 So, good morning. Let us continue our discussion on effectiveness factor. In the last lecture, we looked at how to get a concentration profile based on the boundary condition that we have. And then the concentration profile which was derived for first order reaction, intrinsically first order reaction that was used further to get the expression for the effectiveness factor. Now, I told you effectiveness factor is a factor that normally lies between 0 and 1 for isothermal reactions. Non-isothermal it can go beyond 1. We will look at that separately. But then it tells you how important is the pore diffusion. There was one more factor that we defined earlier that was Thiele modulus. Now, Thiele modulus can take a value from 0 to infinity. It does not have any limits as such. So, but then very large Thiele modulus means more pore diffusion resistance and we need to consider the pore diffusion effects while deriving or looking at the rate of the reaction. And small value of Thiele modulus is where the pore diffusion resistance is negligible. So, there is a inverse relationship between Thiele modulus and effectiveness factor. So, let us look at the effectiveness factor equation or expression that we derived in the last lecture. Eta is equal to 3 divided by phi 1 square y 1 because we are talking about a first order reaction now into phi 1 cot h phi 1 minus 1. And you know what phi is? Phi is the Thiele modulus for the first order reaction, which is nothing but or let me call this as phi 1 square, which is nothing but k 1 into s a rho c. You know what all this is? This is reaction rate constant per unit volume of the catalyst pellet or particle into r square divided by d e. So, I have this relationship, which tells me how effectiveness factor varies with respect to phi that is Thiele modulus. Now, I can plot this relationship and see how it looks. So, for a first order reaction and a spherical particle, remember we are doing all the exercise for a spherical particle. We are assuming particle to be exact sphere, but it may go away from sphericity and you have or it may go away from a spherical shape and we need to incorporate those effects also shape factors. You may have a cylindrical pellet, you may have a slab. So, the shape also matters because all the equations that we have derived are for a spherical particle. So, the effectiveness factor that we have derived is applicable to spheres and for the first order reaction, very important. Just remember what we have done. So, before we go ahead, let me quickly revise what we have learnt because this procedure that we have followed is very important that can be used for deriving or getting the expression for the effectiveness factor for other geometries. As I said, you may have a cylindrical pellet, you may have some other geometries a slab. So, in those cases, how to get a effectiveness factor? We have to follow a certain procedure. So, what have we have done here? Same procedure has to be followed, but of course, applicable to that particular geometry. Now, what we have done is we have taken a balance, component balance which is the form of a differential equation, component balance, a differential element in the geometry. Remember, for sphere it was a differential element. Once we have that, so you get a differential equation with appropriate boundary conditions. Try and solve this if possible. Why we want to do that? Because effectiveness factor is equal to flux into area, external area divided by the rate at external conditions and this flux is nothing but of course, this is proportional to dc by dr or dx, whatever is a gradient. Of course, that is multiplied by diffusivity with an appropriate sign and so on. But the gradient is important and if you want to get a gradient at r is equal to capital R or at the boundary flux at r is equal to capital R for a spherical particle, but in the case of a slab, it would be at x is equal to l. So, at the external surface, where I am sitting and observing the molecules going inside a pellet, inside a particle and a rate at which they are going inside, so flux into area, flux is always per unit area. So, now look at this equation and need this and that is the reason I am getting the, I am solving the equation to get a concentration profile. Then substitute it here and whatever you get is the expression for the effectiveness factor. For simple cases like spherical geometry and first order reaction, it is easy to get analytical solution on expression for effectiveness factor. For nth order reaction or for some complex geometry, it will not be possible to get a expression for the effectiveness factor as such. It will be a quite complicated or rather you have set of equations to be solved simultaneously to get a value of eta that is effectiveness factor. So, this is a general procedure how to get eta. We may, if time permits, probably we look at different geometries or one more geometry to get a value of effectiveness factor following this particular procedure. So, it will be clear and of course, while writing these equations, the coordinates may be different right now using spherical coordinates. But in the case of slab, it will be what? It will be along the length cartesian coordinates. In the case of cylinder, it will be cylindrical coordinates. So, the procedure remains same, but then of course, we will have to take into consideration the geometry fine. Now, let us look at a relationship how it looks phi versus eta. I am plotting a log log plot. So, I have this Thiele modulus. Value varies from say here you have 1 then you have 10 somewhere then you have 100 somewhere. So, it is a log log plot. It is a log and again here you have say 0.1 effectiveness factor 0.2, 0.4 and so on. Finally, it will be 1 here. Now, it is easy. The relationship is inverse is going to go like this. Now, at very low values of phi, let us talk about a first order reaction that we have derived expression for. At very low values of Thiele modulus, the value of effectiveness factor is going to be 1. So, 1 here. Maximum value for isothermal reactions and it will remain 1 over a range for very small values of phi depending on what is the rate constant. So, so many other factors how these two things play with each other or there is a interplay of reaction and diffusion. Now, as we go on increasing the value of phi, effectiveness factor is going to go down and this is what you are likely to see the relationship between eta and phi. Remember, it is not going to go to 0, it is 0.1 is a log log plot here. So, it is very clear that as phi increases the effectiveness factor goes down and what is effectiveness factor? Effectiveness factor tells you what is the resistance of pore diffusion. So, if the pores are open, then my rate of reaction is R is equal to K C A S that is surface concentration raise to N. But at C A S is not seen by all the catalytic sides if there is a pore resistance because inside a catalyst particle the concentration drops down. And that is the reason the reaction rate drops down if your pore diffusion resistance and that is why the effectiveness factor goes down and I get a value say 0.1, 0.2, 0.4 whatever. If it is 1, then you have you are comfortable. It means no pore resistance. Now, this is for the first order reaction. For the 0 order reaction it would be somewhat like this. Why? Effectiveness factor will be more for the 0 order reaction compared to first order for the same value of Thiele modulus probably because the sensitivity towards the concentration is relatively less. It is a concentration that matters because the concentration goes down inside a particle. Sensitivity is less and that is why the effectiveness factor is slightly higher. For the second order reaction it is sorry it should anyway start from 1. It is this way. So, this is for the second order. This is for the first order and this is for the 0 order. We have derived equations for phi. Remember, we have derived equations for phi for 0 order, first order, second order, any order or nth order, phi n. So, this phi is for 0 order, first order, second order. So, let me write phi 0 is equal to R into k. S a rho c d e c a c a s. C a is coming down here. Why? Because it is c a s to n minus 1 inside a root. So, n minus 1 a is 0. So, it is minus 1. C a is coming down. First order phi 1 is equal to R. This is what we have already derived k. S a rho c. Let me write k 0 because it is rate constant for 0 order reaction. This is k 1 d e and now n minus 1, 1 minus 1 0. So, there is no concentration dependency. There is no concentration dependency. Phi 2, you can write the expression for phi 2. What will be the expression for phi 2? Phi 2 is, if you make phi 2 here, then instead of k 1, I have k 2 here and you have to multiply it now by c a s because now it is n minus 1. That is 2 minus 1. C a will appear in the numerator. So, phi varies if the order of the reaction varies. We have already seen that. So, this is the relationship. For different orders, the relationship will also change. The equation that we have derived, that is 3 by phi 1 square into phi 1 cot h phi minus 1. What was that relationship for? It was for the first order reaction. Remember, do not get confused. That is not a general relationship. For second order relationship will be different. We do not have expression for that. But, I can always calculate by iterations by solving those equations simultaneously and get this plot. Get these plots. For first order anyway, I have the expression. This is how it looks. Now, so you have a general plot. You have a general plot showing the relationship between phi and e t. I am not again going to show the coordinates and all. Now, there are two distinct regions in here. There are two distinct regions here. So, there is one region here, where eta is almost constant. Now, this is the region where reaction is not affected by pore diffusion. This is kinetically controlled. Pore diffusion effects are insignificant. Eta is almost 1. Constant means 1. Now, there is another region where eta is changing drastically with phi. What is happening here? Because, value of phi is increasing. The pore resistance is increasing and eta is going down. That means, the effectiveness factor is going down. Resistance is much larger. Now, here the control is taking place. It is taken over by the diffusion and not a reaction. So, in this part, it is the diffusion control or diffusion limited. So, it is very important. See, if you are in this region, then there is no point in working on those parameters which increase the intrinsic reaction. Like for example, temperature. Temperature will have, will not have that impact on the rate. If you are here, it will have some impact. We are going to see that. But, it will not have much impact on the rate compared to what it would be when you are here. Other way round, suppose you are here, then do not work on reducing the effect of diffusion. Why? Because, diffusion is already it is friendly, it is cooperating. That means, there is no issue here as per diffusion is concerned. So, do not try to increase the pore volume. Do not try to increase the effect to diffusivity. Do not try to reduce the particle size. It is very important particle size. I will talk more about it a bit later. So, here in this region where reaction is controlling or when it is kinetically controlled, then try and increase the temperature. Increase the temperature, your rate will increase drastically. Arrhenius equation or law. So, this plot tenses a lot. Diffusion control, reaction control. What about this region? Both of them are important. Now, in this particular region, now your relationship gets a bit simplified. If it is a first order reaction, what we had looked at was eta is equal to 3 divided by phi 1 square phi 1 cot h phi minus phi 1 minus phi 2. This was the relationship that we derived. Now, what happens is that if phi is very large, eta becomes equal to 3 divided by phi. Eta becomes equal to 3 divided by phi. It becomes a very simple relation phi 1 of course. It becomes a very simple relationship. Now, when I substitute for phi, phi 1 rather, eta is equal to 3 divided by phi 1 is equal to 3 divided by r root of de k 1 s a rho c. I am substituting for phi 1 for first order reaction. So, look at this eta effectiveness factor. Keep this in mind. I am not saying you should buy at it, but then always like sometimes, you can interpret the things based on the equations. If you remember those equations, they help you sometimes to come up with a quick estimate or interpretation of what is happening. Fine. In this particular case, eta is larger for smaller radius. Eta is larger for smaller radius or for a smaller particle. Now, you have two particles. Who will offer more resistance? This particle will offer more resistance. Why? Because the molecules have to travel longer path to reach all the sides from this to this. Whereas, the sides inside the small particles are accessible better than what it would be here, because there is not much distance to be traveled. So, that is the significance of it. Like, if your particle is smaller, the value of eta is large. Poor resistance goes down. Overall poor resistance. De effective diffusivity, it is quite obvious. If the defective diffusivity is very large, value of eta is very large. Simple. And look at the denominator. It is a rate constant. So, the intrinsic reaction. Intrinsic reaction, that means when the molecule is seeing the catalytic side, intrinsic reaction, adsorption, surface reaction and desorption. Three steps. Remember. So, the combined equation that we get for this, when a molecule is near the catalytic surface. From that, I am getting this intrinsic rate. If you neglect desorption and adsorption, then it becomes surface reaction. And this is the k 1 is the reaction rate constant that we are talking about. So, this intrinsic rate constant is also important. So, if the reaction is intrinsically very, very slow, what will happen? Eta will become very large. Why? Because reaction is slow and diffusion becomes relatively fast. So, diffusion does not have much impact on the reaction. If intrinsic reaction, that is the value of k 1 s a rho c is very, very large. It is instantaneous. In that case, your problems. Problem in the sense, the poor resistance will play an important role. Because reaction is becoming very fast. So, all these integral factors are D e and intrinsic reaction rate constant together. All these three factors are very important as far as effectiveness of a particle, catalytic particle is concerned. So, it is better to keep this relationship in or equation in mind. Although this is applicable to first order reaction. For any n-th order reaction, you know, there will be some other terms related to surface concentration as well. The first order reaction, that is the only order or reaction in which effectiveness factor is not a function of concentration, surface concentration. Both effectiveness factor and Kille modulus. So, this is the significance of effectiveness factor. And for very large values of phi, it becomes a simple relationship. Eta is equal to 3 by 5. Now, what happens if you have n-th order reaction? A general n-th order reaction. If diffusion effects are significant, again you can derive equations. It will be slightly complicated. And then if you make simplification saying that value of phi is very, very large. What does it mean? Value of phi is very, very large means diffusion effects are significant. If you see that particular plot, I am towards a zone where intraparticle diffusion controls the overall rate. In that case, they have come up with the rate. They have come with the expression, a straightforward expression for effectiveness factor which is given by. Now, for first order reaction, we already know it is phi 1. But then, for n-th order reaction, it becomes 2 by n plus 1 raised to one-half, that is square root of 2 by n plus 1 divided by, sorry, into 3 by phi n. So, this is the definition of effect. This is Thiele modulus for n-th order reactions. Can I expand it further or rather elaborate it further? 1-half, 3 by r, root of D e k n s a rho c c a raise to or c a s raise to 1 minus n. If you take this out, 1 minus n by 2. If you take c a s out, it will be 1 minus n by 2. This is for phi very, very large. Very important. I mean, zone where particle diffusion is controlling, not in the intermediate zone, not in the zone where reaction kinetics is controlling the overall rate. It is the diffusion that is controlling. The value of phi is very large. Value of eta is relatively small. So, this equation is very important. I am going to come back to this. So, we have talked about the relationship between eta and phi. We have spent so much time understanding the significance of eta. We spent time in understanding the effect of parameters like effective diffusivity, radius of the particle and intrinsic rate constant, how they impact, how they make an impact on effectiveness factor and so on. Now, what is the objective? Objective is to design a reactor. Remember, I told you in the last lecture also, say for example, I have a CSTR. I want to design this CSTR, solid catalyzed reaction. It can be a fluidized bed reactor. It can be a spinning basket reactor. It can be a plug flow reactor, whatever. So, I have a performance equation. Say for example, I have a CSTR performance equation is r A w is equal to 0. Now, what is the r A? So, entire exercise here is meant for getting a right equation for r A or expression for r A. In a normal case, it is case A. If you come, you take into consideration the effect of adsorption, desorption, then I have Langmuir initial wood, Euler ideal, whatever. But then, if there is effectiveness factor coming in picture, there is a pore diffusion that is coming in picture. It becomes now very simple how to consider the effect of pore diffusion in the design equation. All I need to do here is f A 0 minus f A plus eta into r A intrinsic is equal to 0. What is r A intrinsic? It is when there is no pore diffusion resistance. Remember the denominator term in the expression for effectiveness factor. When the pores are opened up, all the sides are exposed to the external environment. So, that is nothing but the rate for calculated at the external surface. But because there is a pore diffusion resistance, the overall rate or the actual rate will become equal to eta into this. Remember eta is equal to actual rate divided by rate at external surface. Now, I am interested in the actual rate. As far as design is concerned, I am interested in the actual rate. So, this is actual rate. So, actual rate is eta into the rate calculated at the external concentration, external surface concentration, because here we have external concentrations. I hope it is clear. So, instead of r A i, now I can say r A s. Why? Because this is calculated at the external surface. And this r A s will be in terms of C A s. And this C A s is nothing but or f A is nothing but volumetric flow rate into C A s. So, I can write this as if it is a first order reaction, then K C A s, of course, there should be a negative sign here, because it is a reactant. If it is product to be positive, right. So, fine. Now, I can solve this equation. Sorry, it should be w here. I can solve this equation. If w is given, I can calculate C A s that is conversion. If conversion is given or C A s, that is outlet concentration is given, I can design the reactor to know how much catalyst is required, provided I know eta, provided I know eta. And I have told you how to calculate eta based on the properties of the catalyst, right. For the first order reaction, eta is equal to 3 by 5. If it is pore diffusion control, strongly pore diffusion control is equal to 3 by r, root of what is it, T e into K c. K 1 rho C s A, right. So, this will come here for the first order reaction, fine. So, this is how I solve a design problem, this is how I deal with the reactor. So, all the exercise last two lectures that we have spent or two or three lectures that we have spent on effectiveness factor. From reactor design point of view, it is only this factor that is important and comes from this expression. This was first order. Of course, for other orders, you will have different equation, alright. Effectiveness factor. Now, effectiveness factor will change your kinetics and it may give you an impression that reaction is behaving in a very different way. Suppose, you ignore inter particle diffusion and try and generate some data in laboratory, you will get some effects which are probably not expected. For example, say activation energy. Will it reduce or will it increase in the presence of pore diffusion resistance? If you have more pore, higher, larger pore diffusion resistance, the activation energy is going to go down. Why? Because the sensitivity to the change in temperature of the reaction rate will reduce if you have more pore resistance. Arrhenius law is applicable to intrinsic chemical reaction and not with the pore diffusion. Pore diffusion, that sensitivity to temperature will be relatively less. It is not, it does not come in the form of Arrhenius law then. So, there is a problem then. If you have pore diffusion, then you will see the activation energy which is reduced drastically and you would not expect that low activation energy. So, such things are possible. Order will change. In the sense, the apparent order or observed order in the experimental data will be much different or likely to be different rather than the actual order. So, we are going to see if there are pore diffusion effects, what are the observations possible when I do experiments in laboratory. Now, what kind of experiments I am going to do in laboratory? You have already seen the different laboratory reactors. A fixed bay reactor, a differential reactor. There can be a possibility of slurry reactor, having a slurry reactor, a batch reactor. Specially for catalytic pallets, there is something called as spinning basket reactor which is quite common in laboratory to generate a data for such reactions. What is spinning basket reactor? This is the reactor, the normal reactor. Let me draw it this way. You have stirrer in a normal reactor. Now, this stirrer will have basket attached to it. Now, why basket? There will be another basket like this. And when I stir this or when I give motion to this shaft, the entire basket rotates. So, this basket is attached to the stirrer. Now, what is the purpose of this basket? It holds a catalyst. It holds a catalyst. These are catalyst pallets. Now, what is happening here is instead of putting the catalyst pallets outside, I am putting them inside a basket. Why? Because it is likely that because of intense agitations, these particles or pallets may disintegrate or other get may get converted to fines, small particles. And you know the importance of radius here. You know the importance of particle size here. If it gets converted to fines, then the pore diffusion if it is important, the reaction rate will get will increase because of particle sizes come down. So, I do not I want to see the real effect. What exactly is the pore diffusion effect? I do not want particles to get converted to smaller particles because of some mechanical action. I put them inside a basket. So, they remain intact. At the same time, I need agitation to overcome the other resistances, especially the external mass transfer resistance. We are going to learn the effect of external mass transfer resistance later. But then that effect is also nullified or other it is reduced here or almost eliminated here by having very high speed of agitation. So, this is called a spinning basket reactor. Very common. That is why I spend some time on this spinning basket reactor. So, I am doing experiment in laboratory. So, I am using a spinning basket reactor. It is in a CSTR mode say that is reactant going in product coming out. And I do some experiments and determine the rate and from this rate, I back calculate the reaction kinetics. What does it mean? In the first part of reaction engineering course for normal homogeneous reactions, you have spent some time estimating the kinetic parameters like rate constants, order of the reaction from the laboratory data. So, from the laboratory data, how to get a rate constant and all? You already know about it. The same thing I am doing here. Now, instead of using a simple reactor there, I have a spinning basket reactor. I told you the purpose of that. And from CSTR, from inlet outlet concentration, I know what is the rate at which the reaction is taking place or you may use a differential reactor. And then I calculate rate at different concentrations. I calculate rates at different concentrations and if I plot them, what do I get? I get a relationship like this. C A S that is external concentration and a rate. You can take the modulus of that like if it is a reactant then it should be minus. So, the quantity would be plus. If I have these, I go on changing the concentrations and at different concentrations, I am calculating the rate or measuring the rate, not calculating. I am measuring the rate, observing certain rate. I am going to see something like this. Now, this is a log log plot. Sorry, I forgot to tell you. This is a log log plot. So, if I get a straight line, what is the slope? If I have rate is equal to k minus k C A raise to n, k n rather a. If I take log of both sides, of course negative, negative and make it positive and take log of both the sides, then the slope is nothing but an order of reaction. Now, if poured, if you judge the slope reaction is dominating and I generate this data, I am going to get some slope. Now, the question is whether this n that I am getting is the actual order or not. If the diffusion effect was not there, my earlier exercise reaction engineering part 1, n is equal to the order of reaction as simple as that. But now, I am working with pellets. I have a spinning basket reactor and it is possible that particle size is large or diffusivity is relatively small. Then that case, pore diffusion effects are going to play an important role. In that case, the question is whether the n that is a slope that I calculate here is going to be the order of actual order of the reaction or not. The answer is no. The answer is no. Why? Look at this. If you have actual order n, if you have actual order n, then because of pore diffusion effect, the order that you see is different from n. So, what is that? The rate observed is equal to eta into the intrinsic rate. So, this is equal to pore diffusion effects are significant. Then what is the value of eta? Eta is for any nth order. You already looked at this. C s, I have taken it out of the square root. That is why I divided by 2. Now, look at the order. This is eta and then you have K m. See, this particular expression is for eta. So, I have written this for eta. Then r a is equal to K n. You will have a negative sign here. K n into C s raise to n. Then you have some number here, which is function of D e. Then K save r, whatever n and then C s raise to n plus 1 by 2. I am just looking at the term that involves C s, because C s raise to n into C s raise to 1 minus n by 2 together gives you n plus 1 by 2. You can write here all this with K n come out of the square root. Coming in the numerator and so on. So, what is happening? Now, this particular term or this says that the order that I am going to see, when the pore diffusion effects are significant, when the pore diffusion effects are significant is n plus 1 by 2. Whereas, the actual order is n. So, this is called as falsified kinetics. That means, the actual kinetics is called as falsified kinetics is different, but in the presence of pore diffusion effects. If I do experiments in laboratory, try and interpret that data. I am likely to get an order, which is different from what it is actually as far as the intrinsic kinetics is concerned. So, the apparent order, apparent order becomes or n observed becomes n plus 1 by 2. So, you are going to get this. So, this is going to be a slope from the experimental data. And if you want to calculate the actual order, it is going to be 2 into n observed minus 1. And this is my actual order. I hope it is clear how the order will change, if the pore diffusion effects are significant. And I am doing experiments in laboratory. What will happen to activation energy? What will happen to activation energy? I will come back to this expression. I will come back to this expression. Look at this. I have this. Now, where is the activation energy coming in picture? Activation energy is because of the effect of temperature on K n. Now, again there will be some constant. But now, look at K n. Now, I will separate the terms of K n. K n will appear in the numerator like this, root K n. So, even if the reaction is first order, the dependency of rate on K n is not straight forward. See, if this was not there, sorry, so it was not there, you have rate is directly proportional to K n. Now, because of this factor, what is happening is, rate observed is proportional to root K n and not only K n or not K n rather. It is root K n. K n raise to 1 by 2. Now, let us look at activation energy. What is activation energy? Arrhenius law. Arrhenius law. K n is equal to K m is equal to A, that is, frequency factor exponential minus delta E by RT. How do I get activation energy? I have told you 1 by T versus ln K and the slope gives me delta E, right. This is how I get it. So, what I am going to do in the boundary, I will do experiments at different temperatures. I will get a value of rates. From rates, I will calculate the rate constants or initial rates will directly give you the rate constant because you know the initial concentrations for batch reactor. For continuous reactor, if I know the concentration from concentration, I can calculate the rate constant if the rate is known. So, ln K versus 1 by T, I calculate delta E. Now, if the pore diffusion effects are significant, then what is going to happen? See, the rate is proportional to root K n and not K n. It is proportional to A into exponential minus delta E by 2 RT. Now, as far as Arrhenius law is concerned, I have this term coming in picture. If I do the same exercise, I am going to get this value delta E by 2 whereas, the actual activation energy for the intrinsic reactions delta E, right. Because of this square root, the dependency has changed. In fact, dependency has become less sensitive to temperature or rather the rate has become less sensitive to temperature. The activation energy has come down. Simple, what is activation energy? Activation energy is something that shows how the reaction is sensitive to temperature. If the activation energy is large, the reaction is very highly sensitive to temperature. If the activation energy is low or small, the reaction is insensitive or very less sensitive to temperature. And that is what is seen is being observed here. In the presence of pore diffusion effects, the sensitivity has gone down. So, delta E observed is equal to, this is observed entire thing which is equal to delta E actual divided by 2. I am going to see this. I am going to get a slope equal to delta E by 2. May be noted here that we assume the effective diffusivity to be independent or insensitive to change in temperature. It can change with temperature, especially in the gas phase reactions. It is a bit sensitive to temperature. In that case, they will be combined effectively. That means, the temperature will make influence on not just the intrinsic rate constant, but on the diffusivity as well. So, this is again one more example of falsified kinetics. Falsified kinetics because of pore diffusion effects. So, kinetics is normally done for intrinsic reaction because all these, but all these effects like pore diffusion later on we are going to see external mass transfer. So, they are going to make an impact, their impact and they will change the values of apparent or observed, change the values of activation energies or the observed activation energies would be different from the actual activation energies. That is what we have seen. So, the order and activation energies they are likely to be different if or they are going to be different if pore diffusion effects are there. You have to be very careful. Falsified kinetics. Now, can effectiveness factor be greater than 1? Now, from whatever discussion we have before, effectiveness factor has to be less than 1 because the concentration goes down inside a particle diffusion resistance increases. So, the overall rate that we are going to observe is rate calculated at the bulk conditions. So, effectiveness factor which is observed rate divided by actual rate, actual means the one that is calculated at the external surface. If that is a ratio then it has to be less than 1, but in some cases this effectiveness factor is likely to be greater than 1 and that happens in the case of non isothermal reactions. If the heat effects are very large and heat dissipation is not that good. So, I am going to explain it qualitatively just in 2 minutes now and later on we will see which are the factors which are important. So, if you have a particle then the concentration reduces as you go inside, but since the reaction takes place inside there are thermal effects associated with it, heat is evolved and because heat is evolved the temperature inside will go up and if the conductivity of the particle is not good enough or not sufficient enough in that case the temperature will not reduce or heat dissipation rate will not be significant. So, the temperature inside a particle inside a particle T i and this is T. So, T i is greater than T, T i is going to be greater than T that is inside a particle the rate is likely to be higher because rate depends on what? It depends on concentration and temperature. So, concentration is going down, but a temperature is going up. So, some there is some combination of this concentration and temperature that it is possible that a rate inside is higher than rate calculated at the external surface. Rate inside is higher than rate outside that means effectiveness factor which is ratio of these 2 is likely to be greater than 1 and that happens for non isothermal cases if the heat of reaction is very large not just that, but the heat dissipation also to the outside surface is restricted because thermal conductivity of the particle is not that high. So, this is the non isothermal case and it is quite likely that you may have effectiveness factor being greater than 1. So, if you do some experiments and get a value of effectiveness factor observe value of effectiveness factor to be higher than 1 do not get perplexed do not get confused. See where your reaction is exothermic or not the reaction is exothermic this is quite likely to happen. Thank you. So, we will we will discuss this further and solve some examples if possible. Thank you.