 Good morning. So, let us continue with our problem like we had stated a problem last time if you remember the is a spherical particle that we have and there is some data given about a concentration. So, the diameter of the particle is 2 into 10 raise to minus 3 centimeters and the external concentration C A S is equal to 0.001 moles per dm cube right. Then it is said that the diffusion coefficient or effective diffusivity is 0.1 centimeters per second the data given and moreover it is like also given that at half way distance that is r is equal to capital R by 2 right. The concentration is or C A by C A S is equal to 0.1 right. So, from this they have asked what would be the concentration when r is equal to 0.0 by capital R is equal to 0.7 or in other words if the distance is 0.3 into 10 raise to minus 3 centimeters from the external surface. The radius is radius is 1 into 10 raise to minus 3. So, 0.3 so remaining would be 0.7. So, r by capital R is 0.7. So, what is C A by C A S and we have solved this problem and we arrived at a answer C A is equal to 2.36 into 10 raise to minus 4. How did we do that? Because we know the concentration relation for the concentration profile. We know the relation for the concentration profile that is C A by C A S which is denoted as psi is equal to 1 by lambda right sin h lambda into phi phi that is Thiele modulus sin h phi ok. So, from the given data we can calculate phi which came out to be 6 and then again we substituted for the distance lambda dimensionless distance and we calculated C A. So, C A was equal to or C A was equal to 2.36 into 10 raise to minus 4. So, that is a quick revision of what we have learned last time. Now, the next part of this problem is that what should be the diameter to understand this? What should be the diameter so that eta that is effectiveness factor is 0.8. So, it is very. So, let us stand and learn from this just solving a numerical problem is one thing, but at the same time let us understand what they are trying to say. We can manipulate the diameter of the particle and get effectiveness factor that we want. What is the importance of effectiveness factor? Effectiveness factor is going to decide a rate because actual rate is nothing but the effectiveness factor into the intrinsic rate ok. Now, if the effectiveness factor is close to 1 that means there is no pore diffusion resistance. So, I can manipulate I can design the catalyst particle have a catalyst particle size in such a way that the diffusion resistance is negligible that means effectiveness factor is close to 1. In this case they are asking no do not go for 1 that will be too much I will tell you why, but then 0.8 is good enough. Let us see what happens, but before that we need to understand what is the effectiveness factor otherwise like for the given diameter. For the given diameter what is the effectiveness factor? If the effectiveness factor is already 0.9 or 0.95 then there is no point in bringing it down ok, but anyway they have asked it then we have to solve it all right, but it is quite likely that effectiveness factor is much lower compared to 0.8 and that is why we need to play with the diameter ok. So, in the first part of the question we just looked at a concentration right from the concentration profile. We calculated the Thiele modulus ok, but let us try understand what is the effectiveness factor in the first part itself and then we will see how to manipulate or how to get a value of effectiveness factor to be 0.8. So that there should be a required diameter for it all right fine. So, what is the effectiveness factor in the earlier case when the diameter is 2 into 10 is to minus 3 centimeters which is given rather ok. So, dp is 2 into 10 is to minus 3 centimeter is earlier case phi is 6. So eta i is equal to 3 divided by phi 1 square phi 1 cot h phi minus phi 1 minus 1. We know this ok, we know this. Now, if you substitute phi 1 is equal to 6 here and this is the value ok. You are going to get a value which is much less than 0.8 you can verify that ok. It would be of the order of 0.4 to 0.5 ok that low I am not calculated it, but you can you can do it yourself or it can be somewhere close to this range right. Now, this value is lower. So, the rate is lower, but I am not happy with this rate look at a particle diameter 2 into 10 is to minus 3 centimeters right 2 into 10 is to say 1 centimeter is this much ok 2 into 10 is to minus 3 ok 0.1 centimeter is 1 millimeter. So, this is going to be less than millimeter small it looks like powder ok, but still the effectiveness factor is 0.4 to 0.5 which is something that I am not happy with ok. So, I need to play with a particle diameter can I play with a diffusivity. Now, I have prepared a catalyst ok diffusivity depends on the pore diameter, it depends on the porosity, it depends on the tortuosity, it depends on the constriction factor remember. So, with this now I cannot play with that ok I have prepared a catalyst. Now, only thing that I can play with is a diameter I can grind that catalyst I can grind that catalyst ok, but I cannot change the internal morphology. So, diffusion coefficient cannot be changed the only thing that I can change and get higher effectiveness factor is the diameter ok. I always remember that diameter is something that you can quickly play with right. I can always grind the catalyst if I want to reduce the value of diffusion resistance or increase the value of effectiveness factor that is what we are learning through this problem ok. It would not take much time to numerically solve this problem, but I am trying to convey like what is the meaning behind all this that is more important right. Now, if I want to take this value of effectiveness factor to 0.8 this is d p I need to vary fine. So, in order to vary d p I need to vary d p what do I need to do? First of all what is phi? phi is r root k 1 rho c s a divided by d e. I do not need to buy out this equation I know the proportionality like if you increase r resistance is going to increase if I increase diffusivity resistance going to decrease and relatively if I increase the value of intrinsic rate constant again the diffusion is going to be important ok diffusion resistance is going to be important. So, that is how I remember this equation and what is this numerator? Numerator is numerator is nothing, but the rate constant per unit volume ok. It is per unit volume we already looked at that right. So, this is nothing, but a rate constant let us not get into detail this is the rate constant fine. Now, let me just denote this as r k 1 v divided by d e ok why v because per unit volume the numerator what do I know in all this 6 is equal to 1 into 10 raise to minus 3 k 1 v divided by the diffusivity or diffusion coefficient that is 0.1. So, I get a intrinsic rate constant I get a intrinsic rate constant which is 3 6 0 0 0 0 0 ok you can do this numerical calculation and check ok. What would be the unit? It is a first order reaction unit is per unit volume ok is a volumetric rate constant. So, it is going to be second inverse we already seen that. So, I have determined the value of intrinsic rate constant which is not going to change this value of rate constant is not going to change even if I change diameter ok. So, if I want higher values of effectiveness factor I am going to change the diameter, but rate constant is going to remain same remember that ok fine. So, let us go ahead. Now, I want effectiveness factor to be 0.8 which is nothing but 3 by phi 1 square phi 1 right cot h phi 1 minus 1. So, for 0.8 phi 1 becomes is going to be less or more is going to be less why because effectiveness factor is increase of phi 1 value is going to go down ok right. So, phi 1 becomes 2 just take that ok when I substitute when I do trial and error just assume the value of phi 1 get this see whether it is equal to 0.8 or not if it is not then go and change the value of phi 1 is trial and error method ok and get a value of phi 1. Now, with this value of phi 1 now again I go back to the expression for phi 1 is equal phi 1 is equal to r root of k v divided by d e. What is that I do not know in this of course, this is a new value of phi 1 earlier it was 6. So, I say phi m new value ok. So, I will have a new r these 2 are not going to change because d depends on the catalyst morphology pore diameter and inside structure and this is again something to do with intrinsic rate constant ok which may have adsorption desorption also like what you have seen before. So, these 2 that is k v and d e are not going to change only r is going to change such that I have a new value of phi 1. So, for the new value of phi 1 if I substitute for k v d e and all this is not r raise to n this is a new new r just do not get confused this is superscript and not the index. So, the new value of r you have substitute for phi and k v that I have calculate 36 whatever 1000 lakh order and diffusivity I am going to get r is equal to 3.4 into 10 raise to minus 4 centimeter which says d p is equal to 6.8 into 10 raise to minus 4 centimeter. So, this is my answer. So, what is happening look at this diameter and the one that was given earlier the diameter given was 2 into 10 raise to minus 3 2 into 10 raise to minus 3 now the diameter that I have calculated is 6 into 10 raise to minus 4. So, diameter has gone down further and that is the reason I am able to increase the effectiveness factor. So, that is the meaning of it. So, I am playing with the diameter to increase the effectiveness factor. In general if I just do a reaction at different diameters and calculate a rate for a given concentration then the graph is going to be like this in general I am not talking about this now the problem is solved. Now, I am talking about a general case a diameter of the cactus particle and a rate at a given concentration. So, say initial rate or inlet rate you know how to measure this rate. I am doing experiments in laboratory say I am using a differential reactor I am probably using a spinning basket reactor you already spend some time discussing what is spinning basket reactor I may be doing experiment in a slurry reactor and calculating the rates at given concentrations throughout and just changing the particle diameter. So, what kind of behavior I expect here you plotted on log log or possibly what will I get a what kind of relation will I get this relation is going to be quite similar to the relationship between eta and phi. Why because the diameter has something to do with the phi and the rate observed has something to do with the value of eta. So, the relation is going to be like this. So, at lower diameter you are free of internal mass transfer diffusion resistances. So, I get maximum rate and as I go on increasing the diameter the molecules have to travel or diffuse over a long distance long path because of which the rate is going to go down resistance would increase that is why it is going down that is the meaning of it. So, if somebody does experiment and gives you the data like this you should know the meaning of that and of course, remember all this rate is for given concentration and this is what I have observed for different diameter you are not changing anything else only the diameter. In that case I see something like this means definitely there is a possibility that diffusion resistance is playing an important role I am using a word possibility. Now, what is that with why will the rate get affected because change because of change in diameter here you know the reason. But is that only reason if I change the diameter and the overall rate is changing is there any other possibility. Now, what have we looked at so far adsorption chemical surface reaction desorption and internal diffusion. Of reactant and product apart from that there is something else that will play an important role in the overall rate or govern the overall rate possibly you already mentioned about it, but not discussed so far in detail is the external mass transfer resistance external mass transfer resistance which has nothing to do with the particle morphology nothing to do with the pore diameter nothing to do with the internal structure of the particle. It depends on something that happens at the exterior surface and in the bulk. So, we will see that in detail now, but then before that just remember probably I have not stated it explicitly that I am talking about this rate constant every time intrinsic rate constant say here or wherever I have used the expression k c a and all that k here I have just used it say for first order reaction r a is equal to minus k c a or if for nth order reaction can be c a raise to n. Now, this k or this expression I have used it for as a normal rate equation, but sometimes it can be complicated or in the sense it can have something in a denominator as well why that will take care of the adsorption term that will take care of the lamur initial would mechanism illeradial mechanism. So, I am not considered it deliberately because then our derivations would become difficult and the interpretation is a problem. So, I have considered very simple case where the rate is given by k c a and that happens even if you have lamur initial adsorption predominant there when does that happen I told you like if you have very high temperature if you have very high temperature then the adsorption is insignificant. So, the denominator becomes 1 and then the expression boils down to a form of a normal rate expression that is r is equal to minus r a is equal to minus k c a raise to n. So, always remember that whatever analysis we have done here is but considering a simple equation, but there is a possibility that you may have adsorption constants desorption constants of the species involved in this particular expression if you really want to get very close to reality right now the purpose is to understand it. So, that is the reason we have use a very simple equation. So, there are so many problems in assignment shades or even at the end of text books like Fogler, Levenspiel and all you should attempt and solve this problem. This is where we are closing this chapter of internal diffusion, but anyway we will be referring to that because not that like one studied and then forgotten because it will come again and again later in our discussion whenever we try and design a reactor. So, we will talk about external mass transfer now it is a different phenomena or other the step than reaction adsorption desorption and internal diffusion what is happening external as a name says the external mass transfer. So, you have a particle you have a particle and fluid is flowing over it. This is a catalyst particle now what is likely to happen at external surface of the particle the fluid dynamics knowledge transfer phenomena tells you that is a boundary layer here there is a boundary layer slightly non-spherical in nature because of flow patterns here it is slightly thicker compare to of course, this can be equal here whatever now how do I define a boundary layer how do I define a boundary layer I define a boundary layer is a thickness of a layer in which the velocity is changing the velocity here near the solid surface is ideally 0 there is no slip and the velocity would go on increasing and will become almost equal to the velocity of the bulk that point and that velocity and that particular thickness at which it becomes a say 99 percent of the external because getting it exactly equal to be a problem. So, getting 99 percent equal what is the length that is a length or thickness of the boundary condition we define it for momentum balance momentum transfer based on the velocity. Now, there is another boundary layer for mass transfer the similar like your velocity is changing here in the case of component concentrations there will be a concentration changing here why is it happening because this layer is a stagnant or close to a stagnant layer that is why the concentration will change the concentration here may not be same as concentration here and that has great implications on the reaction rate. So, far we have been assuming that a concentration at the exterior or external surface of the particle is same as a concentration in the bulk C A B is equal to C A S right. So, far we are not really looked at a difference between these two, but now we are going to look at a possibility that there is a resistance offered by the external surface the boundary layer is thick enough. So, that the concentrations are different C A B is not same as C A S C A B is bulk concentration C A S is a surface concentration this is because the resistance offered by that boundary layer. So, the concentration is going to increase. So, typical profile in the boundary layer of the concentration is this this is the concentration boundary layer earlier I mentioned about the velocity boundary layer momentum balance right. Now, I am talking about the concentration boundary layer now let us make certain assumptions and try and quantify the resistance. Now, this boundary layer thickness is not very large in fact, it is very very small compared to particle diameter most of the times right and this boundary layer if you look at a particle particle is very big and I have a boundary layer small boundary layer around it. So, at any given point a curvature can be neglected as far as this boundary layer is concerned. So, I can say that I have a particle and there is a boundary layer now it is quite possible that you have something like this the curvature will start. So, this is this is surface I can assume at a given point I can assume it to be flat and this is a boundary layer this is a boundary layer. Now, there is an assumption that this boundary layer is stagnant enough and whatever resistance offered by the entire this is where you have bulk right and this is where you have solid. So, whatever resistance on this side for the mass transfer to take place mass transfer to take place is offered by this bulk the conditions here may be velocity whatever ok then all number and all that. So, that particular condition is going to offer certain resistance and that resistance we are going to assume that that is concentrated in this boundary itself or it is called as film ok. So, all the resistance is is concentrated in the film this is a very hypothetical case that I have a film that takes care of all the resistance. So, in terms of a concentration profile now it looks quite it becomes very easy for the analysis you have a film in this the concentration is going to change how it is going to change we will see, but there is one concentration here and there is one concentration here. Now, for a reaction to take place say A is going to B. So, A is sitting here in the bulk now A is going to diffuse through this film and go here and B is going to come out of course, inside you have pore diffusion adsorption desorption that is occurring and then B is coming out. So, if A has to transfer from bulk to here the profile has to be like this and for B it would be other way round. So, let us consider a reaction A going to B alright let us consider a reaction A going to B. There is a reason why we are considering this reaction as a first step if you have non-equimbal reaction you have some different things to be considered, but let us now talk about a simple single reaction which is a symbolization reaction, equimbal reaction A going to B. So, A has to diffuse through this layer and go there to the surface inside we already seen what is happening once it is here this is your CAS this is your CAS and this is CAB. Now, I need to know the concentration profile inside what is the nature of this concentration profile whether it will be a straight whether it will be linear non-linear. So, let us let us stand look at this and it is simple we already done this exercise for inter particle diffusion. So, let me consider a differential element like what I did before also, but now it is going to be little easy because I do not have spherical coordinates and working Cartesian coordinates. So, this is at x is equal to 0 and this is at x is equal to say delta. So, delta is the thickness of the film. So, this is x plus dx and this is x and diffusion is going to take place in this direction for A for B it will be opposite. Now, can we write equations here for the flux. So, flux at x flux at x say d A minus d C A minus d C A minus d C A by d x at x minus d C A by d x at x plus d x flux here minus flux here by minus sign because d C A by d x is negative flux has to be positive. So, flux here minus flux here is equal to what earlier what did we do in the case of inter particle diffusion if there was some reaction that was occurring here. So, coming in minus going out plus is the plus reacted is equal to 0 right, but now we have coming in going sorry going out coming in is no reaction reaction is taking place here and not here is only non reactive mass transfer that is taking place. So, right now if you simplify this for dx going to 0 you what you get is a very simple equation a second order differential equation which has a solution and now you will determine A and B by applying proper boundary conditions. What are the boundary conditions x is equal to 0 C A is equal to C A as x is equal to delta C A is equal to C A B right. So, determine the value of A and B first of use this boundary condition x is equal to 0 C A is equal to C S. So, B. So, that implies B is equal to C A S. So, let us calculate the value of A. So, C A minus C A S is equal to A x for x is equal to delta C A is equal to C A B. So, C A B minus C A S by delta is equal to A and if you substitute now this A and B if you substitute here what I get is C A minus C A S divided by C A B minus C A S is equal to x by delta. So, what does it say it is a linear profile inside it is a linear profile inside and we know the boundary conditions and this is a delta the slope is known right. So, it is a linear concentration profile that makes my exercise quite easy. So, it is you have C A B C A S right. Why I am doing all this because I want to get a rate I want to get a flux see mass transfer and diffusion if I want to get a overall rate I have to go through the flux equation flux multiplied by a cross sectional area is going to give me a rate for reaction we already know R is equal to K C A and all that is a rate, but that depends on the unit of K whether it is per unit volume per unit area and all that there is so many possible units. So, be careful about units every time flux is equal to D A right I am not using effective diffusivity here or is a normal diffusivity in the bulk diffusivity into D C A by D C A. So, this is going to be linear. So, substitute for this. So, this is going to be so this is going to be D A by delta right into C A B minus C A S right no delta x is equal to delta right and delta C A is equal to C A B minus C A x linear. So, this is a flux what is the unit moles or kilo moles per meter square second per unit cross sectional area area for flux which area are we talking about by the way we talked about area before also in inter particle diffusion, but do not confuse that area with this area not the area inside it is the area here which is almost equal to the external surface area for the mass transfer to take place. So, this is that area cross sectional area for the flux flux. So, here that area is nothing but the particle external surface area because that is where the mass transfer is occurring, but let us keep the same units right now and go ahead. So, before we go ahead now this flux is equal to normally it is denoted by J A is denoted by J A is one important assumption that we have made another assumption rather we have right we call this assumption because we say that this is A going to B reaction that means it is an equimolar reaction it is an equimolar reaction that is why we have written the equation for flux as flux as J A is equal to D A D C A by D x this is fine, but we have equate it equal to W A normally this happens when you have the many situations possible now you have a surface on which A is going there is a reaction that is taking place B is formed so B is diffusing back what is a net flow it is a equimolar reaction. So, molar flux the net molar flux net molar flux is 0 here very important because how much ever going in same amount is coming out in terms of moles right that is why this is equal to this very important you might have learned that in basic theory of mass transfer W A is equal to J A otherwise W A is equal to J A plus plus what Y A the mole fraction W A plus W B plus whatever total this net flux if it is going to be significant or finite then W A is not equal to J A, but in this particular case for equimolar reaction A going to B right this net flux there will be signs for this plus minus whatever and then there are no other components W A becomes equal to W B, but negative sign. So, this becomes 0 and that is why I have W A is equal to J A this is very important again I want to avoid complication that is why I am considering A going to B reaction, but once you understand this well it would be easy for you to rather extended further for more complicated cases here idea is to rather how to incorporate external mass transfer resistance, how to show the important external mass transfer resistance in the overall rate equation that is the main motivation that is why we are considering a very simple reaction this will happen W A equal to J A will happen for other cases also even if the net flux is finite even if the net flux is finite if Y A is 0. So, you have a very dilute solution then I can neglect this right. So, W A is equal to J A happens in the case of equimolar reactions it happens in the case of dilute solutions as well, but otherwise for non equimolar reactions or suppose your medium is stagnant otherwise no equimolar diffusion that is taking is for only mass transfer say no reaction taking place in that case we have to be careful always remember that fine. So, let us go ahead we have J A or W A is equal to J A is equal to D A by delta C A B minus C A S constant in the bulk the concentration varies only in the film hypothetical film a stagnant film that is what we assume. Now, this particular parameter D A by delta is called as mass transfer coefficient K C let us denote it as K C alright. So, W A is equal to K C into C A B minus C A S again this is flux moles per meter square second and this is called as mass transfer coefficient you have you already learned this probably mass transfer coefficient depends on diffusivity bulk diffusivity and delta film thickness. So, let us spend some time understanding what is this film thickness and all this is the film thickness right. Now, this film thickness depends on which factors this depends on what is the condition here if you have intense agitation the boundary layer thickness will go down the value of delta is going to go down thickness is film thickness is going to go down intense agitation it depends on the properties also say viscosity. Density and diffusivity as well. So, we will talk about the correlations for calculating K C later let us first complete our analysis when the reaction is taking place how to incorporate effect of K C right. How to calculate K C we will see later or look at a correlations and all later, but right now let us try and put this in our rate equation in the overall perspective where you have so many things happening right fine. I have a catalyst particle there is a resistance here I am not going to neglect it now like what I did before and a is going here. Once it reaches here what is the rate the rate is given by eta into intrinsic rate. Once it reaches here, but before it reaches here is the experiencing some resistance. So, these two are again the steps taking place in series and relative importance is going to decide which is the governing step which is the rate determining step and all again the same logic fine. So, let us assume that eta is equal to 1 let us assume that eta is equal to 1 again but we can always incorporate effect of eta later is not a big thing I just multiply it to the rate. And I say adsorption and desorption they are not really important or probably that we are doing it at very high temperature entire surface is available for reaction most of the sides are vacant. In that case what I say is r is equal to say k now I am using k r I will tell you why I am using k r here into C A again a first order reaction r A is equal to minus k r C A. Now, I want to bring these two effects together. So, this is a reaction when this goes inside. So, that is why it is expressed for the concentration at the surface that is C A S. Why C A S is going to be same everywhere why because eta is 1. So, wherever the reaction is taking place it is at C A S concentration inside a particle. Now, this C A S is not same as C A B and we are going to incorporate the effect of external mass transfer. So, this is when the component reaches the external surface. Now, what is the unit of this? This can have many units. So, one unit would be the normally that we use it is per unit volume of the particle provided this is per unit volume. This can be per unit surface area provided by the particle everywhere then in that case this is per unit surface area per unit surface area. This can be this can be per unit weight of the catalyst. So, this would be per unit weight of the catalyst. Now, we are defining one more unit this is per unit area, but not the internal area per unit external area external surface area moles per meter square per second which is same as the unit is same as the external mass transfer. So, that I can marry these two things they are the same platform. So, this is this unit is moles per meter square per second this is external surface area do not confuse this area with the internal area S A that we have used before. So, this is S A, but it has nothing to do with this this is the external surface area which is for a spherical particle it is pi d p square. So, now this rate is expressed in the same unit as that of mass transfer flux that is why I am using k r here. So, k r is defined as the rate constant per unit external surface area for that I need to know the shape of the catalyst I am assuming it to be spherical here external surface area. So, that I can now bring this same platform as that of the external mass transfer. Now, at steady state at steady state r is equal to or say I will now r a is nothing but w a now because I am talking about a flux here per unit external surface area is equal to minus k r C A S and is equal to what minus k c right C A surface minus C A B also. So, this is the surface area sorry it should be other way round should be B here and surface here both are negative negative the magnitude would be equal. So, what I get here is k r C A S is equal to k c C A B minus C A S look at this these two are equal now right. Now, I get these why they are equal because there is no accumulation at the external surface see at the surface surface is two dimensional accumulation is 0. So, whatever going to the surface is leaving the surface how much ever is going to the surface and is same as that is there is no accumulation there is no source term at the surface there is no sink term at the surface. So, that is why is the surface whatever is going external surface whatever is the mass transfer is getting reacted. So, both are equal that is the meaning of this why we are equating now because mathematically we want to go further and get a simpler rate equation in terms of the external or bulk concentration because it is the bulk concentration that I am going to deal with as far as the reactor design is concerned can I can I measure the surface concentration can I sit on the external surface and measure it no I am aware of the bulk concentration only ok. So, my rate equation the overall rate equation should be in terms of the bulk concentration again the same logic what did we do for intraparticle case there again the internal concentrations I did not know them or rather I cannot measure them rather that is why I finally got a expression for the in terms of the external concentration at external surface. Similarly, for adsorption desorption do I know the concentration I adsorbed species I do not know the concentration do not know and say I cannot measure them ok. So, I always express them in terms of the bulk concentration the same logic here ok. So, at the external surface I cannot measure concentration and since I know that this external surface concentration is not same as the bulk concentration I need to do this I need to express this external surface concentration in terms of bulk concentration ok right. So, let us go ahead from this what I get is C A S is equal to K C C A B divided by K C plus K R K C plus K R ok. So, I got expression for C A S now I go ahead and substitute for C A S in the rate equation R A is equal to minus K R C A S is equal to K C K R C A B divided by K R plus K C and what is the unit here again do not forget moles per meters per second meter square per second and which is same as W A. So, this is my rate equation this is my rate equation look at this rate equation this is again something similar to what we have got for adsorption desorption case of course, the expression does not look similar, but the procedure is quite similar that I eliminated the concentration at the interface at the external surface ok and I have got this final expression which is in terms of bulk concentration I am able to measure this. So, again similar to what we discussed before as well now if I have a C A S T R I want to design a C A S T R I am dealing with the bulk concentrations. So, F A minus F sorry F A 0 minus F A plus R A W is equal to 0 and if I convert this into concentrations say and R A is again a function of concentrations these concentrations are bulk concentrations and that is why I need this expression R as a in as a function of C A B and not C A S and that is why I have got this equation. This equation I am going to use this equation in the design now look at this equation this is so much that we can learn from this equation relative importance of K C and K R the smaller one will govern the overall rate ok right I can I can convert it to this form R A is equal to C A B divided by 1 divided by K R 1 divided by K C same one. So, K C K R I have just taken it down now you can see if K C is very large this becomes 0 sorry negative sign yeah K C is very large this becomes 0. So, it boils down to a normal kinetics for K C very large R A is equal to minus K R C A B intrinsic kinetic control that means which is nothing but C A B is equal to C A S this situation no external mass transfer resistance K C is very large diffusion resistance is negligible mass transfer coefficient very large either diffusivity is very large or delta is very small ok. Other way round K R is very large now come relatively always related large means what related to K C ok K R is very large related to K C then R A is equal to minus K C C A B in that case what is C A S? C A S is almost negligible 0 here C A S is equal to C A B whereas in this C A S is equal to 0 you can think about it will continue this discussion in the next lecture, but this is the meaning of the rate equation. If K C is very large then mass transfer resistance is negligible I can ignore it right if K C is very small compared to K R that case I have to consider it ok that reaction becomes instantaneously is a overall or intrinsic reaction becomes instantaneously overall reaction is controlled or governed by the external mass transfer. So, we will continue this discussion in next lecture.