 We continue with gas solid reactions. This quickly recognize what was said little earlier in gas solid reaction. Let us say we have coal reacting with oxygen giving you CO 2 or say wood reacting with oxygen giving you carbon dioxide and energy. Our interest is to see that the energy is available in coal or in wood is fully utilized. Therefore, we do not want to make we want to make a good use of the material that is with us. So, complete consumption is of concern to us and we would like to understand conditions under which we have the complete consumption possible. So, we talked about film diffusion control under which we said if there is a particle and then this particle the resistance to supply of gas is in the external film. Therefore, the rate at which chemical reaction occurs is essentially determined by this concentration all the concentration drop occurs in the external film. So, the rate at which chemical reaction occurs is determined by the resistance in the external film. We have considered this case and we have found out how the time for complete time for chemical reaction is governed by T the relationship we have got is T equal to T by tau f equal to 1 minus r c cubed by r cubed where r is the particle size and r c we mentioned last time is r c is the unreacted core size radius of the unreacted core r c. And since x b which is the extent of the action 4 by 3 pi r c cube rho b divided by 4 by 3 pi r cube rho b 1 minus this is. So, that is why this becomes simply x b this case we already discussed. And we understand therefore, that for the case of film diffusion control the time for complete consumption divided by the time for the time for chemical reaction time for complete consumption is given by this. Where tau f we showed is simply rho b r divided by b k g time c a g where c a g is the concentration of gas which is outside the pellet and does not change as the reaction proceeds is showing that this solid is in contact with the large quantity of gas. Of course, this is an approximate situation we have considered we will relax all this as we go along. The second ways we considered is the case of reaction control. Here we have solid and then we have the solid in contact with large quantity of gas. And we said this unreacted core this is the unreacted core of radius r c the radius of unreacted core is r c radius of the particle is r. Then we said that the time for complete consumption divided by the time for complete consumption time for reaction time for complete consumption we showed is equal to t by tau equal to 1 minus of r c by r. We showed this yesterday where tau r is given as rho b r divided by k s c a g is to one moment I yesterday I missed this term 3 here this 3 I missed b k c a g. So, this case here you have the gas coming in at some concentration and this is the unreacted layer. So, the real drop from the sense that drops here. So, that all the reaction takes place at the unreacted surface this is c a g versus this is distance. These two cases we considered yesterday third case we were talking and then we could not complete and we were that is the case of ash layer diffusion control. Now here what we have is we have a solid and this is the solid unreacted solid. Now what is happening is that as the solid is reacting this is the this is the unreacted layer this is the reacted layer this is the reacted layer. Now as the reaction proceeds the extent of the reacted layer keeps changing and ash layer diffusion control means that you know the rate at which chemical reaction occurs depends upon the resistance due to the ash layer. Therefore, this resistance will keep changing as the reaction occurs which we must take into account this is what we would like to do. So, let us recall that the rate at which chemical reaction occurs d n b by d t is r b s something that we have been saying for a long time. And we also recognize that r a s which is this our reaction please recognize our reaction is a gas plus b b solid giving you products this is the reaction we have taken. So, the rate at which reaction occurs is a rate at which the solid is supplied the rate at which the material is supplied to the unreacted surface multiplied by 4 pi r square this is at r equal to r c. Now the point that we would like to understand here is that in the case of ash layer diffusion control the rate at which chemical reaction occurs is d del c del r 4 pi r square. That means if this is the unreacted surface the rate at which material is supplied to the surface is the rate at which chemical reaction is occurring. And we also recognize that this this interface keeps changing as time proceeds which we must take into account as the reaction proceeds. So, similarly therefore, if you want to say what is r b s then we can say it is b times minus of r a s this comes from stoichiometry. Therefore, this is equal to b times d del c del r 4 pi r square at r equal to r c. Recognize that we have left out this minus sign from the fix law of diffusion. The reason why we have left out this minus sign is because in this case the direction of diffusion is opposite to the direction of positive r that is why we have left out the negative sign. Now that we know that the rate at which chemical reaction occurs essentially is determined by finding out what is c how c changes with r essentially we need to find a way by which c changes with r. So, this what we would like to do now by looking at the differential equation that governs the variation of concentration inside the particle. Let us see how to do this. Let us quickly recognize the situation once again our situation is this. Now, we have the concentration this is the unreacted this is r c this is unreacted unreacted. Now, what we saying is that this concentration remains I will write here concentration and then between this point and this point. So, I will say here this concentration drops this is c h e this is distance. Now, how do we take into account this variation of concentration drop here as the reaction proceeds. So, that the thickness of the reacted layer keeps changing to account for that I will write the diffusion equation input that is input minus of output plus generation equal to accumulation this is the diffusion. So, what is the input 4 pi r squared d del c del r this is r plus delta r 4 pi r squared d del c del r at r. There is no generation and there is accumulation which is del c del t multiplied by the 4 pi r squared d r times epsilon showing that there is porosity that is where the fluids will accumulate. 4 pi r squared del c del r is input at r plus d r and then 4 pi r squared del c del r at r. So, this is what is the input and output and generation accumulation. So, this is the material balance for a differential element. We can simplify this we can simplify this and take the limit as you take the limit as r tends to 0 this becomes our diffusion equation simplifies as r squared del by del r r squared d del c del r equal to epsilon del c del r where r equal to r c equal to c naught we know this r equal to r c we say that c equal to 0 because of reaction. So, this is the problem that we have to solve that to find out just once again let us just recall the problem that we want to solve. We said just now if you want to understand what is the rate at which the particle is reacting we must know what is the rate of chemical reaction. We said the rate of chemical reaction is given by v times d del c del r multiplied by 4 pi r squared at r equal to r c. On other words the rate of chemical reaction depends upon del c del r and we need del c del r through a proper understanding of the process. To understand this what we have done is that we have set up the diffusion equation in an elemental volume of the spherical pellet and then we have simplified this by taking the limit as delta r tends to 0. And then found that this is the diffusion equation that governs the process and the conditions are that this particle at the surface of the particle concentration is c 0 or we can call it c a g whichever is a appropriate for us. And r equal to r c there is c 0 which means the material has been consumed that is the point that means all the drop in concentration occurs in the ash layer. So, how do you solve this now this c equal to when you say c equal to c 0 I mean concentration 0 which means that is an irreversible reaction. Now if it is irreversible reaction then there will be an equilibrium condition and we will look at this as we go along. We can define some non dimensional very to make it a little easier t by tau d where tau d is the diffusion time the time characteristic time for diffusion we can understand this as 4 by 3 pi r cubed rho b. So, this is the total amount of material that is with us and b times d c 0 by r by 2. So, this is this is the maximum rate of diffusion divided by the average diffusion path multiplied by 4 pi r squared this is the maximum surface area over which the diffusion takes. So, this is some way of characteristic diffusion time. So, we can simplify this and write this is rho b times r squared divided by 6 b d times c 0 sometimes we call this as c i g this is we mean the same thing. Now you can ask how is it that I made a definition of tau d in this form. Now actually since I know the final result I found it convenient to write it in the form, but if it does not matter even if you do not write it in this form the answers may not change. So, what we are saying is that if you look at the time coordinate in terms of theta which is non dimensionalized with respect to the diffusion time which is defined as total amount of material divided by the maximum rate of diffusion. This is the maximum rate of diffusion this is the diffusion path. So, this gives the diffusion the moles of material that is coming in. So, this gives you total amount of material of moles of material per unit time gives you a idea of the diffusion characteristic time. And if you non dimensionalized with respect to diffusional characteristic time we get some field for the time constant of the process. So, with this non dimensionalization our equation diffusion equation can be simplified. So, let us do that now. So, let us simplify the diffusion equation. So, please recognize that I am going to replace this r in terms of y replace time in terms of sorry this is time this is time in terms of theta. So, when I do that we get 1 by r square 1 by y square del by del y within brackets of r square y square called y square d by r del psi by del y. Now, I also put one more dimension psi equal to psi equal to c divided by c 0 equal to on the right hand side we have epsilon by tau d del psi divided by del theta. So, please I mean this is fairly straight forward there is nothing new being said here. So, this r square I have written as r square y square del this r I have written as r y r times y this r square I have written as r square y square this r I have written as r y the d remains as such. So, this c I put as c by c 0 of non dimensionalized on both sides. Therefore, I have put it as del psi del psi on both sides. So, nothing nothing new has been done. So, you simplifies this simplifies as 1 by y square del by del y of y square del psi by del y I put a d here equal to epsilon by tau d del psi by del theta. Now, I replace this tau d in terms of what we know. So, let me just write this is del y y square d del psi del y equal to epsilon I just replace this tau d we already done that we already done that tau d tau d is where is tau d which is rho b r square by 6 b d c 0 I just put this put this here rho b r square then denominator 6 b d c 0 del psi by del theta. I am just replacing this tau d here using whatever definitions of tau d we have got before. Now, it simplifies some terms go away. So, that this diffusion equation which describes the diffusion of the reactant gas through the ash layer. So, it simplifies like this let me let me just write this once again just for your sake. So, equation our equation looks like this 1 by y square del by del y of y square d cancels off. So, del psi del y equal to 6 epsilon 6 epsilon c 0 by rho b r square this r square is also got cancelled as where is where is this r square please let us just look once again this r square stays there that is why this r square stays there. So, this what cancels off sorry about that. So, it is rho b del psi by del theta. So, this is the diffusion equation that describes the diffusion of our reactants. Now, we notice here that c naught which is typically of the order of 0.05 moles per liter and then rho b which is the bulk density of solids typically 10 g mole per liter this is g mole. So, that and epsilon let us say about 0.3 typically. So, you put all these numbers here let me just calculate for your sake 6 epsilon is 0.3 c naught is 0.05 divided by rho b is 10. So, this becomes a 0.05 0.15 0.0 how much is it 0.015 0.0 about 0.009. So, this term this whole term is a small quantity. Notice here on this left hand side y has the order of magnitude is 1 order of magnitude order is 1 and then psi also order is order is 1. And then you have on the right hand side theta also order of magnitude is 1. So, everywhere the order of magnitude is 1. So, this is a small quantity in view of that we can delete this term. So, this approximation is called as quasi steady state approximation what it means is that the accumulation of material inside the pores of the solid is not very important from the point of view of understanding the diffusion and how the material gets consumed. And therefore, we might as well look at the left hand side and solve the problem. So, under the assumption that quasi steady state approximation is satisfactory. That means we can as well look at 1 by y square del by del y of y square del psi by del y equal to 0. We might as well solve this equation under the conditions that phi equal to 1 at y equal to 1 and psi equal to 0 at y equal to y c. What are we saying? What we are saying is that to understand the ash layer diffusion problem under the quasi steady state approximation wherein the accumulation of material inside the pores of the solids are not very important from the point of view of understanding the material balance. Therefore, the diffusion equation can be I mean the solution to the diffusion equation can be approximately gotten by looking at the left hand side alone where del 1 by y square del by del y square del psi del y equal to 0. This is the quasi steady state approximation. Now, this is a fairly simple differential equation and the conditions are also phi equal to 1 psi equal to 1 at y equal to 1 psi equal to 0 y equal to y c. What does this mean? It means that all the reaction occurs in the ash layer and there is no material left as the unreacted layer is reached. This is the meaning of this condition. So, let us solve this. It is a fairly simple problem to solve. So, we have when we integrate this I am just integrating this right now. So, maybe I write it once again just make it easier 1 by y square del by del y of y square del psi by del y equal to 0. Let me integrate once when I do that I get y square del psi by del y equal to a constant. We integrate once again del psi del y equal to a by y square. I integrate this when I integrate this I get psi equal to minus of a by y plus b. I put this our conditions are psi equal to 1 at y equal to 1 psi equal to 0 at y equal to y c. So, we can put these two conditions so that we get here psi equal to 1 equal to minus of a by plus b and then that is at y equal to 1 and then we have 0 equal to minus of a by y c plus b that is at y equal to y c. So, you can find out what is a and b which and substitute. So, it is very simply I just write the solution here there is no need to go through this once again 1 by y c minus of 1 a is this and b equal to a by y c. Therefore, we can substitute for a and b and the solution to this differential equation we can get. So, I will write the final form. So, what do we get let me just run through this once again without losing the. So, under the quasi steady state approximation the equation that we have to solve is the diffusion equation. Now, this is the boundary conditions psi equal to 1 at y equal to 1 psi equal to 0 at y equal to y c. This condition implies that all the reaction takes place in the ash layer and as the unreacted layer is reached the reaction stops because there is the diffusing material has been completely consumed. So, that now our solution we just write the solution psi equal to 1 by 1 minus of y c 1 minus of y c by y this is the solution. We can write it in a slightly more comfortable form psi equal to 1 by 1 minus of y c within brackets of 1 minus of y c by y. Now, having said this let us just quickly recall what we have said little earlier what we said little earlier is that the rate at which the reaction occurs. We said this earlier that the rate of chemical reaction we started actually when we started we said we want to understand what is the diffusion that is that is taking place in the unreacted layer. And we said that the rate of chemical reaction which is r b s is given by b times d del c del r 4 pi r squared r equal to r c. So, what we said at that point is that once you find out what is del c del r then we know what is the rate at which chemical reaction occurs at a given r equal to r c. What we have done is that we have got by doing a quasi steady state approximation we found out what is psi in terms of y c we can put it in terms of r c equal to r and all that. So, let us do that right now. So, we want to now find out what is b times d del c del r 4 pi r this is what we are interested in because that is what tells us the rate of chemical reaction let us try to do that now. So, what is d del c del r 4 pi r squared at r equal to r c we want this what is this. Now I want I am writing this in terms of the non-dimensional coordinates please see whether I have got it right 4 pi y squared r squared c naught del psi del y divided by c naught by r. Let us see we have got all the numbers right d is here now del c del r I have written del psi del r I put a r here. So, they have taken care of that 4 pi r squared after 4 pi y squared r squared at r equal to r c. So, that is because y equal to y c all right now let us take this further this 4 pi 4 pi where is this d I have forgotten the d c this is here 4 pi d c naught y squared this r square. So, it is r del psi del y at y c is it all right what I am saying is our interest we know that this is the term we are interested in d del c del r 4 pi r square which we must calculate. We know what is psi therefore, I put everything in terms of psi by converting r to y. So, I have done all that 4 pi r squared put all those terms now we have got everything y now this whole thing at we want to calculate y equal to y c. So, what is del y del psi del y we can calculate del psi del y from here and then evaluate this at y equal to y c it is fairly straight forward now you only have differentiate this and put y equal to y c. So, let me do that it is not doing anything very complicated. So, please let me just restate what I have said already that we have got psi in terms of y c this is r b s we already said that this is r b s we already said this is r b s. So, we have put r b s in terms of the numbers that we know then I is only have differentiate psi with respect to y and evaluate this whole thing at y equal to y c then we will get rate of chemical reaction inside the ash layer that is what I am trying to do now please help me now. So, I will differentiate and substitute and then I will. So, we get d we will not go through all these details this fairly straight forward 4 pi r squared. So, this is what we want at r equal to r c now we have said that is also equal to minus of r b s. So, I am differentiating and then putting all this together. So, we get equal to minus of r b s equal to b 4 pi d c naught divided by 1 by r c minus of 1 by r. What I have done I have differentiated psi with respect to with respect to y put y equal to y c and then the replaced y c we know that y c what is y c y c equal to r c by r. So, I have replaced y c in these equations in terms of replaced y c as r c by r and simplified and I have got now the rate of chemical reaction under the case of ash diffusion control is given by. So, r b s r b s this is the surface area equal to b the minus b 4 pi d c naught divided by 1 by r c minus of 1 by r is this clear. Now, if this is clear the rest is value straight forward we have r b s let me write r b s equal to b times 4 pi d c naught divided by 1 by r c minus of 1 by r. Now, we know that d by d t of n b by definition is r b s what is meant by d by d t of n b rate at which the moles of n b changes with time what is r b s r b s refers to the rate at which chemical reaction this is r b is rate per unit surface area multiplied by surface area which is relevant to this reaction to this kind of controlling regime. Now, left hand side is what d by d t what is n b 4 by 3 pi r c cube rho b we know this what is r b from here minus of b 4 pi d c naught divided by 1 by r c minus of 1 by r is this clear. So, what we have done we have looked at the case of a spherical particle undergoing chemical reaction under ash diffusion control. We also recognize that the thickness of the ash layer will keep changing with time therefore, we wrote the diffusion equation for the ash layer. We recognize that as the reaction moves towards the unreacted surface the reaction would stop because the reagents diffusing reagent gets consumed therefore, we looked at the diffusion problem the unsteady state problem we said under the unsteady state if you look carefully under the non dimensionalization we understood that the right hand side which is the accumulation term becomes unimportant if it is a gas solid reaction therefore, we deleted that term and said that this is called quasi steady state approximation. Now therefore, we looked at the diffusion equation only the left hand side equal to 0 we solve the diffusion equation and then we understood that the rate at which chemical reaction occurs can now be given by the expression which is given as 4 pi b d c a 0 divided by 1 by r c minus 1 by r. Now therefore, this is the rate at which chemical reaction occurs and therefore, we can look at the material balance for the solid which is d by d t of n b equal to r b times s therefore, we have replaced r b s in terms of what we have derived and then we have an equation which tells us how the unreacted core changes with time. So, this how we have simplified the problem let us go further now we have now d by d t. So, on the left hand side we have on the left hand side can see here on the left hand side will differentiate this. So, we get d by d t. So, I will differentiate this we get 4 pi r c square. So, I have differentiated this 4 pi r c square rho b d r c by d t that is a left hand side equal to minus of 4 pi b d c 0 divided by 1 by r c minus of 1 by r or I will put it in this form rho b 1 by r c minus of 1 by r r c square 4 pi I can cancel off equal to b d c naught d t is it. Now, you have got 1 by r c minus 1 by r there is a minus sign there is a minus sign here d I have taken on the terms. Now, I can integrate let me integrate now rho b if I integrate this it becomes r c square by 2 this will become r c cube by 3 r you are going to go for the integration is from r to r c equal to minus of b d c naught t is it clear to all of you what I am saying what have we done what we have done please nothing complicated has been done using the quasi steady state approximation. We have derived an expression for the rate of chemical reaction in terms of the unreacted core radius and the initial radius and so on. And, writing the material balance for the solid we have found out how r c changes with time. And, this is how r c changes with time now we have to do the integrations let us do that now. So, now I have to do the take the limits from r to r c. So, when I do that. So, let me write it once again rho b within brackets of r c square by 2 minus of r c cube by 3 r going from r to r c equal to minus of b d c naught of t c naught sometimes we call it as c a g and both are the same. Now, let us go through the limits rho b within brackets r c square by 2 minus of r c cube by 3 r minus r square by 2 plus r square by 3. Now, let me take r square rho b outside equal to minus of b d c naught of t. If I take this out r square out what happens I will remove the sign also here. Therefore, it becomes 1 by 2 minus of 1 by 3 is that now this becomes minus of 3 r c square by r plus 2 r c cube by r cube equal to b d c naught t is that let me see we have got it right. So, it is this is 3 minus 2 this is 1 by 6. So, r square rho b this is 1 by 6. So, let me let me just do this again not happy with this I will do it again. Let me just write that step once again rho b times r c square by 2 minus of r c cube by 3 r minus of r square by 2 plus r cube by 3 equal to minus of b d c naught t. So, this is what we have. Now, let me simplify this let me simplify this by rho b. So, this is r c square by 2 I retain it like that then is r c square by 2 I retain cube by 3 r I retain it like that. So, this simplifying these 2 this r c square only r c cube sorry r c cube by 3 r. So, this becomes r square by 6 equal to minus of b d c naught t. Now, the r cancels off. So, 3 this is no problem. Now, I will take r square rho b I will simplify this by taking 6 denominator. So, and then this sign also I will take it away. So, this becomes r square minus of r c square by 3 r c square correct plus 2 r c cube by r cube. So, this is by r just a minute this is alright r c cube by r sorry by r I have got it right. I have taken 6 outside. So, it becomes this is this term is. So, I can term is with a minus sign. Therefore, 3 r square this term is this becomes plus and therefore, it becomes twice r c that is fine equal to b d c naught and t. Now, let me put this in the right form. So, what I am doing now is that I am going to take this r square common. So, I will get rho b r square by 6. So, because 1 minus of 3 r c square by r square plus 2 r c cube by r cube equal to 6 d c naught and t or 1 minus of 3 r c square by r square plus 2 r c cube by r cube equal to sorry 6. So, it becomes 6 d c naught t divided by rho b r square. So, for the case of ash layer diffusion we have that the time dependence of the unreacted core radius is given by this relationship. Now, we notice here that at time t equal to or when r c is 0 what is r c equal to 0 means that the particle has been completely consumed which means at r c equal to 0 we have at r c equal to 0 equal to 0 t equal to tau f tau tau d or it is called as the time for complete consumption of the particle. So, let us just calculate at r c equal to 0 the whole thing goes off. Therefore, tau d becomes rho b r square 6 d c naught rho b r square 6 c d naught is it all right. Therefore, you can substituting for this we get 1 minus of 3 r c square by r square plus twice r c cube by r cube equal to t divided by tau d. This is the point that we wanted to mention. So, I mean going through a reasonable amount of simplification. So, on what we have been able to do is that we have been able to find out how the unreacted core radius changes with time for the case of actually at diffusion x b by definition is 4 by 3 pi r c cube times rho b divided by 4 by 3 pi r cube rho b 1 minus. Therefore, r c cube essentially what we trying to say is that r c cube by 1 minus equal to r c cube by r cube r 1 minus of x b it to the power of 1 by 3 is r c by r. Therefore, these terms r c by r r c cube can be put in terms of x b which is a measurable quantity. See what in an experiment it may be more difficult to determine the core radius unreacted core radius. But the extent of reaction can be determined by various chemical methods. So, r x b is a more easily measurable quantity. Therefore, it is more convenient to express this in terms of r c by noticing that r c by r is actually 1 minus of x b to the power of 1 by 3. Therefore, we can represent approximately substitute for r c by r in terms of x b. Therefore, this whole expression can be put in terms of x b. Therefore, the left hand side can be determined experimentally. So, for the right hand side t by tau d can be determined and that is how you can actually quantify what is going on in the chemical reaction. Having said this a question that arises is that if we have in a general case we may have let us just look at the general problem and this state this whole thing once again we have d n b by d t equal to r b s equal to b times r a s this is the reaction a gas plus b b solid going to products. Now, we know from r x what we have done tau reaction is rho b r divided by b times k s c a g. We know that tau f equal to b r divided by 3 times b k g c a g tau d we have just now derived this rho b r squared divided by 6 d c a g or c a 0 whichever you mean this is what we are so far shown. Now, what we want to recognize let me just write here just so that I do not lose the thread r b s r b s equal to minus of b k s c a g into 4 pi r c squared this is for reaction control equal to minus of b k g c a g into 4 pi r squared for film diffusion control equal to minus of 4 pi b d c a g divided by 1 by r c minus of 1 by r this is for ash control this is ash this is film this is reaction. So, what are we saying what are we saying is that we have considered so far cases in which only reaction is controlling only film diffusion is controlling or only ash diffusion is controlling. Now, in a given situation you might have more than one situation under control therefore, we must be able to handle the general case of all the reactions mechanisms being operating how do we handle that. Now, we handle that by recognizing that resistance resistance equal to we can call this omega equal to potential divided by rate. So, in this case potential is c a g and rate is r b s and we have our potential is c a g on other words c a g divided by r b s. So, let us write so the resistance reaction equal to c a g divided by minus of b k s c a g into 4 pi r c squared which is equal to minus of 1 by divided by b k s times 4 pi r c squared is that clear. Therefore, the resistance can be written as c a g divided by r b s therefore, that is equal to minus of 1 divided by b times k s times c a 4 pi r c squared that is what I have written here this is for reaction omega film is c a g divided by minus of b k g c a g 4 pi r squared that is equal to minus 1 divided by b k g 4 pi r squared. Similarly, diffusion is c a g divided by you have I will write it like this 4 pi b d and this term comes on the numerator. So, 1 by r c minus of 1 by r this how the term comes out c a g will come here. Therefore, this comes out as the minus sign. So, minus of 1 by r c minus of 1 by r divided by 4 pi b d and this term pi b and d. So, what we have said here what we have done is that based on whatever we have done so far we have been able to determine what is the rate at which chemical reaction occurs we have done that already r b s expression we have already got. Therefore, we are able to define what is called as the resistance the which is defined as c a g divided by r b s that means, potential divided by rate. And therefore, we get an expression for resistance for the 3 cases. Now, if in a situation we find that more than 1 resistance is important we simply have to add the resistance. So, in a given situation let us say where all the controlling regimes are operating which means d n b by d t d n b by d t which is r b s in the presence of all the reaction all the controlling mechanisms are operating which is omega 1 plus omega 2 plus omega 3 or omega this is film and this is reaction. So, in a general situation where all the all the reaction film as well as diffusion is important you simply have to add these resistances which means what the rate at which chemical reaction occurs. So, now let me replace that slightly messy algebra, but let us do it properly now. So, d n b by d t we know this d n b by d t is what the left hand side is what 4 by 3 pi r c cubed times rho b I have to differentiate this that is the left hand side the right hand side is c a g divided by reaction film and diffusion. So, we can do that now 4 by 4 pi r c squared rho b. So, I have just r c squared rho b I have just differentiated this time. So, 3 cancels of 4 pi r c squared rho b rho b d r c d r c also I should write and we should multiply this by we should multiply this by these resistances which you already calculated. So, I am just going to do that term by term. So, the first resistances let me write it again it is not coming out nicely 4 pi r c squared rho b multiplied by 1 by b k s 4 pi r c squared plus 1 by b k g 4 pi r c squared r squared plus 1 by 4 pi d b within brackets of 1 by r c minus of 1 by r d r c equal to minus d t times c a g. I hope we have not see this is just go through this once again because nothing new is has been said. Please recognize that d n b by d t is r b s and these are the terms omega reaction omega film omega diffusion all that we already derived here you see and simply I have just cross multiplying. So, cross multiplying. So, that is what we have done 4 pi r c squared rho b 1 by all this terms have been taken up problem minus sign is also there. Now, we can integrate and our integration goes from our integration goes from what capital R to r c this is clear what we are saying or in other words you see can notice here the each of this terms is exactly the term that we went through when we did the whole analysis independently. Now, they all occur together now we can integrate and then get the answers. Now, when we integrate we find let me integrate this. So, I am just going to integrate this each of this terms going from r r to r c r to r c r to r c. Now, since we have done all this before. So, I will not go through the whole thing once again instead simply do the integration quickly. So, what I get on the left hand side is rho b r divided by b times k s times 1 minus of r c by r that is 1 term rho b r divided by thrice b k g into 1 minus of r c cube by r cube this is the film diffusion. The next term is rho b r square divided by 6 b d 6 b d into 1 minus of thrice r c square by r square plus twice r c cube by r cube equal to c a g times t. Now, you can put the c a g here and recognize that if I put c a g here then it it simply this the term that multiplies let me do that I will put c a g here I will put c a g here I will put c a g here notice here that this whole thing can be written as tau reaction 1 minus of r c by r plus tau film 1 minus of r c cube by r cube plus tau diffusion 1 minus of thrice r c square by r square plus twice r c cube by r cube equal to tau. What are we saying now? What we have said is that if you have a particle in which all the resistances are important then the time that is required for a given extent of reaction is simply the additive of the effects of each of the three reaction film and diffusion that means whatever previously when we had only one of the resistances operating. So, we had only this term equal to t when it is reaction control or this term equal to t when it is film diffusion control or the third term equal to t when it was diffusion control as diffusion control. Now, all of them simply it is an addition of the three resistances. So, what we are trying to say is that when the problem is linear you can have this additive effects. So, just to cut this long story short what we have said is that if you have all three of them controlling then the results you will get is the results you would have gotten for the individual cases but they are all added together. Thank you.