 We have done the lumped mass system for modeling an explosion and as per the theory what we develop we found that there are two parameters which govern whether an explosion is possible or not. The first one is the ratio of the activation energy divided by the value of let us say the initial temperature that means this is so much joules per mole, this is also the ambient temperature T ambient into the universal gas constant and the other parameter is the ratio of the heat released kg per mole divided by we can say the specific heat per mole that means joule per mole Kelvin into T ambient. We said if both these parameters are large an explosion is more likely. We said based on this the activation energy must be large or the ambient temperature must be small. What does this non-dimensional term represent? It represented the tendency for spurting that means a small value of the chemical energy released time rather we said if I represent the chemical energy by the value of qc dot that is dq by dt and if I put the time over here well initially the reaction takes some time to take place but by the time the actual energy gets liberated it is in the form of a spurt and this happens when the value of the activation energy is large. Equalantly when I say well if the activation energy is large I have this spurting technique and the actual time of chemical reaction is therefore a small value. Well I say this is the chemical reaction time, this is the induction time td and therefore any explosion is associated with the small value of the chemical reaction time compared to the heat release time which we have seen in detail in the last class. Therefore what is this? Therefore activation energy must be large. We said for most explosives the value is around 80,000 to 160,000 joules per mole. The second point is if the ambient temperature is high what is going to happen? If the ambient temperature is high well the reaction starts off because the ambient temperature is high the reaction starts off at a fast rate and therefore the spurting tendency gets diminished therefore an explosion is more likely when the ambient temperature is small. Now this is the non-dimensional parameter which characterizes the spurt in chemical reactions or spurt in heat release after an induction time. This term just tells you what is the magnitude of heat release vis-a-vis the initial internal energy of the medium. If I have more energy release well an explosion is capable of forming a strong blast wave. These are the two parameters. What I do in the class today is maybe we will try to apply it for different chemical reactions one or two typical cases or maybe one typical case such that we know how to apply it. Like for instance let us go back to our starting point. You know we always took this example we said well I have a container in this container I have hydrogen gas I have oxygen gas which are mixed together. We told well hydrogen and oxygen can be in the container for years together let us say it is at room temperature it is at let us say 30 degrees centigrade which is the room temperature. It can be there at 30 degrees let us say one atmosphere pressure for years together nothing is going to happen. But if I want to start a reaction we said well I must put an electrical spark in it that means I deposit some energy at some point and maybe the reaction takes place. I am still within the confinement of overall what we said is uniform temperature uniform concentration therefore I am still looking at the lump parameter approach. I put some energy release and what does the energy release do? It takes the hydrogen maybe the hydrogen is takes place it decomposes the hydrogen or it breaks the hydrogen into H and H atoms 0.1. Because I am releasing some energy a hydrogen and oxygen maybe it breaks it up into H and H atoms we will see what this is. It also takes the oxygen molecules and breaks it up into O and O atoms and also maybe the hydrogen and oxygen since they are mixed together it could also break it up into hydroxyl OH and OH atoms. Now the H atom, O atom, OH atom are all active you know see this is something which is nascent stage therefore these are all active radicals and these are known as chains. That means what you do is active chains therefore what we do is the act of putting some energy release in it activates the hydrogen molecule into hydrogen chain H chain O chain and OH chains and therefore these three reactions I can call as let us say the chain initiation or active radical that is I say well I have chain initiation which takes place when some energy is released into the medium. I have let us say OH being formed it could react with H2 and what could happen? I could form H2O and H2O and 1H that means for every chain of OH another chain of H is being formed we call this as chain propagation that means the chain continues to propagate. It is still possible that I can have maybe the hydrogen molecule which is available when it when one of the chains like O comes and reacts with it I could form H and OH and similarly I could also have may be the O2 which reacts with the hydrogen chain which is formed here to form let us say OH and O O plus OH. Therefore in this case what happens one chain or one O radical or one chain forms two chains over here therefore we call it as chain branching therefore what has happened? The act of releasing some energy starts these reactions may be the chain initiation reaction chain propagation reaction chain branching reactions and what it does it produces a lot of these chains and you will recall we also talked in terms of may be that this axis represents the progress of a reaction we said that the energy could be plotted on the y axis well initially hydrogen and oxygen may be at 25 degrees it is at almost 0 may be at 30 degrees the energy content is over here and what happens they act I want to initiate the reaction therefore I put some energy it gives some activation energy takes it up to a transient stage over here and then what happens it reacts and forms the products and forms the product which is H2O and we know that the heat of reaction of the products H2O we will recall we had delta H2O heat of formation and the standard condition was something like minus 296 kilo joule per mole therefore this corresponds to the product may be lower energy level and therefore this is the heat of reaction which is given and therefore and this is the activation energy therefore what did activation energy do when we want to activate a radical or a reaction what we do is we are essentially forming these chain carriers and therefore what happens is I form all the chains which are in the transient state over here that means I form H I form O I form OH because of this is from this is all not stable it is all chains all active radicals and these things later on combine and what happens let us put this may be H plus let us say H may combine with OH to form may be H2O or let us say H plus OH you know whenever it combines like this may be sometimes the third radical also influences this but let us say well H and OH can combine to form a stable molecule or rather I can also form H plus O2 also can combine let us say HO2 but since these are all helped by the molecules which are available that means molecules which help such a reaction to take place mind you whenever we say chemical reactions the H atom and O2 atom must fit against each other if there are neighboring atoms which are also our neighboring molecules which also help this hitting and breaking the structure we call it as a third body I can also write it as HO2 plus M. Similarly, here H plus OH giving H2O I can write it as M giving H2O plus M I can form H2O by different means H2 plus O plus M that is a third body it could be OH it could be H2 it could be O2 which facilitates this reaction giving me H2O plus M well these are the recombination reactions I say these are the recombination reactions and when recombination reactions takes place well the HO and OH which are formed here come back to the products and I get the completed products of combustion. Therefore, we say well the reaction takes place through something like a chain initiation step followed by chain propagation and chain branching and ultimately the energy is released in the recombination why did I say energy being released in the recombination may be I have to look at these reactions a little more closely let us say what is the heat generated in this reaction because ultimately let us not forget this and what is it I am working towards we want to find out whether this reaction between H2 and O2 I do not know the proportion I presume it goes through these steps of as a function of time if I were to plot the heat release I for an explosion to occur I should have something like a phase during which not much energy is liberated an induction period followed by a spike and this is what I expect in a chemical reaction to take place and therefore let us keep our tendency for this spiking and therefore maybe it spikes and then it drops off over here therefore is it possible how do I calculate this let us go back and see what is the heat release in these reactions to be able to get the heat release it is necessary for me to know the heat of formation of the reactants and heat of formation of the products we tell ourselves well I go back to the literature and look at heat of formation I find that the heat of formation under standard condition for H is equal to the value of heat of formation for H is equal to 218 plus 218 kilo joule per mole the heat of formation of oxygen atoms heat of formation standard for oxygen is equal to plus 248 and the heat of formation of OH is equal to plus 2 no it is plus 39 kilo joule per mole you know in other words to form H I have to supply heat that means H2 has to be heated to form this therefore the heat of formation of H atom O atom and OH over here HF of OH are all positive this being 39 and this using this I can calculate the energy released in this reaction if I were to put it just below this what is the net value minus of heat of formation of H well heat of formation of H is 218 I have 1 plus 1 2 radicals 2 into plus 218 minus H is a naturally occurring element that means H occurs and it is at standard state minus 0 therefore the heat of formation of this is minus 436 kilo joules for this reaction so also for the reaction O2 is equal to O plus O we have heat of formation of O is plus 248 therefore it is minus of heat of formation of products I have 2 of O 2 into plus 248 minus anyway O2 is 0 therefore this is equal to minus 496 kilo joules if I take a look at this reaction these are stable things I have OH over here heat of formation is minus 2 into 39 minus 0 which is equal to minus that is minus 78 in other words kilo joules in other words the heat of reaction over here is minus 436 kilo joules heat of the reaction O2 dissociating into O is minus 496 the heat of reaction is minus 78 that means these reactions are endothermic that means they absorb energy no heat is being liberated in the three chain initiation reactions if we take a look at the heat of reaction corresponding to OH plus H2 giving H2O plus H2O what is it I get minus of H2O well we know by now yes heat of formation of H2O let us say heat of formation at standard condition for water as a liquid is equal to minus 295 kilo joules 295 296 kilo joules per mole therefore I put the value I have one mole of this therefore 295 and the value of H is but mind you you know 295 it is minus that means minus 295 plus H that means H is positive plus 218 that is the value of the two products H2O and H this is the heat and for OH well OH is minus 39 minus of plus 39 this is for the products and if I add this up the heat of reaction of this particular propagation reaction comes out to be equal to plus 28 kilo joules. Similarly I can calculate the energy for the chain branching reactions now I know the heat of formation of H I know the heat of formation of OH I know this I put together similarly I get the value for the chain branching reaction as H2 plus O giving me something like 70 kilo joules in a similar manner and for the last one O2 plus H giving O plus H the energy released in this reaction is 8 kilo joules well we find yes these two reactions and this one these three the chain propagation and the chain branching reactions are exothermic but the exothermicity is small how do I say it is small let us work out one or two cases let us find out the exothermicity of let us say a reaction like H2 plus O plus M is equal to H2O plus M if I take this well the heat of this particular reaction is equal to minus of H2O that is minus 285 and I have O over here we know heat of formation of water is plus 248 therefore I have minus 1 into 248 therefore the heat of this recombination reaction H2O plus H2 plus O plus M is equal to H2O plus M comes out to be something like plus 534 kilo joules similarly if I were to write the heat of formation of this may be H plus O2 plus M H O2 plus M we said well the heat of formation of H O2 if I go to the standard literature tables and plot it out well H O2 has a value of plus 2 kilo joule per mole and therefore the energy liberated in the reaction H2 plus H plus O2 plus M is equal to H O2 plus M is equal to we are talking of may be minus of plus 2 that is the product we have two M and M is the same over here minus I have H over here the value of heat of formation of H is equal to plus 218 minus 218 which is equal to something like 216 plus therefore because this is a small positive value over here it is plus 216 kilo joules therefore these recombination reactions have high value of exothermic heat release whereas the chain propagation and chain branching reactions have very small values 8 or 28 and all this whereas the chain initiation reactions have hardly they absorb energy they are endothermic therefore you know what is going to happen let us say I dump some energy I start this chain initiation chain propagation and chain branching reactions and as more chains and more chains are being formed I have the recombination reactions taking place and what is it I find I find well the energy is essentially liberated in the recombination reactions and hardly any worthwhile energy gets liberated here with the result what I can now tell is well during the initial phase of the reaction when I activate the reactions that means I look at may be the temperature gain what is going to happen the temperature hardly changes it just is endothermic which is balanced by the energy release and once the recombination reactions take place the temperature increases and I have a signature like this in other words initially there is hardly any value of chemical heat release and which is followed by as per therefore in this period I have chains which are being formed chains being initiated followed by a period in which the chains recombine to give heat therefore we tell ourselves well during the initial phase I have chain initiation chain propagation and chain initiation chain branching reactions that means initiation propagation and branching reactions for the chains and the chains combined in the period later on to give energy that means Tc corresponds to recombination and during the initial phase I just do not get much energy release well this matches with whatever we were telling for high activation energy and this is how a reaction progresses in practice having studied this you know why I am why are we going into these details see tomorrow we want to control a reaction or control an explosion if you want to control a control an explosion it is necessary that I do not allow this to happen may be I should somehow absorb the chains during the reaction and this is what I will do subsequently but to be able to be clear on what little we have been doing so far let us put it together in a plot of let us say the last plot could be may be as a function of dt by dt that is the rate of temperature rise as can I work it out as a function of time which means as a function of the temperature itself because we said as time increases the temperature increases therefore as a function of time I could be able to do this I want to take a look at this therefore we again say well my constraints are the following I also have a lumped mass system I have a volume I have a density of the explosive which is rho the CV of the explosive that is the specific heat of the explosive is CV I have the ambient temperature T ambient the temperature at any time is T the initial temperature could have been T0 well what did we tell ourselves well I have the mass of the system rho V into the specific heat that is the thermal mass of the system into I have dt by dt which is and this gives me the rate in so much kilograms or kilo joules per second or joules per second and where does this come from it comes from the rate of the reaction that is so much moles per cc second into the the value of volume that means moles per second multiplied by q so much joules per mole it also give me joule per second and this is what I am balancing I want to see how the value of dt by dt by dt that is d capital T by d small t changes and therefore now I write the value for the reaction I say well reaction given given by ACN exponential minus activation energy by r0 t into I have Vnq over here and you know you will also recall we had an expression for c we derived one expression for c these are well I can write the expression for c is equal to c0 into temperature maximum minus temperature at any point in time when the concentration is c divided by Tm minus the initial temperature we plug this in and then we say we can get the value of dt by dt is equal to well in this particular expression we need not be taken into account I have aq aq over here well I have divided by rho cv and then we have taken these terms over here therefore now we have only the value of cn therefore I can write c0 raise to the power n where we are still considering an nth order reaction or we say the rate of a reaction is proportional to c to the power n n could be 1 it could be it could be even 0 it could be 2 for a second order reaction and we said most of the reactions are n is equal to 2 and we said is equal to Tm minus t divided by Tm minus t0 to the power n AC0 into we have the exponential term coming over here minus ea by r0 into t into the value of q over here well this gives me the expression for the rate of temperature increase and if I look at the rate of temperature increase and let's say I first plot it with respect to temperature that means on the x axis I have temperature what is it I get well when the temperature is initially let's say the ambient temperature have Tm minus t0 Tm minus t0 well this is a strong term well it's a small number over here and as temperature increases and reaches the final maximum value it becomes 0 that means ultimately it becomes 0 over here and in between if my activation energy is large like what we have been talking earlier well the temperature increase is like this it goes it goes after a long induction time let's let's put it down more reasonably we tell ourselves initially the temperature increase is small because the activation energy ea controls it after some time well it peaks like this and it comes down to 0 over here this is the type of dt over d time curve and what is the temperature how it how does it change well initially the the temperature is t0 over here that is and it reaches the value and therefore if I were to now change this x axis instead of temperature I plot the time what is the type of relation what I get let's let's put everything together such that we can now interpret the reaction taking place and the rate at which the temperature build up takes place I have temperature versus time I have dt by dt I want to plot it on the axis again I also want to plot c as a function of time well let us first plot this well it starts with a temperature t0 and it goes to a maximum value tm this is the value of t let us now plot the value of the rate of temperature increase well it's it starts at a low value maybe somewhere over here the peak occurs well it goes like this peak occurs and comes back over here this is the value of dt by dt if I were to plot the concentration and if I denote the concentration c at any time well initially the concentration is c0 the final value of concentration is 0 because all the fuel is consumed and therefore maybe if this is initial value it starts like this and comes over here well these are the concentration plots the temperature increase plots the rate of temperature increase plots and we tell ourselves well based on whatever we have discussed for hydrogen oxygen and it's true for any hydrocarbon or most of the substances we can tell ourselves it is during this period that the chain initiation chain propagation and the chain branching reactions are there and avalanche of chains are being formed and these during this period recombination reactions take place during the actual time of heat release that is recombination of chains to give me the heat and these are essentially the chain formation stage we should know what must I do to control supposing I do not want an explosion and all of us are looking at at ways to control an explosion or to stop an explosion what must I do to to to stop the chains being formed maybe I should add something such that the chains do not get formed but before I do this let's quickly take a look be the chain initiation reactions we told ourselves well h2 forms h plus h you know the formation of chains is is is sort of facilitated when I have other molecules maybe I have something like maybe a wall maybe I have something like an energetic wall which can help in the reaction or I have maybe the oxygen molecule also provides a collision therefore we say the third body helps in the reaction and this third body does not take part in a reaction it helps in the reaction but it is it is available to me after the reaction as the third body itself therefore we say h2 plus m is equal to h plus h plus m and the rate of a reaction we called as omega so much moles per cc second and that is equal to 2.08 into 10 to the power 15 exponential the way we have been telling second order reaction therefore concentration of hydrogen into concentration of whatever is available as a third body over here we also find that there is a weak dependence on temperature coming over here you will recall when I got the value of omega I did say well it could be a small function of temperature and when we do experiments we do find it is a weak function of temperature in addition to temperature coming in the exponential term we do have temperature coming over here similarly the chain initiation reaction of o2 forming o plus o chains is also helped by a third molecule and the rate of reaction is given by in this case again temperature is quite weak over here explicitly a combination of the oxygen and the third molecule over here this is the reaction rate over here similarly for the chain proper chain you also could have chain initiation reaction in which I have h2 plus o2 forming o h plus h this is also chain initiation well in this case the temperature effect is not there the rate of reaction is given by this I can I knowing the rate at which the chains are there I can write the value of omega is equal to d hd of h2 by dt is equal to omega into using the law of mass action as concentration of h into concentration of of the third body I can solve for the rate at which the chains are being formed that means the rate at which h2 is getting depleted or dh by dt is equal to omega into h2 in plus m over here and of course I have two of them therefore two of dh by dt is equal to this and therefore I can find out the rate at which the chains are being formed this is for the chain initiation reactions I get a similar one for the chain propagation reaction o h plus h2 is equal to h plus h2o the rate of a reaction is given in terms of the concentration of o h and h2 and we have the rate of a reaction is there similarly the chain branching reactions h plus o2 giving o h plus o and the chain branching reaction that means for each chain I form two more chains or o plus h2 forming two more chains the reaction rates are given by this these are somewhat temperature dependent reactions in addition to temperature coming under the exponential sign over here therefore we have the chain initiation chain propagation chain branching reactions and when we talk in terms of chain termination reactions I write these two h plus o2 plus a third body mind you you know when we talk of these recombinations and a third body over here you know the collisions you need a number of h number of o2 because three bodies are reacting together at a point this happens essentially at high pressures because at low pressures you know you do not have the feasibility of three molecules colliding with each other therefore these three body reactions essentially happens at higher pressures and we have h o2 plus m we have h plus o h plus m giving h2o plus m and the rate constants of these two are given by this and we just now showed on the board that the energy release in these chain termination reactions are very much greater than the endothermic heat release or the energy absorption in the chain initiation and the small values of energy released in the chain propagation and the chain termination reactions therefore it is possible for us using these known values of experimentally determined rate constants to or the rate of a reaction to be able to plot to be able to determine the values of let us say h and o h and let us say o which are being formed in the reactions and find out the period or the time duration during which the chain initiation takes place and the recombination reactions take place having said this let us come back and take one last look at the chemical explosions from the point of view of the following well we said well hydrogen oxygen will stick to this because it is quite illustrative we said well I have let us say hydrogen and oxygen which is available can I say well all along for all conditions of pressure and temperature that means I want to take a look at the explosion diagram for pressure as a function of temperature in other words I have a mixture of hydrogen and oxygen in a container I have hydrogen and oxygen let us say I vary the pressure and temperature I want to find out well does the reaction does the explosion diagram look quite the same for all pressures and temperatures or are there some regions of pressure and temperature in which no explosion is possible are there some regions of pressure and temperature in which explosion is possible and that is what I want to take a look therefore we just do whatever little we have discussed so far let us say I am I am interested in pressure let us let us consider a pressure limit of let us say 10 kPa which is let us say 0.1 of an atmosphere and let us consider a temperature region let us let us now restrict ourselves to a temperature region around let us say 500 degree centigrade or let us say 400 degree centigrade because you know you will recall we said that the auto ignition temperature of hydrogen oxygen mixture is around 440 degree centigrade or so therefore let us let us keep this in mind let us say in Kelvin I am interested in a temperature of around let us say 700 Kelvin this is my region of interest which I am looking at therefore if I am basically interested in the low pressure region that is sub atmospheric pressures that is I am interested in this particular region can I can I say under what conditions will an explosion occur in this particular region well I I know that when the pressure is large that means when pressure is large that is of the order of let us say somewhere over here around 8 or 10 kPa pressure is large therefore the the collisions between the molecules H2 and O will be higher and therefore we expect more change to be created when the pressure is higher when the pressure is lower well no chains are created but if the temperature is higher well the molecules have higher energy and therefore I can say chains will be formed something like this that means when I talk of very low pressures and high temperatures I can say well chains are created in this particular region that means more chains are possible in this region as the pressure is increases at lower temperature more chains are being formed and therefore I tell myself this is the region in which more chains are being formed and therefore this corresponds to a lot of chains and these chains can recombine and generate heat therefore this shaded region that is the hatch region corresponds to an explosion in this region what is happening some chains are definitely formed but they go back they are absorbed by the wall because the chains are so few and therefore this region corresponds to no explosion this is in the region of sub atmospheric pressures let us now increase the pressure and still discuss this diagram further now in the region let us say I go to something like like like 100 kilo or let us say yeah let us say I go to 100 kilo Pascal or a little higher let us say 200 kilo Pascal then what is going to happen you know in this region well chains are being formed but we also found that the reaction H plus O2 plus M forms something like HO2 plus M which does not really give that much of heat but what happens the chain H gets consumed and maybe at higher pressure and higher temperature more and more chains are getting consumed and HO2 is being formed and HO2 you know is something which can can be kept as it is for some amount of time that means at intermediate pressures and temperatures it can form products and therefore this fellow continues to form HO2 that is the chains get depleted here we also find yes at higher pressures more and more of these things are getting formed chains are getting consumed therefore I can think in terms of the situation something like this wherein maybe if the pressure exceeds this limit well the chain gets consumed here since chain gets consumed over here no explosion is possible in this region that means I put that graph over here I say in this region no explosion is possible but in this particular region well I have the chains which are being formed and recombination reacts reaction generates heat but in this region pressure is high as temperature increases this reaction takes place no explosion is possible and ultimately therefore I say well this is one limit where in no explosion and in this region I have explosion taking place and now if I increase the pressure still further what happens well I have HO2 being formed and this HO2 can react with H2 to form H2O2 plus hydrogen that means chain is getting formed again and this H2O2 I am sorry HO2 plus H2O2 plus H over here and the H2O2 can again at high pressures heat against each other and form OH plus H therefore at very high pressures I again start getting the chain carriers which can go which can produce a lot of chains and these chains can combine to recombine to form heat release like H2 plus O giving H2O or H plus 2H plus O giving me water and therefore the curve goes back over here therefore in this case I have explosion taking place over here and in this series this continues over here no explosion therefore what is it I see well this corresponds to the second phase wherein the chain termination reaction is not allowing the chains to take place and I have no explosion over here but however I get an explosion here I have no explosion here no explosion here beyond a certain value of pressure I do get an explosion coming over here and therefore this gives me an explosion diagram this is the limit where we say is the first explosion limit wherein the vessel absorbs the hydrogen and I have no explosion in this case I have the chain termination reaction H plus O2 plus M which absorbs all the chains and does not allow an explosion to take place and well in the third one well because of the high pressure effects more and more chains are being formed I have an explosion taking place this is the explosion diagram and this this is called the third one wherein I form chains readily I illustrate this in in one of the slides which I will show now and what is it we see in this particular slide I just show the three explosion limits well this is the first explosion limit at low pressures this is the second explosion limit wherein the chain termination reaction robs of the chains over here and I have at the higher pressures the third explosion limit therefore we have a first explosion limit only on the right side which is shaded you have explosions you do not have explosion in this particular region over here and this is how we interpret explosions for the different explosives well this is all about it let's let's spend a couple of minutes on what I told earlier namely we wanted to find out whether I can control my explosion supposing I don't want an explosion what is it I do let's let's see what could be done in other words I want to do something to my explosive let's again get back to the starting point namely I have a vessel let's say I have a container containing hydrogen and oxygen now to prevent an explosion I must not allow my chains H OH may be O to sort of increase in large number and these chains should not react and form the chain termination reactions or given spurt in energy release in other words I want to control such that I do not create these chain chains at all and for this what is it we do well we use some substances which which are like methyl bromide we say CH3 methyl bromide we say methyl iodide or we say methyl chloride CH3 Cl you know these are halogen based compounds they can in general be denoted by Rx where x is the halogen and R is the organic radical the halogen compound Rx reacts with the chain carrier H that means it is robbing the chain carrier from the explosion forming Hx that is the hydrogen halide and the organic radical you know to be more clear about this let's take the specific example of let's say a bromine compound that means we are talking in terms of R halide being bromine RBr the RBr reacts with hydrogen that is the hydrogen chain to form R plus hydrogen bromide this hydrogen bromide so form reacts with more of the hydrogen chain that means it absorbs or robs more of the hydrogen chain from the reaction to form hydrogen molecule plus the bromine atom and therefore now you have effectively taken out H from the reaction that is the chain carrier from the reaction the bromine ions act with bromine ions to form the bromine molecule that happens in the presence of the third body that means the other atoms and molecules in the reaction and therefore bromine forms the bromine molecule and the bromine molecule further reacts with the hydrogen atom that is the hydrogen chains to form hydrogen bromide and bromine atom and now what is happening is see our again robbing hydrogen from the reaction or from the explosion that is the chain carriers are being further removed and this bromine atom again robs further chains of H to form H plus Br in the presence of the third body to form hydrogen bromide therefore you see a series of steps by which the hydrogen chain carrier is being removed and also we find well hydrogen bromine also our hydrogen bromide gets further regenerated in the reaction which effectively promotes further robbing of the chain carriers therefore the regeneration of HBr effectively removes the chain carrier H and in this way you are removing the chain carriers and when we remove the chain carriers well you don't have much of the chain carriers to prolong your induction time and also accumulate chain carriers by which you have a spurt or a peaking in the reaction in the absence of chain carriers therefore you do not therefore get an explosion namely a long induction time followed by the spurt in the reaction and this is how the halogen compounds like Rx are effective in removing the chain carriers and inhibiting the chemical reaction which could lead to an explosion we must remember that phosphorus compounds similarly react with chain carriers and eliminate them and therefore also inhibit the explosion process you use phosphorus compounds and these phosphorus compounds could be mono ammonium phosphate the formula being NH4 H2 PO4 and the second being trisodium phosphate formula being Na3PO4 therefore adding such inhibitors to our explosive substances will arrest the formation of chains and will not allow the explosion to occur well this is all about the theory of explosions and it's time we go to the next chapter but before I do that let's let's take one small model problem such that we revise the total what little we have done you remember we talked in terms of different types of explosions and I think it's necessary for me to tell one last thing you know supposing let's say I have a leak I have a leak of let's say propane C3 H8 or let's say I have a leak of methane such as it happened in Ural mountains in 1989 in which I formed a huge cloud of methane air mixture in because of some pipeline burst and supposing let's say I have this methane air which has which is accumulated as a huge cloud you know mind you this cloud is at atmospheric pressure the ambient is also one atmosphere pressure therefore it is just an unconfined cloud as it were it is not that I have a container or a bomb in which this particular gas mixture is available I can understand when I can contain a mixture in a bomb well pressure builds up on the bomb burst but this is an atmosphere itself in atmosphere itself supposing I have a cloud of methane or a cloud of propane with air which is formed can it result in an explosion the question is well it's not confined but I do realize that maybe propane methane or hydrogen cloud goes through the process of having let's say I plot it now the heat release from this particular cloud what happens initially I have an induction time followed by induction time I have a spurt in the energy release because I have this spurt in energy release after this induction time and this chemical reaction time during which actual heat release takes place is very small what happens is because of this spurt blast waves are generated and you can have an explosion even in an unconfined environment it's not necessary that I put my explosive only in a bomb or in a container and have an explosion it is possible to have unconfined explosions and this is what we will be considering subsequently we will talk in terms of confined explosions and we find even in unconfined conditions that is open unconfined it is possible to have an explosion let me quickly go through one last example may be a worked out problem in which I will now consider may be the example of the Texas city disaster we have done this problem already for the heat release what did we say in the case of the Texas city disaster we have ammonium nitrate and this ammonium nitrate was contained in bags several bags of them something like 7700 tons of ammonium nitrate this happened in 1947 and what is there you have the hull of the ship in which all these bags of ammonium percolate where ammonium nitrate I am sorry we are all packed one against the other you know this represented something like a well insulated system and little bit of energy release we said well under all conditions I can have some at even at room temperature some element of chemical reaction takes place there is a small amount of temperature increase taking place and therefore the small amount of temperature led to higher and higher temperatures and higher energy release and it resulted in a in a very drastic explosion now I want to find out what is the preheat required to cause this explosion when we look at ammonium nitrate well from literature I know that the activation energy is something like 90,000 joules per mole and therefore we say let on that day explosion occurred let the temperature be 30 degree centigrade therefore now I know from what we studied is Tc that is the critical temperature minus T ambient is equal to we have the value of r naught T ambient squared divided by the activation energy and therefore I get the value of the critical temperature minus the ambient temperature as equal to 8 universal gas constant into T ambient T ambient is 30 that is equal to 303 Kelvin divided by 90,000 and this number works out to be equal to 8.48 Kelvin degree centigrade therefore the value of the critical temperature is equal to something like 30 that is equal to 38.48 degree centigrade you find that the preheat temperatures are still small the critical temperature is 38.48 which Kelvin comes out to be equal to plus 273 is equal to the the value is 311.48 Kelvin I add 273 to it now I want to find out the rate at which energy gets liberated and we told yes I must evaluate it at Tc and there are some experiments which are conducted on the rate of reaction for ammonium nitrate and it has been shown by Heiner in one of the combustion symposiums that the rate of reaction of ammonium nitrate is given by 10 to the power 13.8 into exponential of minus 80,000 divided by r naught T second inverse that means it is something in which even the concentration is not playing a role it is this is the rate of a reaction and therefore how do we evaluate the heat release we find out the rate of a reaction under the critical condition that is Tc that means temperature is going to be 311.48 we evaluate so much second inverse we already know if I have ammonium nitrate how much energy is liberated we say well NH4 NO3 we have the products we find out the heat of formation of the products minus heat of formation of reactants we know what is the energy which is liberated per unit mass therefore for 7700 tons what is the energy which is liberated and this energy liberated into the rate at which it is being liberated so much per second that is second inverse gives me the rate at which energy gets liberated in this particular reaction and we find that the time constant of this at Tc is a small number compared to the heat characteristic time and therefore the explosion takes place this is how we do any problem relating to an explosion well this is all about the theory for explosions but mind you we have made a drastic assumption the assumption is we assume a lumped mass system for the explosion in the next series of classes we will try to see what is the effect of this lumped mass system and whether we can really remove this assumption and proceed to interpret an explosion and then we go back and look at confined and unconfined explosions well thank you an announcement please in the lectures 13 to 19 the energy release rate of energy release and the thermal theory for explosions have been covered in detail references for further reading connected with these lectures namely 13 to 19 and a set of homework problems are given in the downloads of this video course