Lecture 31: Reacting Multi-phase Systems

Key Takeaways

all right so today is the last lecture of new material we're going to talk about reactions between gases and condensed phases so we're going to talk about that and then we'll make it specific to oxidation and then on Friday I'll do a extended um uh practice problem I suppose on on reactions between gases and condensed phases so in general some reaction I'm going to write this really generally AA plus b b going to c c plus d d and these are uh not just for ideal gases so the last time we've seen um reaction like this in o2o it's been for ideal gases reacting and we're going to move away from that and do this more generally so let me just quickly remind you of of how we get to um reaction coefficient we have a general expression for the change of Gibs free energy with these unconstrained internal variables and we can reexpress this in terms um in terms of the reaction extent remembering that dni over new I is a constant for all I reaction extent I don't need to have an i there just the C okay and remind us that mu of I the chemical potential of a given species is its reference chemical potential plus um is activity so this is still completely General we um take that expression and we write D of g equals let's see mu I KN new of I plus r t log a of I to the power new of I and this is all times d c and uh this is DG KN plus RT log product a i new of I again whole thing times d c and of course um this is zero at equilibrium because at equilibrium gives energy is optimized so the coefficient is zero and um we get at equilibrium this uh product a of I new of I which you know we call the equilibrium constant equal e minus Delta g/ RT so is a reminder um but uh this is why I wanted to take three minutes for that because we're going to make this specific to metal oxidation that's what we're doing so equilibrium constant for metal and what you're going to find is that it simplifies a lot so that's good metal oxidation as we started the last lecture we have some metal reacting with a mole of oxygen gas going to m z O2 remember Z here has to do with the fact that we don't know um what the oxidation state of that metal is so potassium will be one iron has a couple different oxidation States um copper would be one or two I mean you know so there's different oxidation States so this is to do this generally so so all right um now we're going to write the equilibrium constant for this reaction so we have the activity of the oxide uh m z O2 sorry the writing is getting a little bit small there the activity of the oxide over the activity of the metal to the power Z time the activity of o2 gas um right okay right the products on the numerator reactants on the denominator um the metal is a z moles of metal so we have that activity to the power of Z straightforward to write down everyone with me so far right this is for this oxidation reaction at equilibrium we're going to come back to that this is the metal and oxygen and the oxide all coexisting at equilibrium okay now come the approxim we're going to treat O2 as how are we going to treat oxygen anybody how will we model oxygen constant pressure like atmospheric pressure not constant pressure but we have a we have a as a gas as a what kind IDE yeah because it's the only model for gases that we know you know2 so you know what else can we use so an idea gas the activity of an ideal gas is just this partial pressure by atmosphere for one atmosphere reference State good that's easy and we're going to assume that condensed phases are pure that's big if we assume the condensed phases are pure right that's equivalent to saying that their composition is unchanging right that's saying that this metal uh does not dissolve any oxygen and it's saying that this oxide is what what kind of a compound is always at fixed C geometry like perfect Crystal sorry like perfect Crystal no well not perfect no not a perfect Crystal actually be something else we learned this uh just two days ago L yes right it's a pure oxide meaning it St ometry is ideal it's a line compound it doesn't deviate from that so if we can assume the condensed phases are pure what is the activity of a pure phase you got to think back what's the activity of a component in its reference State one right thanks that was that's that's recalling material from the middle of the semester so it goes back a little bit but the activity of a material in its reference state is one so now I have um some really handy approximations the activity of oxygen gas we're treating it like an ideal gas so it's just P2 by atmosphere and the activity of the metal and the activity of the oxide are both one so this is equivalent to assuming that there's Vanishing oxygen solubility in the metal and that the oxide is a line compound so with those approximations we have the equilibrium constant is P2 by atmosphere to the minus one and this is going to be equal to the E minus Delta G over RT so you see this simplified quite a lot this simplified quite a lot all right I want to share some slides on binary metal oxygen systems Tak convin you that uh they do tend to be pure materials so let's let's look here um I grabbed a couple okay so here's the tin oxygen system and you can see that oxygen well it's just gas here um tin down here is a metal until it becomes a liquid in the liquid phase it dissolves some oxygen you can see that right the the liquid phase here has a solution region so you can dissolve oxygen in liquid tin but what about in solid tin right no oxygen dissolved so you see salatin will just coexist with this line compound to an oxide so this is what we're talking about a line compound of an oxide and a metal that doesn't dissolve any Oxygen here's another example copper oxygen again copper it's written here as a parentheses as if it's a solid solution but you know really there's vanishingly small oxygen dissolution in Copper there's some there's some but it's it's so it's so small as to be invisible on this PL I'll tell you that when you're making a components for cryogenics or for space applications you want oxygen free copper you can buy it from McMaster and uh you want it because oxygen is magnetic when it freezes and and so a lot of times you want non-magnetic copper so have to get oxygen free copper but you know it's there at the parts perm milon level it doesn't show up at all on this plot so again oxygen not doesn't not dissolving in Copper and these oxides being line compounds here's more examples this is a little counter example this is manganese manganese actually does dissolve some oxygen so you see it dissolves one or two% of oxygen so maybe the activity of Manganese should be not quite one right maybe it should be 099 um and Mangan these oxid similar they're not quite line compounds they have some solid solution regions so that's sort of a counter example um another example titanium oxy oxygen system here's titanium uh dissolves a lot of oxygen so this is an even counter counter example this is this is titanium's a funny system dissolves a lot of oxygen into solid titanium but its oxides here are line compounds so um you know what to make of this it's good to understand where the approximations apply and where they might not questions on these phas diagrams before I go back to the board this is this look dissolving all like all2 the molecule or just oxygen General I'm sorry they didn't quite could could you repeat a little like going back like to the solutions is this like dissolving oxygen as O2 form or other for oh it almost certainly disassociate so this would be dissolving oxygen atoms well I mean you you how do you dissolve oxygen atoms I mean you expose a material to oxygen gas and there's some catalytic process by which the oxygen gas splits into oxygen atoms at the surface and the oxygen atoms diffuse into the metal it's almost certainly not dissolving o2 molecules um metals that do dissolve oxygen are are um the basis for a lot of interesting Technologies but we won't have time for that so let's move back to there uh sorry any other questions on these phase diagrams and interpreting them or applications and so forth before I move back all right so so we have this really simple expression for the equilibrium constant and now we want to evaluate Delta G equal Del Delta h knus t Delta s for metal oxidation so we're going to evaluate that so we're going to start with eny enthalpy Delta H not equals Delta H not let's say 298 plus 298 to uh temperature D dummy temperature and you know you can have a Delta CP for the reaction so this is General but here's here's the you know what's coming is an approximation so I'm going to tell you for most metal oxides Delta H knot let's say 298 is large these are energetic reactions why would the enthalpy of formation of an oxide be large does somebody have the sense for why that should be give you a hint it's large negative it's it's large in the negative Direction what reaction are we talking about here oxidation oxidation so what what kind of reaction is an oxidation reaction other than just you know you could say it's an oxidation reaction but what's actually happening maybe somebody who hasn't replied yet today what's happening on the atomic level during an oxidation reaction Does it include like a favorable transfer of electrons yeah yeah this is due to so thank you to to the two of you due to exothermic exothermic nature of electron transfer during metal oxygen Bond formation and I could uh I could say ionic right so it's ionic bond formation you have very Electro negative element that's oxygen and less electronegative elements those are metals and the oxygen the that that bonding mechanism is electron transfer and that's very exothermic so as a result Delta hes tend to be very large and negative and this allows us to neglect temp dependence neglect temp dependence we'll just say that it's whatever it is at 298 because that's a convenient reference point and these are tabulated the the enthalpy of oxidation reactions at standard conditions those are tabulated okay so that takes care of the enthropy what about the entropy entropy Delta s is Delta s at some reference temperature say 298 plus integral okay so who know who has a guess of what we're going to do we're going to approximate this as what what am I looking for I want to justify what approximation aren't entropy values usually much smaller so if the enthalpy is already so big we could assume entropy is zero I see so you're right that entropy values tend to be much smaller but the thing is they're multiplied by temperature so they do matter quite a lot because the temperature number can get large so we're not going to we're not going to neglect it completely but I'll tell you I'm sorry look think get temperature dependent yeah we're going to collect it's temperature dependence so I'm going to um I'm going to write this before I write this which is the why we get to do that so let me write neglect the temperature dependence and now somebody please tell me why we might be able to do this and it's not just that we really like to make our lives easier it turns out to be a really good approximation in many cases what about this oxidation reaction um what's the dominant entropy change what's the dominant contribution to Delta s could it be when the phas change happens so when the temperature is constant the phase change meaning from metal to metal oxide y that's true but maybe tell me what what part of that when the O2 pH condens it yes thank you that's what I'm looking for the reaction entropy is dominated by and see you said that but thank you so much by condensation of o2 from the gas that's it so here's here's metal here's a chunk of metal right and then I have O2 gas right gas is very high entropy as you know right O2 and this reaction pulls the oxygen out of the gas right and makes a metal oxide so the reactants have much much higher entropy than the products simply due to the fact that the reactants included a mole of gas and the products don't include any gas and in almost all cases that change dominates the reaction entropy and since we're modeling um since we're modeling oxygen as an ideal gas the entropy of this gas is temperature independent it's um configurational entropy it's the the thing we've been talking about pretty much the whole semester which is the entropy of randomly configured gas molecules good thank you so now we have we have waved our hands although we've done it scientifically scientific waving of hands has happened and we figured out that we can neglect the temperature dependence of the enthropy and we can neglect the temperature dependence of the entropy and this is going to make our lives even easier than they already were we can solve for this thing we can solve for that oxygen partial pressure P2 by atmosphere equals e the Delta H KN r t e minus Delta s/ R this is the oxygen pressure at which a metal and its oxide coexist at equilibrium okay right so you've got a metal we've got it oxide right and this is the oxygen partial pressure so there's some oxygen here this is the oxygen paral pressure at which this kind of situation is at equilibrium what happens if what happens if the oxygen partial pressure is higher than the equilibrium value what do you think happens spontaneously in this system think principle I have the system in equilibrium and then I shove more oxygen gas in I increase the oxygen partial pressure what do you expect to happen how will the system respond spontaneously oxidize one more time I'm sorry I didn't quite hear that spontaneously oxidized yes fantastic metal spontaneously so you're going to have spontaneous conversion of more metal into oxide we've already established that the oxygen can't dissolve in the metal and the oxygen can't dissolve in the oxide they're pure components so all we can do is convert metal to oxide metal to oxide metal to oxide excellent thank you what about at lower at lower oxygen partial pressure if we're below this value what do we expect to happen spontaneously we will see the reverse reaction happening so more O2 gas will form how does it form through the oxide decomposing back into a metal and O2 gas exactly thank you I'm going to use this term the oxide is reduced but your your expression of spontaneously decomposing is exactly the right thing so if we pull down the oxygen pressure below the equilibrium value the system will respond by converting oxide into metal and oxygen gas and you can think of this in terms of the shot to the a if I add more oxygen the system will try to counteract that by condensing the oxygen into more oxide so it'll take more metal convert it into oxide right pulling some of my exess oxygen out of the atmosphere if I if I pull down the oxygen pressure the system will try to counteract that by giving off more oxygen gas from the oxide right the oxide will decompose into more oxygen gas and more metal that's right okay so um so that that's what happens so for example let's imagine TI oxidizing it's a nice reaction at 298 Kelvin I looked it up the P2 at this equilibrium is anybody you have a guess forget to guess I it's it's it's completely nutty 10 to the minus 150 atmospheres a completely silly make believe number right it's zero it's it's zero might as well be zero what does that mean about titanium metal it's very favorable for it to oxidize it's very favorable for to oxidize titanium metal if you see it in the air will actually be titanium metal with a thin layer of its own native oxide on top of it because if you were to remove the oxide and wait a split second the oxide would spontaneously reform and so this spontaneous oxide formation process is really important for a lot of reasons it's important for microelectronics it's important for stainless steel that's how stainless steel remains stainless you have spontaneous formation of a passivating or you can think of it as a protecting oxide layer and the engineering of those materials continues it's all interesting stuff all right so that is enough to take us to Richards and ellingham diagrams so we're going to do this I can't really motivate this because it sounds kind of random what we're going to do so I'm just going to tell you what it is and at the end tell you why it's useful we're going to plot Delta G as a function of temperature for metal oxidation reactions as you know it's like this and so um this is a pretty simple thing this is a line with slope minus Delta s knot and intercept Delta H knot right this is a a really really simple uh um line so here's temperature let's say in kelvin here's zero here's zero here's Delta G knot and we have a line right we have a line with a slope and an intercept so slope intercept that is a Richardson Ellen diagram now why did we do that give you a couple reasons let's start by talking about the P2 scale what's that the equilibrium constant is P2 by atmosphere to the minus one which is equal to e minus g/ RT so I'm just going to rearrange that and write Delta g equals RT log P2 by atmosphere this is a line on Delta G by T plot with slope R log P2 and zero intercepts let's draw that right here is temperature and Kelvin zero here's zero here's Delta G KN I'll draw the oxidation of a given metal just to have a something on there and now I'm going to draw these P2 Contours right Delta G not equals RT log P2 over atmosphere so I'm going to draw a series of lines all with zero intercepts and varying slope so there's a series of lines with zero intercept and varying slope and as I move in this direction I'm increasing increasing P2 so that's that's what a an ellingham plot is now I get to tell you why it's useful all right so if I rush through the explanation what it is it's because it's just like I have to tell you what it is but now I can show you why it's useful why is this useful it's useful for materials processing let me illustrate that let's consider two metals and I pick these just because they they rep they appear nicely separated on the plot let's consider tin right and manganes all right if I um look up the ellingham riches ellingham diagram for tin and manganes um I could do that but I'll just sort of draw qualitatively what it looks like there's temperature and Kelvin there's Delta G knot and um I'm just here to tell you that uh the the Tim line sits well above the magnes line so that's just a fact about these Metals this here is n 10+2 going to 102 that's what that is and this is two manganese plus O2 going to two manganese oxide okay and here are my P2 Contours what is this plot telling me it's telling me that at a given given uh I I didn't I have a type on my scan lection not sorry sorry at a given temp at any given temperature P2 for the manganese oxidation reaction is lower than for the tin oxidation reaction that's a useful fact how do I see that let's say at a given temperature at a given temperature let's say that temperature here I have this point on the tin oxidation line and this point on the manganese oxidation line well the P2 Contour that runs through the manganese oxidation line intercepting of that temperature corresponds to a lower partial pressure than the po2 Contour that runs through the tin oxidation line at that given temperature again at this given temperature the P2 partial pressure Contour for manganese oxidation is a lower partial pressure than the Contour for tin oxidation what does that mean manges metal will reduce tin oxide so if you were to put tin oxide let's say an ore into a furnace in the presence of Manganese metal and it up the manganese will convert to manganese oxide and the tin oxide will convert to 10 metal why is that manganes has higher affinity for oxygen than does 10 or saying the same thing the enthalpy of oxidation is more negative for manganese oxidation than for tin oxidation you can see that from the plots The Intercept of these plots is Delta H so you can see that the inter intercept for the manganese oxidation that's this point down here is lower let me extend these lines right The Intercept of the manganese oxidation reaction is lower than the intercept of the tin oxidation reaction the eny of Manganese oxidation is more negative or if we speaking a little bit um Cally manganese metal pulls in quotation marks oxygen out of tin oxide so everything I just wrote these are all saying the same thing about Metal oxygen bonds all right these are all equivalent statements they're just ways of interpreting metal oxygen bonding and the implications of that for materials processing right if you're going to be processing manganese in the presence of tin you need to know which one is going to have a higher P2 at equilibrium for oxidation because because it's going to determine your process outcome okay I want to uh share with you some pictures of Alan diagrams so um there's lots of Alan diagrams out there here is uh one representation this is a little busy but you can see lots of things here you can see um many metals which are all represented let me grab my laser pointer sometimes the laser pointer is stubborn so here's iron uh oxidation reaction here's a nickel reaction copper As you move down you're going to metals that are have higher affinity for oxygen so copper is uh sometimes considered Noble right so it's um it doesn't readily form an oxide you remember we talked about roof flashing takes a long time for it to turn green as opposed to calcium right if you have calcium metal around you should you should look out because it's going to explosively oxidizes so down here you have calcium magnesium aluminum aluminum is a very energetic oxidation reaction but we know it forms a passivating oxide so you don't have exploding aluminum all over the place you have um almost instantaneous formation of aluminum oxide on the surface of aluminum titanium silic and manganese chromium so um this is a this is a limited Richardson diagram El diagram because it only shows about a dozen or so metals of course you can get very very busy plots I want to point out one more thing which is the um the po2 scale P2 and you see it's a series of numbers running from one atmosphere to 10 oh sorry to uh 10 Theus 200 atmospheres and each of these numbers has a little tick mark and the tick marks all Point towards the origin you see these tick marks they're they're different angles why is that because it's asking you to imagine a series of straight lines connecting the origin to those values those series of straight lines connecting the origin to uh let's see I've never done this before can I draw the line so here's you know oh there we go look at that so this would be a line of 10 Theus 50 atmospheres of oxygen so for example you might say that uh zinc and its oxide are in equilibrium at 400 degrees C and 10us 50 atmospheres of oxygen that would be one way to read this plot and you also see that the Delta h of formation here is correlated to the electro negativity so this idea that metal oxygen Bond formation is energetic well we know that but it's more ener itic for Less electronegative Metals right so um we expect Noble Metals like silver and copper to be fairly electronegative they like their electrons 1.9 1.93 and we expect alkaline alkaline earth metals let's say alkaline earth alkaline metals alkaline earth metals like magnesium right they're they they are happy to give up their electrons to oxygen and know that has a lower electr negativity calcium is down here at one um okay so just some some solidate chemistry here there's one more thing I want to tell you about ellingham diagrams and that is the effective phase transitions so let's consider melting of metal or it's oxide right Metals melt oxides also melt causes it causes discrete jumps in those values so for example let's let's consider um solid metal plus O2 going to M2 solid it's reaction mon this has standard entropy one it's negative it's negative because we're pulling oxygen gas out of the out of the gas right solids solids now consider case where metal melts at lower temp then2 so Metals having a lower melting point which is true for most but all oxides and raise the temp until the Metal Solid becomes liquid metal so that's what we're going to do we're going to consider that and now we're going to consider the oxidation again except now we're oxidizing liquid metal so the oxide is still solid but the metal is now liquid and we'll call this reaction to have Delta S 2 all right well we know that the liquid metal has higher entropy than solid metal we know that so what that means is that Delta SN 2 is going to be smaller than Delta s 1 and they're all less than zero in other words this is going to be even more negative on a plot it looks like as follows there's temperature this is Delta G knot and we have a kink in the plot the Kink happens at the melting point so here is metal melting at that temperature the lower curve is a solid oxidation reaction solid uh Metal Solid metal oxidation the upper curve is liquid metal oxidation this intercept is Delta H knot for solid metal oxidation and this intercept is Delta H knot for liquid metal oxidation so let's go back to some real plots so now we can understand why there are break points in these there are break points in these curves because uh at these transition points the standard enthropy and entropy of these reactions changes so you know here this melting this this melting point here of aluminum it changes the slope very slightly you can't really see it can't really see it but there is a slope change then up here the the the aluminum boils so so there's a you know there's another break in the slope all here manganes melts here the manganese melts uh here zinc melts and zinc boils so forth and so on and if you want to uh really lose your mind um you can look at more complete inham diagrams so this this is posted on the website right and this is this is a very very very thorough ellingham diagram now with with so many elements and melting points and boiling points and you know anyway and so forth and you can you can start to look at these and learn what the melting points and the boiling points of the metals and the oxides are and you can see how they change the slope of these curves there are cases where the slope actually tips downward for a little while it's typically a melting point of an oxide and so forth and if you uh really like to explore this a little more which I encourage you to do it's a nice way to learn um go to uh the do it Pals page for ellam diagrams for those who don't know do it poms uh you should know do it pumps it is a really excellent resource for learning Material Science Concepts it's maintained by University of Cambridge and um you know it's a nice complement to things like open course Weare and and MX and let me let me share that just to show you where you might go to uh to um play with this a little more get a feel for it so here is uh the do it Palms page for ellingham diagrams um and if you click on oxides right you there's a whole bunch of oxides and so let's let's go random Cobalt oxide see alham diagram so it now tells you shows you Alem diagrams for Cobalt uh for its two different oxidation States and it'll give you data you can you can Mouse around um get the actual numbers uh change the temperature it gives you free energy formation and so forth and it's a nice learning tool um so now I will uh call it a day and stop recording e