 So we take it as axiomatic that atoms and molecules cannot react unless they collide with each other and we are primarily looking at chemical reactions you try to emphasize this all the time so essentially we are looking at electronic exchanges and not going all the way down to the nuclei course there are exceptions to situations like where we have to have molecules and atoms collide with each other you could have thermo the thermo reactions that are sensitive to temperature or in thermal response like decomposition thermal decomposition or you could have photosensitive reactions and so on okay. So let us now look at what is called as a law of mass action will just proceed systematically here so as an example if you now take the hydrogen oxygen reaction so we could now think about a change in the number of moles of hydrogen in terms of a decrease and therefore we have a negative one could be equal to a change in the number of moles of oxygen divided by a negative half because for every one mole of hydrogen that decreases you have only half a mole of oxygen that decreases and that could be equal to then the change in the number of moles of water divided by plus one to indicate that this is a gain as opposed to a loss for the other two so the sign and the value in the denominator for these are actually derived from whether they are reactants or products and you could also now look at what is the what is the loss relatively speaking one mole of hydrogen is lost with half a mole of oxygen that is lost to produce one mole of water so that is what that is what we write so if you now consider a general reaction like an algebraic form that we have considered before except we now write nu i single prime script mi gives sigma i equals 1n nu i double prime script mi so previously we had a ni single prime and an ni double prime we have switched to nu i suddenly we became Greek yeah there is a reason for this we will now get into the difference between what is called as a order of the reaction and molecularity of the reaction so if you are now thinking something like half it refers to something like number of moles but if you are looking at a reaction that is happening at the molecular level you cannot have like half a molecule react with another molecule given you are looking at molecules and atoms colliding with each other right so half a molecule cannot react as if it is half hearted or something like that right so a molecule reacts in molecule reacts so we will try to make the distinction over there but as far as what we are doing now since we have a half here you could stick to our notion of number of moles for the coefficients here okay which could be now taken care of by a nu i single prime and a nu i double prime as before which was done by ni single prime and ni double prime and if you want to now write a set of equations like this this is actually a set of equations is actually a set of three equations for three reactants three three species there so similarly you could for N species you could now write N N N equations right and the way we would do is since we have actually accounted for all the N species on either side with a nu i single prime and a nu i double prime and we know that if something is a reactant the nu i single prime is non-zero and the nu i double prime is zero and if something is a product it is nu i single prime is zero and nu i double prime is non-zero so taking that into account you simply have to take its double prime minus its single prime to get you the right sign as well you can get a negative sign if this is zero and that is non-zero okay if you get a positive sign if this is non-zero and this is zero it does not have to be exactly zeros as well you could have like excess reactant excess reactant that goes to the product side which will be of a lesser value in terms of stoichiometric coefficient then on the right hand side then you get a positive well so negative value here as well okay so all that stuff will be taken care of by this notation and then what do you do with all these things that are equal to each other okay so we now say all these things that are equal to each other could finally be equal to a elemental change in a in a in a quantity called psi yeah so we now say this is equal to d psi where psi refers to sky refers to psi refers to degree of advancement advancement of the reaction right so it tells you how much of the reaction has proceeded depending upon how much elemental change in the number of moles has been affected for each of the species okay so now you divide the divide the above equation by volume that means you now look at whatever is happening for unit volume we right so N over V is equal to C which is a molar concentration molar concentration therefore dn over V is equal to dc and look at the time rate of change right that is to say dn over dt for unit volume is essentially dc over dt right now this is very significant here ultimately what we are looking at is the rate of change the time rate of change of concentration of a species of any species okay so of course when you are now saying N it means the total number total number of moles but we could if you put a subscript I that refers to the number of moles of species I yeah so you could do the same thing for any of those therefore you can now do this for all of these that means you divide the entire set of equations by volume and then take the time rate of change in each case then you now get dc1 over dt divided by nu 1 double prime minus nu 1 single prime equals dc2 over dt divided by nu 2 double prime minus nu 2 single prime etc for the general case of the ith species dci over dt divided by nu I double prime minus nu I single prime up to dcn over dt divided by nu n double prime minus nu I single prime nu n single prime so what happens finally right so you now have a dc over dt divided by volume right so you now have okay. So this is what we would like to call as omega which stands for the reaction rate in that sense the reaction rate is a term that is common to a reaction all right it is you now give a you are given a reaction that means it will have a reaction rate okay what is it the rate of what are we trying to measure out of this actually it is something that is common to all the species strictly speaking what we should be asking or be interested in is what is the rate of production or depletion of a particular species okay that is given by dc1 by dt dc2 by dt etc and therefore dc1 by dt will be nu 1 double prime minus nu 1 single prime times omega and so on for each of those species right so that is such so you have like a common quantity reaction rate from which you will now get the rate of production or depletion of the individual species that is how that is how we try to go about doing this. So what does this really mean it is essentially a volumetric time rate of change of the degree of advancement of the reaction so that the reaction rate essentially means a the rate of change of the degree of advancement of the reaction per unit volume so keep in mind there are two things about a reaction rate this is very very important for you to think about all the time whenever you are bringing in reactions into picture the reaction rate is coming it is going to come into picture and correspondingly the rate of production or depletion of individual species it is always per unit time per unit volume okay it is volumetric rate volumetric means per unit volume rate means per unit time okay so these two things are always there and typically in SI units if you are interested in putting this in things like mass balance and so on we are always looking at measuring this in terms of kilograms per second per meter cube right that is what we are looking at we are beginning to look at at the moment of course we are now having a omega here which is essentially looking at the change in the number of moles per unit time so this moles per unit time per unit volume it is not yet kilograms right so that means we have to now throw in a molecular weight so you cannot do that for a common reaction rate the reaction rate that is common to all the reactions you could do so for the individual species because individual species will have individual molecular weights so you can throw in molecular weights for individual species to convert the rate of change of number of moles of those species to the rate of change of mass of those species right that is what we will do ultimately but I am just trying to tell you what what is all coming ahead and what are the important things that we have to keep thinking about so so so if you now keep that along with that line so in general we can say dci over dt if this is actually omega i this is the rate of production or depletion should in fact we should say net rate of production depletion of species i right species i net rate of production of I will explain why we are saying net in a few minutes essentially what you are thinking is not just one reaction here we will now try to have a sequence of reactions in which a species could be produced in one reaction it could be consumed in another reaction okay so there is like a net rate of production if the depletion so that that is what that is what the word net would mean and we will see this a little bit more explicitly but essentially what we are now saying is therefore omega i is nu i single prime double prime minus nu i single prime times omega okay so where are we we are supposed to in this framework we should now be able to write out the law of mass action. So the law of mass action the law of mass action states that right I am not going to write long sentences like you go through in high school there is a difference now states simply that omega equal to k pi i equals 1 to n ci to the nu i single prime this is what you have actually learnt in high school okay now it is all put an algebraic form so this simply means that the so long as you use a nu i single prime as the exponent over there to ci what it simply means is the the rate of reaction the common quantity for all the species that are participating in a reaction yeah is equal to a what is what is called as a rate constant or a specific rate constant case it is not a constant we will now expand on that a little bit more time but it is like a proportionality constant so effectively you are saying the rate of reaction is proportional to the concentrations of the reactants raised to their stoichiometric coefficients right so when you say reactants that means you are looking at nu i single prime keep in mind you are taking like a product of all the all the species so if some of the species are actually products the nu i single primes for them will be 0 therefore they do not really contribute you now have ci to the nu i single prime for them to be equal to 1 and therefore you do not you do not have to worry about it yeah so it is it is it is pretty general expression there and the next thing that I would like to point out is K here K is equal to a e to the – e over or ut sometimes there is a subscript a to e here so to indicate its activation energy but it is essentially activation energy there so e to the exponential e to the exponential to the negative e over or ut or u stands for universal gas constant right in the Arrhenius so this is the this is what is called as Arrhenius Arrhenius law the Arrhenius law essentially gives you the temperature dependence of the rate constant or this rate constant is also called specific reaction rate okay it is essentially the reaction rate for unity concentrations of the reactions the reactants so that is why it is called specific reaction rate but it is also called the rate constant it is not really a constant it depends on the temperature and the Arrhenius law essentially gives you the temperature dependence you can also write this if you want to now include many times many times K equals B T power M e to the – e over or ut is also used is also used that means there is a polynomial dependence of the rate constant on temperature that is T to the M and there is a exponential dependence on temperature okay so the exponential dependence obviously is a lot stronger dependence when compared to the polynomial dependence so sometimes you could you could tend to ignore it and say let us not worry about the polynomial dependence let us simply say take it as a e to the – e over or T right and that there you are taking in only the exponential dependence the exponential dependence is the biggest problem with combustion so what we are actually now going through is in some sense like the is one of the biggest tumbling blocks in solving combustion problems the exponential dependence is a highly nonlinear dependence so essentially this is what and depends depending upon the value of e and typically when the value of e is larger and larger the nonlinearity dependence of K or the reaction rate in general on temperature becomes more and more nonlinear okay so it becomes very very sensitive temperature that means it does not change over a fairly wide range of temperatures but you now cross a certain threshold temperature it just goes boom and what that means is we should now be able to capture this variation within that very very very small change in temperature because because the suddenly changed until then you do not have any reactions happen significantly speaking right and all of a sudden you have all the reactions happening okay so there are very quick changes that keep happening in space or in time depending upon where you are looking at unsteady problems or or both okay so you typically also have a spatial very spatial resolution issue because of these things and this poses a what is called mathematically what is called as a stiff situation because when you are now looking at convection in a flow and then you have a flame and then suddenly when you go to the flame you are now having a sudden increase in temperature right and so long we are not been having such a high variation in temperature so you are okay with resolving the temperature over larger distances and then suddenly over here where the flame is you now have to resolve very very close and now begin to keep track of these fast changes in temperature if you were to do the same thing over there it is sort of like there is hardly much change so you are wasting your time doing that okay so like for example if you do computational methods you would want to take like a very coarse grid relatively speaking over there but when you get to the flame you want to now suddenly have a very fine grid but you are supposed to find out where the flame is and therefore you should be you should adapt your grid and stuff so it is difficult right so typically you now have a much different length scale for combustion than for reactions so if in the same equation as we will see both the reactions and the convective processes occur and they are happening at two different length scales it is very difficult to resolve both of them for the food with the same scale right and that is typically what is called as a mathematically stiff problem and you will get into these stiff problems primarily coming out of this yeah right so we are just beginning to get into the gory details of combustion if you will yeah there okay so let me just box this for you to make it look good one of the things that I would like to say is whenever you are dealing with the Arrhenius law you are always using the universal gas constant never make a mistake never have any confusion about this okay so typically when you are busy writing an exam you are like they use the specific gas constant or universal gas constant what do you do now no Arrhenius always uses the universal gas constant okay so this is something that you are always keeping your mind when you are doing like back of the envelope calculations exam problems and so on okay second point in this course we are now take going to take these as laws that means we are not going to ask how we got these okay how are you going to how did you get these did experiments yeah I guess yeah this is somebody actually took beakers poor things stirred up and started measuring things is it out is it out they figured yes that is that is true they did that okay plus you could also have a quantum mechanical basis for this for example this is what is called as a or B okay so now you can call this pre exponential exponential factor pre exponential factor right so from quantum mechanics we will be able to find out that this actually depends on probability of what is called as effective collisions okay so what is meant by effective collisions what it really means is you need to have molecules that collide actually come in at a particular orientation with respect to each other so it is it is like you have a molecule that means it has a bunch of atoms in a particular arrangement you have another molecule of atoms its atoms in a particular arrangement when you now have a reaction that is going on it is one atom of this molecule and another atom of the other molecule that are the ones that are actually exchanging electrons and creating bonds there and then disrupting other bonds there right that means when they collide you need to have this particular atom of this molecule collide with that particular atom of that molecule that means they should be oriented in particular ways right and the probability with which they would orient in those particular ways is less when compared to if you did not have to worry about the orientation right so all that stuff is buried in the pre exponential factor okay so you can derive these things from quantum mechanical basis and a good graduate course would do this right or we would have a we have another course like physical gas dynamics where you can actually go through all these things yeah so that that is typically taught in the odd semester that is like the ongoing semester now so you can sit in a course there where you can learn all these things but in this course we are now going to keep on adopting the continuum approach that means at a particular point you have a temperature that is defined you have a concentration that is defined and based on the concentrations the reaction rates are going to be determined and based on the temperature there the reaction rate constant is going to be determined by given given by these laws in reality at every point in space where you have reactions happening you have like millions of molecules that are bombarding each other and going through reactions we are not resolving that level of detail yes we do not want to define that in this course right so one of the things that we go through in chemistry is something like a hill that you have to cross and then there is like a height of the hill that is activation energy we do not want to worry about all that stuff okay so because of this mathematically speaking you will also face a problem in thinking about ignition because if this is your think about this if this is your K you now plug this in K and then you now get your Omega right and plug the Omega in here you get a Omega I which is the rate of production of a particular species right now let us suppose that we have a methane oxidation reaction that you are thinking about and let us do something very sinister right this is something that I always do fill up this room with methane yeah and of course we have some oxygen otherwise we will not be breathing here right what is going to happen are there reactions going on or not I just filled up this room with a decent amount of methane are the reactions going on oxygen is there right according to this yes right so of course we have conditioning and all those things so the temperature here may be like about what 20 degree C right so you are looking at like little plug in our 293 K over there right you will now get some value for different values of AE and so on and you plug in plug it in there I will tell you what the concentration of the methane is what the concentration of oxygen is plug it in there and then you know the reactions the new I double prime and new I single prime then you can find out the rate of production of carbon dioxide or rate of production of water or rate of production depletion of methane all these things right you will find some values which are non zero what do you do the means you just fill in some methane in here reactions are going to go on hey that is dangerous what is happening I have not it ignited that means we do not have a flame that is propagating we are not burning yet right and we are still at 293 Kelvin 20 degree C so but there are reactions going on any idea of this number like what would be your omega that would be extremely low 10 to the minus 14 minus 15 that kind of thing for room temperatures okay so from a mathematical point of view what this actually means and we will relate this to what is called the cold boundary difficulty okay that means we really strictly speaking cannot identify reactants as completely reactants okay because even as they are just being reactants they strictly speaking in principle should be reacting and producing products at a very very very miniscule rate according to this right it becomes numerically significant when the temperature rises past a value that is like related to E over R or you okay that is what this means then you begin to have values that are significant the heat release that is significant it begins to hurt us okay and then we get burnt okay so keep this in mind we are not going to worry about what E means to us except to point out that the value of it is going to dictate the nonlinearity of the reaction rates dependence on temperature okay that is what we want to worry about as far as we are concerned okay typically when you have a full-fledged reaction that is going on and you think that you have a flame there the reaction rates are very high of the order of it could be even 10 to the 10 right so we are now going from one end like 10 to the minus 10 to 10 to the 10 so that is another thing so as I told you we are not only having like a very narrow region over the steep price happens but the steep price happens over a huge range so you have to capture this numerically you know that the values that is very difficult that sets the biggest problem as far as combustion is concerned the chemistry that is thrown in there the exponential dependence of the reaction rate on temperature is the biggest will in chemistry okay that makes the whole problem highly nonlinear and very stiff so I guess I cannot emphasize enough about this but we will just move on and then we now say for a system of reactions for a system of reactions so how do you write a system of reactions you now say sigma i equals 1 to n nu i k single prime mi gives sigma i equals 1 to n nu i k double prime mi k equal to 1 to m so capital M indicates the number of reactions that you are going to have right in a scheme of reactions and each of those reactions then the kth reaction will be identical to any of the single reactions that we have seen before okay except we now want to identify the stoichiometric coefficients with the kth reaction as well as the ith species so we now say it does two subscripts new nu i k single prime and new double nu i k double prime these now become like matrices okay so this then means that you are looking at a multi-step reactions reaction sequence why would you worry about a system of reactions or a multi-step chemistry the reason is as I told you sometime back a species that is produced in one reaction could be consumed in another reaction so the net production the net rate of production of depletion of a species will now be dependent on how much is produced in each of the chemical reactions right so that is what we are going to keep track of so we have for a system of reactions notated by this we have omega k we now have the common quantity reaction rate now specific for the kth reaction okay so the kth reaction the reaction rate for the kth reaction is equal to kk p i equals 1 to n C i nu omega i k single prime okay and that is equal to a k e to the minus e k over Ru t p i equals 1 to n C i nu i nu i k single prime right well if you want to do one more bk t t to the mk e to the minus e k over Ru t p i equals 1 to n C i to the nu i k single prime so look at what is going on we are now having a sudden explosion of the number of parameters that we have to consider okay so previously I said these are going to be given to us okay so in a now have a reaction you know what the nu i single prime and nu i double primes are so that was given to you so you are looking at what this was and that is given by this reaction that means we need to be supplied with these two values a and e okay if you now want to further take the polynomial dependence of on temperature then you need to have b m and e okay so for a given reaction you need to be given b m and e these are what is called as a kinetic parameters okay and the information on nu i single prime and nu i double prime is essentially giving you what the reaction is okay it is a capital N and what these script MIs are the nu i and nu i nu i single prime and nu i double prime is what is going to give you what the reaction is in the system of reactions your capital N capital M nu the script MIs and the nu i single primes nu i k single primes the nu i k double primes all this information gives you what is called as a reaction scheme the reaction scheme is nothing but a list of reactions okay so when you see the list of reactions you should be able to find out what is the capital N what is the total number of species okay and I going from 1 to n what are the species script MI and how many reactions are there capital M okay I k going from 1 to m what are the reactions that is given by that that will be identified by nu i k single prime and nu i k double prime or we should be able to deduce all those things given a scheme of reactions but that is only one part of the story that is only telling you what is the chemistry right this species took reacted with this species to produce that species and that else and then some of those species reacted with this blah blah blah and then so you have a scheme then there is another part of the story so one thing is to actually figure out the chemistry the other thing is to actually figure out bk mk and ek for each of those kth reactions these are the chemical kinetic parameters okay so chemical kinetic parameters many times are quite difficult to measure and obtain even if you were to figure out the reaction scheme you see so what we are and then most of the time the combustion expert or the combustion scientist tries to bank on a chemist to get give this information so you have to actually get this information from a chemistry person first of all find out what the chemistry is I mean what is the reaction scheme that will give you the nu i single k nu i k single single prime nu i k double prime script M is for I going from 1 to capital N what is capital N and what is capital M all these things that is only the scheme in addition to this we have to get bk mk and ek these three parameters for each of the reactions that is that is that is a killer many times you get into a trip problem there so here just to record whatever I am saying this is the specific reaction rate of kth reaction and right now omega i remains the same as what we had before right omega i is the net rate of production or depletion of species I net rate of production or depletion of species I this will be equal to sigma k equals 1 to capital M omega ik so where omega ik is rate of production or depletion of the ith species in the kth reaction that means in every every reaction you have some rate of production or depletion of each species right that is omega ik you sum over all the reactions to find out so in one reaction omega ik could be positive another reaction omega it could be negative it is negative that means it is depleting there is positive it is being produced there you know add up all that stuff algebraically right you should now get the net rate of production of that particular species and that is what we will care about the reason we will care about that is because just like how we normally do a mass conservation okay in your fluid mechanics in addition to doing mass conservation for the mixture of species we will also now do a mass conservation for every species now obviously each of the species mass does not remain constant so to conserve does not mean so it does not means it is going to be a constant okay what it means is whatever is the rate of change of mass will be equal to the net rate of production or depletion that is what that is what we will do as we go along so we will factor this in into the mass conservation equation of individual species as an extent so this is going to be very very important to us yeah so omega ik then is equal to nu ik double prime minus nu ik single prime omega k that is like the multi step version of this over here right we did we did this only for one reaction we now do this for the kth reaction and a multi reaction step now we have to plug this back in here add over all the reactions to get the net rate of production of particular species right so therefore omega i equal to this is nothing but dci over dt that remains the same there you see omega i is dci over dt right so dci over dt that is equal to sigma k equals 1 to m nu ik double prime minus nu ik single prime omega k and omega k is vk t to the mk e exponential minus e ek over Ru t pi j equals 1 to n just using a different running variable just to avoid confusion cj to the nu jk single prime right what do we have here what do we have here you now say dci over dt is equal to a lot of things all now put together so we just started learning something and then pretty soon we now have a all of them collapse together right in one line and finally you now have a dci over dt going all the way you have a pi j equals 1 to n cj to the nu jk single prime okay which means this contains a product of all the species which are reactants right in the kth reaction and you now sum over all the k reactions so pretty much all species concentrations are now going to figure in here right as product in one term and then you are going to have many of those terms added up for each of the reactions right so you now have dci over dt for the ith species equal to one term plus another term plus another term and so on each of those terms is for each each reaction and each term contains products of species concentrations of the same species for which we wrote the dci by dt so two things are emerging here one the rate of change of concentration of species I depends on the concentrations of all species practically for in general that means you cannot evaluate the concentrate rate of change of concentration of species I independent of all other species you have to know the concentrations of all species to be able to find out what is rate of change of this right this means you now have for this box you should be able to write I equals 1 to n that means this is actually n reactions for each species so n equations for each species right and all these equations are simultaneous set of equations they are all coupled to each other because the rate of change of this is going to depend on all of the concentrations to find out the rate of change of so if you want to find out C I you now do a time integration of this right see if I were to give the concentrations that are initially present for a set of reactions right from there I should now be able to do a time integration so I now integrate this in time to go to the next next step next time next time okay during this time lot of species would have got consumed lot of species would have got produced all those things are actually present over here that means I have to solve the next DC by DT for the next species and so on so all of them should be solved simultaneously so it is a simultaneous set of first order ordinary differential equations which we have to integrate in time simultaneously for all of them right this is the this is only the starting they are only scratching the surface here now look at the nature of these equations the nature of the equations is first of all these are highly non-linear not because of the hue and cry that I made about the temperature dependence that is the non-linearity in the temperature dependence that is showing up here okay in each of these terms that is bad enough that is something that I will come to next okay the non-linearity is because you now have each of these CI's is an unknown you have to solve for simultaneously and these unknowns are actually showing up as products right so anytime you have products of unknowns you have non-linear equations right and not only that even if they did not show up as products they are raised to possibly a non-unity exponent right so anytime you have like a square or cube or minus 0.1 anything other than 1 is non-linear okay so linear is only one thing which is like 1 okay non-linear means anything else right that is going to screw up your day and making make it non-linear right so you already have a non-linear set of first order ODEs here and then on top of this we will have to find out in reality in real combustion applications we would not know what the temperature is okay and the temperature keeps changing from place to place or time to time or both place and time that means we have to now have another equation that governs temperature and that equation also will be coupled with this equation because in order to solve that equation you needed to know what was the heat that was produced by these reactions to change the temperature and we will see that as we go along and in order for you to solve this know this solve this equation you need to know the temperature so in addition to this n equations that are getting coupled you will also have an energy equation that gets coupled with this okay because the temperature is coupled to the species and the species are coupled to the temperature okay but if you are given temperature it is possible for you to let us say you are now taking like a bath that is now maintained at a particular temperature and you wanted to know what should be the concentration of species that are produced given an initial set of species this is an initial value problem because it is a ODE in time right that means you can integrate in time given initial conditions for the concentrations of the species you should not be able to advance in time so given temperature the energy equation gets decoupled from this set of equations but in a real practical problem you will have both of them coupled and this becomes a nasty set of equations as it is without even worrying about convection and diffusion of species yet. So we are now beginning to begin to look at combustion like right it just becomes gory and gory by the minute from here on see you tomorrow.