 In this final lecture on surface reactions, we will discuss today complex formation and the phenomenon of penetration into monolayers. We will also look at some aspects of thermodynamics of penetration and adsorption from vapor phase. This will be followed by a little introduction to mass transfer in the interfacial systems. So, we go back a few slides looking into penetration. If under a monolayer, we inject another solution of a different molecular species, a number of different things can happen. The simplest happening could be interaction between two types of molecules mainly electrostatic and that would lead to changes in the surface potential, but not much change in surface pressure. That could be for example, done when you inject very dilute solutions of long chain salts under monolayers. This gives you the pictorial field for what may happen. We might have on one hand anchoring of the injected molecules under the monolayer and there may be interaction between these injected molecules and monolayer molecules, but the molecules only can interact weakly with the monolayer molecules. They cannot actually penetrate into the monolayer. As opposed to this, we see the penetrating molecules over here, which make way into the monolayer displacing some of them, some of those monolayer molecules to find a place in the surface space. And if the energy of desorption is not very high, then it is possible that a mixed monolayer may be made to eject the additive from the surface pressure, from the application of surface pressure. It might be this situation after penetration, but if we apply a compressive surface pressure, these additive molecules may be made to eject into the bulk of the liquid underneath. And generally, the penetration within the monolayer will be characterized by large changes both in surface potential as well as surface pressure. Now, you might have additivity for the contributions of the individual molecular species, the two types, to both surface pressure and surface potential, which means there is an independent interaction leading to the contributions to surface pressure and potential. And we may say that this is like just a physical mixture of the two species in the surface phase. Therefore, you get contributions coming from both. Such may be case for astacine as one species and oleic acid another, spread on 100 normal HCl. You would find that low pressures, there is independence and we may say just a mixture of the two species will be present in the surface phase. But then there could be cases where such an independence is not formed. There by implying that now additivity or superposition of the contributions is not possible. And we have what we call akin to chemical reaction formation of a molecular complex between the two types of molecules in the surface. This kind of complex formation would be of course a function of several things. The polarity of the head groups and the characteristics of the hydrocarbon tails including the lengths and shape. And if we take this example of injection of dilute solutions of substances of this general structure C H 3 C H 2 11 times X under monolayers such as of proteins and cholesterol present at air water interfaces, we would find this order of reactivity. The ammonium, if X is ammonium ion we get the greatest reactivity. And for quaternary ammonium compounds like N C H 3 plus whatever we may have derived from this as the least reactivity in between we have sulphate sulphide and carboxylic contributions. So, we move into the subject proper for today's discussion. We expect a strong interaction between the hydrocarbon and carbon chains especially if these molecules contain long saturated hydrocarbon chains. However, you would expect that strong interaction cohesive interaction would be weakened considerably should there be a double bond in these chains. You know that the double bond would correspond to a significant dipole moment which will lead to a repulsion among the molecules. So, things like oleic acid containing a double bond between 9th and 10th carbon atom in the chain that would correspond to a weaker interaction compared to the corresponding comparable molecule steric acid which does not have a double bond. So, under certain conditions we might have stoichiometric complexes forming. Some examples could be given digitonin and cholesterol and sodium setyl sulphate and cholesterol. These form one is to one kind of complex whereas digitonin and setyl alcohol form one is to two kind of molecular complex. So, there are definite ratios in which these monolayer species will be present in the monolayer and they would not contribute in a linear fashion to the surface potential or surface pressure. We then broach the topic of thermodynamics of penetration. We need to think a little bit about this. In this slide you think of the factors being numbered from here. So, when a molecule in bulk makes its way into an uncharged monolayer then we expect several energy changes to occur. So, the first factor that we may think about is the energy of removal of the hydrophobic part of molecule from water. After all it has moved from bulk of water into the monolayer. So, they must be associated energy change of taking this hydrophobic part from bulk of water into the monolayer because once it goes into the monolayer it will flip away from the water and therefore, we have energy change associated with this removal of hydrophobic part. That is the first contribution. This is what also makes the spontaneous adsorption of an amphiphilic molecule favored energetically because it is taking it away from the structured highly hydrogen highly associated hydrogen bonded water. This chain is taken away from that unfavorable environment into the vapor phase or air in contact with the liquid. The second energy change that you may think about is the energy to make a hole in the monolayer compressing the monolayer molecules into the surrounding areas. So, penetration will mean necessarily creating that space for this particular molecule which has moved from bulk into the monolayer and therefore, that will cause certain energy that is the second factor that I would like you to consider. The third one is the energy of interaction of the penetrating hydrocarbon chains with the molecules of monolayer. Now, you can visualize this may be accompanied by a complex formation. Should there be a possibility of complex formation that should also be included in this term corresponding to energy of interaction of the penetrating hydrocarbon chain into the monolayer and therefore, resulting interaction with the monolayer molecules. Then we have the fourth term which is the dipole interactions between the head groups. Head groups will be associated. There is a uncharged monolayer that we are talking of. It will be associated with certain dipole movement and the penetrated head group will have a dipolar interaction with those head groups. Sir, one day we are going to change for the hydroglyph part as the hydroglyph part is asounded in the bulk from all the direction. It will come. There will be a associated change, but the hydroglyph part is generally the head group is generally small. It would come to the surface. It will be surrounded of course asymmetrically. So, there will be associated change of moving from the bulk to the surface. So, that will be reflected in the last factor and it will come under the category of electrical factors, especially if we have long chain ions penetrating into the monolayer. Then there are two energy terms. One is energy of formation of ionic double layer. Once the ionic chains are penetrating, then at the interface there will be a double layer forming. The energy of formation of that ionic double layer will be one contribution. The other contribution will come from this energy and especially entropy. This is what we have been talking of because of asymmetry in the environment of the partial dehydration. Because once you have a head group coming from the symmetrically placed position in the bulk to the surface, on one side there is no water now. So, that is a partial dehydration. That will mean there will be energy change and associated entropy change. That is also important. This is important for ionic head groups. This happens because of the lower dielectric constant that now we see around this head group. Now, if the original monolayer is ionized, that is another situation. We started with uncharged monolayer, but should the original monolayer be an ionized one, then we will have the fifth term corresponding to energy of alteration of the ionic double layer. Ionic double layer will already exist there, but if you have a penetrating molecule, energy of that ionic double layer will be modified. In practice, this factor 3, energy of interaction of the penetrating hydrocarbon chain and the molecules of monolayer can be appreciable. Now, think of another scenario. Think about the energy of desorption for the penetrating compound. It could be actually higher if a monolayer pre-exists compared to a situation where that monolayer is not present originally. This would be because now the desorption will have to overcome the energy of interaction of this penetrant with the existing monolayer molecules which otherwise would not be present in the comparable situation. To get some idea about the numbers involved, if the original monolayer is closely packed, the energy of desorption per CH2 group could be as high as 800 calories per mole. Therefore, one may be able to compress certain penetrated films to a higher surface pressure than would be possible if the penetrating substance were alone present in the surface space. Similarly, we could argue for complex formation. Although, in case of complex formation, it might be very difficult if not impossible to get the complex penetrant to disengage from existing monolayer molecules. I hope this point is clear to you. Let us say we have air and water and we inject certain molecular species with the which can make its way into the surface. So, we might have eventually these molecules moving from the bulk position into the surface. Now, if this is the only species present at the surface coming through diffusional process after being introduced into the bulk, we can think of having such a system in a Langmuir trough where one barrier is fixed, another one can be made to move exerting different extents of surface pressures. We might be able to compress them so that they come into close proximity by overcoming whatever is the film pressure of this monolayer. And by compressing it further, we will be able to get these back again into the bulk, desorption. You will require certain amount of energy to squeeze out the penetrating molecules which would spontaneously make way into the interface to force them back into the bulk. You will require certain amount of energy here. As against this, you think of a pre-existing monolayer and the same species is a monolayer. It is injected underneath and eventually makes its way to form a mixture with the pre-existing monolayer. Then again, you can do the same thing between a fixed and movable barrier. You can compress this mixture and eventually force penetrating molecules to go back into the bulk. Because of the additional interaction here with the existing molecules, we expect the pressure required to force these penetrant molecules back into the bulk to be greater. This would happen if there is a physical mixture, just a simple mixture of the two species, existing monolayer molecules and the penetrating molecule. Should there be a complex formation, then you probably will never be able to get the penetrating molecule out in the bulk unless of course, you can overcome the energy of desorption for the reacted surface active species. Then you will be able to force the monolayer molecules together with the complex one into the bulk. But it might not be possible to just get the penetrant molecules alone once they have formed the complex. Or it could be a situation in between, the application of surface pressure or imparting higher energy to the surface molecules here, you might be able to split the complex back and then in that case, these penetrating molecules can be forced back into the bulk. In any case, we expect we will require higher energy expenditure for desorption of the penetrating molecules, should there be already a monolayer present. That is what that is the point I was trying to make here. There could be another situation, if the penetrating molecules have only a short chain such as maybe C 4 or C 5 species is the hydrocarbon chain, then the lateral cohesion among the monolayer molecules is so reduced because of the weakening of the nearest neighbor interaction by virtue of the original long chain molecules or long chains, then the film could become almost gaseous. So, this is a situation there could be otherwise expected strong cohesion or moderate cohesion among the penetrating molecules, but because now existing monolayer has chains which separate these penetrating molecules, the surface equation of state for the penetrating molecules may behave or show a behavior like a gaseous monolayer. So, this is another situation that could give you a different picture about the surface behavior of the penetrating molecules. The importance of the second factor energy to make a hole in the monolayer could be assessed from the low rates of penetration which are observed experimentally. Originally, the monolayer may be expanded five times and we may have energy of activation which will permit completion of the penetration rapidly. We may be able to compress this film subsequently again. The number of adsorb molecules which have penetrated the original insoluble monolayer of any substance I at constant area can be found from increase in film pressure. The equation which would be relevant will be this dou ln A by dou pi at constant temperature equal to A bar I where this I you should read as a subscript for A bar partial molar surface area of I. That brings us to the last topic in context of reaction that is penetration from vapor phase. Instead of the penetrating molecules coming from the underlying liquid phase they could be adsorb from the vapor above. We could take example of a monolayer of steric acid. Monolayers of steric acid could be easily penetrated by such vapors as of benzene or hexane even at very high film pressures. Even if we have steric acid compressed to a high surface pressure, if such a surface is brought in contact with a vapor like a benzene or hexane then the monolayer gets penetrated. It might appear that this discussion is very theoretical and probably not of much significance in practice. But to end this lecture I will give you an example which has been taken from industry and shows a dramatic influence of low very low or ultra low vapor pressures of adsorbable or penetrable species on certain practical circumstances. Let us discuss the theory first. So, we have this steric acid monolayer compressed to high surface pressure and yet if the vapor of benzene or hexane is present then you see the vapors actually making way into the monolayer. The adsorption of normal hexane from vapor remains between about 3.6 to 4.8 into 10 raise to 14 molecules per centimeter square for a range of molecular areas of steric acid and corresponding range of surface pressures. The result will be if you maintain the area constant as a result of adsorption of normal hexane there will be increase in surface pressure. Or if you maintain constant surface pressure then there will be a resulting increase in surface area. We could use Gibbs's equation to interpret the results like earlier and we may find that the chemical potential of the monolayer of steric acid itself may be changed because of these penetrating molecules. Will be interesting to see what happens if you continue to have such penetration to a large extent. Initial heat of adsorption of hexane on dilute steric acid monolayer is a high value about 14 kilocalories per mole. That corresponds to effectively a two dimensional solution of steric acid in hexane ok. Because we are talking of now the hexane and steric acid and when hexane is getting into the steric acid we will have this mixture. So, it is a two dimensional mixture or solution of steric acid clusters in hexane. However as the penetration continues from the vapor and we have much more of the surface covered with hexane, the value of heat of adsorption will fall to about 7 kilocalorie per mole from 14 kilocalorie per mole at low extents of adsorption. And that would be the opposite it will be close to the heat of condensation of pure hexane. When lots of hexane is there it is equivalent to hexane condensing in itself later. So, it may be looked at as the heat of condensation of pure hexane in itself near the end. So, in between of course, you will have a range of heats of adsorption. That brings us now to the introduction to what would be the subject proper for the next module and that will be devoted to mass transfer across interfaces. It is evident that much of chemical engineering practice may see introductory undergraduate courses where concepts are simplified, but here since we are focusing on interfaces we have to look at a more detailed picture. So, you are surely going to go ahead of your undergraduate days and many of you beyond the master's phase. So, let us think of what we have here mass transfer means we have one species moving into another and the another species may be a mixture of a number of components. So, we will have transfer of a substance across an interface from one phase into another phase. This part of crossing the interface is a must when we talk about mass transfer across interfaces. So, when a molecule actually passes across the gas liquid interface for instance it would encounter a series of resistances. First it will encounter resistance in the gas phase moving from gas to the interface. Then there will be possibly the resistance offered by the interface and third there will be resistance in the liquid phase. That is what is being indicated by the total resistance here R which is the sum of R G, R I, R I and R I. R G is the resistance due to diffusion in the gas phase, R I is the resistance for diffusion across monomolecular region which makes up the interface and R L which is the resistance to diffusion through the layer of liquid below the interface. So, we might just show this diagrammatically. Let us say for simplicity I show an interface over here, gas phase over here and a liquid over here and you have some species moving from the bulk of gas to the bulk of liquid. This arrow is only supposed to indicate the direction of transfer, transport from gas into the liquid across the interface. So, the three resistances we are seeing are the resistances which may be represented this way that is your R G. The resistance which is offered by the interface R I and the resistance which is offered by the liquid here R L those are the three resistances. Now, what exactly will be this resistance in the gas phase? The one at interface and liquid will depend on how we want to look at it. Experimentally of course, it will depend on number of conditions. Those conditions may include the hydrodynamics of the gas phase and the liquid phase and what is happening at the interface. One may go on to say that hydrodynamics of interface also will play a part apart from the compositions. So, to give you some flavor for the basics, in general we may say that the bulk of gas and liquid they are very well mixed. So, we may represent that by this agitator. The gas phase is kind of mixed with this agitator. This may not be physically present as an agitator, but the mixing may arise because of the flow itself characteristics of flow, but we symbolically represent the mixing in the bulk by this agitator. Similarly, in the liquid they will be mixing for whatever reason. So, in the bulk of gas as well as liquid we have agitation or mixing, but at the interface we might have relative stagnancy. They might be a certain film near the interface in the gas phase and in the liquid phase. As a rough indication we may show these films where the resistance is contained. We might actually say that this is a hypothetical thickness of gas phase and this is the hypothetical thickness of the liquid phase which offers the entire resistance in the gas phase what is indicated here or in the liquid phase. As a simplest film theoretic picture at the interface we will have resistance depending on the hydrodynamics there. One would be interested in knowing what are these resistances and how one could possibly obtain predictions for these resistances for certain representative cases at least. The diffusion in the interface will be possibly across monomolecular thick region. What are the thicknesses of the liquid and gas films would depend on the conditions corresponding to the mixing or hydrodynamics in the gas phase and liquid phase. This is the same thing shown. We have the resistance in the gas phase, resistance in the liquid phase and the resistance at the interface. This is the unstared layer within the liquid and similarly there will be a region here which will be unstared liquid layer, unstared gas layer above the interface. Now, let us get the broad idea about these resistances. First you should remember that it is the liquid phase resistance which is generally the highest molecular diffusion of solute through a non turbulent liquid layer adjacent to the surface is therefore the slowest step. The greatest resistance lies in the liquid R L is highest but you can have interesting exceptions. R L can sometimes be made so low that either R i or R g may become higher than R L and therefore either R i or R g the higher resistances could become controlling rate controlling in comparison to the interface. For getting a broad field remember that these resistances could be imagined to be the resistances to the flow let us say of water through a pipe. Each resistance is linked with a certain wall. The wall which has the greatest resistance will become controlling even when fully open it just does not permit whatever flow rate of water as is permitted by other two walls. So, that becomes the limiting one. The greatest resistance will be the resistance of that particular step which is highest. So, generally the liquid phase resistance is highest but you might be able to tailor make situations where either the interface or the gas phase could become rate controlling. Let us consider the orders of magnitude of these resistances R g, R i and R L and also the units in which they may be expressed conveniently. We later consider some specific examples. Now the first basic differential equation describing transfer of a component across any plane is this equation 2. d q by d t is equal to a k delta c where q is the moles of the material transferring q is the moles of material which is transferring from one phase to another maybe from gas to liquid. T is the time in seconds a is the area perpendicular to the direction of transfer and so we are looking at the flux of a or rate of transfer of a across this area which is perpendicular to the direction of transfer and delta c is the concentration difference in the units of moles per centimeter cube and delta c generally would be the concentration difference in that region for which k is the permeability coefficient in units of centimeter per second. We could identify k to be mass transfer coefficient as we will see later. Reciprocal of k is a measure of resistance in seconds per centimeter. You can look at this equation slightly differently. We are saying the rate of transfer of moles of species is d q by d t and that is equal to the total area across which transfer is taking place a permeability coefficient into the mass transfer driving force concentration difference. Let me give an example. If this is the interface and that is your liquid phase if the greatest resistance as is commonly encountered is in the liquid. So, R L is the controlling one the greatest resistance in the liquid is in the liquid and the solute is moving in this direction. If we look at the concentration of this solute it will be highest somewhere here let us say c sub i and in the bulk it will be lowest which will be c sub b and that may be the concentration profile for this species which is moving. Then this difference c i minus c b is your delta c interface concentration minus bulk concentration that is your delta c. Corresponding to the liquid phase resistance we will have a certain film which may be presumed to offer all the resistance to transport in the liquid and corresponding to that you have the permeability coefficient k in units of centimeters per second. Now I would like you to think of mass transfer as just one of the many possible rate processes so that generalization is important for a proper understanding. For any rate process whether it is transfer of mass from one phase to another or it is transfer of heat or it is the flow of a liquid or it is the flow of charge like electrons through metal and so on including chemical reaction every rate process will have a characteristic rate and the rate will depend on two generic factors one is the driving force and the other one is the resistance. For heat transfer the driving force may be delta t temperature difference for mass transfer delta c is the driving force for flow of a liquid or a gas it may be delta p the pressure difference for a chemical reaction it may be the difference in existing concentration minus the equilibrium concentration equilibrium concentration could be zero if the reaction is irreversible and for transfer of electrons or flow of electrons to a conductor it may be the potential difference and so on. So the driving force is responsible for that rate process to occur rate of that rate process will be equal to the driving force by resistance and this is the generic equation that you always should keep in mind. Looking at mass transfer specifically what it means is that we can identify d q by d t as the rate that is your rate on the left hand side delta c is the driving force so keep it in the numerator and this term a k will be the resistance for the total system. If you in particular have area equal to unity then it should be this right delta c divided by 1 by a k if it is to be equated to be if it is to be equated to a k delta c when you take a k in the denominator it will have to be 1 by a k correct. So, this goes in that position if the area is unity then 1 by k is your resistance if the area is the total area 1 by a k is the total resistance. If it is mass transfer then for total area 1 by a k is the total resistance if the area is unity we are talking of flux rather than the rate rate of transfer and therefore the resistance that we see here for unit area is k raise to minus 1 because k is centimeter per second this will be seconds per centimeter you could look at this restriction of unit area over here if area is unity then k is like a rate constant and that rate constant is reciprocal of resistance that is 1 by k inverse that is k right. So, that is a picture you can have for understanding the concepts in a bigger picture of rate processes in general. We could return to the slides and therefore conclude that k inverse is the resistance r in seconds per centimeter. So, depending on which phase you are looking at this resistance we will have corresponding possibility of relating k inverse or r to the corresponding transport property transport cohesion it could be diffusion cohesion in the liquid or diffusion cohesion in the gas or it could be corresponding reciprocal of resistance of surface resistance corresponding to the surface phase that is what is shown what I have already written k is 1 by r and if this transport is in the liquid phase diffusion through the liquid will be characterized by diffusivity D and a thickness of the liquid layer delta x you can show that k is D by delta x. So, if you say that we are looking at d q by a by d t we will have k times delta c and what we see on the left hand side is the moles transferred per unit time per unit area that is the flux the flux of a whatever species not the area a is a species which is diffusing. So, flux of a is moles of a per unit time per unit area and that will be if diffusion is the controlling mechanism minus d d c by d x or we could write d delta c by delta x where d is the diffusivity delta c is the concentration difference delta s is the thickness of the liquid layer across which we have all the resistance in the liquid that is offered. And therefore, what you have found here is that by comparing the diffusional expression with the generic rate expression in terms of rate constant you find that the rate constant is nothing but D by delta x k is equal to D by delta x that is what is shown in this slide as equation 3. Reciprocal of resistance corresponding to diffusional resistance in the liquid is k and that is equal to D by delta x or in terms of the rate we write d q by d t is equal to a times D by delta x delta c or a d into delta c by delta x this is the concentration gradient which is responsible for the observed mass transfer. I may end the lecture here, but not without adding the following that in general you might be safe to say that is the concentration gradient which is responsible for mass transfer, but not always the more fundamental understanding should be it is the chemical potential gradient for a particular species not the concentration difference, but the difference in chemical potentials which is responsible for mass transfer. Otherwise you may not never be able to dissolve a solute from a gas phase at a very low concentration in the gas phase moles per volume into a liquid which has a high solubility for that component because liquid phase concentration in any case will generally be incomparably high relative to the gas phase concentration, but it is a chemical potential which is decreasing when you have this solute species moving from the bulk of gas to the interface. So, we could stop here for today.