 This looking at advanced reaction engineering today will continue and look at time dependent operations. Now, we said at earlier class that there are several situations where we have to deal with time dependence and the time dependence we are going to be looking at today is catalyst deactivation. So, the context is catalyst deactivation because there is lot of literature on the subject and the example I take here is just to illustrate how we can determine the catalyst activity variation dependent or what is called as the rate function for catalyst deactivation as an example. This is an example situation we are looking at we will be looking at vinyl chloride monomer production from acetylene and HCl. This is a gas this is also a gas it gives you sorry C2 H3 Cl this is called vinyl chloride. This reaction takes place over mercuric chloride catalyst catalyst which is supported on activated carbon. Let me let me tell you how this catalyst are prepared by making a solution of mercuric chloride because mercuric chloride very poisonous you have to handle it great care and so on and you dissolve it in water and then you put so many grams of activated carbon and then allow it to undergo the adsorption and it absorbs very rapidly actually do not have to wait very long and then you filter of the solution and you can determine the un adsorbed mercuric chloride in the solution you can do it in various ways by if you are starting with analog grade mercuric chloride you can just do chloride analysis, but a better thing to do would be to do mercury analysis. There are atomic absorption techniques that we can use by which you can actually determine how much of mercuric chloride has been adsorbed and therefore, you can determine what is the loading of mercuric chloride on the activated carbon. Now, typically this is 10 percent weight by weight this is what is typical of course, this is a well studied reaction and it is also known this catalyst undergoes deactivation and as a result as a result of deactivation the catalyst loses activity and so on and it seems to a good example to illustrate how we can determine deactivation kinetics. So, what we are trying to do is to determine deactivation kinetics first and then use that kind of result to understand how we can design operate processes in which there is catalyst deactivation. Now, determination of deactivation kinetics. So, what we are looking for what we are looking for is we are looking for a rate function which we think will depend upon activity sum to the power of I will just put the power of m it may also depend upon we have three components here if you recall if you recall in our monomer reaction please look recall that if I call this as a I will call this as b call this as a call this as c. So, it depends upon a b and c it could we do not know what is the dependence what is saying is that our intention in this is to determine what is the function that determines catalyst deactivation. So, we expect that the r d the catalyst deactivation function would depend upon activity itself it could also depend upon the compositions of a b and c with some exponents say p q r. So, what in other words what we are saying is that k d m p q r are parameters are parameters of deactivation deactivation rate function and the object of what we are trying to do today is to find the way by which we determine what are m p q r and k d and more importantly how k d for example, depends upon temperature. So, therefore, we are interested in the whole range of things that will help us to understand the deactivation rate function how is it done we do this in this case by recognizing that we have let us say a stirred tank it is also called a stirred basket reactor stirred stirred basket reactor. Why is it called stirred basket reactor we have this terror then you have these impellers these impellers and your basket is mounted like this and your catalyst is sitting like this. And your rotating this and of course, this is a tank and then it is called a closure and there is a valve here through which gases come in and there will be a valve here through which gases go out. And this is spinning at a rapid rate so that we are able to keep the compositions around the catalyst uniform we are able to keep the temperature uniform compositions uniform and so on. So, that whatever data we get the compositions and temperatures are well known in the environment of the catalyst. Now, when we talked about this a little earlier we said our activity a is actually defined as at time t divided by time 0. We recall we said this we recall this we just put in the context we said that when we make a plot I just plot it here plate plot of r a versus time and we can do this for different we can do this for different residence times we have done all these. So, it is not new to you therefore, and if you just extrapolate this if you extrapolate this. So, this gives you this is 0 time. So, you get reaction rate at 0 time and reaction rate at any time. So, at one residence time you get this kind of curves and otherwise you get this kind of curves otherwise you get this kind of curves or in other words what we are trying to say is that we are we can operate this equipment at different we can do this at different C A 0's you can do this at different flow rates at different flow rates at different flow rates. Therefore, different residence times and so on. Therefore, we are able to determine since we are able to measure reaction rate at any time divided by reaction. Therefore, you are able to measure activity of the catalyst in this way. So, you are able to find activity now let us say this is at some temperature t you can do it at some temperature t you can do it at different compositions at C A 0. So, this and then we may also do it at different V 0. So, these are the three variables at which we can do these experiments on other words if you are looking if you are looking at this rate function after all our intention is what is this function. So, this function if you want to determine this function r d what is which we would say that d by d t of a we expect that a to decrease with time equal to r d. Therefore, this is equal to k d which is a function of temperature a to the power of m C A to the power of p C B to the power of q C C to the power of r. On other words now this d by d t of a is given by the right hand side now what is it that we have determined this right hand side is r d. So, we have determined this activity a as a function of time. Therefore, you know in a sense we have data to determine what is k d what is m p q and r. We have data to determine m p q and r now we also said one more one more thing for example, let us say let us say we make a plot of a versus time. Can we do this answer is yes because we have determined a which is r a at any time t divided by r a at 0 time. So, you can find out what is 0 time r a what is r a at any other time therefore, this divided by this is the activity corresponding to this kind of operation. We can get activity for different conditions for every condition that we for which for we get data the compositions a b c this compositions here of a b and c could be quite different. Let us say our data looks like this for one set of one residence time now we do it for another residence time. Let us say at this point we got data and again we get another curve. Let me draw it by another another color let us say previous data I am just putting it the previous data. Now, when we do this data let means for a different residence time let us say our greens are here and let us say one more set of experiments in which we again do the equipment for another residence time let us say our data looks like this. What we are trying to say here is that if it so happens that a versus time data actually collapses into a single curve collapses into a single curve irrespective of the choice of residence time. On other words for every residence time the composition a b and c would be different it is quite obvious because depending on residence time the extent of reaction would change and therefore, the compositions would change. Therefore, every choice of residence time we get different compositions and then we find that the activity as determined by this function r a t at r a at 0. We get let us say the black dots and then for another residence time we get this green dots and for another residence time we get this red dots all of them seem to be collapsing into one curve suggesting that this form of deactivation is concentration independent. What are we saying what we are saying is that if you look at this deactivation rate function what we find is that when we change compositions a b c by an appropriate choice of the residence time compositions of a b and c change, but in spite of that we find that our activity time relationship collapses onto a single curve showing that the concentration dependence is not there in this data we talking about this particular data we do not know different data may be different accordingly these functions could be different these exponents could be different. What is being said here is that if you find that this concentration dependence is not there it means that p q and r it means that p q and r of this rate function r 0 that is all it means and in a particular case if you find this dependence is there and you will be able to determine what the value is from your data. So, in this particular case the example I have taken is that p q and r seem to be 0 let us say all right having said this let us just go further. Let us just take the data that we have I have got some data here it is just write down the data this says temperature is 180 c and data is serial number 1 serial number it is 1 2 3 4 5 6 7 I have got 7 sets of data y a 0 this is comes as 0.1 this is 0.1 this is 0.2, 0.28, 0.39, 0.28, 0.22, 0.17 and 0.22 the data is available like this and time in hours we talking about 0.0 time in hours 2.5, 12.5. 14.0, 17.0, 25.0, 38.2 activity is defined as r t at r 0 I am just putting it as 1.0, 0.86, 0.63, 0.56, 0.41, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.36, 0.24 this is a data at 180 c. Now what is y a 0, y a 0 is mole fraction, y a 0 is mole fraction h c l in this case component a and then pressure is 1 atmosphere temperature in this data the temperature is 180 c. What have we done? Now let us just recall how this experiment was done. It is quite interesting to know how experiments like this are done. Now how do you do this experiment? You set the temperature as 180. How do you set this temperature is 180? You will have this particular reactor nicely covered with let us say heating tapes or jackets. In this particular case it is a heating tape and well insulated so that the heat losses are small. Therefore, by adjusting the heating rate you can get the temperature that you desire. So, and then compositions how do you get let us say y a 0 as 0.1, 0.1 to whatever you have a rotameter let us say which is all calibrated. So, adjust the flow of the rotameter of hydrogen chloride gas as well as acetylene gas and adjust it appropriately as per your calibration. So, that you get the mole fraction of hydrogen chloride gas is 0.1, 0.1 to whatever. So, this composition of you have both a and b similarly C b 0. This is in this case C a 0 is H c l C b 0 is acetylene and the gas goes in undergoes reaction and this is kept spinning typically at about 1500 rpm. Let us say so that the mixing is satisfactory and then product gases come out and what happens to this? This product gas generally goes through what is called as a absorber of H c l absorber. So, you have to remove the H c l before it can go for chromatography goes to a G c or gas chromatograph where the compositions are continuously monitored. Is there understand is this clear what we are saying? So, the gas goes into H c l absorber to remove the H c l and generally absorber is the sodium hydroxide in a lab typically it will be sodium hydroxide whatever or potassium hydroxide and then the H c l free gas goes to gas chromatograph for analysis. Once again gas chromatographs are calibrated there are elaborate procedures for calibration with internal standards and so on. So, you know as you put your gas through the chromatograph you get a we get a response and you can compare this response with a standard. So, that you can get your compositions. So, what we are trying to say here is that you can do a measurement of the activity at a given temperature. You can do this activity measurements by and to determine whether this concentration dependence is there or not by doing this experiment is various compositions that is what we have done. And the data that is in front of you this data is front of you tells you that this is H c l a is H c l a is H c l. So, it data tells you that we have changed the compositions from point 1 to 2 various compositions in that range to find out what happens to the activity is it clear. So, we find the activity to be this. Now, you can do the same kind of experiments at other temperatures say let us say t equal to 210. So, I will just write down the data here serial number 1, 2, 3, 4, 5, 6 and 7 then you have y a 0 it is 0.12, 0.28 please bear with me because this is important that is why I am just writing it down. So, do not be impatient and we will be with you shortly just few minutes and I will be with you y a 0 and then you have how much of you know total molar flow f t 0 is moles per per minute that is also given let me write that down 0.071, 0.032, 0.023, 0.032, 0.041, 0.051, 0.512, 0.041 is it all right. Then, you have temperatures activity you have time sorry time is 0, 0.4 and then 10.2, 3.4, 10.8, 7.9, 15.8 activity is 1.40, 0.4, 0.5, 0.0, 0.96, 0.35, 0.70, 0.32, 0.43, 0.19. Now, please bear with me, in the previous data I had not mentioned this F T 0 I forgot to write down. So, let me just write down 0.07, 0.07, 0.071, 0.032, 0.023, 0.032, 0.041, 0.051, 0.041. So, I have got this moles per minute, moles per minute. So, I have got data and then I have calculated activity using the relationship r of t divided by r of 0. So, this is how calculated. Now, we have data now at 180, we have data now at 210, we have got 2 sets of data. In this particular case fortunately, we have little bit more data and the context of having telling this is that, see you need to be able to determine the activation energy, you need minimum data at 3 temperatures. So, this is an instance where you have no choice about to do experiments minimally at 3 temperatures. So, we have serial number, you have Y A 0, you have F T 0, activity and sorry time and activity. So, 1, 2, 3, 4, 5, 6, 7, 0.12, 1, 2, 3, 4, 3, 4, 5, 6, 2, 0.28, 0.39, 0.28, 0.22, 0.17, 0.22, it is 0.071, 0.032, 0.023, 0.032, 0.041, 0.051, 0.041. Now, 240, time 0, 0.9, 1.9, 5.8, 3.8, 1.5 and 0.6, activity is 1, 0.84, 0.62, 0.30, 0.44, 0.70, 0.86. Now, what is it that you have got? You have got from our experiments, data, data at 3 temperatures. So, we have data at 3 temperatures, 1, 2, let us say 3, this is T is 180, 210 and 240. Now, when you make a plot of A versus time, you get a plot like this, plot like this, plot like this and interestingly, interestingly, interestingly, interestingly, interestingly, the concentration dependence is not to be seen. On other words, even though we have done these measurements at different flow rates and different compositions as you can recognize from the data given that activity versus time data at the 3 temperatures does not seem to reveal any serious concentration dependence. So, first thing we conclude is that in this data, concentration dependence of activity seem to be very, very small or P, Q and R equal to 0. It seems like that, it seems that P, Q, R are 0. Is that clear? Now, if moment we recognize that P, Q, R are 0, therefore all you have to do is to, all we have to do is to solve. Now, the only thing that we have to do is that minus of D A times D T equal to K D times A to the power of M, A to the power of M and then C A to the power of P, C B to the power of Q, C, C to the power of R, all these dependences are not there. Therefore, you have to only solve this equation. Now, if you look carefully at the data, carefully at the data A versus time, you notice that this dependence, this dependence, this dependence, this dependence is exponential. It is observed that the exponent M is nearly 1, showing that A equal to X power of minus of K D times T. So, what we are saying is that from the data that is given, it appears, from the data that is given, it appears that activity decays exponentially, it activity decays exponentially. How did we come to this great conclusion? We first plotted activity versus time and we found that it was not showing much of temperature, concentration dependence. So, then we decided that let us take the data and then determine what is the value of M, which is the exponent of activity. So, we simply assumed it to be 1 and then therefore, we plotted A versus time and then we find, when we plot A versus time, we find that our data looks something like this at each temperature. You understand what I am saying? What I am saying is that the data seems to suggest that the exponential decay is valid and this rate constant K D for different temperatures can be observed from the data given. So, you can determine K D, K D equal to what? K D 0 e raised to the power of minus of E D by R T. So, if you have data for K D at 3 temperatures, then you can plot them and find out what is the activation energy for deactivation. So, let us see how it looks like. So, we can make a plot of 1 by T versus ln of K D, ln of K D. So, what is the plot we would expect to see? What shall we expect to see? K D should be, so this should be aligned at 3 temperatures. Is it correct? So, the slope gives E by R. So, the slope gives E by R. So, the slope gives R. Is that clear? What we are saying? So, when we plot the data of K D versus 1 by T, where T is in Kelvin, T is in Kelvin. So, then we get a straight line, reasonably good straight line and slopes we can call this as E D. E D turns out to be approximately 10000 calories per mole. So, what is observed from the experiment? This is clear. Now, you can compare this E D, 10,000 calories per mole with various other kinds of data. A simple data which people would like to compare is the heat of vaporization of mercury chloride. Now, the fact remains is that the heat of vaporization of mercury chloride is much, much higher than it is 10000. And therefore, we should not be unduly perturbed by the fact that our activation energy is so different from a thermodynamic property. Please recognize that activation energy is a kinetic property. Therefore, it only gives you kinetic information. It does not give you equilibrium information. Therefore, do not try to compare the activation energy data with heats of vaporization. The heats of vaporization of mercury chloride etcetera are a thermodynamic property. And here, even though the deactivation may be very strongly related to evaporation of mercury, but the activation energy for that process may be quite different from the activation energy that you see in what is called as evaporation and condensation. Evaporation and condensation involves heats of vaporization and condensation, which is around 17, 18, 19, 20,000 or so. So, this is very different from heats of vaporization. So, this is the point I wanted to get across to you. Having said this, let us look at a more challenging problem. The more challenging problem in front of us is the following. You have a reactor. It has got a catalyst. Now, it goes to a separator. This is separator. So, it goes to a separator. From there, it comes out. Now, frequently, what is of interest to us is that we do not want to waste this product. So, it is recycled. It is recycled. Is it clear? Now, what is being said is the following. Let me just write down. You have some data. The reaction is A goes to B goes to A. So, k value equal to exponential of 8.8 minus 5000 by T, k rate constant k. k D equal to exponential 13 minus 2500 by T. Equal to constant k equal to exponential minus 19.5 plus 10000 by T. So, the question in front of us is the following. First one is questions. Obtain design equation process. Is that clear? What is meant by design equation for the process? It should relate conversion to the residence time to temperature to various features that is in our hand when we do a design. So, first what we want is that obtain a design equation for the process. Having done that, this is the first question. Second question is since catalyst deactivates. What is the reactor temperature of operation? Is this clear? We have two questions. First question is what is the process design? And then since catalyst deactivates, what is the reactor temperature of operation? After let us say after say we put some time after 30 days, some numbers. So, because you should know how you should run the plant with time. So, this is the idea. How do we do this? How do we do this? This is reactor. So, I will call this is 0, 1, 2, 1, 2 and 3. So, sorry. This is 0. This is 1. This is 2. This is 3. This is 4. I am sorry for that. Now, what we are now trying to say is that what is the process design means that we should be able to relate conversion to some variables which is in our hand. Let us first do that and then address various questions that might arise. First, let us do that. We want to relate conversion to various operating variables. Let us do that first and then all the rest will become quite easy. To do this, I will draw this once again. You have a reactor. You have a reactor. You have a separator. So, it is 0, 1, 2, 3, 4. A goes to B. This is the reaction. This is pure B. Therefore, C is pure A. We understand this. Why this is pure A? Because here it is pure A. Due to reaction, you will find a mixture here. But you record all the pure B. Therefore, only pure A comes here. Therefore, the concentration here must be C A 0. This is quite obvious. F A 1 is F A 0 plus F A 4. F A 1 equal to F A 0. F A 4 is same as F A 2. That is equal to F A 0 plus F A 1 times 1 minus of Y 2. Y is constant conversion defined with respect to position 0. Please, let us just understand this fully. What are we saying? We have this process and it is a time dependent operation. That also we have explained. In this time dependent operation, what we find is that we have to approve. This is the reactor. We have to find a way by which we will operate the reactor. This is separator. The whole time dependent activity here is that how do we change the temperature of the reactor so that we can get what we want? This is the point that we are trying to get across. Before we start with this problem, what we said is that we will do some elementary stoichiometry and set up all the relationships so that we can deal with them as we go along. F A 1 is F A 0 plus F A 1 1 minus of Y 2. When I say Y 2, it is meant, it is conversion at position 2. But in this case, conversion position 2, assuming that position 1 is a reference, then we conversion at 1 is 0. That is the assumption. This is only a reference. So, I do not think we should worry too much about this. We have done that. Now, what do we have? Now, we have to be able to write the design equations. After all, we want to write the design equation for the reactor. Let us write the design equation for the reactor. What is it? This is the plug flow reactor. So, what is D F A d V equal to R A? What is R A by definition? Rate of formation of correct. This is rate of formation of A and rate of consumption of A. So, the left hand side becomes F A 0 d X d V. F A 0, notice that F A equal to F A 0 times 1 minus of X. This we know. Now, because of this negative sign here, I will put a negative sign here. The K 2, what is C B? C B equal to F B divided by F B divided by V. So, C B equal to F B is what? F A 0 times Y 2. Correct. Exit. F B is at the exit. What is V? That is important. What is V? V equal to that I have done here. Let me just see what is V. Let me write here equal to V at the outlet. V at the inlet. This is V at the not V 0. It is not V 0. This is C B exit means it should be V 2. Is it alright? See concentration in a reactor, what comes out? The volumetric flow is V 2. Therefore, this is C B at any position. Therefore, V at any position. Therefore, V at any position. If you want C B at any position, it is F A 0 Y by V. Similarly, C A. C A at any position is equal to I have done all these things. Let me just write down instead of doing it again and again. Let me write it down. We will first calculate. What is F A 1 equal to F A 0 plus F A 4? Please recall this figure F A 0 plus F A 4 equal to F A 1. F A 4 equal to F A 2 and F A 2 is F A 1 multiplied by 1 minus Y 2. So, we will write that that is equal to F A 0 plus F A 2 that is equal to F A 0 plus F A 1 multiplied by 1 minus Y 2. So, this becomes F A 0 divided by Y 2 F A 1. Is it clear? Now, what is C A? C A 4 is C A 4 is same as C A 0 because B is fully recovered. Therefore, here we get purely. Therefore, C A 0 C A 4 is equal to C A 0. Now, we said V 4. What is V 4? Please notice here. V 4 volumetric flow at this point V 4 by definition is what? V 4 equal to F A 4 divided by F A 4 divided by C A 4 that is F A 2 divided by C A 4 equal to this is F A 1 into 1 minus of Y 2 divided by C A 4 is C A 0. And what is F A 1? F A 1 is F A 0 into 1 minus of Y 2 divided by Y 2. Now, what is V 1? V 1 equal to V 0 plus V 4. Notice here V 1 equal to V 4 V 0 plus V 4. I mean we say this volume balances are not correct, but if the densities are the same for all the streams this is that is why volume balances are used. Now, this is equal to V 0 V 0 plus just a moment F A 0 C A 0. I forgot C A 0 here. So, it is V 0 by V 0 into 1 minus of Y 2 divided by Y 2 that is equal to V 0 by Y 2. So, we got V 1 which is V 0 by Y 2. So, now therefore, we are now in a position see after all why we have done all these things. We want to say write the design equation for this reactor to be able to write the design equation for any reactor. We should know what is the concentrations at various positions. Now, if you want to write concentrations in any positions then you should be able to understand how to define conversions. So, here we have defined conversion with respect to position 1. Therefore, we need what is concentration at any point? What is C A here? What is C B here? This we should be able to tell. Once we are able to tell that you know we are in a position to write the design equation for the equipment. So, what is let us write everything quickly now. So, without losing too much time. So, we have C A equal to F A divided by V by definition at any position. In this reacting equipment we notice here that V 1 equal to V 2. There is no volume change and V 1 is equal to V 0 by Y 2. Therefore, V 2 is also equal to V 0 by Y 2. Any position the flow is V 0 by Y 2. That we already done. So, this therefore, F A at any position is equal to F A 1 into 1 minus of Y divided by V 0. V at any position is V 0 by Y 2. So, that is equal to F A 0 by Y 2 divided by V 0 by Y 2 multiplied by 1 minus of Y. That is equal to C A by C A 0 into 1 minus of Y. Similarly, C B equal to C A 0 times Y. So, what we have got? We have got the concentrations at this at any point. Therefore, now we can write the design equation and find the process design equation for this reactor. That is what once we know that we know everything. Let us quickly do that. So, we have the process design equation. Please recognize. We are writing the process design equation for this reactor keeping in mind keeping in mind the fact that the concentrations A and B are given by fairly simple relationships. So, we have D F D F A by D V equal to R A. So, F this is written as minus of F A 1 D Y D V. Equal to R A, which is minus K 1 C A 0 1 minus of Y plus K 2 C A 0 Y. Is that clear? Now, F A 1 is we know F A 1 is F A 0 by Y 2 D Y D V D Y D V. So, minus sign goes to the off. Therefore, it becomes K 1 C A 0 1 minus of Y minus of K 2 C A 0 Y or we can write this as D Y D V V 0 by Y 2 equal to we can simplify this and write this as K 1 times 1 minus of beta Y, where beta equal to K plus 1 by K. So, I will not show the details, but it is very obvious take K 1 common it comes nicely like this. So, we can integrate this and then the integrated form I just write down the integrated form. So, that we can go back and then look at the problem of our interest because we are going away from and then doing some. So, the integrated form the integrated form of this equation I will write it here itself by going to K 1 V equal to minus of B naught l n 1 minus of beta Y divided by Y 2 times beta. So, this is the this is the integrated form I will just box this. So, that you know we know what we have done. Now, the rest is fairly straight forward and the context also we cannot be forgotten and so on. So, what we have got here is that we have a deactivating catalyst and therefore, we have got the design equation for this reactor and then it is separator and so on. Now, the question that is in front of us is the following it says this catalyst deactivates which means what this particular K 1 which contains the activity. So, frequently it is better actually that is why I have we can put an alpha here we can put an alpha here which is the activity. So, that this alpha. So, basically what we got here is this term is this term is K 1 times alpha where K 1 is the rate constant and alpha is the activity. Now, what did we say we said that this activity decays and this activity decay function we have determined that is what we said you know in the example that we illustrated. We found out from our experiments how activity decays with time and then we said what is this at dependence on concentration in that example we said the concentration dependence is not there. Therefore, we determined the deactivation rate constant K d as a function of temperature. So, what we got that we just put it in the context what we got was what we got was the following. So, we got from our experiments K d equal to some K d 0 exponential of minus of E d by R t this we got from our experiments. So, essentially what we have got all the data we have got from our experiments and what is being asked of us is that now how do you run this process. How do you run this process on other words we want this product to be produced at a steady rate which means that as this catalyst deactivates even then this amount of product B product B does not change with time. How do you manage this we said yesterday we said this yesterday that if you look at this rate function r a which is typically it is a reaction velocity and there is an activity function which is a function of time. And of course, our rate function will have a composition what we pointed out yesterday is that we want to see that this term which is function of time this should not change as this activity keeps on going down. And how do you manage this we manage this by ensuring that this product this product does not change with time. How do we ensure this we ensure this by looking at what is a t a t we said it is some exponential of minus of k d times t where k d is given by this equation let us say star. On other words if I call this as star star and star. So, I have to keep this function this function let us say function of temperature this should be constant. So, as temperature changes this should be equal to a constant value. How do we do this we do this by changing the temperature at which we will run the process that means we will change the temperature at which we will run this process. So, that this product this product is always a constant. And if you do that if you do that this function remains a constant. And therefore, the composition at position 2 does not change. And this is how we take care of running processes that involve catalyst deactivation. So, in a sense what we have try to do today is that we took an example of a catalyst which deactivates we actually got the data from an experiment. We looked at the data and then we said this is a data in which concentration dependence is not significant. And therefore, it is an exponential d k because we found the exponent is m equal to 1. And that way we found that the rate function d m r d is simply k d times a with a minus sign. And therefore, it is exponential d k and therefore, we are able to incorporate this effect into the rate function r a. And therefore, we recognized that by incorporating that rate function into this r a we recognize that k d times a t can be kept constant in a process. How do we keep it constant by changing the temperature of the process or actually we increase the temperature? This temperature is increased because this is decreasing. As this goes down we keep on increasing this. So, that this product is kept constant. And that way we can ensure that this process runs. So, that the amount of this product that is produced at position 2 does not change with time even though this is a time dependent operation. The reactor is time dependent, but we adjust the temperature. So, that the time dependence is not seen at position 2. And therefore, it is not seen in our separator. Therefore, our process runs as though it does not recognize the effect of this decrease in activity. Thank you.