 Let us summarize what we have learnt in the last lecture. So we have learnt how to, if there is a heterogeneous catalytic reaction, we have learnt how to deduce the rate law from experimental data, deduce the rate law. We have learnt the second step, how to find mechanism, how to find the mechanism behind a certain catalytic reaction and we have found the, for a particular example, we have found the rate parameters and then we initiated discussion on how to design a reactor. In particular, we considered the example of hydro demethylation of toluene on solid mineral catalyst containing a certain material called clenoptilolite and we looked at the experimental data for that particular reaction and then we tried to deduce the mechanism and tried to estimate the rate parameters. We found that for the toluene reacting with the hydrogen leading to benzene and methane formation, so for that reaction, we deduce that the rate of the reaction can be represented as the rate of consumption of the toluene can be written as some rate constant k multiplied by the partial pressure of hydrogen multiplied by the partial pressure of toluene divided by 1 plus the adsorption constant for toluene multiplied by the partial pressure of toluene plus the adsorption constant for benzene multiplied by the partial pressure of benzene. So then we use the experimental data in order to perform a linear regression analysis and estimate these parameters k, estimate the parameters k, estimate the parameter kT and also kB from experimental data. Then we initiated discussion on the design of a reactor using this rate law. So while designing the reactor, we wrote a mole balance and found what is the balance equation that represents the that captures the conversion in the reactor as a function of the weight of the catalyst and so that turned out to be dx by dw, x is the conversion and w is the weight of the catalyst in the reactor equal to minus RT prime divided by FT0 where minus RT prime is the rate at which toluene is actually being consumed in the reaction and also FT0 is the initial feed molar rate of toluene species which is participating in the reaction. We also found that the partial pressure of toluene can actually be expressed as the partial pressure of toluene at the inlet PT0 multiplied by 1 minus the conversion into y which is the mole fraction and similarly for the partial pressure of hydrogen which is expressed as PT0 into the feed ratio of hydrogen to toluene minus the conversion into the mole fraction and for benzene it will be PT0 into x into y. So we will continue from there today and so the first exercise is to find out what is this expression for what is y that is what is the mole fraction and that can be obtained what is the relationship of mole fraction and the other and the conversion inside the reactor and that can actually be found out using the differential form of Ergen equation differential form of Ergen equation and which is which can be written as dy by dw that is equal to minus alpha by 2y multiplied by 1 plus epsilon x into T by T0 where alpha is the pressure drop parameter alpha is the pressure drop parameter epsilon is the epsilon is the total fractional change in the number of moles and x is the conversion and T is the temperature at that location and T0 is the inlet temperature. If we assume that it is isothermal conditions then T by T0 is equal to 1 if we assume that it is a isothermal the reactor is actually reaction is conducted in isothermal conditions then the temperature in the reactor is same as the inlet temperature of the fluid stream and therefore T by T0 will be equal to 1 and for the chosen reaction the epsilon which is the total change in the number of moles because of the reaction is equal to 0 because the net stoichiometric coefficient is 0 and from here we can solve the differential Ergen equation and obtain the expression for y as a function of the other parameters which is P by P0 equal to 1 minus alpha times w to the power of half where alpha is the pressure drop parameter and w is the weight of the catalyst. So the mole fraction at any location in the reactor will now depend upon how much catalyst is present till that location and the corresponding pressure drop parameter. So now we can actually plug this into the model equation and then we can attempt to find out what is the conversion as a function of the weight of the catalyst. So therefore the partial pressure of toluene is now given by Pt0 which is the total partial pressure of toluene at the inlet multiplied by 1 minus x which is the conversion into 1 minus alpha w to the power of half and similarly that of partial pressure of hydrogen is given by Pt0 into 1 theta H2 which is the feed ratio of hydrogen versus toluene minus the conversion into 1 minus alpha w to the power of half and the partial pressure of benzene is given by Pt0 into conversion into 1 minus alpha w to the power of half. So now we have all the information that we need to solve the equation and so let us look at how to let us put them all together. So the model equation that relates the conversion with the weight of the catalyst is given by Ft0 multiplied by dx by dw that is equal to minus RT prime which is the rate of generation of toluene and we know from the rate mechanism that the rate of generation is given by K into partial pressure of hydrogen into partial pressure of toluene divided by 1 plus Kb into partial pressure of benzene plus Kt into partial pressure of toluene. So now what we need to know is what are the limits of the integration? We know what is we know the expressions for partial pressure of hydrogen in terms of conversion. We know the expression for partial pressure of toluene in terms of conversion and we know the expression for partial pressure of benzene in terms of conversion. Now once we know this in order to integrate the only piece of information that is required is what is the total weight of the catalyst in the reactor? Weight of the catalyst in the reactor. Now why we need this because we need to in order to integrate this expression here in order to integrate the expression which is present here we need to know what are the limits for the limits for W which is the weight of the catalyst. So of course at the inlet at the inlet the weight of the catalyst is 0 but at the outlet let us assume that there is W is given by some maximum weight of the catalyst. Now how do we find this maximum weight of the catalyst? So we know the expression for the mole fraction as a function of the weight of the catalyst. So what we can do is suppose if we assume that the reactor the fluid stream exits the reactor at atmospheric pressure. So suppose the pressure at which the fluid stream leaves the reactor is 1 atmosphere. Then we will be able to estimate what is the maximum weight because we know the relationship between the mole fraction y which is P by P0 that is equal to 1 minus alpha times W to the power of half. So from this expression we can now substitute the pressure at the exit of the reactor and find out what is the weight of the catalyst in terms of the other parameters. So therefore 1 by P0 equal to 1 minus alpha W max to the power of half. So this means that the maximum weight of the catalyst if the pressure at the outlet is 1 atmosphere that will be equal to 1 by alpha into 1 plus 1 by P0 whole square. So now we have all the required information in order to integrate the model equation. So let us look at let us put them all together. So we need to solve this equation F t0 into dx by dw equal to minus RT prime with the rate of the equation being rate of the generation of species toluene being K into partial pressure of hydrogen partial pressure of toluene divided by 1 plus the equilibrium constant for benzene multiplied by the partial pressure of benzene plus the equilibrium constant of toluene multiplied by the partial pressure of toluene. And we know that partial pressure of hydrogen is equal to Pt0 into theta h2 minus x into 1 minus alpha W to the power of half. And then the partial pressure of benzene is given by the inlet partial pressure of toluene multiplied by the conversion into 1 minus alpha W to the power of half and the partial pressure of toluene is given by Pt0 into 1 minus x into 1 minus alpha W to the power of half. And the maximum weight of the catalyst is given by 1 by alpha into 1 plus 1 by P0 to the power of 2. So this gives you the this captures the model equations and all the others necessary relationship between the partial pressures and the conversion in order to find the conversion as a function of the weight of the catalyst. So remember that the weight of the catalyst at the inlet of the reactor is 0 and at the outlet of the reactor is the maximum weight that can be packed maximum catalyst that can be packed inside the reactor. So let us look at how the solution is going to look like. So if we have the numbers and if we have the quantities all the required quantities then in principle we should be able to integrate this expression and find the profile of conversion as a function of weight of the catalyst. So let me just sketch here I will sketch the profile of the conversion as a function of weight of the catalyst. So suppose in the x-axis I have the weight of the catalyst going from 0 to the maximum weight. So remember that the maximum weight is given by 1 by alpha into 1 plus 1 by P0 the whole square. And if I plot x which is the conversion then the conversion as you go into the reactor as the fluid stream sees more catalyst you see that there is an increase in the conversion as a function of the weight of the catalyst. While if you look at the pressure drop inside the mole fraction inside the reactor the mole fraction y the overall mole fraction y actually goes down. So this is y equal to P by P0. So that actually goes down because the reaction is being conducted and the reaction is happening as the fluid stream goes through the reactor. We can actually get much more information than this. So we can actually look at how the partial pressure changes with the weight of the catalyst. So if I plot that is the maximum weight of the catalyst that is the exit of the reactor. So this is the inlet stream and this is the outlet stream that is the outside location of the reactor. And so if I plot the partial pressures of toluene partial pressure of hydrogen and partial pressure of benzene for typical set of parameters we will see that the partial pressure of hydrogen has a profile which looks like this. And then we have a toluene partial pressure which actually decreases because toluene is being consumed and then the benzene partial pressure this is the kind of profile that is observed. So the reason for so this is for this is the partial pressure for hydrogen partial pressure for toluene and the partial pressure of benzene. Now the reason why the partial pressure of benzene goes through a maxima is because we said that the outside pressure is the pressure of at which the stream leaves the reactor is one atmosphere and so there is clearly a pressure drop and the pressure drop actually causes the partial pressure of benzene to go through a maximum and then towards the exit of the reactor it starts falling down. So this sort of captures systematically the method for designing a reactor starting from experimental data. So let us look at what happens if we conduct the reaction in a fluidized CSTR. So suppose if we take a fluidized CSTR so CSTR is essentially a tank and there is a feed that goes into the CSTR and it is well mixed and there is a fixed volume and there is an outlet stream. Generally for a fluidized bed the inlet is at the bottom and exit at the top just for sake of convenience we assumed other way here. So the question is how much catalyst weight weight of the catalyst or how much catalyst that needs to be packed which directly correlates to the volume of the reactor. So how much catalyst has to be packed inside the CSTR in order to obtain the same conversion. So if FT0 is the molar flow rate of toluene at the inlet stream of the CSTR and FT is the molar flow rate at which the toluene leaves the reactor then a simple mass balance can be mole balance can be written in order to estimate the amount of catalyst which is required for obtaining the same conversion in a fluidized CSTR. So the simple mass by mole balance just that the rate at which the stream enters the reactor minus the rate at which things leave plus the generation should be equal to the accumulation. So now what the rate at which moles enter the CSTR is given by FT0 for toluene minus the rate at which it leaves is FT plus the rate at which it is generated is given by RT prime which is basically the moles of toluene consumed per gram of catalyst per unit time multiplied by the weight of the catalyst and that should be equal to 0 if it is a steady state. So we can now rewrite this equation and we can say that the weight of the catalyst is given by FT0 minus FT divided by minus RT prime. So from here we know that FT which is the rate molar rate at which the molar flow rate at which the species leaves the reactor is related to the conversion because FT is given by FT0 into 1 minus x. So therefore we can rewrite this expression as FT0 into x divided by minus RT prime. So if we know what is the conversion in the packed bed reactor then we should be able to use that conversion and if it is under the same condition that is the same molar flow rate at which the species enters the CSTR then we will be able to estimate what is the weight at which the same conversion can be obtained if you use a fluidized CSTR and this provides an important parameter in terms of designing the reactor. So let us next look at a specific case study. So we will look at the chemical vapor deposition process, we will take the example of chemical vapor deposition process which here after I will refer to as CVD. So chemical vapor deposition is actually an important process in the microelectronic fabrication and it actually plays a strong role in manufacturing the tandem or the multi-layer solar cells. So the chemical vapor deposition is essentially depositing a particular material on a given substrate. So it is very similar to the heterogeneous catalysis process and it has several applications in microelectronic fabrication and it has applications in manufacturing of tandem solar cells. So we will take a specific case of chemical vapor deposition which is applicable to micro electronic fabrication and see how it can be viewed as a heterogeneous catalytic system and how it can be modeled in order to obtain certain design parameters. An interesting example of that is growth of germanium epitaxial film. Germanium epitaxial film is actually an interesting example of, so while manufacturing in micro electronic fabrication, it is important to have different layers between the gallium arsenides and silicon and so germanium epitaxial film serves a very good purpose of acting as an interlayer between the silicon layer which is present and on top of it will be the gallium arsenide. It also acts as a good contact layer, so the growth of this epitaxial film is actually obtained by using this process called the chemical vapor deposition. So the mechanism that actually governs such a deposition process is actually very similar to that of the heterogeneous catalytic reaction. So if you look at the mechanism, it is very similar to that of the heterogeneous catalytic systems. So it involves the classical three processes. One is the adsorption, the adsorption of the reactants onto the surface of a substrate and followed by a surface reaction, followed by a surface reaction and then followed by a desorption. In addition to this, there may be gas phase reactions, so it is the adsorption of reactants onto a substrate surface followed by a surface reaction and then followed by desorption depending upon the nature of the reaction. In some cases, there may not be a desorption and in some cases, there may be a desorption step. It is very similar to the heterogeneous catalytic processes that we have learned so far and so we will use those principles in order to model this epitaxial film growth process. So the mechanism involves three steps. The first step is actually four steps. First step is the gas phase dissociation, gas phase dissociation step. Now in the gas phase dissociation, what happens is that the germanium chloride GECL4 GECL4 in the gas phase. Now that reversibly gets converted into the germanium Cl2 in the gas phase plus Cl2 plus chlorine and the second step is the adsorption process where the germanium chloride GECL2 in the gas phase, it adsorbs onto the catalyst site and it is an equilibrium process. If Ka is the specific constant corresponding to that and GECL2.s which basically represents the adsorption of germanium chloride onto the catalytic site. Now here the germanium chloride actually reacts with the hydrogen and so in this particular case the hydrogen gas in the gas phase also gets adsorbed to the catalyst site. So it actually interacts with two catalysts, two vacant catalyst site and so if the specific constant for that particular forward step or adsorption step is kh then it leads to so one molecule of hydrogen actually binds to two sites, two vacant sites which is available. And so there are now three sites which are actually, so there are three catalytic sites are actually involved, three vacant sites are required, three vacant sites are required for this processes, this catalytic process and so then the next step is basically the surface reaction. So the next step is the surface reaction process where it is the surface reaction process where the germanium chloride which is now adsorbed to the catalyst site interacts with the hydrogen atoms which are actually adsorbed to two vacant sites, two sites leading to formation of germanium in the solid form which is actually deposited on the surface of the substrate plus two HCl plus two vacant sites. Now so basically this is the one which is deposited, the germanium which is formed because of this reaction is the one which is deposited on the substrate. After the reaction it may appear that there are only two vacant sites which are actually left out, it may appear that there are only two vacant sites. Now that is not true because the germanium which is deposited also becomes active and so subsequently the deposited germanium in its solid form also acts as an active site for binding of germanium chloride or the hydrogen atom itself. So therefore in addition to the two vacant sites which are created the germanium deposit also acts as an active site and so totally there are three vacant sites, three active sites that are actually created because of the reaction. So really there is no loss of active sites on the catalyst because of the reaction. So now if I assume that the surface reaction is the limiting reaction then I can write or I can find out what is the rate of deposition of germanium. So the surface reaction is GCl2 which is adsorbed to the surface plus the two hydrogen atoms which are adsorbed to the surface leads to deposition of germanium in the solid form plus 2 HCl plus two vacant sites. So now based on this reaction we can actually write a rate of deposition of germanium and so the rate of deposition of germanium must be a function of the total number of active sites which are actually present, total number of active sites on which the germanium chloride is adsorbed and also the total number of active sites on which the hydrogen atoms are adsorbed. So therefore we can write this reaction rate as R deposition is equal to the specific reaction rate multiplied by the fraction of sites which are occupied by the germanium chloride multiplied by the fraction of sites which are occupied by the hydrogen atom and the square of that. So the reason why there is a square here because for every reaction, every surface reaction which is assumed irreversible there are two catalytic sites are actually involved in the surface reaction, two catalytic sites containing adsorbed hydrogen atoms are actually involved in the reaction and therefore the rate of deposition has to be some constant specific reaction rate multiplied by the fraction of sites on which the germanium chloride is adsorbed and multiplied by the square of the fraction of sites on which the hydrogen atom is adsorbed and the units of the rate is actually nanometers per second. So that is the rate this can actually be converted into the molar units by multiplying the molar density and this is the specific reaction rate and the units are nanometers per second and this is the fraction of sites occupied by germanium chloride and this is the fraction of sites occupied by hydrogen atom, hydrogen atom. So this sort of captures the rate of deposition of germanium on the substrate. So now the exercise is to find out what is the fraction of germanium chloride which is adsorbed onto the surface and what is the fraction of the hydrogen atom which is adsorbed onto the surface. So this can be obtained by looking at the expression, looking at the adsorption process and so the adsorption steps, so let us look a little more deeply into the adsorption steps. There are two adsorption steps, one is the germanium chloride which adsorbs onto the vacant sites and the hydrogen adsorbs onto two vacant sites. So the first adsorption step is GeCl2 in the gas phase plus a particular vacant site, reversibly it is an equilibrium process and if Ka is the corresponding specific constant it gets adsorbed and leads to the adsorbed, germanium chloride adsorbed to the active site. So the rate of adsorption is given by Ka multiplied by the fraction of vacant sites Fe multiplied by the partial pressure of the germanium chloride in the gas phase minus the fraction of sites onto which the germanium chloride is adsorbed divided by the corresponding equilibrium constant. Now because we assume that the surface reaction is the limiting reaction, the rate of adsorption has to be approximately 0. So therefore the rate of adsorption divided by Ka is approximately 0 from which one can estimate that the fraction of sites on which the germanium chloride is adsorbed is given by Ka into fraction of the vacant sites multiplied by the partial pressure of the germanium chloride in the gas phase. Similarly, we can look at the other adsorption process which is the adsorption of hydrogen which is the adsorption of hydrogen and so this the reaction that corresponds to the adsorption desorption step is hydrogen plus two catalytic vacant sites, if Ka is the corresponding specific constant leads to two sites in which the hydrogen atoms are bound. So now the rate of adsorption of hydrogen is given by a specific constant K s multiplied by the, so it needs two vacant sites, so the rate will be given by Fe square multiplied by the partial pressure of hydrogen minus the fraction of sites that are actually occupied by the hydrogen atom F h square divided by the corresponding equilibrium constant. Because we assume that the surface reaction is a limiting reaction, we can actually say that the rate of adsorption of h2 divided by K s is approximately 0 and from here one can estimate that the fraction of sites to which the hydrogen atom is bound is given by the vacant sites Fe multiplied by square root of the equilibrium constant for hydrogen adsorption multiplied by the partial pressure in the gas phase of hydrogen in the gas phase. So now the total number of sites which is present in the catalyst is actually a constant because it is a conservation property, so therefore that can be written, the conservation property can be written as Fe which is the total fraction of vacant sites plus the sites on which the germanium chloride is bound plus the total, the fraction of sites on which the hydrogen atom is bound that should be equal to 1 because the total number of sites which is present in the catalyst is a constant and from here we can express that F v equal to, so we can find, we can substitute the fractional sites on which the germanium chloride is bound and the fractional sites on which the hydrogen is bound and from this conservation property we can estimate that the fraction of vacant sites which is present is given by 1 divided by 1 plus K a into partial pressure of germanium chloride in the gas phase plus square root of the equilibrium constant for hydrogen adsorption multiplied by the partial pressure of hydrogen in the gas phase. So now we can plug all these expressions into the rate of deposition, so the rate of deposition is now given by K s into F v into K a, F v is the vacant site number of vacant sites which is present multiplied by the equilibrium constant for adsorption of germanium chloride multiplied by the square of the number of vacant sites which is present multiplied by the partial pressure of hydrogen in the gas phase into the equilibrium constant for hydrogen. So now we know the, we know what is the relationship between the vacant sites which is available and the partial pressure of the germanium chloride in the gas phase and the partial pressure of hydrogen in the gas phase, so we can relate, we can plug in this expression in the reaction rate, deposition rate expression and so we can rewrite the, we can rewrite the deposition rate expression as our deposition is equal to K s, K a, K h multiplied by the partial pressure of germanium in the gas phase multiplied by the partial pressure of hydrogen in the gas phase divided by 1 plus K a partial pressure of germanium chloride in the gas phase plus square root of K h into P h 2 whole cube. So now we can plug in, we can actually collate these three constants, we can collate these three constants together and we can write it as 1 constant K prime, so that will be K prime into partial pressure of germanium chloride into partial pressure of hydrogen in the gas phase divided by 1 plus K a into partial pressure of germanium chloride whole cube. So now when we, the germanium chloride is actually in the form of G C l 4, C l 4 and not in G C l 2 which is the one which gets absorbed on to the surface. So now there is a gas phase reaction which actually is responsible for the conversion of or transformation of G C l 4 into the C l 2 form and so the gas phase reaction is given by G C l 4 in the gas phase C l 2 plus chlorine, so that is the gas phase reaction. Now one can write an expression relating the gas phase partial pressure of the germanium chloride G E C l 4 and the germanium chloride G E C l 2. So that can simply be written as because it is an equilibrium conditions we can write it as K p which is the corresponding equilibrium constant and that should be equal to partial pressure of G E C l 2 multiplied by partial pressure of chlorine divided by partial pressure of germanium C l 4. So from here we can find out what is the partial pressure of germanium G E C l 2 which is given by C l 4 divided by partial pressure of C l 2 multiplied by K p. So this is required to complete the problem because what is practically measurable quantity is actually the partial pressure of G E C l 4 and the partial pressure of chlorine. So therefore it is required to express the partial pressure of G E C l 2 in the gas phase in terms of the measurable quantities. So plugging in these expressions you can find that the net rate of deposition r d net deposition rate is actually given by K prime K p into partial pressure of germanium chloride G E C l 4 into partial pressure of hydrogen divided by partial pressure of C l 2 into 1 plus equilibrium constant K a multiplied by partial pressure of G E C l 2 plus square root of K h into partial pressure of hydrogen the whole cube. So we can easily rewrite this expression as with a little bit of algebra you can rewrite this expression as K which is basically K prime into K p. So we can club these two constants here into one constant K multiplied by the partial pressure of G E C l 4 multiplied by the partial pressure of H 2 into partial pressure of C l 2 square divided by the partial pressure of C l 2 plus constant K into partial pressure of G E C l 4 plus the square root of K h into partial pressure of H 2 divided by partial pressure of C l 2 cube the whole cube. So now suppose if the partial pressure of hydrogen in the K h into partial pressure of hydrogen is very small suppose if this quantity is very small then we can rewrite this expression as we can rewrite this expression as r deposition the deposition rate is given by K into partial pressure of germanium chloride 4 multiplied by partial pressure of hydrogen multiplied by partial pressure of chlorine square divided by partial pressure of chlorine plus modified constant K multiplied by the partial pressure of germanium chloride 4 in the gas phase whole cube and this is valid only when only square root of K h P h 2 is much smaller than 1. So let us stare at this equation so we will see that now we have expressed the rate of deposition as a function of the gas phase partial pressure of germanium chloride which is a measurable quantity and then the partial pressure of hydrogen which is again a measurable quantity and then partial pressure of chlorine which is again a measurable quantity. So from here we can find out what is the expression for the rate of deposition of germanium on the substrate. Now there is another process which can actually occur along with this the germanium which is deposited on to the surface because of this reaction can actually react with the germanium chloride in order to form 2 germanium Cl2. So this has actually been ignored in the current framework but in principle this can actually occur in reality. So let us summarize what we have learned today so what we have seen is we have we completed the reactor design that we initiated in the last lecture so we looked at the design of a packed bed reactor and then we looked at how the conversion can be expressed in terms of the weight of the catalyst which is packed inside the reactor and then we also looked at what should be the performance equation for a CSTR in case we want to find out what is the weight of the catalyst which is required to obtain the same conversion as that obtained in the packed bed reactor and then we went on to describe a case study of the germanium epitaxial film growth. So germanium epitaxial film growth so which also is can be modeled as a heterogeneous catalytic process and so with these 2 examples of hydro demethylation of toluene and the germanium epitaxial growth what we have learned is a systematic way to start from first principles by determining what is the mechanism of a particular catalytic reaction find out what is the surface react what is the limiting step in both these cases we assume that it is the surface reaction which is the limiting step and in principle the adsorption or desorption also can be a limiting step. So it is important to note that the only the 75% of the heterogeneous catalytic reactions are actually surface reaction limiting and so in principle the same procedure can be followed if the adsorption and the desorption step were to be the limiting step. So using these rate law for hydro demethylation of toluene we estimated the rate parameters and then we went ahead to design a simple packed bed reactor model to find the relationship between the conversion and the weight of the catalyst catalyst that is actually packed inside the reactor. Thank you.