 In this eighth lecture in the module 4 on monolayers, we look at couple of more properties of monolayers, the indices which characterize their strength related to the action of shear and with respect to when the flow can begin. And we also take up the leftover topic of how possibly fibers may be produced from monolayers and then we will connect it with a couple of examples in nature with perhaps a conjecture about how the primary material for living cells might have come into being in the very first instance. Then we move on to the generic topic of surface reactions to which we alluded a few times earlier in passing in this lecture. Before I do that, I did want to make a couple of comments on the action of wind on slicks of hexadecannol on wind ruffle lake surface. By very definition the slick is an isolated patch of a floating material. Now if it is a monolayer like of hexadecannol and this patch is floating in the middle of a lake, it is clear indication that the action of wind drag will be perceived in the normal sense as more and more momentum is transferred in the direction of the wind. The slick of hexadecannol will tend to accelerate until of course it reaches a location closer to the shore where a continuous monolayer exists between the region we are looking at and the shore and then only the action of the back surface pressure will come into picture. If we do not have any floating or adsorbed films then turbulent ripples would dissipate most of the energy that is transferred from the wind to water. So first let us look at the shear elastic modulus of a monolayer. The picture here will be reminiscent of the rotational torsional methods for measurement of surface viscosity. But having said that I will urge you to think of the differences right. You do not have to respond but keep that at the back of your mind. So we are here thinking of a monolayer which is undergoing a shear and this monolayer is contained between two concentric rings lying in the plane of the surface and that means now we have a constant area over here. Now if the applied stress is to be calculated you will have to take into account the applied torque which would be constant throughout the annulus. Now a little bit of imagination will make you understand why we have this applied stress equal to L by 2 pi r square. You have to think in terms of the geometry. You have an annulus we are looking at a certain portion of this film which is under the action of shear and the torque applied is the constant torque L. So that has got to be this L by 2 pi r square. You could think of 2 pi r square as a split 2 pi r times r. Then it will become at once clear what area we have in the denominator. In any case in order to understand the concept of this shear elastic modulus we need to look at a diagram. So let me draw a diagram for you. Let us say we are looking at an annulus indicated by this sector the radii r r 1 and r 2 and we look at an element over here which is undergoing shear. So that radius is r, dr is over here and if the angle subtended here is d theta then we have r d theta giving you the measure of distortion under the action of the shear. So now you can look at the action of the torque which is constant over this area and now focus on a location between r and r plus dr. So that gives you the stress the applied shear stress as L by 2 pi r square. Now the corresponding shear strain will be given by r d theta by dr that is the distortion here by dr. Now the ratio of this applied stress on an element by the shear strain of the element which is this L by 2 pi r square by r d theta over dr is the shear elastic modulus E s. And when we simplify this we have E s is equal to L by 2 pi r cube dr by d theta L is a constant. So upon integrating this we will be able to write E s as 1 by L by 4 pi theta times 1 over r 1 square minus 1 over r 2 square. Now if one ring is stationary and another is attached to a torsion wire and is made to oscillate then we can substitute for L in terms of the moment of inertia total moment of inertia which should be I this would give E s is equal to pi I 1 over t square minus 1 over t naught square where t is the period of oscillation of this ring attached to torsion wire when the surface is covered with a film and t 0 is the corresponding period of oscillation for a clean surface that into 1 over r 1 square minus 1 over r 2 square r 1 is the inner radius for the sheared monolayer r 2 is the outer radius. You could look into this formula a little bit further if we have a monolayer covered surface and we look at the period of oscillation that is t if it is a clean surface the period is t naught. So that together with the relative magnitudes of r 1 and r 2 would make it clear that E s should be this way. Now experimental methods are based on measuring the movement in the surface of the circular ring which is weighted and attached to a torsion wire and head and it is that film between the ring and the outer ring or the outer circular dish which is subjected to the shear and damping could be measured. We may want to look at next how the monolayer will respond to an effort to move it or make it flow. For every monolayer there exists a certain surface yield stress value that we represent as y s with the same units as for surface pressure dynes per centimeter. This is the magnitude of the tension that a monolayer will sustain before it starts flowing. Now, when the molecules are far apart there is hardly any interaction and it is a ideal kind of state of a film which is properly described as a gaseous film. We should expect y s equal to 0 it should not be able to resist any stress or tension within the surface it will start flowing at once. However, if we have polymers or long chain compounds forming a film or the surface which is capable of resisting this flow you should get appreciable magnitudes for yield this yield value. In a sense yield value could be looked at as parallel to the yield for the solid or the kind of yield that you have in the Bingham model for the non-Newtonian fluids it does not start flowing. The analogy with the yield of solid is a bit far-fetched because that is where the solid becomes plastic it yields to the stress there and for solids we really should not be talking of flows, but the analogy with the Bingham plastic may be reasonably good in this two dimensions. So, what are the magnitudes of y s? What is this monolayer? The monolayer could be in any measuring system like a Langmuir trap and you have a stress applied in the surface and we watch until the film starts flowing. So, if the monolayer is made up of molecules which are exhibiting the gaseous surface state then they would not be able to resist this it will start flowing at once. But if we have condensed monolayers polymeric then we expect there will be certain resistance and that could be small values apparently in dynes per centimeter, but when translated into three dimensional equivalents those will become appreciable. So, you could also think about how this yield value will change as we increase the number of carbon atoms in the molecules making the monolayer. It turns out this yield stress or yield value will increase proportionately with the number of carbon atoms in the molecules making the monolayer alcohol is an example. So, measurements reveal that for this mixture of sodium laurel sulfate and a long chain alcohol at one twentieth of a percent and one two hundredth of a percent very small values the yield value increases linearly with the number of carbon atoms in the alcohol. Perhaps you should be able to remember something related to this that we had looked at long time back where else did we come across a linear dependence on the number of carbon atoms making the molecules in the monolayer cohesive pressure remember that m value comes here. Now, you could imagine that you have a monolayer which you are trying to make flow before it does so it will have to overcome these intermolecular interactions. So, longer the carbon chains higher will be the cohesion and greater will be the opposition to flow. So, yield value is expectedly increasing linearly with the number of carbon atoms. Let us look at some magnitudes if we have sodium laurel sulfate with no laurel alcohol at tenth of a percent with the gaseous film yield stress is 0 dynes per centimeter. The same solution of laurel sulfate with 0.005 gram per 100 ml of laurel alcohol will give you yield value of 0.05 dynes per centimeter with increase in the laurel alcohol there is an increase in the stress if you have only laurel alcohol from saturated solution the value is appreciable about 0.12 dynes per centimeter. For spread protein films about tenth of a dynes per centimeter and polymolecular compounds like adsorb film of saponin can give yield value between 0.3 to 1 that can be regarded as substantial. Next, we come to this topic I had mentioned last time to fibers from monolayers. We will talk about the fibers once again when we go into the reactions a few lectures later. But for the time being you can think of a protein monolayer may be on a Langmuir trough compressed so that this monolayer occupies a very small area. If you have such compressed protein monolayer we should be able to bring it out in the form of a thin fiber by withdrawing a fine glass rod through this compressed monolayer. The monolayer forms an adhering layer on the glass rod and when it comes out it continues to build this fiber. You will have to of course take care that this glass rod is removed slowly. This way it is possible to take out fibers of proteins between about 1 to 100 microns. So very thin fibers actually can be made by simply withdrawing the monolayers continuously. Now how easily such fibers would form would be a measure of the molecular size and the extent of crosslinking which might have occurred in the monolayer. Another factor which may come into picture is whether the liquid is coad soft along with the molecules in the surface because that will determine the interaction among these molecules and their ability to form a continuous fiber. If other factors are to be taken into account we could visualize the case of oval albumen. You know that oval albumen when treated with heat UV radiation or chemicals like urea would tend to lose their ability to form the elastic fibers because you expect the protein to get denatured to an extent. So if a protein like oval albumen is treated by these external agents before spreading then either you cannot form any fibers or the fibers are very fragile they will break easily. Now it is this notion which is believed to be one of the theories for forming the components of first living cells. In the primordial soap you had large organic molecules coming together forming monolayers and the action of wind could actually convert these monolayers into fibers which could have been the basic building blocks for living cells. This is one of the conjectures of Goldakar in 1958. That brings us to a couple of interesting examples to think of. First we take a case from nature a certain species of water snail found in South America uses its own foot as a Langmuir trough. What it does is it concentrates the protein present in the water bodies perhaps as monolayers by creating this kind of geometry. So you can think of a tank wall or any floating object and we are looking at the surface which is covered with monolayer. If the snail has to take these proteins forming the monolayer it will be inevitable that it will take together with these protein molecules the water around and that would mean energy expenditure. To obtain this food it has to take in lots of water. So what the snail does is something smart. It forms a little Langmuir trough by anchoring its curved foot onto the tank. So this is the Langmuir trough that the snail has formed and by up and down movement of its foot it pushes the monolayers material here through this little Langmuir trough. When it is compressed and forced through this little space it comes out as a fiber. Having now used the cohesive pressure among molecules these surface molecules now are formed into fibers which are then far easier to ingest. This is what it does. So in the normal manner it would have had take a lot of water in but here it is able to concentrate the monolayer into fibers concentrated in proteins. Rim of the funnel is placed in the surface of the water. Perhaps we will be keeping our discussion incomplete if we did not take the example of a tadpool. You would find it fascinating that a tadpool in one day engurgitates its own body weight of protein taken from monolayer. But this is not a smart way of doing it because it has to take in a lot of water. So the nature has different species at different levels of intelligence. We try to understand from them how best to utilize that knowledge in our everyday life. We next run into the surface reactions. So we could ask a few questions in comparative sense for the surface reactions and ordinary bulk reactions. The first question is about the activation energy. Would the activation energy be same for a surface reaction and a bulk phase reaction? If your answer is correct then we will have to see how to justify that under given circumstance. So expect the activation energy to be different. More we are opposite. Second question is related to the potential. When we are talking of reactions we got to think of now species which can exist as monolayers and if you have to compare them against the bulk phase reactions we have to visualize that they are brought into the bulk and there the question will be will there be differences because of orientation and corresponding effect on configuration. So first thing which will strike is that with respect to potential I will come to activation energy a little later. For the potential only for surface phases we will have equi-potential state because the monolayers will have oriented molecules. All the molecules will be in equi-potential state. In case of bulk phase reaction the orientations are all random that is for the monolayer this is for bulk. Secondly we could control the potential and also measure it only for such surface reactions. The reason is the orientation could be controlled by having barriers in the surface. By adjusting the spacing between these barriers we can have the surface monolayer in different states of compression therefore different orientations. We cannot do anything like that for molecules which are dissolved in the bulk phase. So we have a scope to control and also measure the potential for surface reaction in bulk phase we do not have. Now if that is clearer let us talk of activation energy. But before I go into activation energy you might have to think a little more about the structure of the molecules which are in the surface. For reaction to take place we got to have certain reactant in the liquid phase. So let us say our reactant in the liquid phase is B and this monolayer molecule is A. It will be a certain part of this molecule which can react with B. So if it is a case like oxidation of a double bond unless the double bond is in contact with the permanganate solution which is trying to oxidize it you cannot have reaction ok. So depending on the symmetry or asymmetry of these molecules the surface pressure or the compression may be able to bring that reactive part in contact with the other reactant. Depending on the structure we might be able to force the reactive part of this molecule to come in contact with B or take it away. So by doing that we will be able to alter the reaction rate ok. Now activation energy will be based on the measurement. If we can maintain the same proximity of the reactant group within the molecules with the other reactant in the solution we expect activation energy to be not different ok. So depending on the structure of molecule and the conditions under which you carry out the surface reaction we might be able to see variation in activation energy or no variation. So that is a factor the structure of the molecule and its orientation can play a part in determining the activation energy ok. The third question you could ask will be with respect to catalysis can you get strong catalytic effects for reactions which can take place in the surface space. You do get on account of the different potentials that you can generate at the surface a considerable degree of catalysis. Kinetic factors or salting out factors can result and we would be looking at a number of these effects in course of time. Because this is an introductory lecture on surface reactions let us think of what all varieties of surfaces you would be interested in studying the reactions at. Of course the model experiments could be in a Langmuir trough but that is more for the primary interest of studying what the surface reaction rates are and what the catalytic effects or what the activation energies are. But in practice the surface reactions will be important mostly in the disperse phase systems. We will be talking about disperse phase systems later but to give you an idea you could carry liquid-liquid reactions in a variety of ways perhaps in the form emulsions where a large interfacial area is created by dispersing one liquid another and therefore the two reactant species from two phases can be brought together with very high areas and emulsions can be of different kinds depending on the required areas that is specific interfacial areas and the factors related to stability and phase separation at the end of the reaction. When you talk of emulsions it will be desirable that the interfacial area be as high as possible and how do we increase the area by splitting the disperse phase into finer droplets that will mean energy expenditure. In order to create finer dispersions you will have to put in energy that is the mechanical energy. The same thing could be achieved by lowering the interfacial tensions. By incorporating surface active agents which can lower the interfacial energy we will bring down the energy required to achieve the same levels of interfacial areas. If we could be luckier and if the reacting system permits we might be able to reduce the interfacial area, interfacial tension to 0 or even momentarily negative values in which case we will have spontaneous emulsification. Then the mechanical energy required will be minimum but there will be other factors. Talking about emulsions let me put this in perspective of the practice. Ahead of time let us say we are talking of saponification. If a vegetable oil is reacted with an alkaline solution and the practical systems will involve oil and alkali at about 150 degree centigrade could be made to react to produce soap. That is the common saponification example. You would want to have oil and this aqueous alkaline solution to have as much interfacial area as possible as large interfacial area as possible. So, you would want this oil to be present in the form of tiny droplets. In this case you might begin with the mechanical energy required to split oil in the form of droplets but as the reaction proceeds the soap itself is capable of lowering the interfacial tension and thereby reduce the mechanical energy requirement. But it is here that you might have another factor coming in. The soap molecules because they have an affinity for the interface they will also compete for the reaction space. Oil has to come in contact with alkali in order to react but the soap molecules now tend to accumulate at the surface. So, while they aid in increasing the interfacial area they are at the same time competing for the space at the reaction surface. So, the soap will have a detrimental effect on reaction unless you are able to effectively remove it continuously. So, the picture will be that a magnified interface between oil and alkali will have the soap molecules accumulating here thereby reducing the effective interfacial area available for reaction. Not just that as the soap layer builds up over here as more and more oil is reacted it will add to the resistance for the penetration of alkali or oil before reaction can occur. So, the soap layer here will cause an additional resistance slowing down the observed reaction rate which means apart from reaction the mass transfer effects will come. One will have to remove this soap layer effectively. Going beyond emulsions you could have micro emulsions. The micro emulsions are different from emulsions in the sense that the domain size for oil and aqueous phase is much smaller, smaller than about quarter of a wavelength of light. If droplets are present they will be smaller than quarter of wavelength of light that is one picture where it is still droplets disperse disperse in a continuous phase, but the droplets are very small and therefore, you can have oil in water or water in oil kind of micro emulsions. But there is also a special structure which is bi-continuous wherein the two phases aqueous and oil phases are in a transiently interlocked fashion and they keep changing their special locations. So, if you take in a bi-continuous structure a particular point a geometrical position at one instance you might find oil there at another instance they will be the aqueous phase. These domains keep on changing continuously. So, one simple way of explaining the bi-continuous structure is that within the bi-continuous structure where these domains are varying over time one may be able to move from one point to another by just moving on and waiting till a particular phase appears. So, it would appear like if in an instance if you have to move from this location one to location two it will be necessary to cross the oil barrier. It might be possible if these lines indicate the presence of oil and if one has to move from this point one to point two immediately in one moment then it will be necessary that you have to cross the interface sometime. And there may be other way of travelling you might go from location one to location two waiting for the position where oil domain is now to be replaced by water or aqueous solution. So, you might be able to go from one point to another point without ever having to cross the phase boundary or interface. One interesting thing about micro emulsions unlike emulsions is that they are thermodynamically stable which means unless you change the state variables oil and water will never separate whereas in emulsions phase separation will take place over time. So, all the products based on emulsions will show some kind of creaming over the time of storage and there comes the factor which is the keeping factor of a product. Every emulsion based system will be thermodynamically unstable micro emulsions however are thermodynamically stable and because of this fine scale there will be one visible difference the micro emulsions are transparent. You can see through a micro emulsion, but the emulsions will be translucent or opaque. Looking at the color of an emulsion we might be able to tell what is the domain size depending on the size of the droplets. The emulsions of two clear liquids a clear oil and clear water could show different hues. So, with experience you can probably guess the droplet size just looking at the color of an emulsion and because the domain sizes are so small in micro emulsions obviously you have interfacial areas very high by continuous structure. We are moving from first point to second point and they are in the same area or different area. That is what I am saying. If you want to move from point 1 to point 2 all the time through oil they were having to cross the boundary with water you can do so. You start with position 1 when oil is there then keep moving towards to you will have to stop. Let us say this is oil the clear region is oil you move from here up to the phase boundary then wait until the water is replaced by oil and then you can go through only oil from 1 to 2. Same way you can do with water you can move from one point in water to another point in water. Whereas if these were permanent steady structures then it would mean that these barriers are always there and we will have to cross that. So in bi-continuous structure there are these transient patterns. Reaction is taking place at interfaces and interfaces are swapping positions. So it is a dynamic picture and interface might be existing between oil and water now here at another instant it will be elsewhere. But the magnitude of the interfacial area that you get will depend on the domain sizes. Bi-continuous are very small so interfacial areas are very large right. So what is it that is making bi-continuous? What is it that is making bi-continuous? It is the first the ratio of the phase volumes you know if you have very little of oil and most of water or the other way around very little of water and most it is oil then it is unlikely that you will have the bi-continuous structure. So there are ways to characterize when you will get the bi-continuous structure. And in these systems the interfacial region is the bulk of the phase I mean most of the volume is in the interfacial region. This is shown only as a representative diagram with these domain sizes become very small the total volume contained in the interfacial region itself is the very significant portion of the total volume right thereby you get very large areas. So interfacial areas and resistances arising out of the product accumulation are now clear to you. So you can visualize now in this saponification reaction how you would expect to find the reaction rate if this is a surface reaction. In the early part the reaction rate is expected to be rapid right later as the soap accumulates there is less phase for reacting molecules at the interface. At even later stages in last stages soap would have accumulated to such an extent that the reactant can react only by penetrating and diffusing through the soap layers. So you expect the reaction rate actually to drop down as a function of time. Before I conclude this lecture maybe I will cite one last example of a reaction in the surface where the rate will be directly affected by the orientation. Maybe I can put that as a question to you first where else do you expect? The orientation and our control over orientation of the reactant molecules could become important in governing the rate of reaction. Think of a type of reaction where the orientation not from the perspective of accessibility of the reactant groups or reactive parts of molecules to be able to come in contact with the other reactant but perhaps some other reagent or some other agency. Maybe some zero order reactions where the rate does not depend on concentration of any of the reactants but some other factor. In photochemical reactions ability of a molecule to absorb a quantum of light will be dependent on the position of the chromophore the part of molecule which can intercept and absorb a photon. Our ability to control the orientation of molecules and therefore the orientation of the chromophore will allow us different extents of capture of photons for photochemical reactions. Such specialized cases will be important in say photopolymerization. Depending on the orientation we might be able to alter the quantum yield which is basically the number of molecules reacted per quantum absorbed. So we might have special effects coming into picture depending on the orientation at the surface. One last question before I resume in the next lecture if we have to track a surface reaction how would one go about contrast this against a bulk reaction. If it is a bulk phase reaction what you do is probably take a reaction vessel have the two components over here either in homogeneous fashion or in a system where interfacial transport of one of the species allows for bulk reaction in the second one. So it will be a combination of bit of surface and bulk reaction. So in such systems how will you track the reaction? You need to measure concentration let us say of one of the reactant at least versus time. So you take samples out at different times and analyze. Our problem here is now you have let us say a Langmuir trough. The reaction is occurring between two kinds of molecules in the surface. How will you obtain the rate data? How will you measure the concentration as a function of time in the surface layer? If you are thinking of the same thing take out samples may be by siphoning out small portions with a capillary attached to a pump that probably is not a good idea because that will mean that we are disturbing the interface. But they may be another way in which you can track how the concentration changes with time. Any ideas? Yeah, but that will be a bit difficult because depending on what the product is and depending on the activity of the surface molecules we may or may not get significant change in the interfacial tension or the surface pressure. But something more promising will be information from another module in this same course. Think of that interfacial potential. That is a powerful way in which we can relate the changing concentrations in the surface to the time that we require. So it is possible to express all the interfacial concentrations in terms of interfacial potentials and then work out the kinetics. So we will look into all those in the next lecture.