 We look at recycle effects in process in this lecture. We will take some examples to illustrate and go through the numbers to illustrate how we can take care of recycle and what kind of effects they will produce. We have talked about design equation for recycle reactor where we said that the reactor volume V is given by this integral where d x by minus of r a r a is the rate function. While doing this we also said that when r value of r is 0 then we get a P of r or plug flow recycle reactor. When r is tending to infinity it is very large we get a CSTR and therefore, r in between 0 and infinity you are able to approximate various types of devices which lie between a P of r and a CSTR. If you look upon CSTR as a very well mixed vessel and P of r as why there is no mixing recycle is one way of approximating any type of reactor which lies between a P of r and a CSTR no mixing to a very large amount of mixing. So, that is the great advantage of CSTR that you can get various degrees of mixing and therefore, appropriately take care of the requirements of a process. Now, what we would like to do today is to look at some examples where recycles become important. Say for example, if you have a reaction a going a plus b going to C and C is a condensable gas you would condense this gas and then collect this C and then recycle and reacted material back into the process. We may not this stream C may be there may not be there. Now, the fact that this happens that the certain amount of condensate comes out and the recycle is affected by the condenser we will have to take into account how the reactor design takes into account the effect of the condensate which you will take an example to illustrate. Now, there could be another situation that we will encounter in fact this is the most common kind of situation where you have a reactor where let us say reaction a going to b plus whatever let us say a plus b going to C whatever might be the situation. Now, we have a separator where we remove the product and then we recycle the unreacted. And this reactor need not be a recycle I mean it can be a CSTR. This reactor can be for example, a CSTR or it might be a PFR or whatever might be the situation of your interest. So, what we would like to do is to take some examples of these two situations and then go through the numbers to understand what are the new features that they bring into the process. First example I have taken here is a plus b going to C reaction take place at 570 C at pressure of 1 atmosphere. And the vapor pressure of C is about 0.2 atmospheres showing that you know it has a finite vapor pressure at 45 this is at 45 C. On other words you have to condense means you have to cool before we can condense the. And the example I have taken is the desired conversion is 0.5. So, and then we recycle part of it where the recycle ratio is given as 4. There could be examples in which stream 3 might not be there. So, that we actually recycle a substantial part to be able to look at the numbers we will have to go through elements of stoichiometry which you all understand. So, at any position inside the reactor which means we have this reaction equipment this is the reaction equipment. And then at any position in the reaction equipment we say that the if the recycle ratio is r r plus 1 times f a 0 multiplied by 1 minus of x a would be the molar flow at this point. Since a and b is taken as equimolar in this particular problem a and b is taken as equimolar. And therefore, b is also r plus 1 f a 0 1 minus of x a and c the amount of product formed if x a is the extent of reaction r plus 1 f a 0 x a is the amount of product that is formed. So, if you add up all this you will get the total molar flow at any point in the equipment is as what I have written here 2 times r plus 1 you can add these 2 here 2 times r plus 1 f a 0 and then r plus 1 f a 0 x a. So, this is the total molar flow at any point in the equipment. Now, what we have said is that component c is a condensable gas therefore, it would condense. And we can find out when the condensation occurs by putting this condition that at the point of incipient condensation that means at the point of incipient condensation we have the mole fraction of component c because the total pressure is 1 atmosphere at p is 1 atmosphere. And vapor pressure of c p star of c is 0.2 atmospheres therefore, when it becomes mole fraction equal to 0.2 it would condense therefore, we have based on the stoichiometry that this relationship should hold over x s is the conversion at which the condensation begins to take place. So, if you solve this and find that x s equal to 0.33 at x s equal to 0.33 condensation would begin. On other words beyond the value of x s equal to 0.33 you will find that all the c condenses and therefore, in the gas phase the value of c will not be more than x s that is the important point which we must take into account in the design of the equipment. So, let us see how it shapes up as you go along. So, what is our stoichiometric table for x less than x s there is no condensation therefore, a is r plus 1 f a 0 1 minus of x b is r plus 1 f a 0 1 minus x because a and b are equimolar. And all the c that is formed stays in the gas phase therefore, r plus 1 1 a 0 x all of them are in the gas phase. Therefore, the total molar flow becomes r plus 1 times f a 0 times 2 minus of x this is for x less than x s. Now, when x becomes greater than x s what happens we have here a and b are the same and c what happens is that c r plus 1 f a 0 x s and beyond that c does not go into the gas phase. So, the highest amount of c in the gas phase is given by r plus 1 f a 0 x s therefore, the gas phase beyond x s equal to 0.5 condensers therefore, it does not enter the gas phase. So, the total molar flow for the case of x greater than x s is given by this representation f t equal to r plus 1 f a 0 2 minus of 2 x plus x s. Now, what we want to do is to do is to is to recognize is to recognize that reaction a plus b going to c where c is a condensable gas and c is a condensable gas and in the reaction rate constant of 200 liter per gram volt second. So, we have a catalytic reaction catalyst here and then we are recycling it and recycling after removal of the condensate. This is condensate c once again recognize that the condensate stream 1 till it is still it starts to condense till it becomes saturated it does not condense basically that is the point that we have to take into account. Now, we have said all these. So, we quickly recognize that for the case for the case of condenser for x a equal to 0.5 for x a equal to 0.5 conversion we the by gas law we get v by v 0 is this that the total molar flow all those details we have done. Therefore, we can get the total the volumetric flow is given by this therefore, the concentrations are given by simple expressions like this. On other words if our rate expression is minus of k c a c b now we can substitute for c a and c b in this case c a equal to c b because the feed is equal molar feed is equal molar. And therefore, we can substitute for c a and c b in this rate expression and then integrate to find out what is the size of the equipment that is required. We have done this I am doing this again just to bring your attention an important feature. That means, we can find the residence time we can we find the molar flow rate at the inlet is 138 cubic meters concentration of the inlet is 3.6 10 minus 3. Now, if the reaction rate is 200 liters per gram mole second if the reaction rate is 200 liters per gram mole second then we can find out and based on this calculation that the residence time is 1.15 seconds. That means, if the reaction rate is as high as 200 liters per gram mole second then the residence time becomes 1.15 seconds. On other words if for this flow rate if f a 0 f a 0 equal to 1 kilo mole per hour and then kilo mole per hour and f b 0 equal to f a 0 if this is the situation then we can find out that there is the residence time that is required to process this is 1.15 seconds. Therefore, the size of the equipment the size of the equipment that is required to process 1 kilo mole per is 45 liters. What are we saying what you are trying to say here is that what you are trying to say here is the following. We have a reaction a plus b going to c and then this c is a condensable gas and as it is and then as a result of which you find that this stream the recycle stream only after it is saturated that condensation begins. And that effect of that condenser we have taken into account through the procedure that we have set out. When the reaction rate constants are as high as 200 liters per gram mole second we have just now shown that the equipment the size of the reactor required to the both 45 liters. Now, the point that perhaps is relevant to recognize is the following that is I mean this whole example has been taken being inspired by what happens in the sulphur dioxide plant. Sulfur trioxide is a condensable gas it is the vapor pressure I mean at 45 degrees significant amount of vapor pressure and so on. Now, if you are looking at a commercial catalyst I mean just making an estimate whose race constant let us say is 200 liters per gram mole second. Then to produce something like 3200 tons per day of sulphur trioxide we can easily calculate based on the numbers that we have taken it will require about 75 cubic meters of catalyst. On other words what we are trying to say is that a plant producing about 3200 tons per day of sulphur trioxide will actually use up about 75 cubic meters of catalyst. This is the kind of numbers that we have I mean that we estimate of course what happens in reality and so many other things may take place. But, if the rate constant is 200 liters per gram mole second to produce 3200 tons per day we will need about 75 cubic meters of catalyst and that is the point we are trying to get across. The second example which you would like to look at is also very common as you see in the process industry. The example I have taken here is a stirred tank this is a CSTR stirred tank. And after the reaction it goes to a separator where the product the reaction A goes to B it is reversible rate constants are given equilibrium constant is given. And in the separator product B is completely recovered and the unreacted is recycled. But, since product B is recovered the concentrations here may not be satisfactory therefore, you make up the concentrations in this stream and make it up to 100 kg per cubic meter as for specified in this exercise. And this recycled so that the concentration at position 1 is also equal to what we have have in the fresh feed stream. Now, the point we noticed here is that as we try and put the unreacted material back into the process there is a cost associated with separation and therefore, there is a cost of separation that goes back into the process. Therefore, you will find in general that the effect of this would be to give you a certain optimum at which you must operate. Now, this optimum might be defined in various ways in this particular exercise we have taken optimum with respect to cost of running the process. There would vary other ways in which you can define the objective function to do your optimization in this particular case just to illustrate what we might be able to do is to take the example of cost. So, accordingly in this particular problem what is specified is there is a process in which 12 cubic meters per hour of component A is entering the process. And it is mixing with the recycle stream which can which is suitably made up. So, that the concentrations are the same and then it is a mix after mixing it enters the equipment the reactor of volume 60 cubic meters is given as a stirred tank. And the reactor after reaction it is entering a separator where product B is completely recovered and the unreacted A is recycled after appropriate make up of the concentration streams. There is a stream 6 which is discarded there is certain amount of stream 6 is discarded and what is said in this problem is a fraction y of stream 4 is recycled not all of it is recycled a fraction y is recycled. And the idea is to find out what is the fraction y which would be optimum from the point of view of cost of running this process. So, this is the exercise that we want to do the one part of the reason why this example has been taken is that it illustrates how we understand the interaction between the reaction equipment and the separation equipment. Because every process involving a reaction you will find that you have to separate your product and the cost of separation and the cost of reaction and they have to properly balanced. So, that your process runs satisfactorily. So, let us try and understand this problem in some. So, what we have said to be able to solve this problem what we have done is we have done a material balance at position 1. So, what is happening at position 1 f A 0 if I call this stream A f A 0 is entering there is no B in the product. So, f A 0 is coming in and it is mixing with f A 5 that is what is written here f A 1 is f A 0 plus f A 5 statement of material balance. Now, since we are dealing with recycle in this we have defined conversions that take place in this equipment with respect to position 1. We find it convenient when you define conversion with respect to position 1 f A 2 f A 2 as you can see here it becomes f A 1 times 1 minus of alpha. That means a fraction alpha based on f A 1 has been consumed that is what is stated here. So, fraction alpha of f A 1 is consumed therefore, what emerges from the reaction equipment what emerges from the reaction equipment is the unreacted A which is f A 1 times 1 minus of alpha. Therefore, the amount of reacted material is f A 1 times alpha that will be the product B in this. We have defined conversion alpha with respect to position 1 this is with respect to position 1 that means with respect to f A 1. Now, what is also given is that f A 3 there is no product there is no pure B and therefore, A is 0 here it is given. What is f A 4? f A 4 would be whatever is present in f A 2 would come at f A 4. So, it is written as f A 1 times 1 minus of alpha and what is f A 5? f A 5 by definition fraction y of f A 4 is recycled. So, y times f A 4 is recycled and our object is to find out the fraction y that would optimize the operation of this process. So, what are we saying now? What we are saying is that f A 1 f A 1 is f A 0 plus f A 5 that is what is mentioned here. Now, f A 5 is fraction y of f A 4 is recycled therefore, y times f A 4 is f A 5 and we also said that f A 4 whatever is coming at 4 is what is present at position 2 which is f A 1 times 1 minus of alpha. Therefore, we have put here f A 1 times 1 minus of alpha. So, from this statement of material balance what we find is that f A 1 is f A 0 divided by 1 minus of y times 1 minus of alpha. So, we are able to determine what is f A 1 in terms of the of the process variable. What are the process variable? y and alpha see y and alpha are the variables of the process which we have to find out what is the optimum value of y that would run the process well. Now, we have C A 1 what is C A 1? Fee is coming at C A 0 and this stream 5 as per the problem statement it is made up. So, that the concentration here is same as concentration of the fresh feed stream. So, fresh feed stream concentration is 100 it is given as 100 kg per cubic meter and this stream is also made up. So, that it becomes 100 kg per cubic meter this is stream 5 this also 100 kg per cubic meter that is C A 5 is also 100 kg per cubic meter. So, that at position 1 our concentration is also 100 kg per cubic meter. So, that is what is being said here. Since, C A 1 is same as C A 0 what is V 1 or first principle F A 1 by C A 1 is V 1 F A 1 we have just now found out is F A 0 times 1 minus of alpha. Therefore, you can replace C A 1 in terms of C A 0. Therefore, V 1 becomes V 0 divided by 1 minus of y times 1 minus of alpha. Interesting point here is that we are able to relate every variable of interest to us in terms of parameters that we want to determine y and alpha we do not know why we do not know alpha. So, our interest is to understand how y and alpha affect our process and in fact, every process that you run you will find that these are the kinds of decisions that you have to make to get your process to work at the most optimum point. So, once again we hide the molar flow rate as a volumetric flow at 1 is given now in terms of volumetric flow rate at 0 divided by 1 minus of y times 1 minus of alpha, but y and alpha are variables of our interest. Now, we also need various other things to be able to put the process equations in the form that we can solve. So, I just put them down just to put it in the context. V 1 we have just now said is V 0 divided by 1 minus of y times 1 minus of alpha. Similarly, what is C B 2 concentration of component 2 is F B 2 divided by B 2. F B 2 we have said based on our understanding of the stoichiometry of reaction is F B 2 what is at this position 2 is simply whatever reacts F A 1 times alpha is F B 2. So, it is what is mentioned here F A 1 times alpha is F B 2 and what is the volumetric flow this particular reaction A goes to B there is no volume change. Therefore, whatever is the volumetric flow at position 1 we should expect that will be the volumetric flow at position 2 because there is no volume change in the reaction. So, therefore, V 2 equal to V 1 therefore, C B 2 becomes C A 1 times alpha and we have said just now that C A 1 is because C A stream 5 is made up to concentration 100 kg per cubic meter same as C A 0. Therefore, position 1 concentration is C A 0. So, we replace C A 1 as C A 0 therefore, we get C B 2 as C A 0 times alpha. What is C A 2? C A 2 is F A 2 by V 2 which is F A 1 1 minus alpha V 1 once again it is C A 0 1 minus of alpha. Notice that our effort is put every variable of interest in terms of parameters process variables which we are interested in so that we can do our optimization properly. What we have done so far is that we have expressed we have expressed concentrations at position 1 concentrations at position 2 volumetric flow at position 1 volumetric flow at position 2. That means, whatever affects the process the reactor all that we have expressed in terms of alpha and y. Therefore, we are now in a position to write the design equation for this reaction equipment. So, that we can understand how they come together to affect the process. So, this is what is written here it says C S T R equation says that V A the volume of C S T R is F A 1 times alpha divided by the rate at which the reaction occurs at position 2. This here that C S T R operates at exit concentration which is position 2 that is how it is written R A at position 2. Therefore, whenever we write the reaction rate expressions at position 2 we should take the concentrations at position 2 to write our rate expressions. That is what we have done here. So, F A 1 times alpha we have already shown that F A 1 is 1 this we have shown that F A 1 is F A 0 times 1 minus or y times 1 minus of alpha. So, we have replaced F A 1 in terms of F A 0 divided by 1 minus of alpha and what is R A 2 minus of R A 2 is the rate of chemical reaction. So, R A 2 is mentioned here. So, it is K 1 C A 2 minus of K 2 C B 2 rate at which this reaction is taking place with a minus sign is given by this. Because R A we all know this R A by definition is K 2 C B 2 minus of K 1 C A 2. This is something that we all know that the first order reaction that is given takes place then this is the rate function which will account for the. So, what we have done now we have replaced R A in terms of K 1 C A 2 minus of K 2 C B 2. This subscript 2 refers to position 2 where K 1 and K 2 refers to the reactions A this reaction A going to B this is reaction 1 and B going to A this is reaction 2. So, we have the size of the CSTR or size of the stirred tank is now given in terms of F A 0 alpha divided by our parameters y and alpha which define the process. So, we can see once again we are able to put everything in terms of the parameters of our interest. So, we can manipulate and simplify this and we get residence time in the reaction equipment is given by this relationship where alpha divided by y minus of y times 1 minus of alpha multiplied by the rate function appropriately modified. So, you have an equation now basically what we have done we have set up an equation for the chemical reactor which takes into account the inputs and outputs are appropriately. And that final form of that expression which relates residence time in terms of parameters which we are interested in our parameters are alpha and y. See that is how it you know the whole balances help us to do. Now, we said just little bit little while ago that whenever we have this kind of recycle there is an optima that exist because of cost cost of operation and cost benefits of producing the product. So, you will see in the optima and in this particular exercise cost functions are also given. For example, in this particular problem it is specified that the profit that you are likely to get from this process is the amount of product F B 2 that we make multiplied by the worth of that product which is beta B is the worth of that product. So, this is the benefit of doing this process and what is the cost of running the process it is given as it is equal to the material that is entering the separator that means volumetric flow entering the separator multiplied by the cost per cubic meter of solution entering the separator. What is the cost of operating this process it says it is equal to V 2 V 2 is volumetric flow here multiplied by beta 0 where beta 0 is the cost of operation of this process expressed per cubic meter of solution entering the separator this is what is given. Now, we have this cost function the profit function F B 2 times beta P and minus of V 2 times beta 0 both are given beta 0 is given beta B is given. Therefore, we have now a cost profit function which very nicely stated and you notice once again that F B 2 is F A 1 alpha where V 2 V 2 is V 1 we know V 2 is V 1 we already mentioned that V 2 is V 1 and V 1 is in terms of V 0 we already done that 1 minus of y times 1 minus of alpha we know all that. So, you can replace everything in terms of numbers that you know in terms of parameters that you know. So, that you finally, end up with an equation that defines the profit that you will get from this process profit that you will get from this process once again this profit function is in terms of two parameters y and alpha. So, recalling what we have sent so far we have done little earlier and recalling once again that what we have done so far this quickly recap because important to understand why we are doing this we have a reactor we have a separator. Now, the fact that the reactor and separator interplay and there is an optimum value at which the process if it operates then the benefits are the highest to quantitate this this problem has been taken. So, what we have done first done is that we have done a balance for the CSTR and that balance told us that the residence time is related to alpha and y by this relation equation 4. Now, then next thing what we did is that we have set up a profit function which says the profit depends upon alpha and y as per equation 5 you have equation 4 which relates to the process equation 5 which relates to the cost or profit. Therefore, now we have to only manipulate equation 4 and 5 to see how we can handle the whole issue. Interestingly, if you look at equation 5 we find that this term 1 minus of y times 1 minus of alpha and appears and the same time if you look at equation 4 this term 1 minus of y times 1 minus of alpha appears. So, on other words it is possible for us to find out the value of 1 minus of y times 1 minus of alpha from equation 4 and then substitute in equation 5. The interesting thing is that moment we do that moment we do that y disappears from our formulation. So, moment we replace the term 1 minus of y times 1 minus of alpha from equation 4 into equation 5 this what I have tried to do here from equation 4 here I have expressed it in terms of alpha and tau and others. So, you find here 1 minus of y times 1 minus of alpha now depend only an alpha and tau tau is a process variable which in this case is specified by the problem statement. So, what we have been able to do is that by appropriately formulating the problem in a hand you are able to eliminate one of the two process parameters which is y. So, that you are able to get everything in terms of one variable which is alpha. So, by substituting for this term 1 minus of y times 1 minus of alpha which is equation 6 into equation 5 what is equation 5 or equation 5 is the profit function. Now, you are able to get the profit function the whole profit function if you eliminate this term 1 minus of y times 1 minus of alpha the profit function now looks like this which is essentially a function of alpha. So, now what we have done we had we had a problem in which there were two parameters y and alpha which determine the process through the design equation for the CSTR we could knock out one of them therefore, your profit function now depends only an alpha. Now, the question problem statement is that find the value of y which maximizes the operational profit of the plant that is the question that we have to solve. So, what we have done we have now eliminated y, but the question is what is the value of y which maximizes p that is the question. So, but what we have done is that we have eliminated y we have an equation which represents p in terms of alpha we are able to maximize profit with alpha as our variable. So, we have to now show the maximizing p with respect to alpha is same as maximizing p with respect to y. Now, to do this it is not very difficult to do, but to do this we have equation 6 which we already set up we have already set up equation 6 as you all know. So, from this equation we can express y in terms of alpha. So, you can I have done this here. So, I have done y in terms of alpha now if you do d alpha let us say we have to show that d p d y I have written it here see we have to do d p d y is same as d p d alpha because d alpha d y is finite. Since d alpha d y is finite if we maximize p with respect to alpha it is same as maximizing p with respect to y I hope it is clear let me run through this once again we spend a lot of time. So, far trying to express our process in terms of y and alpha then we eliminated y and got the whole profit function in terms of alpha we have done that here by putting all the numbers appropriate to the process. Now, the question remains that is the maxima of p or the profit in term with respect to y same as the maxima with respect to alpha the answer is that if d alpha by d y is finite in the region of our interest which it is in this case you can check for yourself. Therefore, d p d y is same as d p d alpha equal to 0. So, if you put d p d y d alpha equal to 0 the point of maxima that you will get is also the same point of maxima corresponding to d p d y. So, what are we now saying to say what we are trying to say is that if we simply maximize p with respect to alpha it is as good as maximizing p with respect to y. So, what we have done here you do d p d alpha when you do d p d alpha what do we get when you do d p d alpha we find that the value of alpha at which p takes the p is maxima comes out to be 0.447. So, on other words what we are saying is that in this process in this process the value of alpha is the fraction of f a 1 which is converted it is 0.447. If you operate at that value of alpha then this whole process the profit of this process being the highest. So, that is what we have been able to conclude by looking at this process. Now, the question that is being asked is given that value of alpha is 0.447 what is the conversion that we can expect from this process and what is the value of y that we can get from this process. What is the fraction y is the fraction of stream 4 which you have to recycle we still do not know y we only know alpha. How do you find y? y we can find from our equation 6 if equation 6 if you know alpha you can find y because this is equation 6 and in this process all the numbers all the numbers are specified you can see here tau is given as 5 hours k 1 is given as 0.4 k 2 is given as 0.1 per hour f a 0 is given as 1200 c a 0 is given v 0 is given. Therefore, you can get the value of y from this equation. So, you can put a numbers and see how it looks I have done that and then it turns out that y is 0.89 this is the number y that is 0.89. So, what are we getting now if you know alpha is 0.447 y is 0.89. Now, if I ask you what is the conversion that we can get from this process? How is conversion defined? Conversion is defined as the amount of product that you make divided by amount of feed that you put into the process. So, how much product have we made is f a 1 times alpha? How much of feed have you put in it is f a 0 and what is f a 1? f a 1 is 1 by 1 minus of f a 1 f a 1 is f a 0 divided by this f a 0 gets cancelled. Therefore, this conversion that you will expect from this process is alpha divided by 1 minus of y times 1 minus of alpha. If you put a numbers alpha is 0.447 and then therefore, y is 0.89 you put all the numbers you will get the conversion to be expected in this process is 0.88. So, just to put this in the context the context is important that if you have a process in which you have continuously a processing in a reaction equipment CSTR and then you are removing the product in the separator and then you are recycling only a part of it because that is operationally more profitable. The fraction y that maximizes the profit of this plant is coming out of this formulation that I have discussed with you just now. So, depending upon the worth of this product and the operating cost how these two come together if they change of course, the value of alpha will change therefore, value of y y will change and therefore, the conversion from the plant will also change. So, these numbers essentially determine at what point you must operate the process. So, this is what we have tried to get across in this exercise of recycle effects in the process. Now, before we conclude let me just draw your attention to one more issue regarding autocatalytic reactions. So, when we have an autocatalytic reaction A going to B and R A is minus of K times C A and C B. We said that if this reactor if this reactor is a stirred tank with this stirred tank which means comes in and goes out then R A is minus of K times C A and C B then it is not necessary to have C B 0. C B 0 can be C A 0 is finite, but C B 0 can be 0 even then the reaction will proceed. We have explained this last time when we met when we said this is because in this when you look at the unsteady state during the start up. As long as during start up you have put in some C B into the process then stirred tank remembers that start up addition of C B and therefore, that process will be able to get initiated because of the presence of even traces of C B in the equipment. In other words, the stirred tank is an instance of a reaction equipment where you do not need feed sorry product to be primed into the feed as for an autocatalytic reaction, but if the same thing is to be done in a in a PFR without recycle. If you have to do in a PFR same autocatalytic reaction because there is no there is no mixing here and therefore, the feed does not contain product therefore, you will find the reaction will not take off. So, PFR is not effective not effective for for autocatalytic reaction autocatalytic. CSTR and recycle reactors are extremely valuable for autocatalytic reactions because you are able to in a in a recycle reactor this is a recycle reactor. You are able to recycle the product and therefore, the reaction gets initiated. Therefore, this very very valuable from the point of view of autocatalytic reactions stirred tank since the recycle because of recycle because this infinite recycle that is what we have understood this you find is very effective from the point of view of autocatalytic reactions. I will stop this this process.