noc18-mm11 Lec 10-Equilibrium Concentration using Defects Reaction

[Music] [Music] so we left it at this point we saw the equilibrium concentration so I will add one more thing over here that this is equal to ks and when you take this this is equal to square root of KS which is 1 by 2 and that is why we get these terms - which also represents that we are that where the Delta H efforts and delta s are related to pair of defects so we get the factor two over here okay so that is Schottky defect now we have looked at Schottky now let's revisit the Frenkel defect here we'll make some more additional we'll have to make some additional assumptions or simplifications you can say so what was Frenkel defects cation going into vacancies so we can represent it as mm + VI VI as I said does not have to be there this is just to keep a track of the interstitials that are being used so the metal iron has already moved to interstitial so this is something their nose chart should have been so the whatever charge it is at the metal site is an additional charge so effectively it has two positive V M double Prime now this rate constant for this now you should be able to predict this at this point so it will be equal to X mi dot dot X these are the fractions VM so we have like this but here remember that the this fraction is with respect to the total number number of interstitial sites this fraction is with respect to the total number of lattice sites where the metals shall set so therefore there will be the denominator in both these cases are different and therefore we can have two different cases if n I is equal to n which is the more simpler case and I remember is the total number of interstitial sites and is the total number of lattice sites and we have seen in the example of I steel Fe carbon that there are two different kinds of interstitials that was that just two examples there can be more type of interstitials so there are two examples that we discussed for tetrahedral and octahedral sites and depending on whether we are talking about BCC or FCC there will be different relation between ni & n but for the sake of simplicity first let us assume that ni is equal to N and therefore KF and KF is given like this and using the assumptions as before that this is the dominating defect Frenkel defect is the dominating defect in this system that we are talking about and the fraction of defects is small we can say that this is approximately equal to 1 and therefore and another assumption or another understanding we already have because this is the dominating defect we can again also say that XM i dot dot is equal to x basically what we are saying is number but this translates to the fraction because ni is equal to n and therefore what we have is again first we'll take a look to take where first we'll just take that product which is X VM is equal to K F is equal to exp we'll say frontal pair to represent that we are talking about this defect reaction this enthalpy change of the formation of central pair and divided by RT now when these two are equal which is again only the correct possible when this is the dominating reaction and there are no other ways of generating these if there are other ways then these two will not be equal so that assumption is important and when that is true we can say this each of them are equal therefore this is square root so each of them will be equal to the square root of this quantity XM I dot dot is equal to X M KF to the power 1 by 2 and when you take square root of this this is equivalent to putting 2 factor over here so this becomes exp delta SF by 2r so for the case when we have ni equal to n we get the relation like this now this will be a little different when we have if you remember already from the earlier derivation for equilibrium concentration that this is different when we have an I not equal to n so let us consider when ni is not equal to n but what we still have is that ni is equal to NV that is number of interstitial sites is equal to number of vacancy sites why because this is the dominating defect domicile which means this is the largest in concentration and therefore whenever a metal cation site goes into the interstitial there will concomitant li vacancy created and interstitial site created therefore what we can write is something like this ni by n i where and i at the upper case and i as we know is the total number of interstitial sites smaller case and i is the actual number of interstitial sites which are being used into NV by n again now the right hand side is similar that we have already used earlier which is delta SF by r so what does this become Ni and V is equal to ni n so there is a factor here and since ni is equal to NV so this whole thing will have to be taken a square root so ni equal to NV equal to n FB we will call it Frankel pair defect so this is or not representing actually a pair but only one of these defect well then it will become ni n 1 by 2 so this we have to take a square root and therefore it will also again have a factor 2 over here or you can say that NFB if you want to if you are talking about let us say n V then you can say n V by n is equal to ni by n to the power 1 by 2 and this whole thing or you can say ni by ni equal to n by n I to the power 1 by 2 so we are now able to get fraction of interstitial as well as fraction of vacancies so you see that even for the case when ni is not equal to n we can get a relation and the relation has symmetry in it when you are talking about XV then it is ni by n when you are talking about X I the fact the pre-exponential factor is n by n I and presumably you will be knowing what is the nth total number of sites and ni which is the total number of interstitial sites given what type of interstitial you are talking about and what type of crystal system you are talking about so we will be able to find these fractions so again you can see that the whole derivation that we looked at over here it becomes so much simpler if we know if you put it in the form of a defect reaction now we will look at next we will move on to intrinsic ionization of electrons so we will try to find and first what we'll do is when it is localized so localized on metal sites and for that we had seen that instead of Fe we'll just simply use em so it is something like mmm electron going on to one of the metal site so it has a negative charge hole going on to the other one so it has a positive charge and it is nothing but saying that something like Fe three-plus going to Fe 2 plus and Fe 4 plus like we in told you earlier and now in this particular case I am representing at that rate constant by ki plus ki is equal to X again so we have to write it like this X mm which is prime X M m dot by X mm square so this 2 is coming from this 2 and like we said earlier the assumptions would be that this is the dominating defect and that would mean that these numbers be influenced by any other factor and the total concentration is very small so the total concentration being very small this is approximately equal to 1 which is 1 square so it still remains 1 and since this is the dominating defect we can again say that X mm prime is equal to XM m dot so we will this time we will directly go and say this is square root of K I which is the rate constant and therefore this 1 by 2 will manifest itself inside exponential factor exponential which will be Delta s formation to R exponential minus Delta H formation for intestine for intrinsic ionisation by 2 RT so we get the equilibrium concentration of metal with negative charge and metal with effective positive charge which are already equal which are equal and if we know the enthalpy change for this and trapeze change for this reaction we will be able to calculate the equilibrium concentration so this is when we are assuming that it's localized now the other case is when it is delocalized now here we will have to move on to next level meaning some more understanding and simplification will have to be done so now we will say this is delocalized electrons why is it will be different okay so let us we will see that in a moment so let us say that this is a perfect crystal giving out electrons plus whole right now this is not localized on any site now classical thermodynamics in the sense that we cannot apply number of sites etcetera we cannot have with respect to number of sites the density over here the fraction cannot be calculated like that so classical thermodynamics and accounting of sites and configurational entropy etcetera when we see classical thermodynamics we actually mean configurational entropy cannot be applied so this is a caution when you are using it for some relation like this however now it is a defect reaction so we can write it like this we will we have put a prime - to designate it is different from it will be used differently than the earlier rate constants so here this e prime which is the electron with negative charge is given by N and since we are talking about concentration but here on the denominator we will have an effective C meaning effective density of states in the conduction band similarly for P which is the hole this is an effective V or the effective density of states in the valence bands this will be given by exp minus eg by KBT depending on whether you are using per electron per mole as I said you will have to use or or KB respectively and why eg because temperature dependence of N and P or the ionization is given by the band gap energy so this EG is nothing but the band gap energy so temperature dependence of ionization is dependent upon band gap energy which is eg and so we get some relation like this which is very similar to what we obtained like for the earlier so just making it clear that this ineffective is effective density of states in conduction band this is effective density of states in the valence band so we have a relation very similar to the relation that we use for the defect reaction when there are lattice sites involved so this is the form of the equation and it as you can see it is very much similar to what we have been using so that gives us a handle to be able to write down equation when both of them are involved which is what we will do in the next example which is non stoichiometric oxides so the relation if you remember we are talking about when oxygen escapes into as oxygen gas so let us say we have au at its site plus and we are over here talking about metal oxide so it will form I can see at oxygen site and initially you can say that it is positively charged so something it will have charges like this the effective charge is like this so the first case for non-stoichiometric we are not yet involving the free electrons are delocalized electrons we are talking about localized ionization so here kvo would be given as and apart from these we also need activity 402 and that activity 402 is given by pressure partial pressure partial pressure of o2 so this reaction becomes something like this so far so already we have added one complication here or you can say next level of difficulty so we have Oh two whose activity is measured in a different way these activity can be directly measured as fraction and this for activity of this gas it has to be measured in partial pressure which is given like this now taking so again the two assumptions would hold the assumptions that this is the dominating defect meaning these are not being generated by any other mechanism primarily primarily the only the mechanism that is operative in creating these defects are this non stoichiometric escaping of gas and that the defect concentration is small so when the defect concentration is small the first thing that we can do is make this approximately equal to one so because they are so small the defects that these are the majority of real that is they are covering the real lattice sites now taking to each oxygen vacancy site is creating two electrons and each of those two electrons are going moving on to two different metal sites so the defect the metal sites with an additional negative charge the earth number would be equal to twice the number of vacancy sites that have been oxygen vacancy sites that have been created so our relation would become something like this and it would mean that if you want to have the concentrate constant for kV oh it will now this is equal to 1 this is equal to 1 and this since this number is equal to twice this number and that would also translate to since it is a mo type of metal therefore it will also translate to XM equal to 2x vo and therefore we will have if we convert this to 2x vo it becomes 4x vo square and therefore it becomes 4x vo cube so 4x v o cube into po2 1 by 2 so you see that this relation again we have mm prime equal to 2 times V 2 times V o meaning each vacancy at oxygen site is leading to defect The effect of negative charge on to metal sites and therefore the concentration of metal sites would be twice the vacancy concentration and therefore the number fraction it being a mo type of solid would be also twice therefore X mm is equal to 2 V o 2x vo and therefore this is so you are taking the square of this will come for X be your Square and into x vo is become 4 X V of Q and therefore what we have over here we can write it as X vo now this is this comes to the LHS so it is 1 by 4 which is equal to 1 by 2 square but we have to take cube root so it becomes 1 by 2 to the power 2 by 3 kvo we have to take 1 by 3 of this we also have to take this on to this side which means it is on minus 1 by 2 and then again cube root of this so it is equal to minus 1 by 6 now like I said mmm prime by n which is the fraction is equal to 2 your dot by n this is equal to what we are doing here is we have taken multiplied it by a factor of 2 so this is into 2 so this 2 to the power 1 minus 2 by 3 is equal 2 to the power 1 by 3 and therefore it becomes 2 to the power 1 by 3 into kvo to the power 1 by 3 into P or 2 to the power minus 1 by 6 now these two terms remain and we can expand this term so this will become so this is kvo which we can expand as Delta s and Delta H therefore we will have exp Delta s and since it is cube root so there will be a factor of 3 over here into exp exponential minus Delta H formation for the constituency oxygen by 3 or T so this gives you the ratio or the fraction of metal site as well as the letter in Seaside so if you are talking about the metal site this is the metal site defect that is one electron localized on to it this is how the relation is and if you are talking about vacancy oxygen oxygen vacancy sites you will have to divide it by two so we are able to obtain a relation for metal sites as well as vacancy at the oxygen site and what we understand from here is that concentration of these oxygen vacancies and also these metal defects these increase as the po2 decreases so when you decrease this you would see that thus whole quantity increases so if you reduce the value of po2 this whole value increases and which means the concentration of V O and M M increases therefore we have an important understanding gained from this reaction or from this equilibrium concentration and that is that concentration of oxygen vacancies and metal defects defects I mean with localized electron increases with decreasing partial pressure of oxygen now this is something we can also absolutely directly see from this relation now if you go back to this relation equation over here what we see is that on the left hand side this is the reaction on the right hand side you have this reaction was not reaction but these components so if you decrease this what it means is that the reaction will have to go forward to maintain the reaction constant and therefore the concentration of these two have to be increased so whenever you decrease the partial pressure of oxygen the concentration of vacancy and the concentration of metal with electron localized onto it has to increase so what we are obtaining or understanding or what we understand from the defect reaction over here is confirmed by the equation which gives you the direct concentration of these defects the vacancy oxygen vacancy defect concentration and metal with electron localized defect so this is the relation reaction when we have localized ionization next we will look at the same reaction but in this in the second case we will assume that charges are non localized meaning instead of two mmm we will take two electrons which are free electrons and you remember that the classical thermodynamics is like we said will not well or we cannot take the site ratio similar to that we take four rackin C and interstitials so and for as in the case of N and P so if the relation will be little bit different now there will be mixing the two electrons will also be their lattice sites will also be their vacancies will also be there so we'll see how that relation comes out in the next lecture so see you next time [Music] [Music] 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