 We will look at gas solid reactions today. Now, there are several examples of interest to us in which we want to react say a gas with a solid which gives us products. Now, we can look at several types of examples. A simple example would be let us say we have calcium carbonate and it undergoes decomposition. It gives us calcium oxide is carbon dioxide or a very common example would be an oxide reacting with carbon monoxide giving you iron plus CO2. We can balance this reaction. Now, whenever we are looking at a gas solid reaction of course, our prime interest is to see that the product that we get is of good quality. We want the product is of good quality or in other words we do not want unreacted material in the product. Let us take one more example zinc sulphide plus oxygen giving you zinc oxide plus sulphur dioxide. Now, these are all commercially extremely important reactions. Millions of tons of products are produced by these reactions. Each of them great value to us in daily life. If you look at reactions like zinc sulphide, zinc is available in this form called zinc blend. It reacts with oxygen to form zinc oxide and sulphur dioxide. Notice here that sulphur dioxide is also a very valuable product. On other words, if this zinc sulphide does not react completely then part of the sulphur is also lost because sulphur dioxide is used for making sulphuric acid. Therefore, ensuring that this reaction actually goes to completion is of great interest to us. Therefore, in whatever we do, we would like to understand how to manage the reaction so that we are able to drive the reaction to completion. Let us look at some simple examples to understand what might be the way to handle all this. Let us say you have a spherical solid. Now, let us say that this solid was of this size when we started and due to reaction it has become of a smaller size. In other words, the material which is unreacted, this is unreacted. So, when you started the whole thing was unreacted. Now, as the reaction proceeded part of it has gotten reacted. We want to understand how this whole thing solves this reaction proceeds. We will take once again, we will just recognize a gas plus b b solid giving you products. Let us for the moment assume that the film which is on the outside of this solid. There is a gas film which is in the outside of the solid on the solid and all the resistance to supply of the gas is in this external film. So, in other words you have I mean I am just draw this external film in a slightly big way. So, I mean it is I mean it is I am just drawing it slightly big. So, that you know we understand that you know it is this is the external film. I will call this as the external film external film. So, this gas concentration let us say gas is coming. So, this is the gas gas is here and in this distance the concentration of gas becomes 0. So, if you start at C A G is a concentration of gas and as it enters the film it concentration drops from C A G to 0. On other words the rate at which the supply of gas to the solid surface is controlled by the resistance in the external film. Now, we know from our basic understanding that as the velocity of gas over the solid changes this external film thickness changes. In fact, it decreases velocity increases to the extent we can control this rate of transfer by managing the velocity of gas over the solid. So, if this is the reaction then we can say A A based on our understanding of stoichiometry by minus of 1 sorry equal to R B S divided by minus B sorry minus minus B that is equal to this is these two are equal because of the fact that it is stoichiometry. And also we say that R A with a minus sign is K G times C A G. How do we understand this that the rate at which this is R it is not written properly this is rate R A. Then is rate of supply the rate at which chemical reaction takes place what are we saying that we have a spherical particle. Now, the rate at which the gas is supplied to the solid surface is controlled by external diffusion that is the case we are considering right now. Therefore, we have the rate at which chemical reaction occurs is minus of K G times C A G. K G is the mass transfer coefficient C A G is the gas concentration. We are assuming that the gas outside is in large quantity therefore, as the reaction proceeds the change in the concentration of gas external to I mean on the outside as the solid does not is not significant. And therefore, we can ignore it at least for the moment in this formulation. Now, we know from stoichiometry that R A times S by minus 1 is R B times S by minus B. So, this basically statement stoichiometry the word this term S refers refers to the surface area of solid which is in responsible for reaction. On other words what we trying to say here is that in gas solid reaction surface area is very important. Therefore, R A is the reaction rate per unit area if you multiply by S it tells you the total reaction rate. Now, if you want to find out what happens to the solid after all our interest is to find out what happens to the solid and how the solid changes. So, we write here our material balance D by D T of N B this is the rate which change must be equal to R B times S we understand this D by D T is R B times S. So, what is R B times S we have said this minus of K G times C A G times S please recognize here. So, R B times S is the rate of chemical reaction. Now, what is R B times S is related to R A times S you have to multiply by this stoichiometry that clear what we are saying. Therefore, the rated which the solid changes that means the rate number of moles of N B the rated which changes is equal to R B times S which is minus of B times K G times C A G multiplied by S which is the surface area which is appropriate to the controlling regime. What is the surface area through which the supply of gas is contact the solid whatever may be the state of the unreacted core. We find that the surface area through which the oxygen the material has to diffuse is the external surface area which is size of the radius if it is R the external surface area is 4 pi R square. Therefore, surface area S that is relevant for this controlling regime is C A G times 4 pi R square. Let us go through this formulation once again what we are saying we are having a spherical particle of radius R. This spherical particle radius R is in contact with the large quantity of gas and it undergoes chemical reaction. And this chemical reaction is controlled by rated which the gas is supplied to the solid surface and that rate of supply of gas is taken as K G times C A G with a minus sign showing that component A is getting consumed because of the reaction. Now, the surface area which is relevant to the supply is the external surface area because that external surface area is 4 pi R square. Therefore, we have on the basis of these that the rated which the solid B changes is R B times S which is minus of B times K G times C A G times S where this surface area refers to the external surface area. And what is N B by definition? N B by definition is the unreacted material whose quantity is this multiplied by density refers to total number of moles. So, what is our model? Our model is we look upon the solid which is undergoing chemical reaction as that of a sphere which undergoes chemical reaction. And as the reaction proceeds as the reaction proceeds the unreacted material which was initially here it keeps on progressing inwards or unreacted course keeps on shrinking. Therefore, it is also called as a shrinking core model. So, what is our model? We have a solid sphere spherical particle which is in contact with the gas of concentration C A G and this concentration does not change because the amount of gas is so large that the change in the gas concentrations are taken as unimportant in this formulation. We will relax all these things that we go along. Now as the reaction proceeds the unreacted surface which is initially of radius R this unreacted surface keeps on shrinking as the reaction proceeds. So, this is the reacted material is this. This is the reacted material this is the unreacted material. So, our interest is to find out how the radius of the unreacted core this is the radius the initially it was R. Now, the radius is let us say small R unreacted core how the radius of the unreacted core changes as the reaction proceeds this is what. So, this equation here now that we know we can substitute for N B we get D by D T of 4 by 3 pi this this is at any time it is small R. So, pi R cubed times rho B. So, this is left hand side equal to minus of K G C A G. Now, we know that the area this that is relevant to this supply of the reagents that is 4 pi R squared. Even if the core radius keeps on shrinking even if the unreacted core which keeps on shrinking the surface area through which is supply of reagents takes place that is 4 pi capital R squared that surface area does not change even though the reaction the core which is unreacted keeps on shrinking. So, what do we get we get that D R D T 3 cancels of. So, we get 4 pi 4 pi R D T rho B equal to minus of K G C A G C A G. So, this is G 4 pi R squared we just check we have got everything right. So, here we get 4 pi R D R C by D T. So, when you differentiate this it becomes R C squared there is a 3 here this is a 3 here which cancels of R C squared here have you got it right rho B 4 pi R C squared D R C by D T minus of B missing here B is B K G C A G 4 pi capital R squared. So, this tells you how the general approach here is to call this is R C to indicate that the core this is the unreacted core. This is the unreacted core and initially the radius of the unreacted core was capital R. So, the capital R is the radius of the particle which, but R C is the radius of the unreacted core which is shrinking because of the chemical reaction. So, we have set up the equations properly here N B is 4 pi R C cubed times rho B. So, I put all the numbers here. So, that the equation looks like this. So, let me simplify this a little R C squared D R C by D T equal to minus of K G C A G R squared. So, this that because you cancel the 4 pi. Now, we can integrate this. So, that the integrated form becomes R B divided by when I integrate this it becomes R C cube by 3 it goes from we started R and goes to R C equal to minus of B K G C A G B missing here R squared and time. Now, put the limits it becomes B times I take the minus sign on this side it becomes R cubed minus R C cubed divided by 3 equal to B K G C A G R squared. Now, I can divide throughout by R here. So, I put R squared here. So, this whole thing can be written as T equal to B K G C A G T equal to rho B R divided by 3 is B K G C A G within brackets of 1 minus of R C cube by R cubed I have got it right that is equal to time. So, it is simply you know taking R cubed outside it simplifies like this. So, let me write it once again just to put it in a nice form. So, time required for complete for reaction of this particle under what I call as external diffusion control is given by T equal to rho B R divided by 3 equal to R C cubed divided by 3 times B which is the stoichiometric coefficient K G is the mass transfer coefficient and C A G is the bulk concentration which is assumed to be unchanging. Now, what happens when R C becomes 0 which means the particle fully consumed or I can call this as the time required for complete consumption of the particle when the external film controls. So, what is tau f time required for complete consumption when external diffusion controls. So, that is equal to rho B R divided by 3 is B K G times C A G. So, that now what it becomes is that time required is simply 1 minus of R C cubed by R cubed or T divided by tau f is 1 minus of R C cubed by R cubed. So, this relationship tells us that if there is external diffusion which is controlling then the time required for complete consumption of particle or time required for consumption of a certain amount of particle is given by 1 minus of R C cubed by R cubed. Now, we also know from our basic understanding is that N B which is a total particle that is present is 4 by 3 pi R C cubed times rho B and we also know N B 0 equal to 4 by 3 pi R cubed times rho B. Therefore, what is this ratio N B divided by N B 0 by definition is 1 minus of X B which is the conversion of the solid that ratio is equal to R C cubed by R cubed. In other words we can also write this equation T by tau f which is 1 minus of R C cubed we can say it here therefore, X B equal to 1 minus of R C cubed by R cubed and therefore, we write T by tau f is also equal to X B. This is a result which comes out of our understanding of the system. What are we saying now what we saying is that if you have a spherical particle which is undergoing chemical reaction which is undergoing chemical reaction under external film diffusion control. If this solid is in contact with the large quantity of gas and this reaction takes place and as a result of this reaction the core keeps on shrinking unreacted core keeps on shrinking. We have our equation which says that the time required for complete consumption to the time required for any extent of reaction the time required for complete consumption is given by 1 minus of R C cubed by R cubed and when we recognize some of these simple relationships we will notice that the extent of reaction X B is equal to T by tau f that is how this relationship nicely comes. So, for the case of external diffusion control we have the extent of reaction X B is simply T divided by tau f. So, if you want to find out a case where there is external diffusion kind of control. Similarly, find the time required for complete consumption time required at any given instant of time that ratio is the extent to which the reaction is occurred. Our next example is we have a particle we have a shrinking particle sorry we have a particle which undergoes reaction. So, what does it mean the particle is undergoing reaction means we have this is the reacted core this is reacted core this is reacted. So, this is unreacted in this case we are looking at the case where this reaction that is taking place is controlled by chemical reaction chemical reaction control. What is meant by chemical reaction control is that let us say at any instant of time at any instant of time the this is the what we call as the reacted core let us say at any instant of time let us say this is the reacted core. So, what we are saying is that the concentration of the gas remains at C A G and as it reaches a solid surface it drops to 0. So, in the same way here concentration remains like this and it drops to 0. So, what we are trying to say here is that there is no drop in concentration of gas as it moves through the external film there is an external film here you can see external film and as it moves through the react this is reacted layer these are all reacted layers you see. So, the assumption here is that this reaction control reaction control means what it means that all the resistance lies in the at the reaction surface. Therefore, there is no drop in concentration as it moves through the external film as it moves through the reacted components and as it comes to the surface it drops to 0. When concentration at the surface drops to 0 means it is an irreversible reaction. Now, if the concentration does not drop to 0 it means that it is a reversible reaction. So, that the surface concentrations will be appropriately described by the equilibrium relationship as of now we are taking the surface concentration to going 0. Therefore, it is an irreversible reaction. So, for reaction control we have d n b by d t d n b by d t as usual equal to r b times s. We also know r a s with a minus 1 equal to r b s with a minus b this also we know because comes to the stoichiometry. So, what is s? What is s which is appropriate to reaction control? We recognize that as the reaction proceeds the surface area that is appropriate to the reaction is the unreacted core surface. So, as the core keeps on shrinking the unreacted core of radius r c it keeps on shrinking. Therefore, the value of s refers to 4 pi r c square. So, here s refers to 4 pi r c square. So, that now we can write what is our rate functions? We have r a r a which is the rate of chemical reaction which is equal to minus of k s times c a g r b equal to minus of b times k times c a g. And we also know r a times minus of 1 equal to r b times minus of b therefore, we get this relationship. So, what is d n b by d t? So, d n b by d t equal to minus what is this relationship minus of b times k s times c a g multiplied by the area which is appropriate to the process. We said the area which is appropriate to reaction control is the area which is the unreacted surface. And at any instant of time the radius of the unreacted core is r c. So, that particular surface area is 4 pi r c square. So, this equation represents the variation of what is happening to component b as a reaction proceeds. So, we get now here the left hand side is 4 by 3 pi r c cubed times rho b. So, this is the left hand side d by d t equal to what is the right hand side b times k s times c a g times 4 pi r c square. This refers to the surface area which is relevant to the controlling surface area. This surface area was 4 pi capital R squared when we talked about external film diffusion control. So, the surface area for reaction control is the unreacted core surface where the reaction is taking place which is 4 pi r c square. Let me simplify this. So, 4 pi. So, it is r c squared into 3 divided by 3 rho b d r c by d t minus of k s c a g 4 pi r c squared. 3 cancels of 4 pi cancels of 4 cancels of. So, you get d r c by d t equal to minus of b k s c a g by rho b. Integrating it is r c minus of r equal to minus of b k s c a g by rho b. So, this r we can simplify this further and write this as r minus of r times rho b equal to minus of b k s c a g t. This t should come here because integrating. So, taking r common 1 minus of r by r rho b equal to b k s c a g t. So, you have rho b r divided by b k s c a g into 1 minus of r by r equal to time. So, what we get here is for the case of reaction control, for the case of reaction control, the time required for consumption to any extent of reaction is given by this. Therefore, when r equal to 0, you have to put r c here, this is r c here to indicate there is a shrinking core. So, r c c equal to 0 becomes t becomes tau reaction control, which is rho b r divided by b k s c a g. So, that what we get here is for reaction control, the expression for the effect of time t by tau f tau s equal to rho b r divided by rho b times b k s c a g times 1 minus of r by r. So, this is reaction control. What we say, if you have reaction control, the time required for any extent of reaction to time required for complete consumption is given by the right hand side. At r equal to 0, we know it is fully consumed. Therefore, the time therefore, t becomes equal to tau s that is equal to rho b rho b r rho b r divided by b k s c a g. So, what we are saying once again, let us just repeat what we have said, we have said that this is the relationship that explains reaction control. So, when r equal to 0, which means the particle is completely consumed, which means the r equal to 0 means then this rho b r by b s k h becomes the time required for complete consumption. Therefore, this whole term can be written as tau s t divided by tau s is equal to 1 minus of r by r. This is what I have written. Notice here that as the reaction proceeds, this point the surface which is in contact with the gas, concentration at that point is taken as 0. If equilibrium constants are finite, then that assumption is not correct and then we have already developed procedures to take care of all that. So, what have we done? We have taken two cases so far. First case is when there was external film diffusion control. The second case is the case of reaction control for which we have both derived what is t by tau a tau f or t by tau f, which is in terms of r c, which is unreacted core and then r is the size of the particle. Now, let us look at the third situation, which is what is called as third situation of interest to us is what is called as ash diffusion control. What are we trying to say here is the following? We have a particle, we have what we call as the unreacted core and outside the unreacted core we think and there is a laminar film, which is the external film. Now, what we expect is the following? As the reaction proceeds, as the reaction proceeds, this is the unreacted layer and this is the this is the reacted layer. Now, the distance between the unreacted and the reactor, this thickness keeps on growing as the reaction proceeds. That means, as the reaction proceeds, the thickness of the reacted layer keeps on increasing. So, and we also say that the reaction takes place, reaction takes place at this unreacted surface unreacted surface, which means what? As the reaction proceeds, that the diffusion path around which this gas must diffuse before reaching the solid surface keeps on increasing. This is an instant, we should keep on understanding. Now, for this case, let us see how to solve so that we can understand, what is the rate of chemical reaction? How do we do this? We do this by recognizing the following. We know d n b by d t equal to r b s, we know this. You also know minus of r a times s equal to minus of r b times s equal to d times del c del r into 4 pi r square. What do we mean by this relationship? What we mean by this relationship is that, in the case of ash diffusion control, the rate at which the chemical reaction takes place would be the rate at which the material arrives at this solid surface, because that is the meaning of reaction control. Control means what? The entire process is governed by this rate of supply. So, in other words, here is an instance, where the rate of supply through the reacted layer is what will determine the rate of chemical reaction. So, we must account for it appropriately, so that we can appropriately put these numbers in our equations. Let us see how we do this. What we said so far is that, the unreacted layer, the unreacted layer is here, the reacted layer is here. So, this driving force through this column of this reacted material is what determines the rate of chemical reaction. So, we must take that into account, that resistance into account to be able to handle this problem. Now, what is the general situation? The general situation is that, this is a case in which we have continuous diffusion of particles, diffusion of gas at the same time shrinking of these particles. We have an instance, where the diffusion of gas through the product layer. Simultaneously, at the solid surface, the reaction is taking place. So, we have to take all these things into account, which we will try to do by writing the material balance for what is going on. Let me put the material balance like this input, minus of output plus generation equal to accumulation. So, we are writing the material balance for this gas in this. So, what are the inputs? Our inputs are 4 pi r squared d del c del r at r plus delta r minus of 4 pi r squared d del c del r at r. So, this is the input. In this, between r plus d r and r, there is no chemical reaction. Therefore, there is no reaction is 0. That must be equal to del by del t of 4 pi r squared d r epsilon times del c del t times c. You have to multiply by c. What are we saying here? What we are saying here is that the unreacted layer, this is the reactor layer. The red one is the unreacted layer. The rate of supply of gases to this surface depends upon resistance here. That is what we have taken into account. That difference is the accumulation. So, simplifying, what we get is 1 by r squared del by del r, r squared d del c del r multiplied by porosity. So, this gives you the accumulation of gas in this system. So, simplifying, we get 1 by r squared del by del r, r squared d del c del r equal to epsilon del c del r, del c del t, this epsilon. Make sure all the terms are correct. 4 pi r squared cancels off, 4 pi r squared cancels off. So, this becomes d del c del r 1 by r squared d del c del r with a minus sign. So, no minus sign r plus d r to r, no minus sign epsilon del c del t. So, this is the point of writing this is the following. What we are saying is the following. What we are saying is that in the case of ash diffusion control, the resistance to all the resistance lies in this layer of in this red layer, all the resistance lies here. As the reaction proceeds, this thickness will keep on changing. We have said and correctly that the rate of chemical reaction is minus of r a s or which is r b s with a minus b with a miss already taken. So, r a s and r b s are related. Therefore, minus of r a s into b equal to minus of r b s. This is something we know comes stoichiometry. So, what we are saying now is that the rated which chemical reaction occurs at this solid surface can be given can be obtained from this equation which tells you the variation of c with respect to r. So, what we have done is that we have looked at an unsteady state problem and made certain simplifications. Let us see what the simplifications are. The simplification we would like to do is to do non dimensionalization r by r theta equal to t by tau d. And tau d you might ask how I came to this kind of formulation. Let me explain how it comes? It becomes very easy d c 0 by r by 2 d c 0 r by 2 into 4 pi r square. Now, how do we understand this term tau d equal to 4 by 3 pi r cube rho? What is 4 by 3 pi r cube rho? This is a total mass of material that is been put into the system. What is the denominator? Denominator d c 0 this term t c 0 is the maximum diffusive flux because c 0 is the concentration at the inlet. Therefore, t c 0 is the maximum diffusion flux divided by what is the path length? The length of the path as you can see here or path length is at best r by 2. This is the r by 2. So, this is how path length of r by 2 is chosen. So, that we get this tau d is the total amount of material divided by the rate of diffusion which is d c 0 r by 2. So, we are able to determine a priori what is the diffusion time. Now, let us look at a simplification. What is the simplification that we want to do is we want to solve this. So, our equation is 1 by r square del by del r by r square d del c del r equal to epsilon del c del t. To solve this, we have what is called as the non-dimensional representation. It makes it a little easier to understand what is going on and tau d equal to 4 by 3 pi r cubed rho b divided by b times d del c del r by 6 d c 0 by r by 2 4 pi r square. Please let us understand what is meant by these terms. Why equal to r by r? This is you know it is a well known thing and theta is tau this is reaction time versus the complete consumption time for diffusion control. So, with this we find that our tau d becomes this and when you simplify what is it become rho b r square 6 b d c 0. Let us see whether all the numbers are correct 4 by 3 pi r cubed by r cubed. Now, let us look at the solution. Now, I want to non-dimensionalize this differential equation in terms of these variables. So, you get 1 by r squared y squared del by del y del by del y which is r squared y squared d del by del y is del by del y times r correct. That is equal to the right hand side which is epsilon by tau d tau tau d del psi by del theta. Please understand this what are we saying now? What we are saying is that we have this differential equation 1 by r squared del by del r r squared del psi by del r. The right hand side is epsilon is porosity this is porosity and this is time and so on. So, we have non-dimensionalize with respect to y equal to r by r and theta equal to t by tau d. So, this whole equation must now be written in terms of the our del y squared therefore, it will be r squared del by del y. Have you got everything correctly c r squared del c del r del c del r r is r squared will not come. So, let me see I have got all the terms right 1 by r squared is correct del by del y is correct del by del r. So, 1 more r will come. So, 1 more r will come here because of this r correct. So, r squared y squared. So, d del c del r. So, it is r del by del y on the other side epsilon tau d del c by del theta this is tau d. Let me simplify this further what do we get going from this to the next one you can see here r squared there is r cube. So, there is r squared here. So, essentially the denominator there is 1 r here there is 1 r here. Therefore, it simplifies as 1 by r squared y squared del by del y of y squared d del psi del y equal to epsilon by tau d del psi del theta. Now our tau d please recognize I will just put it down here tau d we have defined tau d in such a way that it looks like this rho b r squared by 6 b d c 0. Now you can replace tau d here. So, that I am just writing the right hand side only epsilon and then tau d is what 6 b 6 b d c 0 by rho b r squared the right hand side is del psi del theta. Now there are lot of simplifications you can do from. So, that now this equation looks like this r squared cancels of d cancels of and so on. So, you are left with 1 by y squared del by del y y squared del psi del y equal to 6 epsilon c naught by rho b del psi del theta this is the equation that governs what happens inside the particle. See why are we doing all this we are doing all this just to understand an important feature of our problem what is the important feature of our problem. The important feature of our problem under ash diffusion control is that as the reaction proceeds the thickness of the ash layer or product layer keeps on increasing. And our reaction the drop in concentration the drop in concentration comes from it is c a g here and it drops to 0 here. So, it from here now when this when this distance is more that means the second case in this case the second case the drop this distance has increased. That means the drop occurs over a larger distance on other words our mathematical formulation must take into account this increase in the diffusion path has the reaction proceeds. What are we done for this what we have done for this is to recognize that our equation that describes the motion or the variations are given by this equation. Where we can non dimensionalize under certain conditions where we assume tau d to be this which is the total amount of material divided by the average diffusion through this. So, that we define the average time of what is called time of diffusion time. So, with all these simplifications we have reduced this equation to look like this 1 by y square del by del y equal to equal to right hand side 6 epsilon c naught by rho b del side del theta. Please notice that the order of magnitude of y is maximum value y can take is 1 the order of magnitude of psi is 1 order of magnitude of theta is also 1. Because it is non dimensionalized with respect to tau d. So, on other words here is a non dimensional representation of the diffusion equation through the product layer where the multiplicative factor is 6 epsilon c naught by rho b. What is this 6 epsilon c naught by rho b you know from our experience we know the following what do we know we know that c naught. So, let me write this equation once again 1 by y square del by del y y square del psi del y equal to 6 epsilon c naught by rho b del psi del theta. Now, c naught typically is about 0.05 gram mole per liter we know this rho b for solids is typically 10 gram moles per liter. So, what are we saying this and epsilon is about 0.2 to 0.4 shall be same. So, this whole term 6 epsilon c naught by rho b we can calculate 6 multiplied by 0.3 let us say 0.18 this is 0.05 0.18 multiplied by 0.05 divided by 10. So, you can see here this number is a small number on other words this term 6 epsilon c naught by rho b this is approximately I will just put the numbers here you please calculate say 0.3 is 0.05 divided by 10 equal to. So, you can see 0.3 0.3 0.09 something like 9 10 minus 3. So, this is the value it takes is it correct 6 into 0.6 into 0.5 0.3 0.3 0.09 0.009. So, about 10 minus 3. So, this is a small number what you want to say here is that this quantity is a small this is small. So, what is it mean it means that under the conditions of the formulation of the problem it is possible for us to delete the right hand side saying that the right hand side is not very important it goes to 0 or this is what is called also called as quasi steady state approximation. So, that means accumulation accumulation of material inside the particle as the reaction proceeds is small therefore, we can as well assume that the left hand side is approximately 0. So, our problem therefore, what we are trying to say here is that this 1 by y square del by del y y square del psi by del y equal to the right hand side 6 epsilon C naught by rho b del psi by del theta. This is not very important and this is what is called as quasi steady state approximation. Therefore, we might as well solve this problem by assuming that 1 by y square del by del y square del psi del equal to 0. This would be a satisfactory way of understanding diffusion of gas through a solid particle under the conditions of quasi steady state approximation. On other words to understand the rate at which the reaction would take place over the solid during the instance of as diffusion control. We might as well solve a simpler problem 1 by which is essentially the quasi steady state approximation problem which will give us answers which are quite satisfactory. We will take this up when we meet next. Thank you.