 Welcome back. In the last lecture, we looked at the mechanism of mass transfer from a gas to an agitated liquid and we considered two theories which seek to explain this phenomenon. The two theories, the first of which was the film theory and the second the surface renewal theories. These two theories start from very different standpoints and the mechanisms that are postulated for the mass transfer process are also quite different and not surprisingly the mathematics also works out quite differently and we end up with two different expressions for the mass transfer coefficient from a gas to a liquid. Now, in the ultimate analysis neither of these theories is able to predict the mass transfer coefficient from first principles, but as we shall see the main utility of the theories is in their ability to predict the effect of mass transfer, effect of chemical reaction on mass transfer. So, this is what we are going to concentrate on in the present lecture as well as in some of the lectures to follow. So, the situation is something like this. There is a gaseous solute A which is getting absorbed into a liquid B, liquid solution and liquid contains a component B with which A can react. So, we write the stoichiometry of the reaction as A from the gas phase reacting with B from the liquid phase giving rise to C and although it does not really matter all that much we will assume that the C remains dissolved in the liquid phase and B and C are stoichiometric coefficients and we shall assume that this reaction has an intrinsic rate which is first order in A and first order in B. As usual as we did in the case of the physical mass transfer we will assume that there is no gas phase resistance the entire resistance to the mass transfer is concentrated on the liquid side. So, we shall loosen these assumptions towards the later stages of this topic both the assumption with respect to the orders of the reaction as well as with respect to the non-existence of gas phase resistance and we will see the consequences of those at that point, but at this point we will make this make these assumptions in order to keep the focus on the essential aspects and keep the other details simple. We will further assume that the film theory is good enough to explain this process of mass transfer with chemical reaction and while we have said earlier that in view of the available evidence the surface renewal theories are to be trusted more than the film theory. In view of the simplicity of the film theory we will do the development first in the framework of the film theory and later on we will see what differences come about if it is the film if it is the surface renewal theories that we invoke and we shall also give a justification for why film theory is still useful in spite of the fact that the prediction it makes about the way the mass transfer coefficient depends on diffusivity is not borne out by the available experimental data ok. So, we shall proceed with the analysis of this process. So, A is getting transferred from the gas phase to the liquid and on the liquid side when it diffuses into the liquid side it reacts with one of the components of the liquid phase which we have called as B. The concentration of this B in the bulk liquid we take it as C B B and C B B is the concentration of B in liquid bulk that is to say outside the film and C A star is the concentration of A in equilibrium with the prevailing partial pressure and within the film theory parlance therefore, this is also the concentration of A at the interface. So, with those notations having been agreed upon we can write the diffusion reaction equations that govern the diffusion of A into B A into the liquid rather in the following manner D A D squared C A by D x squared equals K C A C B which is the same equation as we had worked with before except for the presence of this reaction rate term. So, what this is saying is that if you look at the film with this is the gas liquid interface that is the gas side and this is the liquid side and that is the film. So, in any differential slice of the film there is a rate at which the solute is entering by diffusion there is a rate at which the solution is leaving by diffusion the difference between these two rates is consumed by chemical reaction within this slice. So, we had a 0 here when there was no reaction. So, whatever was entering was equal to the rate at which it was leaving, but now there is a rate at which the material enters at x and there is a lower rate at which the material leaves at x plus delta x and this difference is consumed by reaction. That is all that is happening because this is a steady state theory and there is no accumulation. So, input minus output equals consumption. So, that is what this reaction is this equation is saying. Now, this equation cannot be solved by itself because it contains the concentration of B and it is possible that the concentration of B also has a profile within the film. In order to describe that profile we have to write a very similar equation for B. It is almost identical except for the fact that we use the diffusivity of B instead of the diffusivity of A in this equation. So, this is input minus output of B into the differential slice and the reaction rate term contains the stoichiometric factor in addition to the rate term that was present in the equation for A. That is because the rate of consumption of B is lower case B times the rate of consumption of A. Consistent with the postulates of the film theory, the boundary conditions can be written in the following manner. At x is equal to 0, we have C A being given by C A star and at x is equal to delta. This is delta C A is given by C A bulk and when it comes to the boundary conditions for B, at x is equal to delta the concentration of B is C B B as we have already assumed. At x is equal to 0, we make the assumption of non volatility that is to say we assume that B has no tendency to pass over onto the gas side by vaporization. What this means is that the flux of B at the gas liquid interface is 0 and if you invoke a fixed law for the flux, what this means is that we are saying D C B by D x is equal to 0 because it is a product of this with the diffusivity that determines the flux and since we are saying that the flux is 0, this gradient must be 0 at the gas liquid interface. So, this is the formulation of the problem that is the diffusion equation that governs the transport of A. This is the diffusion equation that governs the transport of B and these are the relevant boundary conditions with which these equations have to be solved. Now, if you look at these equations, there are concentrations, there are distances, there are concentrations and there are distances and these are in different units and they have different magnitudes and therefore, in order to analyze the effect of this chemical reaction on the mass transfer and particularly in order to understand the effect of chemical reaction as the chemical reaction rate is varied relative to the diffusion rate, it is far more useful to treat these equations in a dimensionless form. Let us do that by defining the following non-dimensional variables. Let us call lower case A as the non-dimensional concentration of A which is obtained by dividing the concentration of A divided by the concentration of A at the interface. Similarly, small B is concentration of B divided by concentration of B in the bulk, maybe what we should do is in order to avoid confusion between this B and the stoichiometric factor B, we will conserve the stoichiometric factor as nu. In other words, we will go back and change this stoichiometric coefficient to nu, the Greek letter nu so that there is no confusion between the two sets of symbols that we are using and we are left with the independent variable that is distance and we shall let zeta, the Greek letter zeta stand for non-dimensional distance which we obtain by dividing the value of x by the maximum value that x can take namely delta. So, what we see in this scheme of non-dimensionalization is that each variable is being divided by the maximum value it can take and therefore, the non-dimensional variables will have a value between 0 and 1. So, they are all comparable. So, earlier in the dimensional form they need not have been comparable, right now they are and substituting these into the differential equations, we can do the non-dimensionalization as follows, d A we have got d C A that is d A multiplied by C A star, this is the second derivative and we have got the second derivative of x which becomes delta squared times d squared zeta equals K C A is C A star times A and C B is C B bulk times B. So, the C A star will get cancelled and we can rearrange this equation to read as follows delta squared K C B B divided by d A multiplied by A B. So, I shall call this group as m and write this equation as d squared A upon d zeta squared equals m A B. So, what is the let us stop for a moment and consider the meaning of this dimensionless group that we have defined here m which is delta squared divided by d A multiplied by K C B B and if you look at this K C B B is a pseudo first order rate constant for the reaction and the reciprocal of the first order rate constant has the significance of a characteristic time for reaction that is the time that is required by the reaction to proceed to a significant extent. So, 1 over K C B B is the characteristic time for reaction and delta squared by d A is the characteristic time for diffusion because delta squared is the square of the distance over which the diffusion is occurring and d A is the diffusivity which has units of centimeter square per second or meter square per second and therefore, this is the characteristic time for diffusion. In other words, this parameter m compares the characteristic time for diffusion with the characteristic time for reaction or it is comparing the reaction rate with the diffusion rate. So, if m is large it means that the characteristic time for diffusion is much larger than the characteristic time for reaction or reaction is much faster than diffusion. If m is small it means that the characteristic time of diffusion is much smaller than the characteristic time for reaction which means that diffusion is a much faster process than reaction. So, by varying m we can look at reactions of varying grades relative to the rate of diffusion. In the literature square root of M has come to be called as Hata number and in some books you will see this being referred to as H A and we shall use the nomenclature M because much of the literature continues to use that kind of nomenclature. So, we have got the non dimensional form of the diffusion equation for A in that manner. We can do the same thing for the diffusion equation for B and here we get D squared B divided by delta squared here and D squared zeta D zeta squared equals nu k C A star A C B B B. So, this is the concentration of A and this is the concentration of B and that is the rate. So, in order to express this equation also in terms of this dimensionless number that we have defined we will group these various quantities that we have obtained here in the following manner multiplied by nu D A C A star divided by C B B into A times B. So, we see that in the equation for B this familiar quantity has made its appearance there is an additional dimensionless group that has arise. Let us see what is the meaning of this additional group that has made its appearance now. So, let us call Q as the reciprocal of that group. So, by Q let us denote this quantity D B C B B divided by nu D A C A star. So, what does this represent? D B C B B divided by D A C A star is the relative rate of supply of B to the rate of supply of A that is because you know the maximum supply of A can be written in that manner which is nothing but the mass transfer coefficient of A multiplied by the maximum driving force that is possible. And the similarly the maximum rate of supply of B would be D B C B B divided by delta. And so it is the ratio of these quantities that going to the numerator and this coming to the denominator that is appearing here that is being divided by the stoichiometric factor B which is the number of moles of B that is required by the reaction per mole of A. So, we can regard this as the relative rate of supply B to A divided by relative requirement B to A. So, because of this definition we can associate the significance of relative abundance to the parameter Q and we will continue to denote by this letter Q. So, with that definition we can now write the equation for B as D squared B upon D zeta squared equals M divided by Q multiplied by A times B. So, we have the equation for A we have the equation for B it is a simple matter to write the non-dimensional boundary conditions x equal to 0 corresponds to x equal to 0 corresponds to zeta equal to 0. And at that point small a equals 1 and the equation for B is D B by D zeta equal to 0 and at zeta equal to 1 which is the end of the film small a is A B which is nothing but the concentration of A in the bulk divided by the concentration at the interface and the concentration of B in non-dimensional terms is 1. So, we have got the equation for we have got the equation for A we have got the equation for B and we have the initial and boundary conditions and now we are ready to look at the implications of these equations on the rate of mass transfer. So, in order to understand the various things that can happen when reactions of various velocities take place in conjunction with mass transfer instead of directly going ahead and solving these equations by brute force method let us do some analysis and see what happens. So, we first note one thing that the value of Q is almost always much greater than 1. This is because if you look at the definition of Q it has D B and D A which are similar in magnitude it has the concentration of B which is rather large B could be if you take an example such as the absorption of carbon dioxide in a solution of monoethanol amine B would be the concentration of monoethanol amine in the aqueous solution that can be very very large and C A star is the equilibrium concentration of carbon dioxide in water and this is usually much smaller than the concentration of B in the bulk. So, this is the case for most gas liquid systems whether you take oxidation reactions in which oxygen is the gas or hydrogenation reactions in which hydrogen is the gas or gases like carbon dioxide or H 2 S or whatever. Usually the concentration of the gaseous species on the liquid side at equilibrium is a much smaller number as compared to the concentration of the liquid phase reactant. There that is to say this in this expression the C B B is almost 100 times or more as compared to C A star and nu. Once again is a stoichiometric factor which is a small number like 1 or 2 or 3 etcetera. Therefore, the net result is that the value of Q is governed by the ratio of C B B to C A star which is usually very large. So, this can be of the order of 100 or even higher and therefore, we make the point that this is usually the case that is Q is much smaller than much greater than 1. So, let us look at reactions in the increasing order of the values of m. In other words in the increasing order of the relative rate relative to diffusion. Let us first take the case of m being small in particular we will take the case where m is much smaller than Q. Q is a number that is much higher than 1. So, it is not very difficult to find values of m which are much less than Q. So, if you if this is the case looking at the equation for B looking at the equation for B where we have the ratio of m to Q and we are saying that m is much less than Q. We can make the approximation that D squared B upon D zeta squared is approximately equal to 0. Now, if this is the case then it means that D B by D zeta is a constant at all values of zeta. We know that at zeta is equal to 0 D B by D zeta equal to 0 because of the non volatility condition and therefore, this is equal to 0 everywhere which means that B is equal to 1 everywhere that is all values of zeta. What we are saying here is that under these conditions where the value of m is much less than Q what happens is that the concentration of B is uniform throughout the film and the bulk concentration prevails right up to the interface. So, if that is the case looking at the equation for A we immediately have D squared A upon D zeta squared. This is the equation that we are looking at and because B is equal to 1 we can write this equation as D squared A upon D zeta squared equals m A. In other words now this equation has become independent of B and this can be solved by itself and what we see is that the reaction is really behaving like a first order reaction because the dependence of B does not matter B concentration being constant everywhere. So, this not surprisingly therefore, this condition that we are considering that is m being much less than Q is called as the case of pseudo first order. So, we are going to be considering first the reactions for which m is far less than Q and these are called the pseudo first order reactions. Under this once again we will further make the assumption that the reaction is slow enough we will call these as slow reactions for which the value of m is not only much less than Q it is much less than 1. So, this is a more stricter condition because Q as we have already remarked is much larger than 1. So, we are now demanding that m be not just less than Q it must be far less than 1. So, in the equation that we had earlier we had D squared A upon D zeta squared equals m A for the general pseudo first order regime. So, this is very very small in this case. So, what are we left with we are left with an equation that is saying D squared A by D zeta squared equal to 0 with the conditions at zeta equal to 0 A is 1 and zeta equal to 1 A is A bulk. If you look at this this is no different from the equation that we had written in the physical mass transfer case except that in the last lecture we did not non dimensionalized equations here we are talking about non dimensional equations. But this is exactly the same equation as obtains for the case of physical mass transfer. So, what we are saying is when the reaction is slow enough that the value of m is far less than 1 the reaction really has no effect on the mass transfer process. It is unable to influence the concentration profiles in the diffusion film and the concentration profiles in the diffusion film are governed by the same equation as if there were no reaction which means that the concentration profile will remain linear. If the concentration profile remains linear then the mass transfer coefficient is given by D A divided by delta as in the case of physical mass transfer and the rate of absorption is exactly same as the rate of absorption in physical absorption case for the same driving force. So, what is it that the reaction is doing in this case as we shall see the reaction has the important task of actually setting the driving force in this case. So, how do we show this in order to see this we make a balance on the bulk. So, what we are going to consider is the role of reaction in slow reaction regime these are called as reaction regimes. So, we shall adopt the same nomenclature the role of reaction in the slow reaction regime is certainly not to influence the processes in the diffusion film rather it is to influence the magnitude of the driving force. How does it do this we have seen that the reaction is slow enough not to take place to any significant extent within the diffusion film. So, where is the reaction taking place the reaction is now taking place outside the film in the bulk of the liquid. So, if we were to understand the role of reaction we have to draw a balance of the bulk of the liquid and say that the rate at which a is being transported into the bulk is equal to the rate at which a is being consumed by chemical reaction there and it is this balance that sets the value of the value the concentration of a in the bulk. Now, before we do that let us take a value of the let us take a look at the typical values of the numbers that we are talking about here. So, we have invoked the diffusivities and these diffusivities usually have the order of magnitude of about 10 to the power minus 9 meter square per second in SI units. We have talked about the mass transfer coefficient K L and this usually in process equipment has magnitudes of the order of 10 to the power minus 4 meter per second which means that the value of delta which is nothing but d a divided by K L in the film theory is of the order of 10 to the power minus 9 divided by 10 to the power minus 4 that is to say order of 10 to the power minus 5 meters that is approximately the thickness of the film. We will also need the to define another quantity this is we call it as a hat and we shall call this as the interfacial area per unit volume of liquid this in process equipment usually is of the order of a few hundred square meters per meter cube of liquid. So, let us say this is of the order of 10 square meter square per meter cube. Now, if you look at the this quantity film volume per unit volume of liquid the volume of liquid that is inside the film as a fraction of the total volume of the liquid then this is nothing but a is the interfacial area per unit volume multiplied by delta is the thickness of the film. So, that it is this and this is of the order of 10 to the power minus 5 multiplied by 10 square. So, this is of the order of 10 to the minus 3 which means that the film volume the volume of liquid within the film is only about 0.1 percent of the total volume of liquid there is. So, when counting the volume of liquid there is in the tank you will be making no significant mistake if you neglect the volume of liquid in the film altogether because that is only about 0.1 percent having made that point now let us draw a balance for a in the bulk. We will assume that the bulk liquid is in a state of quasi steady state. In other words if concentrations and other conditions are changing in the bulk at all they are changing sufficiently slowly with respect to the mass transfer time scales that as far as the mass transfer process is concerned the conditions in the bulk may be considered to be stationary. Under these conditions the bulk is receiving some amount of a from the film and that amount of a must be being consumed by reaction because we are not allowing for any accumulation. And therefore, it is K L multiplied by C A star minus C A B this is the rate of mass transfer into the liquid this is multiplied this is per unit area multiplied by a hat V L. So, this is the total amount of area that is present. So, this is the number of moles per second of a that are being transferred from the gas to the liquid gas to the liquid at the interface and since there is no reaction in the film itself all of this gas is being passed into the bulk as well. So, this must equal the rate at which reaction is taking place in the bulk which is K times C A B divided by C B B per unit volume multiplied by V L. So, this is the equation that governs the state of affairs in the liquid bulk. So, V L can be cancelled and if you rearrange this equation we have C A B divided by C A star the quantity that we have called as A B this is equal to 1 divided by 1 plus K C B B divided by K L A. So, now we see that there is an additional dimensionless group in the works and that is this quantity here. So, as usual we will stop and consider what that means we will call this quantity as P and this is K C B B divided by K L A. Notice that both the numerator and denominator have units of reciprocal time and therefore, P is once again dimensionless K C B B is the reaction rate constant is a first order reaction rate constant for the you know second order reaction considered in pseudo first order terms and K L A is the mass transfer rate constant if you like. So, what this is comparing is the reciprocal rate constant is the characteristic time of reaction time of reaction and reciprocal mass transfer coefficient is the characteristic time for mass transfer. So, this quantity P is comparing the rate of mass transfer with the rate of reaction if the reaction is much faster than the mass transfer process then P has a large value if reaction is much slower than mass transfer P has a much smaller value. So, this is P and therefore, we can write this earlier equation as 1 over 1 plus P. Now, this is where the role of the reaction comes in because the role of the rate of the reaction determines the value of P and we can imagine two extreme situations one in which P is much less than 1 that is the value of reaction rate is much smaller than the value of the mass transfer rate. In this case what this equation says is that P can be dropped from the denominator and this implies A B is almost equal to 1 which means that under these conditions if P is so small then the concentration of A almost attains saturation. The other extreme is P much greater than 1 and under these conditions P will dominate in the denominator and therefore, A B will tend towards very small values. So, there are these two situations. Now, what happens to the rate of absorption in these two situations? The rate of absorption in units of moles per second is nothing but the mass transfer flux multiplied by the interfacial area or moles per centimeter cube per second is mass transfer flux multiplied by interfacial area per unit volume and this is given by K L A writing this equation again writing this equation again. So, this is K L A C A star minus C A B or under the quasi steady state conditions this is equally equal to K C B B C A B. So, this can be written as R A equals K L A C A star into 1 minus A B or this is also equal to K C A star C B B multiplied by A B. So, if P is much less than 1 that is this condition here then A B is equal to 1 A B is nearly equal to 1. So, this part of the equation will tell you that the mass transfer rate is operating with minimum driving force. So, if you want to calculate the rate using this expression it is not going to be very accurate because the driving force is so close to 0 that it is difficult to estimate. On the other hand it is very convenient to use this part of the equation because A B is close to 1 here and therefore, R A can be written as K into C A star C B B. In other words the rate of absorption is equal to the maximum rate of reaction that you can expect in the system because this is the concentration of B and this is the maximum allowable concentration of A under the circumstances. So, the rate of absorption being completely determined by the kinetic rate here. So, this would be considered as a case of kinetic control. On the other hand the other extreme of P being much greater than 1 this situation that we talked about A B is equal to 0 or A B is nearly equal to 0 which means it is not going to be very realistic to calculate the rate from this expression because the reaction driving force is very inaccurately estimated here. It is so close to 0 that it is not very easy to measure and therefore, on the other hand it is very easy to calculate the rate from this expression here. So, we write R A equal to K L A C A star. Now, if you look at this this is the maximum rate of mass transfer that is possible in the system because under these conditions the driving force takes the maximum value it can. Therefore, the rate of reaction being completely determined by the rate of mass transfer here this can be considered as a situation of mass transfer control. So, let us so what happens in the situation where P is neither much smaller than 1 nor much greater than 1 in that case we can calculate R A in general we can calculate R A as K L A C A star into 1 minus A B that is 1 by 1 plus P as we have derived earlier or this is given by K L A C A star multiplied by P times 1 plus P. So, this is how we can calculate the rate of reaction or rate of absorption in the slow reaction regime. So, let us step back and then summarize what we have seen so far. So, we have seen the following first is we are considering cases of M far less than Q not only that additionally we consider M far less than 1 and this we are calling as the slow reaction regime. Under this there are two extreme situations there is an additional parameter that comes about P which is nothing but K C B B divided by K L A P much less than 1 and P much greater than 1 these are the two extreme situations and this is called as the kinetic control or we will use the term kinetic sub regime and we will use the term diffusional sub regime for the other extreme where P is much greater than 1 and therefore, you have got P is much less than 1 is the case where the liquid get saturated that is the kinetic sub regime P much greater than 1 is where the liquid is starved of the gas and that is the diffusional sub regime. So, we have the pseudo first order regime under that the slow reaction regime and under that two extreme situations kinetic sub regime and diffusional sub regime. Of course, it is possible for the value of P to be somewhere in the middle and in that case we have got no particular rate process controlling and we have got this equation K L A P divided by 1 plus P C A star as giving the rate. Now, before we go to other matters we should examine this case of diffusional sub regime in a little more detail because it is rather interesting. So, we shall call it the interesting case of diffusional sub regime. Why is it interesting because we have on the one hand said that the reaction is slow and we have ensured this by saying that M is far less than 1 on the other hand we are saying that the reaction is fast in some sense because we are saying that the value of P is much greater than 1 and P contains the rate constant recall that it is relative to the mass transfer coefficient the value of the rate. So, is this possible are we talking about two opposite things here can the reaction on the one hand be slow enough to ensure that M is much less than 1 on the other hand at the same time can it be fast to ensure that P is much greater than 1. In order to understand this we examine the definition of P which is this and we can write this in the following manner K C B B divided by K L divided by K L divided by K L multiplied by a delta there and multiplied by a delta here. Now, what is delta divided by K L? So, this is P. So, P is delta K C B B and K L is nothing, but D A divided by delta right a delta. So, looking at this is nothing, but the value of M divided by a delta. We have already made the point that this a delta is order of 10 to the power minus 3 in most process equipment because a delta is so small it is possible for M to be also small and still P to be large. In other words if M is small, but still much greater than 10 to the power minus 3 P will be large and you will always have a diffusional submerged. So, as you consider the reactions in order of ever increasing values of M you have a situation where M is so much less than 1 that in spite of being divided by a quantity of the order of 10 to the power minus 3 P remains much smaller than 1 and then that is when you get the kinetic sub regime. As you go to increasing values of M, M becomes somewhat large that M divided by a delta still is fairly large value. So, you have M less than 1 and therefore low reaction is ensured, but P is much greater than 1. Therefore, the reaction is fast enough compared to the KLA process that the diffusional sub regime results. So, because of the circumstance that in most process equipment a delta has this small value of the order of 0.001 you almost always have a situation of diffusional sub regime before the reaction becomes fast enough to affect the processes in the diffusional film. So, this is a statement of some importance. So, what we have said is that in normal process equipment in cases where a delta is of the order of 10 to the power minus 3 you always have a diffusional sub regime before reaction starts influencing film processes. A corollary of this is that what happens in diffusional sub regime? In diffusional sub regime the concentration of A goes to nearly 0. So, what we are saying in as a corollary of this statement is that in practice as you consider reactions of larger and larger values of M or larger and larger velocities they are always or rather the bulk concentration of A always goes to 0 before reaction in the film becomes important. That is to say so far we have neglected the occurrence of the reaction in the film and remember that we solved the equation D squared A by D zeta squared equal to 0. We can continue to do that as the value of reaction rate increases till the reaction rate is fast enough that the bulk concentration is nearly equal to 0. So, if the reaction rate increases beyond this value only then you need to consider the reaction rate term in the diffusional equation. So, what happens in those cases we will see in the next class.