 We are looking at practice problems on gas solid reactions. The object of these problems is to demonstrate to you the order of magnitude of these numbers. That is whether it is reaction velocity constant, whether it is external diffusion coefficient or the internal diffusion coefficient etcetera. So, all these numbers is what is the prime object of these five or six exercises. So, that in a given problem you can understand how important the resistances from the different controlling regimes. Quickly the first one is about combustion and then it gives you some numbers regarding the reaction velocity and temperature and so on. We let us quickly calculate what is the time. Let me see whether I have done this. Here we have twice of Z n s plus thrice O 2 is twice Z n O plus twice S O 2. The size of the particle is 1 mm. It is roasted in 8 percent oxygen. Its temperature is 900 C. Data is given for time required for complete consumption. We have diffusion coefficients. The data is given here. K s is given as twice centimeter per second. D e is given as 0.08 centimeter square per second. Now, the context see please recognize that this is 1 mm particle and zinc sulphide is roasted around the world in fluidized beds. So, you have a fairly small size of particle. That is why calculations are for small size of particle. As zinc sulphide is converted to zinc oxide, there is a unreacted core. Therefore, both the reaction velocity as well as diffusion coefficient in the product layer are important. Notice that the external diffusion is not very important in this problem because of the fluidized bed. There is a lot of external mass transfer is high. Therefore, that point is not taken into account. Let us quickly calculate tau r. We know this divided by B times K s times C A g and what is C A g? 8 percent oxygen P by RT 0.083. Temperature is how much? What is the temperature? 900. So, it is 1173. So, that comes out to be at P by RT is 1 atmosphere 8 10 minus 4 mole per litre or 8 10 minus 7 mole per ml. Now, we can substitute tau reaction equal to what is the density is 0.0425. The size is 1 mm. 1 mm is 0.1 centimeters. K s is 2 centimeters per second 2 and C A g is 8 10 minus 7. So, that turns out to be small b is missing. Small b here is 2 by 3. No, a gas plus b b solid is 2 by 3. So, it comes out to be 2.8 10 to the power of 3 seconds. Please tell me whether this is correct? 2.8 10 to the power of 3 seconds. We all get this 2.8. What is it? 3.9 is it? You all get this 3.9. Minus 10 to the power of shall I say it is 3.98 10 to the power of 3. Let us go for it. Then we have tau d is rho b R squared by 6 b times C A g. Let us put all the numbers 0.0425. R squared is 5 squared. 6 diffusion coefficient is 0.08. What else? C A g. C A g is 8 10 minus 7. Is it? R is 1 mm is it? 0.1. Now, b have we taken b into account? We have not. It is 2 by 3. Tell me what is the answer? 1 6 6. Something like this? Now, please recognize that the reaction time for complete consumption under reaction control is 3900 and time for complete consumption under diffusion control is 1600. Here is an instance where the reaction seems to be having more resistance compared to the diffusion through product layer. The reason is the size of the particle is small. That is why you are not seeing that effect so much. If it is a bigger particle, you would have seen that effect. The learning aspect here is that when you dealing with small particles, you will find reaction control that resistance is important. That is the important point that we are going to get across. Ignoring. See in a fluid bed, the solids and the gas, solids moving rapidly. Gas is therefore, the hydrodynamic effect on the solid surface is such that film thickness is very small. So, it is not very important because it is a fluid bed. Now, look at this. I have taken this from Levenspiel. He always gives nice problems like this. So, there is stock pile which is burning. It says and then surface is in flames and then you are looking at it from a great distance because you cannot go near it because it is burning. Huge quantity of coal is burning. So, you will look at it from a great distance. So, you will try to understand what is happening. He says the linear rate of the pile as measured by its silhout. Silhout means the shape of the burning flame. It seems to be decreasing at the rate of 5 percent per. The rate of 5 percent is not mentioned per hour. Maybe it is per hour. 5 percent per hour. Maybe I do not know. It is not mentioned there, but it is because I might have missed it. That is all. It says 5 percent per day. I am sorry. It is 5 percent per day. Now, how do you do this? 5 percent per day. So, let us quickly understand this. You have this flame and this is what we are seeing. This is decreasing at the rate of it is percent. So, r by r is 0.5 per day. This is what is given. You understand that. Now it says obtain an expression for size of the burning mass with time. So, we can integrate this. So, it is r by r equal to 0.05 plus a constant. Is that clear? At t equal to 0, size is left hand side is 1. So, it is minus. Thank you. So, at t equal to 0, r equal to the left hand side is 1. So, the constant is 1. So, it becomes constant equal to 1. So, what is our solution? 1 minus of r by r equal to 0.05. Is it all right? Yes or no? How long does it take for this to burn out? Sorry. How long would it take for it to burn out? How long? How many days? When does r become equal to 0? So, t equal to 20. 20 days. What is what Professor Levenspiel is trying to convey is that, when you have a burning mass, you have to actually judge all these things from a distance. You see, this is one simple way of judging it from a distance. You cannot go near because it is burning. See, we did this yesterday, but I was not satisfied with the kind of answers you gave me. I would like you to do it again. Let us quickly run through this. So, this is particle size d p. 4 mm and 12 mm and then temperature is 550 C and 590 C. Time for 50 percent conversion. That is given as 15 minutes and 2 hours. Now, if conversion is 0.5, we know that this is equal to 1 minus of R C by R whole cube. This we know. Yes or no? Therefore, R C by R equal to 0.5 to the power of 1 by 3. Whatever that is, tell me what is it? 0.8. We know from this. I will just set this down. 1 minus of R C by R plus tau f 1 minus of R C by R. It is not important because in this case, it is not important. Tau d 1 minus of 3 R C squared by R cubed plus twice R. Thank you. Is it okay? Here, film in this particular case, it is mentioned clearly somewhere. Ignore film resistance. It mentions here. Generally, what will happen is that many of the commercial exercises that we do, the external diffusion may not be very important. In most cases, because the velocities are quite large and so on. What is given is 0.25. Let me say, this is for the case of 550 C corresponding to 4 mm particle. This is 4 mm particle. We have tau R 1, 1 minus of 0.8 plus tau d 1, 1 minus of 3 times 0.8 squared plus 2 times 0.8 cube. So, this is what we have. So, this simplifies as 0.2 tau R 1 plus 0.08. This is what I get. Please tell me whether this is right. Equal to 0.25. So, this is equation 1. Is it all right? I have got it right. 0.2 and this is 0.08 is correct. This is what? 1, 0. Thank you. Now, this is for 4 mm particle. Then, we will do for 12 mm particle. This is at 590 C. So, what I get is 2 equal to tau R 2, 1 minus of 0.8 and then tau d 2. I have just put all the numbers. This is what I get. Please tell me whether this is correct. R C by R is a substitute. 0.1 you are saying. I think it is 0.1. So, you are saying it is 0.1. This is equation 2. Tau R 2 is thrice tau R 1. Tau d 2 is 9 times tau d 1. This we know. Yes or no? Now, we can substitute here. We can substitute in equation number 0.1. 2. I will replace this as 2 equal to tau R 2 is 3 times tau R 1. 1 minus is 0.2 plus tau d 2 is 9 tau d 1 times 0.1. So, this is the equation. I can call it equation 2 b or something like that. So, these two equations, can you please solve and then tell me the values of tau R 1 and tau d 1 please? 0. Is that clear? What we are saying? Quickly solve and tell me the results. Please solve for 1 and 2 and tau R 1 equal to tau d 1 2.08. Good. Let us go further now. Now, the question here is what is the residence time required to achieve in a flow reactor operated at 550 C? When he says flow reactor, he means CSTR. We have not said explicitly. This is what I wanted to draw your attention. This is not explicitly said. I mean flow reactor could mean a rotary kill, where it is a plug flow, which means the answers directly. You can use the single particle results. If it is a CSTR, I mean if it is a fluid bed, then you will have to integrate appropriately. Those results, they are already done in class. So, let us do this for the case of a fluid bed. A fluid bed, where our 1 minus of x b is known to us as 1 minus of x b bar equal to, but fluid bed means, we cannot do it for a fluid bed. No, we do it only for a rotary kill, because fluid bed means more complicated. So, rotary kill, where you can use the single pellet results directly. So, you have 98 percent. So, we have x b equal to 0.98. Therefore, equal to 1 minus of r c cube by r cube. So, what is r c by r? It is 0.8. What is r c by r? 0.98, 0.27. Very good. Thank you. So, now, what is the flow of 550 C? Therefore, T equal to tau r 1 times 1 minus of r c by r plus tau d 1 1 minus of 3 r c square by r square plus 2 r c cube by r cube. Now, tau r 1 is known, tau r tau d 1 is known. You can substitute and tell me the result. Now, please, r c by r is 0.27. Put all the numbers and tell me the result. Tau r 1 equal to 0.2, tau d 1 equal to 2.08, sorry, particle size is what? What is the particle size? 2 mm. So, tau r 1 becomes half of that. So, 0.2 becomes 0.1. So, 0.2 becomes 0.1. So, it is 0.2 divided by 2 is 0.1 and tau d 1, tau d 1 is divided by 4 is 0.5 hours. So, you put these numbers and give me the result, please. So, what is time T equal to? So, let me summarize here 2 mm particle T equal to 0.67. Let us go further. Let us look at this. It is 0.48. See, all of you might know about this 4 H 2 plus F e 3 O 4 4 H 2 O plus 3 F e. Now, see this particular reaction, you see blast furnace is a well known technology for a very long time. But, blast furnace economics is not very suitable for small scale production. So, the sponge ion is considered a very suitable technology for smaller scales of production. In the last number of years, very large number of sponge ion plants have come up in India, particularly in the eastern area, Belair or Kela in Varisa area. What they do here is, hydrogen is reacted with the ion oxide to give you iron and water. What happens is that, because it is a solid and hydrogen reacts inside, when you get your product, the product looks very porous. When you look at the product, you will find that it looks very porous, because hydrogen has gone inside and moisture has come out. Therefore, it looks very porous. That is why it is called sponge ion. It looks like a sponge. So, once again hydrogen diffusing into ferric oxide. Numbers are given and it is important to appreciate that once again, that its reaction velocity and diffusion coefficient product layer, which are important. The external diffusion is not all that important. But, he has given a number. You can see he has given a number. So, when you do the calculation, you will realize what is the relative importance of all these resistances. Let us quickly calculate. Let us calculate this numbers quickly. You have K s is what 1.93, 10 raise to the power of 5, exponential minus 12,000 divided by, what is the temperature? It is 800. So, 873. RTR is 2. Please make sure the units have to be properly matched. So, it is 1.03 is what I get. Please tell me centimeter per second. 1.03, sorry, 10 to the power of plus 2 is what I get. Please tell me or it is about 19.3 centimeters per second is what I get. Please tell me. You have to substitute, sorry, you can see power of 1. I have put 10,873. That is temperature or gas constant because this number is generally in calories per mole. It is not mentioned. It is not mentioned. So, it is in calories per mole. That is why I have taken 2 in the units of calories per mole. Do you all get this? So, what should it be? 1.199. K s. Are you sure this is okay? Please check. I want decimals seems to be not correct. 12,000 divided by 873 multiplied by 2. Is it 19.9? Is this correct? Let us go forward. Now D e, what is D e? D e is 0.03 centimeter square per second and K g is 10 centimeters per second. Okay. All right. Let us. So, t equal to tau r 1 minus of r c by r plus tau f 1 minus of r c cube by r cube plus tau d 1 minus of twice r c square by r square. There is r c cube by r cube. Okay. All right. So, tau r tau f and tau d. You have to calculate. Please calculate and tell me. Please calculate. Let me write 1 by 1 tau r c a g. I am calculating as p by r t hydrogen. So, it is 1 divided by 0.082 multiplied by 873. Is this correct? 873. I get this as 0.0138 10 minus 3 mole per mole per liter or 0.0138 10 minus 7 mole per centimeter cube. Is this okay? Please tell me. C a g minus 10 minus 3, 10 minus 6. Is it all right? Okay. So, let us just calculate tau r all the numbers. Let us calculate tau r equal to rho b r. What is the value of b? What is the value of b? We have to calculate where r is. What is the value of b here? A gas. Very good. B is 1 by 4. And density is how much? 4.3 divided by 225. That is density. Particle size is 5 centimeters. And b is 1 by 4. K s is 199. That is what you said. And then c a g is 0.0138 10 minus 6. So, what is the value of b? C a g calculation is wrong. Is it 0.082 R t? P by R t. Pure hydrogen. Is this number okay? Yes sir. All right. Shall we go forward? 0.01 per liter. Therefore, per centimeter cube, I have written 10 minus 6. So, tell me what is this time for reaction? What is this? What is this number? Okay. There is a problem. What is the problem? This c a g number is right or wrong? See, he is right. My friend is right. Now it is 0.0138 10 minus 3. It is okay. Thank you very much. Thank you very much. So, he is correctly said this is 10 minus 3. Now what is tau r? 13. 138 seconds. Very good. Very good. Tau f equal to tau f. What is tau f? 46 seconds. Anybody else? 46 seconds. 46 seconds is right or wrong? What is so difficult about doing this? Rho b is given. R is given. b is 1 by 4. Kg is 10. 46 is right? Yes sir. Okay. Very good. Thank you. Tau d equal to rho b r squared 6 b d times c a g. Rho b is 4.3 by 225. R squared which is 5 squared 6 1 by 4. Diffusion coefficient is 0.03. And what else? C a g 0.0138 10 minus 3. What is this? 0138 10 minus 3. So, what is the answer? 10 minus 3 is here. Please. What is the time for complete consumption? How much is it? Rough estimate I will make. 10 minus 4, 10 minus 7. Just a minute. Tau d will finish. We will go to tau f. What is tau d? Do this calculation and tell me quickly. What is it? 7 10 raise to 5 seconds. 7 10 raise to 5 seconds. Is it correct? Very good. Very good. Very good. Now somebody is saying tau f is wrong. Somebody says it is 923. Now it means that now see now the moral of the story is that when you have product layer diffusion that is the most important resistance. With ion oxide that is the very important resistance and therefore, your reaction times are large. Therefore, you have to have large equipment to be able to take care of this. So, with this so we have gone through gas solid. So, we have an idea of the relative importance of all the resistances and then we conclude that whenever we have a product layer, the product layer will be an important resistance. We will stop there with that. So, that finishes what I wanted to do in gas solid. So, from today onwards we will look at biological processes. So, biochemical and environmental. So, this is what we want to do biochemical and environmental. Now the context all of us know the context or environment globally is in bad shape. We want to understand the reaction. So, that we will design our systems better and so on. So, let me take some examples to illustrate illustration number 1. So, what is this reaction? You have carbon, hydrogen, oxygen, nitrogen, phosphorous and so on. So, this is this is the biomass. So, it reacts with oxygen in the I mean in moist to give you carbon dioxide water and minerals in energy. So, this is what happens in soil. Soil respiration where organics in soil react with oxygen to give you carbon dioxide and water and in that process generate minerals for plants uptake generate energy for the biological processes of soil. So, waste organics that we all throw out from our homes and all that that is oxidized to release energy for soil processes. Release minerals for plants to take up and release carbon dioxide once again it is an important nutrient for plants. On other words what we are trying to say here is that waste organics actually serve a very valuable purpose. They give carbon dioxide for plants they give minerals for plants at the same time provide energy for biological processes of soil. But this waste organics that is around if it does not go to soil for whatever reasons then it accumulates in the urban environment to create various types of problems. And this is what we face around the world of course, the problems are very serious in cities like Bombay or if you go to a city like Sampolo in Brazil there are very serious problems urban areas are in very bad shape. Now, let us look at the opposite that means carbon dioxide plus potter pulse minerals plus energy is photosynthesis that we know. On other words respiration and photosynthesis are complementary chemical reactions of our environment. And if we can engineer our processes. So, that this complementarity is there in our design even for urban systems then clearly we will not face problems of waste management. The problem is that this synergy between photosynthesis and respiration does not seem to happen in the urban areas of the world. This is the fundamental problem for which we do not seem to have good solutions. But the fact the context to us is this is a gas solid reaction and whatever learn we have learnt so far is that this gas solid reactions we understand. Therefore, we understand how to design such systems if they are required. Let us take one more example of greater importance. Now, what is this reaction you have read in your school I suppose primary mineral reacts with carbon dioxide in the presence of moisture water to give you this mineral which goes into solution and this bicarbonate which goes into solution in that process generate soil or clay or sand. Now, let us put it in the context the context is this is what is called as mineral weathering it is also a gas solid reaction. Now, this reaction this reaction gives number of things in fact production of soil production of clay production of sand. For example, we mine lot of sand around the world including in bombacity and how does it get produced anyway what is this production takes place because of this reaction. This reaction goes on in the environment then primary mineral reacts with carbon dioxide of the environment to give you sand. In some places it has formed over millions of years therefore, we can mine forever. This clay for example, if you go to catch area for example, huge amounts of clay is mined for various purposes you know for industrial purposes and this clay was formed because of this reaction. So, mineral weathering is source of several products for human activity that power the industries of the world clay for example, or sand for example, we make lot of glass around the world using sand for example, because we are able to mine and this is come from this reaction. What is soil? Soil is reaction between primary mineral and carbon dioxide in the presence of water. So, on other words if you ask a reverse question can we make soil in a factory is it possible? Can we make clay in our factory? Can we make sand in our factory because we have run out of sand? Bombay city today has no sand for construction is well known part of the reason why our constructions in bombay city IIT for example, lots of cracks because you do not get good sand in the city. If you go to Jaipur for example, the constructions are excellent very good sand is available. Now, if we have to overcome this problem then we have to make this material in an industry. Can we do this? Can we engineer these reactions? So that we do not have to worry about mining our raw materials for industry from the environment but indeed we make it ourselves. What this reaction says it is possible? See if you can spend our time to study this reaction we can make soil, we can make clay we can make sand. So, it is a gas solid reaction are we short of carbon dioxide? No that is abundant are we short of primary mineral? The largest resource in this country is our basaltic rocks throughout this country billions of tons of rock is around. We are not even looking at this kind of possibilities to solve our problems. You see we must bring carbon dioxide of industry with the primary rock then these kinds of things become possible. Let me give you one more example the reaction is not balanced but what is this reaction nitrogen plus water plus energy gives you ammonia plus oxygen. Where does this energy come from? This energy comes from biological respiration. So, this energy if you can put it in here that means if this energy of respiration in soil can be channeled into nitrogen fixation pathway we get ammonia which we manufacture in our factories. Not only that we get oxygen which becomes a product of this reaction. This oxygen this oxygen can go here see the synergy. So, how beautifully this environment has evolved which we do not seem to appreciate that if this synergy can be established that means the energy of respiration. If we can channel into pathways of soil for nitrogen fixation then this oxygen can support energy respiration. And look at the other alternatives suppose we do not do this. Suppose we do not do this carbon dioxide accumulates our waste products accumulate. So, moment we break the synergy all types of problems starts to mount. On other words environment is about synergy in chemical reactions you know that is the most important message that comes from the environment. It is all synergy you break the chain problems after problem will evolve. So, by and large our design in the future of course our time is up, but your time is start starting your designs must look at synergies with the natural processes of the environment. Then by and large your designs would be long living and it will last a very long time. With this let me ask you few questions first question. Where do you see this reactions happening in daily life? Daily life at our homes I mean what I mean in our daily environment. Where do you see this? Where do you encounter? We encounter this problem in its most serious form in public toilets. It is most serious form and why is it? The reason is that there is not enough water. Because there is not enough water I have got some numbers here which says in households where the water is not abundant up to about 10 grams in 50 liters of ammonia can accumulate. 10 grams of nitrogen in 50 liters and this is what you know then this it cannot hold so much. It simply creates problems ammonia comes out hydrogen sulphide comes out and it is very obnoxious kind of environment. Now in the initially when these flushes were designed you will find that the designs that the flush that we do nowadays it does not hold more than about 4, 5 liters of water. Because there is not enough water around anyway so you cannot put too much. When it was designed 100 years ago it had something like 25 liters of water. So, from 25 liters we reduced it to 5 liters. Now in that process what has happened is that the concentrations have gone up and because concentrations have gone up all these problems have become more serious. So, if you have to solve problems like this for example, if you have to get rid of of I mean obnoxious order in public environment what do you do? We have to increase the availability of water so that you can dissolve these nutrients in large quantity of water then only this problem will go away. On other words solving this problem is that you have to increase availability of water. I mean such a resource where it is not available you have to increase availability of water. On other words we must look very seriously at ways of reclaiming waste water. You see I mean I mean it is one of the most serious problems of the developing world most serious problems of highly you know dense populations like Bombay, Sampolo, maybe Shanghai etcetera. What do we understand from these reactions? Carbon dioxide reacts with water to give you H plus HCO 3 reacts further to give you CO 3 minus carbon dioxide and water to form H 2 CO 3 carbon dioxide dissolves in liquid. So, what is this reaction? I mean in terms of the of what we see in our natural environment what shall we say? These are reactions of carbon dioxide in our environment. Now if the carbon dioxide concentration in our environment goes up for whatever reasons what happens to these equilibria? Suppose carbon dioxide concentration in the environment goes up it increases. What happens to all these products? We expect it will go up. So, we should expect that H plus of our environment will go up. So, you would expect that sea water will become more acidic correct sea water will become more acidic and what will be the effect of sea water becoming acidic in a physical sense when carbon dioxide concentration goes up H plus goes up the what is being said is that as H plus increases in sea water the sea waters capability to hold carbon dioxide decreases. See you will find in the natural environment lot of the carbon dioxide in this planet is in rock it is as carbon it rocks. Now lot of it lot of the carbon is present which is now becoming carbon dioxide due to fuel combustions and so on and therefore the effect of carbon dioxide in our environment appears to affect the holding capacity of carbon dioxide in sea water. Sea water holds close to about 4000 billion tons of carbon dioxide. So therefore, it has this net effect of increasing the CO2 even worse making the sea water even more acidic creating more and more problems for aquatic life apart from increasing the carbon dioxide. So, what we are trying to say here that these reactions these reactions essentially determine how life in this planet will perform and our designs must respect this chemistry. And that chemistry is what that the carbon dioxide concentration in our environment is about 300 ppm because it is this chemistry which has been able to evolve life and the aerobic life that you and I see today is because of the fact that this chemistry that work for billions of years. So, that respect must come. So, this is another feature that we must content in our design. Let me just look at one more example. What is this reaction? What do we make out of this reaction? It only says suppose we have a biomass. This can be a protein you and I consume a lot of protein which contains sulfur and if this reaction takes place in limited supply of oxygen. That means, wherever there is limited supply of oxygen which is waste dump is a good example, oxygen availability is limited. Then you find you produce acid there is production of acidity and then you also produce ammonia and hydrogen sulfide. So, you will find that places of accumulation of waste organics it is acidic environment is acidic and also there is ammonia obnoxious smell because of hydrogen sulfide and ammonia. On other words what we are saying is that if we prevent the aerobic respiration from working. If this does not work that means availability of oxygen is limited then this will happen and we see this in whatever we have done. So, what the what is called as the United Nations panel for climate change and all that what they are saying now is that this particular material is what is creating a huge amount of problem in terms of global warming. While we cannot prevent these reactions what they are saying now let us capture this. If we capture this and burn it then the effect of this see the global warming potential of methane compared to the global warming potential of carbon dioxide methane is much worse may be 6 7 times worse than carbon dioxide. So, they are looking at ways by which we can capture this methane. So, that you know we can capture it burn it perhaps in a boiler perhaps at home whatever then what is involved in this reaction what is involved in this reaction you have to biologically convert this to methane in enclosures where you can capture the methane. So, in what we want to do in biological reaction engineering concerned with environment is you prevent global warming by capturing methane alternatively we will ensure that the availability of oxygen is such that aerobic respiration can take place. Supplying oxygen cost money and aerobic reactions do not put so much demand on oxygen. So, there are certain advantages there are certain disadvantage will come to that shortly if we can do it nothing like it. For example, if you can maintain an aerobic environment in the waste dumps of the world Devanar for example, in Bombay or may be similar dumps in other parts of the world then these problems will not come. But what seems to be happening is that the urban areas of the world there is no space there is no space in the city of Bombay. So, what do you do? So, you keep on piling up waste and therefore, these reactions will occur because there is no space correct. So, what is being thought of now is that alright since there is no space let us keep on putting it in the same area, but capture this methane and we will use it appropriately this is the present thinking. Yes it is methane when you methane and carbon dioxide together people call it as biogas people call it as gober gas various things depending upon the type of raw material. Whether this would solve a problem is something people are battling with the present situation appears to be that the cost at which we are able to capture this methane seems to be not very satisfactory. We are not able to afford the cost of collection of this methane from waste dumps of the world because that collections cost lot of money. But if you have an enclosure in which you have reacted design to capture that methane the cost become even higher. So, unless the economics satisfy it is not worth looking at it. So, these are issues that we are facing, but the fact is in the absence of oxygen this reaction will occur in the presence of oxygen this things will not happen all these productions will not happen hydrogen sulphide will become sulphate it will get oxidized to sulphate nitrogen will get oxidized to nitrate and methane will get oxidized to carbon dioxide. So, in oxygen rich environment you are looking at all of them in its highest oxidation state. And therefore, you get the oxidized product phosphate become phosphate and so on. What is this reaction it is not clear please what is the question what I said was if there is inadequate supply of oxygen what happens in waste dumps of the world anaerobic reaction sets in and methane is produced. Now, since these dumps are not designed for capture of methane this methane goes into the atmosphere since the global warming potential of methane is several times that of carbon dioxide this is very very deleterious effect on the environment. What is the alternative the alternative is that we have designed dumps dumps are designed. So, that you can capture the methane there are many places in the world where they do this it cost money alternatively you this waste material is taken to a biological methanation where we capture the methane and then you take it for whatever purposes. So, waste dumps you have to design it to capture the methane or you take the waste directly to a biomethanation. So, that you can capture the methane both are being practiced, but there are problems of economics as a result we are not able to implement. What is this reaction and what does it have to tell us where do you see this reaction ammonia oxidation nitric acid factory correct nitric acid factory is where this happens all right. Now, we are not talking about factory now we are talking about soil where that is in soil this reaction takes place when you put proteins into soil or you put urea into soil or in other words nitrogen nitrogen to soil this reaction could take place nitrogen to soil this reaction could take place. Now, what happens to this H plus in soil it increases the acidity of soil now what is it that we observe around the world India in particular if you go to for a good example would be Gujarat. You see Gujarat is one state where because of these two rivers Tappi and Narmada the lot of ground water lot of ground water 1940's the lot of data 1940's ground water in Gujarat with the TDS or dissolved solids of about 250 to 500 in the range of 250 to 500 excellent quality water 50 years later it is 10000 50 years why is it is because this H plus reacts with minerals of soil and dissolves those minerals and all those minerals are now accumulating in ground water. The water which was in excellent shape 50 years ago has become unsuitable for drinking unsuitable for agriculture is something that we have done partly due to agriculture partly due to human interference with the natural process a partly due to the huge amount of carbon dioxide that we generate in our industry and so on. So, what is required in the see all these are reactions biologically mediated reactions all these have a technological solution we have to understand this technological solution and this is what we want to do in the next few classes try to see what is it that you and I can do to us understanding the foundations of these problems and to put in measures that will at least you know partly reduce these problems of the environment I will stop there.