this molecule is moving from the external surface inside if I look at the actual distance travelled by the molecule it is this distance but the molecule is going to go this way. So, it has to follow a torches path and the diffusivity the effective diffusivity if I consider the actual moment this then the diffusivity will further go down because of this path that it has to follow. So, for the same displacement or the movement or distance it has to go this way. So, this is another effect we are going to put all these effects in one parameter later then the pores these pores there is no guarantee that they should be of uniform size there are very few catalysts especially zeolites and all that where the pore size is well defined. But, take example of say ion exchange resin you cannot define the pore size it is in the range there is a pore size distribution that you will see. So, if you take a single pore the pore itself will have something like this the wall is not straight and then there is a possibility that the area of cross section will vary as the molecule moves ahead. So, there is again some constriction that it is going to experience and because of that further the diffusivity will go down. So, now the effective diffusivity if I want to determine the effective diffusivity that diffusivity is nothing but the normal diffusivity and the effect of all these factors. So, let us define a term called effective diffusivity. So, we call this as DE effective diffusivity the effective diffusivity is equal to Knudsen diffusivity DA which is a diffusivity inside a pores that means it considers the resistance offered by the fluid inside a pore and the walls of the pores. Now, how do I relate these two? This is definitely going to be larger than this because this is smaller right. So, there will be some effect because there is a space constraint. So, some effect will come because of phi p which is nothing but porosity because some space is occupied by solid that is not available for the movement of the molecule. So, it has to be multiplied by the porosity it will further reduce. So, porosity is always less than 1 then I was telling you about the constriction this is called as constriction factor because the variable area and there is another factor called as tortuosity alright. Sometimes people club these two and they call as tortuosity factor. So, in some books you will see sigma not present there and you have a tau f or whatever less tortuosity factor, but what needs to be understood from this is that this is a normal Knudsen diffusivity that we define for a molecule right. You have phi here taking effect of considering the effect of porosity presence of solid constriction and then the tortuosity ok. This considers the effect of wall anyway right, but these three factors are the factors which are dependent on the solid that we have ok. It is a morphology it is porous structure and so on right. So, these are the factors which will change from particle to particles right solid to solid catalyst right. So, this is the effect to diffusivity. Now, this diffusivity once I have this I have this diffusivity in terms of other factors I can write a normal fixed law type of equation for a solid porous solid ok. So, what this allows me to do is to write mass balance flux equation based on fixed law ok. Only difference is instead of normal diffusivity we have a we have an effect to diffusivity here ok. Now, let us spend some time understanding tortuosity. Now, the tau that we have defined is as I said like the molecule has to follow a path this way because of the porous structure, but actual distance that it travels is only this much ok. So, I am at the external surface and I am going inside, but then it has to follow this path. So, the extra path that it has to follow will result in reduction in a diffusivity ok. Now, the tortuosity is defined as actual distance travelled divided by shortest possible distance between these two points. Suppose this is point a this is point b the shortest possible distance between a and b right. So, it is very easy to define or even easy to calculate the value of tau suppose I know the angle here ok. So, let us say I have I just approximate it like this and this is same 90 degrees is the angle ok. Can I calculate the value of tortuosity? You can do this exercise tortuosity is the actual distance travelled. So, let me say that this is L this is L. So, the actual distance travelled is 2 L if this is 90 then I can calculate this distance this is going to be 2 L divided by root 2 right 2 L divided by root 2. So, the value of tortuosity comes out to be root 2 which is say 1.41 tortuosity is greater than 1. So, always greater than 1 we cannot we cannot expect to have straight pores very difficult. So, at. So, in extreme situation would be tortuosity is 1, but not quite unlikely sometimes the tortuosity can be of the order of 5 6 7 10 it can be that high it all depends on the material ok. And of course, people have control over making a material also like this is separate all together different area where we design catalyst as per our requirement, but then for our analysis let us take this tortuosity as a material property and that will come when the or that can be evaluated when the material is design or it is available. And then we go ahead and get a value of effective diffusivity beds based on the known value of Knudsen diffusivity alright. Now, once we know the value of diffusivity as I said before we can write the fixed law for the solid. So, let us assume that I have a spherical particle I have a spherical particle and then the molecule is going inside ok. Now, this is a center and let us take a differential element and write fixed law around this. So, this is r plus delta r and this is r and the radius is capital R ok. Now, this is a diffusion taking place as I told you this particle you go to have pores inside, but now I am going to assume that the diffusion is taking place straight along the radial direction actual movement is like this, but then now that I have considered that effect of tortuous movement in my effective diffusivity I am free to write equation something similar to what I do normally for diffusion or flux writing mass balance when there is a concentration gradient alright. Now, look at this particle again there is catalysis happening in sense there is a reaction taking place. So, as n when the molecule goes inside it is going to meet catalytic species and reaction is going to take place. So, it is something happening in parallel. So, reaction and diffusion they are taking place simultaneously. So, as n when there is a movement is a reaction also taking place. Now, the overall rate of the reaction will depend not only on the actual rate of reaction actually in a sense the one that we derived yesterday considering the effect of adsorption desorption and chemical reaction whichever is the controlling whatever ok. So, that particular rate and now the rate of diffusion will also matter you can imagine the situation where the reaction is instantaneous very fast. Even if adsorption controls or desorption controls or even the reaction chemical surface reaction controls all these steps are happening very fast. But then diffusion becomes very slow that particles molecules are not able to diffuse inside then the overall rate will go down right. So, diffusion is playing an important role here. If the diffusion is very fast the pores are wide enough diffusivity is very large then it is the chemical reaction or whatever adsorption desorption at the site that will probably control the overall rate. So, try to understand the importance of diffusion now internal diffusion we are not yet talked about external diffusion or mass transfer fine. So, we are going to write the law sorry the mass balance for this the flux balance there is no convection here there is only diffusion that is taking place. So, the flux at r is W a r this is the way I am defining the flux W is the flux of component a that is reactant in radial direction into 4 pi r square what is it the area right at r at r minus W a r at r minus W a r minus W a r sorry a r 4 pi r square at r plus delta r here the flux here minus the flux here can I write this equal to 0 no right this is this is another term that we need to consider here that we need to consider here what is that term that term is reaction if there was no reaction then this would be equal to 0 right, but there is reaction taking place. So, look at this as a CSTR so material coming in going out and plus reacted is equal to 0 right. So, I need to add the reaction term so plus reaction rate r a into let me explain all this later first let me write it into rho C which is nothing but a density of the particle mass per unit volume into 4 pi r m square into delta r. So, this is a reaction term I will explain this detail is equal to 0. So, material at r is going out minus coming in plus generation term right. Now, r a will have its own sign depending on whether we have reactant or product if a is a reactant it will have a negative sign what is this this is volume this is volume 4 pi r m square into delta r then rho C is a density. So, this is volume this is density then density into volume is nothing but mass. So, rate is expressed per unit mass of the catalyst I told you in the last lecture most of times in solid catalyst reactions the rate is expressed per unit mass of the catalyst. So, r a is per unit weight of the catalyst you have to multiply it by the weight. So, this is to be this has to be the weight of the catalyst. So, density into volume right. So, this is the mass balance this is the mass balance. Now, let us go ahead and try and elaborate this term called W a r which is nothing but flux. As I said we can use a fixed law for this with in spherical coordinates to write the flux equation for W a r. So, W a r is nothing but minus d e that is effective diffusivity into d c a by d r. So, this is a flux. Now, if you substitute for this flux in the main equation then what do we get we get and of course like you have r and you have r plus d r. Then just play with the equation assume delta r approaching 0 I can convert that difference equation to the differential equation. And then what I get is this d minus d e d c a by d r into r r square d c a by d r r d y, but this is d by d r of course is minus r square rho c r a dash is equal to 0. So, I have just substituted for W and converted the difference equation to the differential equation and I get this particular equation right. Now, if I want to write the expression for r a dash what is r a dash? r a dash is the rate of reaction, intrinsic rate of reaction when the molecule is present near the catalytic site. So, if r a we can obtain it by rate law then r a can be k into c raise to n. So, if I expand it further what I get is d to c a by d r square just expand this further plus 2 by r d c a by d r square plus 2 by r d c by d r. Let me explain it later, but let me write it first k n into s a into rho c divided by d e look at this d e this d e is coming here I am just dividing both the terms by d e right that is effective diffusivity into c a raise to n is equal to 0. Now, this you would be able to guess what I have done here I have divided both the sides by r square right. So, what is happening is you have d d 2 c a by d r square second order equation d c a by d r 2 by r. Now, I have written expression for r a dash that is the rate of reaction at the catalytic site in terms of these constants. So, let us try understand these constants k n is going to be the rate constant we will see the units then s a is a surface area is a surface area rho c I have already told you it is the density mass for it volume c a is a concentration raise to n this n is nothing, but order of the reaction with respect to a the intrinsic kinetics here has been assumed to be nth order reaction that means the kinetics is a power law model rather than the commonly used models like Langmuir-Hinshelwood and so on for heterogeneous catalysis right. I have a reaction let us say that I have only an isomerization reaction a going to b only single reactant present n. So, k into c a raise to n right and this is nothing, but a diffusivity all right. Now, can you guess the unit of k n I have multiplied it by surface area now. So, now this surface let us understand what is s a s a is a surface area. Now, what is the unit of s a s a is say meter square per gram of the catalyst or k g of the catalyst whatever. So, this is a unit of s a right. Why I have defined this particular parameter here because normally k n would be available in the unit of or unit on the basis of area that is per unit area. See now the different type of rate constants in a normal reaction without catalysis or rather without heterogeneous catalysis rather normally we have the rate constant per unit volume that is why we multiply rate by volume in many design equations. In the last lecture I told you that for solid catalyzed reactions or heterogeneous catalysis the rate is expressed per unit weight of the catalyst. So, that is why we multiply normally to the rate by w that is the weight of the catalyst right. Now, I am defining another rate constant here which is expressed per unit area of the catalyst which area it is a area present in the pore can be very high very large ok. So, per unit area of the catalyst. So, if you have a rate constant per unit area of the catalyst it has to be multiplied by the surface area per unit weight of the catalyst. So, this gives me k n into s a gives me the rate constant per unit weight of the catalyst because k n is per unit area s a is area per unit weight. So, this becomes per unit weight of the catalyst right and rho c if I multiplied by rho c then it becomes my normal rate constant that is per unit volume of the catalyst. So, this is nothing but the rate constant expressed per unit volume of the catalyst. It may be noted that k n in some standard reaction engineering text books like for example, J N Smith. So, this is the constant of the catalyst. The k n is taken to be per unit mass of the catalyst. So, in that case k n that they are using in their book is equivalent to k n into s a that is surface area that we are using in our notations. I hope it is clear. So, this together like all these parameters together is nothing but my normal rate constant per unit volume of the catalyst. Now, what is the actual unit of the rate constant? So, for a first order reaction r a is equal to minus k c a right and if I want to get a unit of k r a is expressed this is a per unit volume of the rate that is moles per second is equal to minus k moles per meter cube. So, what is the unit of k for a first order reaction? It is second inverse. So, this is a k this is a k small k normally. This is per unit volume because I am expressing rate per unit volume, but for a first order reaction the unit is second inverse second inverse that is the unit of first order rate constant when the rate is expressed per unit volume. Now, if I express rate for a first order reaction per unit weight of the catalyst. So, what is the what is the unit of the rate constant? Just look at this. Now, this is for the volume per unit volume. Now, I am going to look at weight of the catalyst. Now, r a is expressed per kg of the catalyst right moles per second moles per meter cube right. So, what is the unit of k here? Unit of k is meter cube per kg second. This is for the volume this is for the per unit weight this is per unit volume this is per unit weight right. Now, what about the unit when I express my rate per unit area per unit area area of the catalyst. 1 upon meter square moles per second minus k moles per meter cube this is for area basis is area of the catalyst. So, what is the unit of k here? Unit of k is meters per second and this is nothing, but our k m that we have just written our balance along the spherical particle the k m right. So, let me summarize I have 3 types of rate constants 3 types of rate constants expressed per unit volume for a first order reaction k has a unit second inverse second per unit mass of the catalyst right. First order reaction k has a unit meter cube per kg second and third surface area of the catalyst first order reaction k has a unit meter per second. So, these are the 3 different units that we probably going to deal with as and when we solve the problems later. So, in our rate equation k m is this right k n is this then k n into s a is this and k n into s a into rho c that is the density of the catalyst particle then it is nothing, but the rate constant per unit volume right. So, let us get back to the equation the original equation that we have derived you have this equation in this you have the entire k n s a into rho c is nothing, but the rate constant per unit volume. Now, why we doing all this we are doing all this to get the concentration profile inside a particle we need to get we need to know how the concentration changes as we go from the external surface to the interior of the catalyst. And why do we want all that because later on I am going to determine how much is the reduction in the concentration and because of which how much is the reduction in the rate overall rate right when the diffusion is important. So, that is the purpose now we have this differential equation all I need to know is the boundary condition first to solve this differential equation and get the concentration profile inside a catalyst. So, what is the boundary condition here I have a particle I have a particle I have it is a second order differential equation I need two boundary conditions. So, at r is equal to capital R I have c a is equal to c a bulk or 0 whatever outside concentration right. Since external mass transfer effects are not present we are assuming that c a s is equal to c a b which is something that I am going to see which is something that I am going to measure I cannot go inside a pore and get a concentration or measure the concentration. I will see the concentration only in the bulk why the concentration inside a particle is lower than what you have outside it has to be lower why because there is a resistance face by the molecules while diffusing from the external surface to the interior. And since there is a concentration gradient there is a flow there is a flux diffusion that is taking place if this diffusion is significant then the concentration gradient is significant even if you have a very active catalyst present inside it is not going to see the concentration outside a particle it is going to see the concentration which is smaller than outside right. And because of which the rate will go down and our aim is to rather characterize the quantifier rather these particular resistance and reduction in the rate because of the inter particle diffusion alright. So, r is equal to capital R c a is equal to c a b and there is another boundary condition that I need is equal to r is equal to 0 since there is a symmetry you have c a is equal to finite or d c a by d r is equal to 0. So, with these two boundary conditions I need to solve this equation alright I need to solve this equation and get a concentration profile once I get a concentration profile my job would be easier I would be able to know the local rates at every point inside a particle and from that I can calculate the overall rate which is going to be the actual rate rather and it is going to be less than what it would be when there is no inter particle diffusion resistance fine. So, I have this equation and I need to further modify this to get some dimensionless parameters which are going to give me some idea about the importance of diffusion. So, let us de-dimensionalize this equation by defining certain dimensionless variables. So, let us say I have I express concentration as psi equal to c a divided by c a s or c a b rather that is what I have used c a b what is c a b is a bulk concentration is at the external surface that is the external surface you can even call this as c a s. So, some books you may see a notation c a s c a b right and then there is another dimensionless variable that I define is lambda and something to do the radial distance r divided by capital R. If I have these two variables defined dimensionless variables dimensionless concentration this is dimensionless radial distance. If I have that then my differential equation gets reduced to a form. So, this is a differential equation that I have I am going to substitute for dimensionless concentrations and dimensionless radial distance what I get is this I am directly writing this you can try it on your own. So, d 2 psi by d lambda square plus 2 by lambda d psi by d lambda minus k n s a rho c r square c a b raise to n minus 1 divided by d e that is diffusivity into psi raise to n minus 1 divided by m equal to 0 equal to 0. So, this becomes my non-dimensional form of the equation psi is a dimensionless concentration lambda is dimensionless radial distance. Now look at what I have got here now this is a term that is going to tell me something. So, let me write that term separately and denote it as phi m. So, this phi is different from porosity this is not phi s this is phi n, but then I will say at phi n square is equal to k n s a rho c r square c a b n minus 1 this term I am just writing it here again. Y square will be clear later to you it makes the solution easier. So, this is a term now look at this term carefully what do you have in this term in the numerator you have the parameters related to reaction what do you mean the rate constant is more the rate of reaction is more surface area rho c all these parameters have something to do the reaction of course, this r is a it is common for all I will just manipulate this equation further let us not consider r square at this moment, but otherwise rest the parameters are showing the dependence or the they are influencing the reaction what about the denominator d e d is something that will affect the diffusion. Now let me just manipulate this other other just play with this equation further phi n square is equal to k m s a rho c r c a b n b divided by d e into c a b see earlier I had c a b c a b raise to n minus 1 now what I am doing is I am writing c a b raise to n and I am taking c a b here minus 0 divided by r. So, what do I see here I see that this is nothing, but the diffusion rate see that concentration gradient distance diffusivity something if you have a linear gradient then this becomes the flux due to diffusion. Whereas in numerator you have something that will affect the reaction actual reaction. So, the 2 rates that I am comparing here, so this particular parameter phi n square is going to give me the relative importance of actual reaction taking place at a site which may have adsorption desorption surface chemical reaction and all and diffusion. So, relative importance of these 2 processes if phi n is very large what does it mean? It means that diffusion is slow diffusion is important if phi n is very small diffusion very fast or not important at all that means it is like an open catalyst there are no pores right or pores are very large. So, this value is going to tell me everything. So, for a given reaction for a given catalyst if I can determine the value of phi n or phi n square the magnitude of this particular parameter is going to tell me whether reaction is important or diffusion is important. If this value is very large then diffusion is important sense is slow and it is going to control the rate and if it is very small then I do not need to worry about it I will just take it as if like there is no diffusion problem. So, this phi n is called as Thiele modulus it is a very important term in heterogeneous catalysis that tells you the importance of pore diffusion this term is called as or this particular number is called as Thiele modulus. So, phi m is Thiele modulus I can write it for different types of reactions first order reaction second order reaction and so on. So, the equation that I have is for a spherical particle because I am talking in terms of the radius of the particle if I have cylinder if I have something else some other shape then I will have a corresponding characteristic dimension of that particular shape we will look at that separately like suppose I have a cylindrical pore how do I write expression for how do I derive expression for Thiele modulus. But, right now I am doing it for a spherical particle and I have the expression given by this that is phi n is equal to k m rho C s a C a b n rho c s a raise to n minus 1 r square divided by d e and suppose n is equal to 1 the first order reaction because very easy k 1 rho c s a r square divided by d e sorry this is phi n square now phi 1 would be r root of k 1 rho c s a b n rho c s a b n rho c s a b n rho c s a divided by d e. If I give a rate constant per unit volume I do not need to write all this it would be r k v d e why k v because it is expressed in terms of per unit volume because this has a unit per unit volume. First order reaction it will be second inverse this is a dimensionless number look at a units r meter r square divided by d e. k second inverse d e meter square per second right. So, if you look at this no units this is dimensionless number everything will get cancelled here. So, remember phi is the dimensionless number called as Thiele modulus which gives you the importance of tells you the importance of pore diffusion effects. If the value of phi is large then the pore diffusion effects are significant I cannot ignore them and if the value of phi is relatively small say 0.01 or whatever then I can safely ignore pore diffusion and treat the reaction as if there are no pore diffusion effects and the reactions taking place on the catalytic surface which is wide open for me. We will go further and try and simplify this and understand pore diffusion effects in detail get some more insight in next lecture. Thanks.