Chemical Science - Lewis Diagrams - Lecture 12
Principles of Chemical Science/n * Email this page/nVideo Lectures - Lecture 12/nTopics covered: /nLewis Diagrams/nInstructor: /nProf. Sylvia Ceyer/nTranscript - Lecture 12/nLast time we were talking about the general interactions that are present in a chemical bond./nThe nuclear-nuclear repulsion. The electron-nuclear attractions. The electron-electron repulsions./nAnd we talked about how the competition of those three interactions are what gave rise to this interaction potential that we drew./nIt is the competition between those interactions that determines what the bond length of a molecule is. It is the competition between those interactions that determines what the well depth is, where the bond strength is in a chemical bond./nWell, today what we're going to do is move on and talk about a specific model for chemical bonding./nAnd the model we're going to talk about is that of Lewis diagrams. Now, interestingly, this is a model that is non-quantum mechanical. This was invented 20 years before quantum mechanics by G. N. Lewis./nSomehow G. N. Lewis recognized, and I don't know how he recognized./nThat's genius, I guess. But he recognized that the key to chemical bonding was a sharing or an overlap of two electrons, one from each atom./nAnd he represented these ideas using these Lewis diagrams in which he took the valence electrons, the total number of valence electrons in the two atoms that were forming a bond./nAnd he redistributed them so that one electron from each atom that was making this bond shared an electron and the rest of the valence electrons were distributed amongst those two atoms such that each atom had an octet of electrons around them./nEach atom had a rare gas electron configuration. And that's the basis of this idea of Lewis diagrams./nThe idea of the Octet Rule. There's no chalk today. I've got a piece. OK, the Octet Rule./nFor example, if you have a chlorine atom with its seven valence electrons here coming together with another chlorine atom with its seven valence electrons to make a fluorine molecule./nWell, the idea is that each one of these extra electrons now on the fluorine, or these unpaired electrons are shared between the two fluorines./nThe result is now, given the sharing, that this fluorine has an octet of electrons around it and this fluorine now has an octet of electrons around it./nThey have this octet configuration which is the lower energy filled shell configuration./nNow, that's the case, of course, for all the elements except for hydrogen. In the case of hydrogen, we're coming in with one atom. Say it's combining or reacting with a chlorine atom. Well, in this case, this electron from the hydrogen is going to be shared with this electron from the chlorine./nAnd now, again, this chlorine has this octet around it./nBut the hydrogen, of course, is just two electrons around it. Its inner gas configuration is that of helium, and so it is satisfied with just two electrons around it. With this sharing idea, Lewis could really understand the Stoichiometry of many compounds./nHe understood why hydrogen gas was H2 and not a hydrogen atom./nHe understood why nitrogen gas was N2 and not a nitrogen atom. He could understand why oxygen was 02 and not an oxygen atom. Now, these Lewis diagrams have absolutely nothing to do with orbitals or wave functions, which we actually know to be a more accurate description of chemical bonding./nWe're going to get to them./nLewis diagrams work 90% of the time to tell you where the electrons are in a molecule. It works so well and it is easy that chemists have used these Lewis diagrams before quantum mechanics and have continued to use them after quantum mechanics./nThey use them because it's easy and it works. It allows you to understand where the electrons are in a chemical bond, on which atoms the electrons are centered, where the lone pairs are, where those electrons that are in the molecule that actually don't participate in the chemical bond./nFor example, here in HCl, these two electrons, these two electrons, these two electrons, they aren't participating in the chemical bond. We call these electrons here lone pairs./nThat's an important definition that we're going to use./nSo we're going to look at this system. Again, it's easy. The alterative is solving the Schrödinger equation, which you don't want to do if you don't have to do because that's hard. You cannot do that quickly./nSo that's the value here. It's going to give us a framework, these Lewis diagrams, to kind of organize a lot of chemical reactions./nTo kind of organize our ideas of bonding and what will react with what./nLet's start doing this. And to that we're going to have a set of rules here. And let me put this front screen down. Actually, on the side screen, could you put my rules up there?/nAll right. The basic idea in Lewis structures is that we're going to take the total number of valence electrons and the two atoms that are going to form this bond./nWe are going to share electrons between the two atoms and then take the rest of the valence electrons and distribute them amongst the two atoms in that molecule such that each atom has an octet configuration./nTo draw a Lewis diagram you're going to have to know what the valence electrons are. In particular, how many valence electrons each atom has./nThis you have to know really off the top of your head./nAnd this is just a quick list of the number of valence electrons. Carbon has four, nitrogen has five, oxygen has six, etc. Let's, as an example, illustrate how to draw these Lewis diagrams. Here are our set of rules./nWe are going to go through them step-by-step./nAs you get better at this, you won't have to go through these rules in such detail. But for now, when we're just learning how to do it, I think this is a very effective way./nSo let's start. Let's take an easy one. We want to draw the Lewis structure of methane. The first thing you have to do is you have to figure out what the skeletal structure is. You have to kind of figure out, as a first pass, what is bonded to what./nAnd now I'm going to give you some hints./nFirst of all, hydrogen, if you have a hydrogen atom. That's always what we call a terminal atom. Terminal means that it is only bonded to one other atom. Chlorine atoms also are usually only terminal atoms./nHydrogen always is. Fluorine is 99% of the time only a terminal atom./nIn the case of methane, which you know has four hydrogens and one carbon, if the hydrogens have to be terminal, well, then there is only one choice here./nYou put the carbon in the center. So that's our skeletal structure. Number two here on our Lewis diagrams, it says count the total number of valence electrons. So let's do that. Hydrogen has one valence electron, but there are four hydrogens./nCarbon has four valence electrons./nThe total number of valence electrons in our molecule here is eight, and I am going to draw those eight valence electrons. Number three here in our rules says count the total number of electrons that you would have to have if each of the atoms had an inner gas configuration./nThese are the number that we've got, eight./nAnd now we are calculating how many we would like to have in the optimal situation if each atom has an inner gas configuration. Let's do that. Well, hydrogen wants two./nAnd we have four hydrogens, so four times two. Carbon wants eight. If each of these atoms had an inner gas configuration we would need 16 electrons./nStep four says to calculate the number of bonding electrons, the way you do that is by taking the total number of electrons that you would have to have if each atom had a noble gas configuration and subtract from it the total number of electrons that you actually do have./nSo we're going to take 16, subtract from that eight. We're going to have eight bonding electrons./nThe way I'm going to represent the eight bonding electrons, or bonding electrons in general is that I am going to draw a red square around my electrons that are bonding electrons./nUp here these are the valence electrons that I've got. It says eight of those valence electrons are bonding electrons, so I am going to draw a box. Those are my bonding electrons. Step five says to assign two bonding electrons to each bond./nWell, let's do that, assign two bonding electrons to each bond./nWell, if we put two there, two there, two there and two there then we've used up all of our valence electrons and all of our bonding electrons in that structure. Number six says if you've got bonding electrons that remain, well, the answer is no, we used them all up./nAnd number seven says do we have any valence electrons left? The answer is no, we don't./nWe've used them all up. So we are done. And you can see the hydrogen has an inner gas configuration around it, the carbon has an inner gas configuration, an octet of electrons around it. Hey, we're done with methane./nBut you could have done that anyway, right? Yes? Yeah. Here is another one./nOh, one other concept here. In this particular structure, carbon is the central atom. What I mean by a central atom is a atom to which everything else is bonded./nSometimes we're going to ask you to write a Lewis structure, and we are going to tell you what atom is the central atom so you know to put it in the center and everything else gets connected to it./nLet's do a higher one./nHere is this molecule. It is going to be hydrogen cyanide, HCn. The first thing you have to do, according to our rule number one, is to draw a skeletal structure. You have a choice here of how to draw that skeletal structure./nYou can put the nitrogen in the center or the carbon in the center./nHydrogen is not going to go in the center because we know the hydrogen is terminal always. What you often do is a first pass when you're drawing the skeletal structure./nYou put the element with the lowest ionization energy in the middle. That is what our rule number one said. Element with lowest ionization energy goes in the middle. That often is true, but not always./nThis is a first pass./nI am going to show you a little bit later some cases where this won't be true and how we are going to determine that the lowest ionization element doesn't go in the center. We're going to have a way to figure this out./nBut for right now, if you had no other choices, that's what you would do, you'd write a structure with the lowest ionization energy element in the center. OK, so we did that./nNumber two, calculate the total number of valence electrons in this molecule./nWell, hydrogen brings one valence electron into the party, carbon brings four and nitrogen brings five. We have a total of ten valence electrons. I am going to draw those ten valence electrons. While you are learning to do this, you might also want to draw ten valence electrons like this./nNext step, calculate the total number of electrons each atom would have to have if, in an ideal world, they had a noble gas configuration without sharing./nWell, the total number of electrons that we would have to have is two for the hydrogen, eight for the carbon and eight for the nitrogen. A total number of 18. Step four says we calculate the number of bonding electrons./nTo do that we subtract the total number that we would have to have for an inner gas configuration, 18, from the total number that we actually do have, which is ten./nThe result is eight bonding electrons. I am going to now draw this red square around the valence electrons that are going to be bonding electrons. Here it goes, we've got eight bonding electrons and two electrons left out here./nWe will see that in a moment, where they go./nNumber five, assign two bonding electrons to each bond. We'll put two there, we put two there, that's great. We've now used up four bonding electrons. We've got four more bonding electrons left. Our next step is do we have any bonding electrons left? We sure do./nWe've got four of them. In that case you start to make double bonds or triple bonds./nA double bond is one in which four electrons are shared. A triple bond is one in which six electrons are shared./nIn general, which atoms do you make double bonds or triple bonds between? Well, in general, double bonds form between carbon, nitrogen, oxygen and sulfur. Triple bonds usually restrict it to carbon, nitrogen and oxygen./nIn this case, we've got a carbon and a nitrogen here, we've got four extra bonding electrons./nWell, what we are going to do is we are going to put these in between the carbon and the nitrogen. We're going to have a triple bond right there. We took care of all our bonding electrons now. Step seven, do we have any valence electrons remaining? We sure do./nWe've got two valence electrons remaining. What do we do with them?/nWell, we assign them as lone pairs. We look to see which atoms don't have an inner gas configuration. Well, hydrogen does. The carbon does now./nIt's got eight electrons around it. But the nitrogen doesn't. It only has six. So we're going to put those two extra valence electrons, as a lone pair, here on the nitrogen. And now the nitrogen has an octet around it./nThese two electrons are called lone pairs./nThey don't participate in the chemical bonding. We did that one. Next one, thionyl chloride. It is SOCl2. How do you do this one? Well, as a first guess, the halogens usually are terminal atoms so they are not bonded to anything else./nThat's usually the case./nNot always, especially for chlorine. We're going to put the chlorines out on the ends somewhere. So then do we put oxygen in the center or sulfur in the center? Well, we're going to pick the lowest ionization energy element and will put that in the center./nThat is going to be sulfur. So we will try this as a skeletal structure./nCount the total number of valence electrons. Chlorine brings seven valence electrons into the party, and there are two chlorine atoms./nSulfur brings six. Oxygen brings six. We've got a total number of 26 valence electrons. Let me draw them all out. Step number three is calculate the total number of electrons that we would have to have for each atom to have an inner gas configuration./nWe've got four atoms here./nEach one of them has got eight. We're going to have 32. That's what we would need for an inner gas configuration for each atom. Let's calculate the total number of bonding electrons. Well, that's 32 minus the number of valence electrons./nThat's six. Of those 26 valence electrons, let me draw my bonding box around six of them./nNext step is to assign those bonding electrons. That's going to be easy. Put two there, two there and two there./nThat means we've just used up our six bonding electrons. Do we have any more left? No, there aren't going to be any double bonds here. However, we've got lots of valence electrons left. We've got 20 of them./nWe're going to have to make lone pairs with those 20 excess valence electrons./nWe're going to have to make ten lone pairs so that each atom has an octet around it. So let's do that. We're going to put six of them around the oxygen, so now the oxygen has an octet, another six, there they go, another six around the chlorine, there they go, another six around this chlorine, there they go./nAnd, finally, we've got two left./nWell, that's going to have to go onto sulfur to make an octet. There it goes. We are done. We've used up all of our valence electrons and we have an octet around each one of the atoms in that molecule./nSo that's the basic procedure for writing these Lewis diagrams. But now I talked about putting the lowest ionization energy element in the center sometimes./nAnd I talked about you've got try this skeletal structure first./nHow do we really know if we've drawn the right Lewis diagram? Well, the way we're going to tell is by this concept called Formal Charge. Formal charge is a measure of the extent to which an atom has gained or lost an electron when it has formed a chemical bond./nSo compared to when it is an atom, the formal charge tells you how much of an electron that has been gained by it or lost by it./nI will make that a little more real in just a moment. How do you assign a formal charge?/nWell, we've got a recipe for this. That formal charge is the following. You count up the number of valence electrons on the free atom./nWe're going to call that V. We're going to subtract from it the number of electrons that are lone pairs on that atom in the molecule, so V minus L. And then the third step is you're going to subtract one-half of the number of bonding electrons that atom has./nOr, one-half the number of electrons that it shares in that molecule./nSo it's V minus L minus S over two. So is the number of electrons it shares. That's the formal charge. Let's illustrate this. Let's take this ion, CN-. I already drew the Lewis structure here, so now what we're going to do is we're going to calculate the formal charge for carbon and the formal charge for nitrogen within this Lewis structure./nFormal charges depend on the Lewis structure./nReady? Let's do it. For carbon, V is four. We have four valence electrons. In this structure L is two. L is the number of lone pair electrons. Look at in this structure./nHere are these electrons, the lone pair that don't participate in the bond./nWe have two lone pair electrons. It's minus two. And then we have to subtract from that minus one-half the number of shared electrons. Well, the number of shared electrons in the case of the carbon is six./nWe've got a triple bond here. That's six over two, that's three. The formal charge on carbon in this Lewis structure is minus one. What we often do is put a minus one somewhere near that carbon and circle it./nThat is the formal charge on the carbon./nWhat that means is that the carbon has gained one more electron than it usually has as an atom. It's a little bit more negatively charged in this molecule than it is in the atom. Let's do nitrogen./nNitrogen, valence electrons five./nThen minus the number of lone pairs. Well, that's two. Here is that lone pair. And then it's minus one-half the number of shared electrons or bonding electrons./nWell, that's three. The result is a formal charge of zero. And so what we would usually do is we'd write a zero with a circle around it near that nitrogen./nSo the formal charge on carbon is minus one, the formal charge on nitrogen is zero in this structure, but now you've got to check to make sure you did everything right./nAnd here is the check. The check is that the total charges here, minus one plus zero is minus one, the sum of the formal charges has to equal the overall charge on the molecule. Minus one plus zero is minus one./nThat does equal the overall charge on the molecule./nSo we did everything right. That is an important thing to check. That's how you calculate formal charge. Now, how do you use it? Well, first of all, before I tell you how you use it, let me just emphasize that formal charge is not oxidation number./nOxidation number does not depend on the Lewis structure./nFormal charge depends on the Lewis structure. You're going to talk about oxidation number with Professor Drennan. If you don't know what it is that is OK./nYou will learn about it in a few weeks. But if you do know what it is, I just want to emphasize that formal charge is not the oxidation number./nHow do we use formal charge to make sure that we've got the correct Lewis structure? That's what you do./n[LAUGHTER] You use the formal charge. What you do is you realize that structures with lower absolute values of formal charge are the more stable structures. Structures which have formal charges on the atoms that are lower are more stable./nStructures where the number of electrons that the atom has gained or lost in forming a molecule is smaller are the stable structures./nThis is really important because you're going to often have a situation where you've got two different skeletal structures. You've obeyed all of the rules for drawing the Lewis diagrams of those two different skeletal structures./nThose two different arrangements of atoms./nBut only one of those structures is going to be correct. And the one that is correct is going to have the lower set of absolute values of the formal charge. So let's illustrate this. Let's take this ion./nIt's the thiocyanate ion, NCS-, as I have written it here./nNow, if you were just told you've got an ion that has nitrogen, carbon and sulfur in it, you might say, well, how do I start? Well, you could start by putting the element with the lowest ionization energy in the center./nThat would be a good place to start. You should do that. But what we're going to do is we're going to look at all three possibilities here./nWe're going to look at a Lewis structure where the carbon is in the middle, where the sulfur is in the middle, and now where the nitrogen is in the middle./nAnd we drew these Lewis structures. I've already drawn them here for you according to our rules posted here on these walls. But only one of these structures is correct. And we're going to use the idea of formal charges to determine which one is correct, which one has the lowest energy./nIf we take this structure and we calculate the formal charge for nitrogen, we will find it is minus one./nThe formal charge on the carbon here is zero. The formal charge on the sulfur here is zero. So I already did these calculations for you. If you sum up the formal charges we get minus one./nThe overall charge on this ion is minus one./nWe did it right. We did everything right. You should check your formal charges like this. Let's do the same thing for this structure. We calculate the formal charge on carbons minus two. The formal charge on sulfur is two./nThe formal charge on nitrogen is minus one. Let's sum these up. When we do that we get a minus one./nIt indeed matches the overall charge on the ion. Let's calculate the formal charges for each atom here./nSulfur is zero. Nitrogen is one. Carbon is minus two. Overall charge minus one. We did that right. Now, which one of these structures is the lowest energy structure? Ready? I'm going to ask you to raise your hand or to scream./nIs it structure number one? Yes./nIs it structure number two? No. Is it structure number three? No. You're right. It's structure number one. It has the lowest set of formal charges. Minus one, zero, zero. Only one atom here has a formal charge other than zero./nThese two structures have formal charges that are minus two or two./nI don't think that there is a structure on earth, well, maybe there is, that has a formal charge of minus two or two that will make it the lowest energy structure. If you've got a structure with a minus two or two, that's very, very likely not the lowest energy structure./nThis is a powerful tool to know if you have drawn your skeleton correctly, so remember this./nIt's really important. On an exam, we're going to tell you to draw the Lewis structure with the lowest energy. You're going to have to try several of them to see which one has the lowest set of absolute values for the formal charge./nQuestion?/nNumber of valence on the carbon is four minus the number of lone pairs on the carbon is four./nThat's two. Four, is that your question? Yeah. Other questions? OK. Here is another use of the formal charge./nThis is also determining the skeletal structure here, but you will see a problem in just a moment. Here is another special case. We have CH3NHO-./nHow do you draw, in this case here, that skeletal structure? Could I have some lights on the board here?/nCH3 here, this is what we call a methyl group./nIt is always terminal since the hydrogens are always terminal./nSo the structure looks like this. And the carbon then is bonded to the next element over. If you see a structure like this where you've got lots of different elements, the first pass just take these elements in a row./nThe CH3, this methyl group is bonded to the nitrogen./nThe next one is hydrogen, but we know we're not going to put the hydrogen between the nitrogen and the oxygen because the hydrogen is always terminal./nSo the hydrogen is either going to go here or the hydrogen is going to go here. And that's the question that we're going to ask here on this slide, is this the correct structure or is this the correct structure?/nI drew the skeletal structures, I drew the Lewis diagrams./nWhich one is right? Well, if I go and calculate the formal charges here, I would find that the formal charge on oxygen is minus one and that the formal charge on all other atoms is zero. In this structure, if I go calculate the formal charge on the nitrogen in this structure is minus one./nAnd then the formal charges on all the other atoms are zero./nWe've got a problem here. We've got two different structures, and they have absolutely the same set of formal charges in terms of the absolutely magnitudes. We have only one atom that has a minus one formal charge, only one atom that has one more electron that it usually does in its unbonded state, in its atomic state./nSo which one of these is correct? Well, it turns out that this structure is correct./nIt is correct because of this. It is correct because we put the formal charge on the most electronegative atom which is oxygen. Oxygen is more electronegative than nitrogen./nThat is the other key here, is that if you've got two structures, same sets of formal charges and you have a choice of putting the minus formal charge on an atom, you put it on the most electronegative atom./nThat is going to be the lowest energy structure. In general here we talked about the electronegativities. The electronegativity of fluorine is greater than oxygen, is greater than nitrogen, is greater than carbon./nThat is something that you will need to know./nSo we've seen how this formal charge now is useful in figuring out whether or not we had the right skeletal structure to begin with. Now, here is another --/n-- utility of formal charge. Let's look at this structure NO3-./nWell, I know ahead of time that we're going to put this nitrogen in the center. If you just had NO3- and you didn't know anything. You would have to play around with some structures and look at formal charges, but I have already done that playing around so I know to put nitrogen in the center./nBut let's do the Lewis diagram for this./nWe count up the valence electrons. We've got six for every one of the oxygens and five for the nitrogen. And now, this one is important, we added an extra electron when we counted up the valence electrons./nThat extra valence electron comes because this is an 03-. We have an extra electron on NO3 so we have to count it in our valence electrons. If this was NO3+, we would subtract one right here./nBut this is NO3- so we've got to add one./nThat's 24 valence electrons. Let me draw them all out. Number three. Let's calculate the number of electrons that we would have to have if each atom had a noble gas configuration. That would be 32 because we've got four atoms./nLet's calculate the number of bonding electrons./nWell, that's 32 minus 24 or eight. So, of those 24 valence electrons, eight of them are bonding electrons. Let's assign them. Well, here is two, here is two, there is two./nThat's great. We used up six. Do we have any remaining? We sure do. We've got two, so we're going to have to make one of these a double bond. So we've got one double bond there between nitrogen and oxygen./nNow, we've got a lot of remaining valence electrons./nWe've got 16 that are remaining. We're going to distribute them around the oxygens in order to make an octet around each atom. There we go. We did that. And so now we've used up all of our valence electrons./nLet's calculate the formal charges. The formal charge on the singly bonded oxygen is minus one./nI put a minus one there. I'm going to put a minus one there. The formal charge on the doubly bonded oxygen right in here, that's zero./nI'm going to put a zero there. The formal charge on the nitrogen, well, that's one. I'm going to put a one there. If I sum up the formal charges, I get a minus one. That does agree with the overall charge on this molecule, NO3-./nSo we did everything right./nBut now you see something a little strange here, right? And that is, for whatever reason, I picked this nitrogen and oxygen to make a double bond. But I could have also made the double bond between this nitrogen and oxygen./nAnd if I had calculated the formal charges, I would have identically the same set of formal charges. Or, I could have picked this oxygen and nitrogen to put in the double bond./nIf I calculated the formal charges, I'd have the same set of formal charges./nSo we've got three different structures all which have the same set of formal charges. What are we going to do here? Which one of these is correct? Well, when you've got this situation here, this is a hint that you've got what we call a resonance structure./nSo if you have the same skeletal arrangement and you've just go three different structures and the electrons are in different places and the same set of formal charges exist, this is a hint that you have something called a resonance structure./nWhat is a resonance structure? Well, here is the bottom line./nIn NO3-, if you look at the structure of that ion, you would not find that one of the nitrogen-oxygen bonds was a double bond and that the other two were single bonds./nA double bond is stronger than a single bond. A double bond is shorter than a single bond. So if you do the appropriate experiment you can tell whether or not you have a shorter or a longer bond./nYou can tell whether you have a weaker or a stronger bond./nThe bottom line is that when you look at this molecule experimentally, you do not see one bond that is shorter and stronger than the other two nitrogen-oxygen bonds. What you find is that all three bonds are equivalent./nAll three bonds have the same strength and the same length./nYou find that the length of the nitrogen-oxygen bonds are a little bit shorter than those of a single bond and you find that the nitrogen-oxygen bonds are uniformly, all three of them, a little bit stronger than a nitrogen-oxygen single bond./nBut not quite a double bond./nWhat you find is that all of these three bonds is like a bond and a third. So what is happening here? What is happening here is that this extra pair of electrons, instead of being localized on just one of the nitrogen-oxygen bonds, those extra pair of electrons are actually delocalized over all three bonds./nAnd we call that a resonance hybrid./nWe denote that resonance hybrid in this way, with three different structures with double-headed arrows. This does not mean, this is important, that the structure is flickering between here, here and here./nThat's not what is happening./nRather, these two extra electrons are delocalized over all three bonds such that the bond is like a bond and a third rather than one bond or two bonds. This is called a resonance structure, a resonance hybrid./nAnd you will be clued in that you've got this kind of resonance hybrid if you're able to write three structures with the same set of formal charges./nYou will know to do this. OK. Have a nice cool weekend./nSee you Wednesday.
Tags // Lewis Diagrams
Added: April 16, 2009, 10:59 pm
Runtime: 2749.20 | Views: 12899 | Comments:0

Chemical Science - Bond Energies / Bond Enthalpies - Lec 17
Principles of Chemical Science/n * Email this page/nVideo Lectures - Lecture 17/nTopics covered: /nBond Energies / Bond Enthalpies/nInstructor: /nProf. Sylvia Ceyer/nTranscript - Lecture 17/nAll right. We are first going to start talking about bond strengths. You know that we have been drawing these energy interaction curves as a function of r. And we have defined here a bond association energy E sub d as that energy difference from the bottom of this well to the separated atom limit./nAnd let me consider the separated atom limit now to be a separated methyl group and a hydrogen atom because I want to talk about the CH bond. This is the bond association energy for CH in methane. But now what we are going to do is change our definition a little bit./nInstead of talking about a dissociated energy, a bond energy, we are going to talk about a bond enthalpy. And I am going to represent the bond enthalpy by delta H. It turns out that in the laboratory we can more easily measure a delta H, a bond enthalpy rather than a bond energy./nWe can do so because most of the experiments we do are at constant pressure. When we do an experiment in the lab that is at constant pressure. And the relationship between delta H and delta E is the following./nDelta H is equal to delta E plus delta PV. Now, where this relationship comes from is a subject of a great discussion in 5.60 Chemical Thermodynamics. You will see that if you take 5.60. At the moment, we are just going to take this as a given./nBut I will tell you that although delta H and delta E differ for gases, the difference is on the order of 1% to 2% for gases between delta H and delta A. They are both an energy but we are going to call it an enthalpy./nThat is actually what we measure typically in the laboratory. For solids into liquids the numbers are essentially the same, but for gases there is a 1% to 2% difference. So if you are doing an exact calculation, you really do need to know that number exactly, you need to get the bond enthalpy./nAnd, in fact, if you go and you look up bond strengths they will, for the most part, all be bond enthalpies and not bond energies. There is a more extensive table of bond enthalpies than there is for bond energies./nSo we are going to talking about delta H for a bond now. And in thermodynamics, as again you will see in more detail in 5.60, we always need a standard state. We are always going to be, for the most part, measuring things relative to one another./nAnd so we need a common ground, we need a standard state. And what your book uses, and what most people now use for a standard state is the pressure one bar. The standard state really refers to the pressure and actually not the temperature./nOne bar is equal to ten to the five Pascals. And a Pascal is a kilogram meter per second squared. Ten to the five kilogram per meter second squared. And the relationship between bar and atmosphere is one atmosphere, 1.01325 bar./nAn atmosphere is 1.01325 bar. I am sorry? OK. And so to represent this standard state here, we are going to put a knot up here. We are going to put a superscript up here. This is going to be delta H knot./nThat is going to represent our standard state. This slide is what we will do now. On the first slide here, what I have is bunch of bond enthalpies. And these are bond enthalpies where all of them are for CH bonds./nThe first one is for the CH bond in methane. The second one for the CH bond in ethane. The third for the CH bond in CHF3. And what you can see is that they are all about 400 kilojoules per mole. They are not all exactly the same, and they shouldn't be./nThey are different molecules. The bond strength really does depend on the individual molecule, but they are all roughly speaking the same. And so what is often done is to take bond enthalpies, such as this, and average those strengths over many different molecules for the same bond so that you have an average bond enthalpy./nAnd that is what you will see plotted, or that is what you will see a graph of. This is a table from your book that gives you the mean bond enthalpy for, for example, CH bonds. 412 kilojoules. It gives you the mean bond enthalpy here for CC bonds./n348 kilojoules. Carbon-carbon double bonds. 612 kilojoules. And so those are the kinds of tables that you will often see, but remember these are average bond enthalpies. They are going to give you that bond enthalpy approximately correctly./nBut not exactly correctly. And that is going to be important, as we are going to see in just a moment. Now, why in the world are bond enthalpies important to us? Well, they are important to us because they determine the enthalpy of a chemical reaction./nIf the bond enthalpies in a product of the reaction, if those bond enthalpies are stronger than the bond enthalpies in the reactants. Well, then the enthalpy of that reaction is going to be negative./nYou are going to have an exothermic chemical reaction. If the bond enthalpies in the reactants are stronger than those in the products, well, then we are going to have an endothermic reaction. So the relative bond enthalpies are going to determine the sign of the enthalpy of the chemical reaction./nSo let's look at this reaction here, a very important reaction. This is the oxidation of glucose, very exothermic reaction. Minus 2816 kilojoules per mole. This reaction is going on in every single cell of your body./nIt is this reaction that you are using right now to maintain your body temperature. To move your muscles. To repair tissue. To think. This oxidation of glucose is important. I mean, this is the reaction./nAnd this is the reason why we eat, this is the reason why we breathe, this is the reason why we exhale and this is the reason why we pee. So what do we have to do to calculate the enthalpy for this reaction? Well, what we have got to do here is take glucose, and that is the structure of a glucose, blood sugar, and that of oxygen./nAnd we have to calculate how much energy it is going to require, how much enthalpy it is going to require to break every single one of these bonds in glucose. What we have to know is how much energy is required to break seven CH bonds./nWe have to know how much energy or enthalpy is required to break five OH bonds. Here are our OH bonds. And how much enthalpy is required to break five CO bonds. Here are CO bonds. And to break five carbon-carbon bonds./nAnd to break one CO double bond. And then finally we need the enthalpy to break six molecular oxygen bonds. So we can calculate that. We can calculate the enthalpy to break all of these bonds by looking up the values from that table that I just showed you./nThe table of average bond enthalpies. And when we do that we find it is a whopping 12452 kilojoules per mole. That is what we are going to need to take glucose and oxygen to the elements, the atomic species./nWell, then likewise we know that although we need to put that much energy in to break the bonds, we are going to get some energy back when we make the bonds. So now we need to know how much energy enthalpy we are going to get back when we make 12 CO double bonds and when we make 12 OH bonds./nAgain, we can use our table of average bond enthalpies to calculate that when we make those bonds we get 15192 kilojoules of energy or enthalpy back. And then the difference between this energy and that energy, well, that is the enthalpy of that reaction minus 2740./nWhat did we do here? Well, what we did to get the enthalpy for the reaction is we took the bond enthalpies for each bond of the reactants and we summed them all up. And we took the bond enthalpies for each bond of the products and summed them all up./nAnd then we took the difference. That's how we got this number minus 2740 kilojoules per mole. But now what I want you to notice here, and this is important, is that we calculated these enthalpies by taking the enthalpies of the bonds of the reactants minus those of the products./nWhen we measure enthalpies for chemical reactions using bond enthalpies, it is reactants minus products. The reason I tell you this is because in a moment I am going to show you another way to calculate the enthalpy for a reaction./nAnd that way will be product minus reactants. You've got to keep this straight, this is reactants minus products. But now you see that I just calculated a number for that reaction enthalpy of minus 2740./nAnd the experimental value here was minus 2816. Some of you are looking at your notes like it is not in your notes. Is that right? It's in the notes, OK. Gee, this is what I told you was the experiment, this is what we calculated./nWhat happened here? Well, remember that what we used to calculate this number was average bond enthalpies. We didn't use the exact bond enthalpies for every bond that we had. And there is a difference./nAnd so how can we calculate this in a more accurate way? I mean sometimes, depending on what you're doing, it is OK to have the average bond enthalpy. You are close enough. It's good enough. But sometimes it isn't./nWhen it isn't, what can we do? Is there a more accurate way? Well, you could have the exact bond enthalpies. But the bottom line is that if you had the exact enthalpies for each bond, for each known molecule, can you image the table of numbers that you would have to have? In fact, such a table doesn't even exist./nI mean just for glucose alone, we would need 23 different numbers, 23 different bond enthalpies. That is a large amount of data that we would need to do this exactly from the bond enthalpies. I am sorry? Absolutely./nThat is your job. That is why you are here, to sit in class and say, gee, that's too hard, let me think of another way to do it. That is what a MIT education is all about, OK? [LAUGHTER] I am serious./nYou're laughing. But we do have a less cumbersome way to handle this. And that is using heat of formation. A heat of formation, what is that? Here it is. I am going to define this as delta H knot sub f./nThat's the heat of formation. It is an enthalpy. It is the enthalpy of a reaction that forms one mole of a compound from the elements in their most stable form in the standard state, which is one bar of pressure here./nLet me explain that. Here is a reaction that produces one mole. That is important, one mole. One mole of liquid water. The enthalpy for this reaction is minus 285 kilojoules per mole. That enthalpy here is defined as the heat of formation for water, because the enthalpy for this reaction forms one mole of water./nAnd it is forming one mole of water from its elements, but the elements here are in their most stable form. The most stable form of hydrogen then makes up this water here, the most stable form at one bar pressure and at room temperature./nAnd I forgot to tell you that I was going to talk about delta Hs at 298.15 degrees Kelvin. All of our delta Hs will be for that temperature. Delta H does depend a little bit on temperature. But we are not going to do that here./nWe are going to do that in 5.60. 298.15 is the temperature at which you are going to be dealing with all of your delta Hs. Anyway, H2, that is the most stable form of hydrogen. And then oxygen here./nWell, that makes up one of the elements in water. And so the most stable form of oxygen at this temperature and pressure is indeed molecule oxygen. And so this equation satisfies this definition, forming one mole of a compound from the pure elements in their most stable form./nLet's look at this reaction. Here we are forming one mole of oxygen. And we are forming it from one mole of oxygen. The delta H for this reaction is a whopping zero. However, this is the definition for the enthalpy of formation for oxygen./nBecause we are forming one mole of oxygen from its elements in their most stable form at the standard state, the most stable form of oxygen is oxygen. So the heat of formation, the enthalpy of formation of molecular oxygen is zero./nSuch is the case for the heat of formation of hydrogen, H2, or nitrogen, N2, or chlorine, Cl2 gas. The heat of formation of those compounds are all equal to zero because they are in their most stable form at one bar pressure and 298 Kelvin./nWhat about this reaction? Well, this reaction here is forming one mole of glucose from glucose's elements in their most stable form. Here is the molecular hydrogen, here is the molecular oxygen, and here is the element carbon in its most stable form./nWhen you see this GR, that stands for graphite. The most stable form of carbon is graphite at this standard pressure and this temperature 298 Kelvin. The enthalpy for this reaction minus 1260, that is the heat of formation or the enthalpy of formation for glucose./nNow we looked at the definition for the heat of formation. How are we going to use the heat of formation to calculate the enthalpy for a reaction? That is what we are going to do now. Here are our reactants, glucose and oxygen./nWhat we are going to do is use the heat of formation to calculate how much energy we need to produce the elements of glucose and oxygen. The elements in their most stable form meaning graphite, molecular hydrogen, molecular oxygen./nThat energy here is minus the heat of formation of glucose. Now I am taking this reaction and going backwards. I am going from glucose to the elements, so this is going to be minus the heat of formation of glucose./nIt is going to be 1260 kilojoules. But then what I am going to do is form the products. I am going to take the elements in their most stable form and put them together to form the products. In this case, I am going to form here six CO2 molecules./nI need the heated formation of CO2. Since this is per mole, we've got six moles of them minus 2361. And then I am going to take these elements and form the other product, water. And there are six moles of water./nSo I need the heat of formation of water times six minus 1715 or something. And then the difference between the sum of these two energies and these energies, well, that is the enthalpy of the reaction minus 2816./nThat is a tabular or energy diagram form. We can also write out the equations. Let's do that. Here is the reaction that defines the heat of formation of glucose, except I wrote it backwards. This is glucose going to its elements in their most stable form./nThe enthalpy change for that reaction is minus now the heat of formation of glucose. Here is the reaction that defines the heat of formation of CO2. The enthalpy change for that reaction is six times the heat of formation of CO2 because we've got a six here./nThe heat of formation is for one mole, but now we've multiplied this whole equation by six so we need to multiply that by six. Then we've got an expression here for the formation of six moles of water, so the delta H for that reaction is six times the heat of formation of water./nNow we are going to add them all up. We are going to add up the reactions first. And what you can see is that there is going to be a lot of cancellations if we did this correctly. This six carbon atom in the form of graphite is going to cancel with this graphite on the left-hand side./nThese six moles of H2 is going to cancel with these six moles of H2. These three moles of 02 are going to cancel with these three moles of 02. And then the six 02 cancels with this six 02. We add them all up and we have the reaction that we're looking for, the oxidation of glucose./nAnd then, of course, we are going to have to add up the delta Hs for all these reactions. When we do so we get minus 2816. That is the enthalpy change for this reaction. What I just did here, adding up these reactions and the corresponding delta Hs for these reactions, well, this is an example of Hess's Law, which I think you have seen before./nWhich says that if you have two or more chemical equations that are added up to give another chemical equation, that is what we did here, you then also have to add up the corresponding enthalpies to get the enthalpy change for the overall reaction./nNow, what did we actually do? I showed you how to calculate the enthalpy for this reaction both in kind of this graphical form and also kind of this mathematical form where we wrote out the equations./nBut the essence of what we did is that. What we did is took the heat of formation of each one of the products, multiplied it by the appropriate stoichiometric number, and then summed overall all of the heats of formation./nAnd then we took the heats of formation for each one of the reactions and multiplied it by its appropriate stoichiometric number. Summed over all of that and subtracted the two to get the enthalpy for the reactions./nFor glucose we took one product, CO2, multiplied that by six for the heat of formation, added to the six times the heat of formation of water, subtracted from that the heat of formation of glucose and that of oxygen./nAnd that is our calculated value for the reaction. And, indeed, that does agree with the experimentally measured enthalpy for the oxidation of glucose. Why? It agrees exactly because we used the heats of formation./nThe heats of formation are specific for the individual molecules that are participating in that reaction. It is exact. And having tables of heats of formation, the tables are much less extensive than what you would need for the bond enthalpy for every bond in every known molecule./nAnd so these tables of heats of formation are readily available to you. And you can get exactly the enthalpy for the reaction. But now the second very important point here is look at how we calculated the enthalpy for this reaction./nWe took the heats of formation of the products and subtracted them from the reactants. In the case of getting enthalpies for chemical reactions from the heats of formation, it is products minus reactants./nFor bond enthalpies or for getting the enthalpy of reaction from enthalpies it is reactants minus products. You can be sure to see this soon, like on November 2nd. It is something you want to remember, something you want to know./nQuestions on that? All right. This is important. And I want to just make one other brief point here about delta Hs and many thermodynamic quantities, but delta H in particular here. And that is that delta H is a state function./nThat means it is independent of path. For example, here I have glucose decomposed to the elements. Not in their most stable form, but glucose decomposed to the elements. And then rearranged to form CO2 and water./nThe difference between this energy and this energy is the enthalpy of the reaction. And we calculated the enthalpy of the reaction in this way, if we had used exact bond enthalpies. But we also calculated the enthalpy using heats of formation./nTaking glucose to the elements in the most stable form and then recombining them. Again, the energy difference is minus 2816. What this is an example of is the function enthalpy being a state function./nIt is independent of the path that we took to calculate it. You can go up here and then come back down here. You're going to get the same number. As you go up here and come back down here. That is what we mean by a state function./nAgain, you will talk about that in more detail in 5.60. Now, that takes care of enthalpies for the moment. What I want to talk about is another concept, and I want to talk about spontaneity. Spontaneity or spontaneous change is certainly something that you are familiar with./nIt occurs by itself without any outside intervention. It has directionality. For example, if you put a rock near the top of the hill and you let it go, you know what is going to happen. It is going to roll down./nIt is not going to roll up. That's a spontaneous process. If you have a gas, say, over here at high pressure and then a stop cock and a bulb here with a vacuum and you open the stop cock, you know that that gas is going to flow from high pressure to low pressure./nIt is not going to go the other way. That is a spontaneous process. You know that if you put a hot object next to a cold object, the heat is going to flow from the hot object to the cold object. It is not going to go the other way./nHowever, although a change might be spontaneous it may not be fast. If you take a bottle of ketchup, in the days when the bottle of ketchup was a glass bottle and you couldn't squeeze it, and turned it upside down, well, the ketchup has the spontaneous tendency to come out but it may not be fast./nWhat we want to understand is what is the key to spontaneity, in particular for chemical reactions? For example, you know that iron rusts and you don't have to do anything about it to make it happen./nIt rusts. It turns out that iron rusting is a very exothermic reaction, minus 824 kilojoules per mole. You know that if your stomach is a little bit acidic that as long as you ingest something that is basic that you can neutralize that acidity./nThat is also a very exothermic reaction, minus 56 kilojoules per mole. Another reaction that occurs in every cell of your body is this transformation or the reaction of adenosine triphosphate. The hydrolyzation, the reaction with water to adenosine diphosphate, this also is an exothermic reaction./nYou don't have to do anything to make it go. It goes. The question is, is delta H the key to spontaneity? Well, let's look at this reaction. This reaction is ammonium nitrate dissolving essentially in water to make ammonium ion and nitrate ion./nAnd the question is, is this reaction spontaneous? This is a reaction that is about 24-25 kilojoules per mole endothermic. Does an endothermic reaction happen spontaneously? Well, to answer that question we are going to have to do an experiment./nAnd the experiment we are going to do is one of these cold packs. And we are going to carry out this reaction and see if it is spontaneous. What you are going to do is, and the TAs can give out a few of the cold packs, take a cold pack and find the little packet inside this cold pack that has water in it./nYou are going to squeeze it to break that packet and then you are going to see if this reaction goes. If it goes, hey, it is going to get cold because this is an endothermic reaction. It is going to pull heat out from the environment./nAll right. And some of you have gotten hot packs. Is an endothermic reaction spontaneous? Yes. Is an exothermic reaction spontaneous? Yes. Is delta H the key to spontaneity? No. So it is not the key of spontaneity./nWhat is the key of spontaneity? What is the key to spontaneity? Delta G. That is the key to spontaneity. The Gibbs free energy. There it is in all its glory. Delta G is equal to Delta H minus T delta S./nWhat is delta S here? Delta S is the change in entropy. What is entropy? Entropy is a measure of the disorder in a system. If you have a positive delta S, if the change in delta S, if the sign of delta S is positive, you are going to have an increase in disorder, the systems for disorder./nIf you have a negative delta S. If the change in delta S is negative, you are going to have a decrease in this order. That is going to affect delta G. It is not just delta H. But the change in delta S is going to affect the spontaneity for a reaction./nThe combination of delta S and delta H is this delta G. And our conditions for spontaneity are the following. If delta G is less than zero then you have a spontaneous reaction. If delta G is greater than zero then you have a non-spontaneous reaction./nAnd if delta G is zero then you are at equilibrium. These are the conditions for spontaneity. These are the under conditions of constant temperature and pressure. This is also important. We won't dwell on this./nAnd the actual origin for these conditions is something, again, you are going to talk about in exquisite detail in 5.60, but let me try to give you a feeling here for what delta G is. And that is the following./nThe bottom line is this. The reason why it is delta G and not delta H for spontaneity is because when you have an exothermic reaction, for example, that releases this delta H. It turns out that not all of that enthalpy, that delta H, not all of that delta H is necessarily released to the outside world to do useful work./nWhen you have an exothermic reaction, it is not necessary that all of that delta H is released to the outside to useful work. Some of that delta H actually gets stuck in the reaction, stuck in what I call the nooks and crannies of the product molecules./nThe amount of that energy that gets stuck is this T times delta S. T times delta S has the units of energy. What do I mean about some of the enthalpies getting stuck in the nooks and crannies of the molecules? Well, it turns out, and we are unfortunately not going to have time for this discussion, that molecules have a way to store energy in their vibrates./nMolecules vibrate. These bonds move. Molecules giggle. Molecules rotate. That is a way in which molecules can store energy. And it is that energy that is not available to the outside system to do useful work./nAnd so the amount of energy that is actually available to do useful work is delta S minus this T delta S, the amount that gets stuck in the products of the reaction. And so that's why delta G here is called the free energy, because it is this amount of energy that is free to do useful work./nDelta H, that is not necessarily all free. Delta G is free to do useful work. For example, let's look at this reaction. This is an oxidation of methane. Methane is a major component of natural gas./nWhen you home for dinner and you turn on your natural gas stove, this is the reaction that is going on. You are burning methane. A very exothermic reaction, minus 890 kilojoules per mole. This is the energy that you are going to use to heat up the water to boil the water so you can cook your pasta./nHowever, not all of this energy is going to be available to heat up that water. The amount that is going to be available is delta G. First of all, we have to know how to calculate delta S for this reaction./nWell, I will show you how to do that in a minute. But the bottom line for this reaction is that delta S is minus 0.242 kilojoules per degree Kelvin per mole. If I put that delta S into this expression and calculate delta G, well, you can see I am going to have a minus 890 plus some number./nThat is going to give me a negative number, and that negative number, that delta G is going to be less negative than delta H. Not all of this enthalpy is available to do useful work to heat up that water./nSome of that energy has gotten stuck in these internal degrees of freedom effectively of the product molecules. On the other hand, let's look at what your body does. It doesn't burn methane but it burns glucose minus 2816 kilojoules per mole./nIf we go to calculate the entropy for that reaction, that entropy is positive 0.233 kilojoules per degree Kelvin per mole. Positive entropy means we increase the disorder when we go from reactants to products./nLet's calculate delta G for this reaction. Delta G here is this delta H minus then a positive number. We are going to get for delta G minus 2885 kilojoules per mole. Look at this. Delta G for this reaction is even more negative than delta H./nIn other words, in this case we are going to get out -- Your body gets out of the oxidation of glucose all of the enthalpy to do useful work and then some. Your body has figured out how to do this./nIt gets even more energy out. And it does so because it has gotten that energy from the nooks and crannies of the reactant molecules, the centrally internal motions of the reactant molecules. There are more ways to store energy in the reactant molecules than there is in the product molecules, and so delta G here is even more negative than delta H./nYou get all that delta H out and then you get even more because you don't have as many ways to store energy in the products as you did in the reactants. And so the energy has got to be conserved. It is going to come out./nAnd it comes out. Delta G is more negative than delta H. This works really well. Now here is entropy. I told you we to know how to calculate that, entropy change for a chemical reaction. If you want to calculate that, you need to do so in this way./nThe entropy for a reaction is the sum of the entropies for all of the products of the reaction, but you have to multiply it by the appropriate stoichiometric number because these entropies here are per mole, minus the sum of the entropies of all of the reactants multiplied by this stoichiometric number./nThese entropies are actually absolute values. That not too often happens in thermodynamics, having an absolute value. These are absolute values. They are not deltas. Those absolute values are a result of the third law of thermodynamics, again, discussed in exquisite detail in 5.60./nSee you Monday.
Tags // Bond Enthalpy
Added: April 16, 2009, 11:01 pm
Runtime: 2844.33 | Views: 24589 | Comments:0

Chemical Science - Chemical Equilibrium - Lecture 19
Principles of Chemical Science/n * Email this page/nVideo Lectures - Lecture 19/nTopics covered: /nChemical Equilibrium/nInstructor: /nProf. Catherine Drennan/nTranscript - Lecture 19/nIn this course, because I am a biological chemist now, I am going to give you some examples between chemistry and biology./nWe are going to teach you these principles of chemistry./nI am going to explain to you why that is relevant and how it applies to the more life sciences or biological areas to try to give you a sense of how I made this transition and being someone who did not like chemistry in high school to somehow who has now been a chemist for the rest of my life./nAnd so this is another important point that I like to make which is never say you really hate a subject, because I said I really do not like chemistry and became a chemist. I am doing chemistry for the rest of my life./nAnd that can happen to you, too./nSo be very careful what you say because you may discover later that that is really what you want to be doing. All right. What are we going to do in this second half of the course? We are going to first start out by finishing chemical equilibrium./nProfessor Ceyer started this. She led you from thermodynamics into chemical equilibrium and introduced you to the chemical equilibrium constant K. And so we are going to take it from there. And we will do a little review of what she covered on Monday as well./nThen we move into acid base equilibrium./nThis is a form of equilibrium. And from there to oxidation reduction reactions. And just from this connection between chemistry and biology, enzymes, which are the proteins in your body that catalyze all the reactions which allow you to walk around and breathe and all those kind of good things, most of these enzymatic reactions involve either an acid based kind of catalyzed reaction or oxidation reaction or both./nSo these are fundamental parts of how living systems work./nThen we cover transition metals, and I am going to explain during that part about metals in biology. A lot of enzymes require transition metals for their activity, so we will talk a little bit about that./nAnd then we end with kinetics in terms of the connection with biology, kinetics rates of reaction. What enzymes do is they work as catalysts to help speed up those reactions./nSo all of these topics are fundamental chemistry topics./nThey also apply to biology. The second half of the course is quite a bit different, really, from the first half of the course. In the first half of the course, you really get a lot of the basic principles, atomic theory, orbitals, bonding, Lewis structures, a lot of really fundamental things which in thermodynamics is what you need to know to be able to do the second half./nThe second half is really more applications of these basic principles./nAnd when you put this together, you get the fundamentals of chemistry that you need to go and study biochemistry, organic chemistry, inorganic biology, life sciences. These are the fundamental principles of chemistry which you get through in this half semester course./nSo it is somewhat of a different transition, and I just like to let people know ahead of time that it is going to feel like a little bit of a different course./nYou have a different instructor, and it is really somewhat different material./nIt builds on what you learned before, but it is really more applications. It is going to have a little bit of a different feel to it, so don't be surprised by that. Are there any questions? I will say the technical aspect of the course is the same./nYou have lecture notes, the way the problem sets are structured, everything like that is going to be the same. Use of calculators, all of that, the rules of the course do not change./nBut it is a little bit different material./nAre there any specific questions? And feel free to raise your hand. Maybe the TAs can help me if I don't always see that someone has a question, and phone extension, but usually email works best. We are doing a fairly smooth transition from the material on Monday to today./nWe're still on the topic in Chapter 9 in the book on chemical equilibrium./nAnd I just want to point out some things about the handouts that you've received. At the top of the handout, I have the chapter and page numbers available, so if you want to refer where this material is in your book that is always going to be listed at the top of the handout./nI also list on each handout at the top, the topics that are covered in that handout or in that lecture./nThis is just useful when you are going back and, say, reviewing for the final to see, oh, yes, on this day we covered these topics./nSo that is going to be listed at the top of each of these. And, as Professor Ceyer told you in the beginning of the course, these are lecture summaries that help you so you don't have to sort of scribble and write all the time that you can sort of sit back and listen more and write a few things./nBut, like Professor Ceyer, there are some blanks. Sometimes I am going to be wanting to ask you questions./nAnd so there are some blanks that you can fill in. So you should have a pen or pencil available because there are going to be places you are going to be wanting to write./nBut we provide these summaries to make it easier for you to not have to scribble constantly, especially with PowerPoint, so that you can really listen to what is going on. That is pretty much the same as what you've experienced./nBut if, of course, don't like those lecture summaries then put them away and you can take notes if that's how you feel that you learn better./nAll right. Chemical equilibrium. You were introduced to this a little bit on Monday./nYou moved from delta G into thinking about the relationship between delta G and K, the equilibrium constant. Basically, the point to remember about equilibrium is that it is a dynamic process. It's not that the reaction has stopped when it reaches equilibrium./nThe reactions are still going, but the rate of the forward reaction and the rate of the reverse reaction are equal so there is no net change in the composition of the material./nReactions are still going but there is no net change. Let's take a look at an example of what that means./nIf we take the reaction N2 gas plus H2 gas going to NH3 gas, a reaction with a delta G knot minus 32.90 kilojoules per mole, and if we consider what happens when the reaction starts, so we look at the concentrations of the reactants and products over time, at the beginning of the reaction you are only going to have your reactants./nYou would have hydrogen which would start and it would react so the concentration will change, it will decrease and then it will level off./nThe same will happen for nitrogen, the other reactant. It will start at a higher level, and then the concentration will drop and will level off./nAnd there will be no product in the beginning, but the product concentration will come up and then it will level off./nI want to make the point from the lecture notes. I cannot tell you how long trying to make the curves in PowerPoint. If anyone is really good at drawing sort of curve lines in PowerPoint you can see me later./nSo, if it looks sometimes like it is not really leveling off, that is a PowerPoint artistic issue./nAnd I will just say in general, for this course, that in terms of your ability to draw things on the exam, the standard for my parts of the exam, exam three, and my part of the final is that you have to draw as well as I draw./nThis is an extremely low bar in terms of drawing. And you will notice that what I went to college to study art was not on the list./nSo, I apologize for some drawing inabilities that may occur during this course./nBut the point here is that with the reactions that are starting at higher concentrations they are being consumed as they make product, product is coming up, but then all the lines flatten out. And what that means is that equilibrium has been reached./nThe reaction is still going on. It's going in the forward reaction, in the backward reaction, so there are arrows going forward and backward./nBut there is no net change so the concentrations are leveling off./nThe rate of the forward reaction is equal to the rate of the reverse reaction. We can think about this in a different way of what is going on here in terms of free energy./nIf we think about plotting free energy versus the progress of the reaction --/nSo free energy, or our friend delta G, versus the progress of the reaction./nIn the beginning you would have pure reactants./nAnd, at this point, delta G is going to be less than zero. It will be negative. So you will have a reaction that is spontaneous in the forward direction./nHere you would start up and it will come down./nYou will have a delta G less than zero until you reach some point where delta G equals zero./nSo you start here with pure reactants, and those pure reactants then will react and form some product. If you, however, start with pure products and you have no reactants, delta G will be greater than zero, so you will be spontaneous in the reverse direction of the reaction./nSo you will use up your products to form reactants./nAnd so here you go this way and here you are going to go down this way. And here you have a delta G that is greater than zero. So you are getting rid of your products. You started with pure products and you're making reactants until you get to this midpoint here where delta G equals zero./nAnd when delta G equal zero then you are at equilibrium./nThere is a relationship between delta Gs and the directions of these reactions. They are all moving toward equilibrium./nAlong with this then, we can think about the fact that delta G is going to be changing as the reactants and products change, as the composition of the reaction mixture changes./nWe start with pure reactants. You are going to have this negative delta G. At some points at equilibrium it will be zero. It is also possible to have a positive delta G if you start with pure products./nSo delta G is going to depend on your composition of your material./nAnd here is just a quick review from Monday. Professor Ceyer introduced you to this equation. The relationship between delta G, delta G knot and Q, which is the reaction quotient./nAgain, delta G is at any particular point in your reaction. Delta G knot has to do with the reactants and products in their standard states./nQ is the reaction quotient. And then R is the gas constant./nAnd T is temperature. She also introduced you to this on Monday, but I just want to review this. I think she actually just talked about the gaseous state, but the same equations and principles apply if you are talking about reactants in solution./nQ, this reaction quotient, the longer sort of definition of Q depends on the partial pressures of the reactants and the products or the concentrations if you are in solution./nShe showed you this equation of just A plus B going to C and D and said that the top of the reaction quotient is products and the bottom is reactants./nYou would have the partial pressure of C over some reference raised to the power of C, because you consider the stoichiometry of the reaction, and then it would be times D, the partial pressure of D over this reference raised to the power of D, again stoichiometry./nAnd then the same for the reactants. And over here, for solution, you're just talking about things in concentrations rather than in partial pressures./nBut then since the reference here would be one bar for partial pressure or one molar for concentration, you often see these with this reference dropped out of the equation, and you see Q as being equal to the partial pressure of C raised to the C, partial pressure of D raised to the D over partial pressure of A to the A and B to the B, or the same form in terms of concentrations./nThis is how you will be, in general, seeing it. And remember what your references are there. Again, she went over this on Monday./nWe will just look at it again. Also on Monday, she looked at this equation here and considered delta G when it is equal to zero, which means you are at equilibrium./nAnd, at that point, when delta G equals zero, Q, that reaction quotient equals K, which is the equilibrium constant. Q equals K, it has a new name when you are at equilibrium./nWe can take this equation then and substitute in zero, because at equilibrium delta G is zero./nAnd then this equation can be rewritten. You just bring the RT natural log of K to the other side. Again, K then is the equilibrium constant. It has the same form as Q./nBut you are only considering the concentrations at equilibrium./nIn the gaseous state, this looks just like Q, but these concentrations or partial pressures are the ones at equilibrium because K is the equilibrium constant. And, likewise, in solution, here you are just talking about the ratio of the products to the reactants at equilibrium./nI have these little equilibrium things to remind you that these are the concentrations at equilibrium./nShe also showed you, on Monday, that you can rearrange this equation. You can substitute in for delta G knot the minus RT natural log of K, because those are equal to each other./nAnd then you can rewrite that equation as delta G equals RT natural log of Q over K./nIf you're interested in this, this is a very useful equation. If you think about at what concentration are the reactants now, how does that compare to the concentrations at equilibrium? If you know where you are now, if you know Q now, you know what K is, you can think about the direction of the reaction./nYou can think about delta G. Is it negative or positive? Are you going the forward direction or are in you going in the reverse direction?/nThis is a very useful equation to predict the direction of the reaction depending on the composition at that moment compared to the composition at equilibrium./nIf Q is less than K, that is going to mean you will get a negative delta G, and you will be spontaneous in the forward direction./nBased on this equation, if Q is greater than K then delta G is going to be positive and you are going to go in the reverse direction./nAgain, this is very useful thinking about the relationship of Q and K and how that affects delta G. Let's look at an example then. We have our same reaction. Now we are given a value for K at 400 degrees C./nAnd we are given some partial pressures of the reactants and products./nAsk now are we at equilibrium here? If not, which direction is the reaction going to go? Let's figure out what Q is because we want to know the relationship between Q and K. That is going to tell us where we are./nAre we at equilibrium? What direction will the reaction go?/nAll right. Q. What is going to be on the top of the expression for Q? What goes on the top? Products or reactants? Products. And so we have one product here, so we are going to put our partial pressure of our product./nAnd then we have to consider the stoichiometry which would give us a two./nOn the bottom we have our reactants, so we will consider the partial pressure of N2 and the partial pressure of H2 cubed, again, the stoichiometry of the reaction./nThat is going to equal 1.1 squared over 5.5 times 2.2 cubed./nAnd if you work out the math there you get 2.1 times 10 to the minus 2, so that is Q. And K was listed as 1.9 times 10 to the minus 4 at the same temperature./nHere Q is greater than K. And so what direction will the reaction go?/nIt will go, as people said, toward reactants, or you could say in the reversed direction. So, NH3 will dissociate until equilibrium is reached./nAnd just one point about doing this, I have noticed people have come to you and said, oh, I knew the answer on the exam./nWhen I said left, what I meant was right. And since I have done that numerous times, I suggest that you say toward reactants or toward products or draw an arrow./nBecause usually then you actually write down what you mean./nMy personal tip is stay away from left and right on exams just in case, unless you are particularly good at that. But I have seen people be heartbroken when they knew the answer and just wrote the wrong direction./nHere we have an example where we were able to predict what direction the reaction would go by looking at Q and K./nLet's think about what else we can learn from K. The size of the equilibrium constant, the largeness of that value tells you something about the reaction./nWhat does K tell us?/nK, again, is equal to the products over the reactants at equilibrium. The size of K, whether it's greater or less than one, is going to tell us about the ratio of products to reactants at equilibrium./nWhen K is large, or we can say really what we're talking about is a K that is greater than one, you are going to have more products --/n-- than reactants at equilibrium, or high products./nThe products are going to be greater. If K is a big number, it has got to mean that there are more products than reactants at equilibrium, so you already know something when you know the value of K./nWhen K is small, when you have a K that is less than one then you are going to have low products or more reactants at equilibrium./nLet's look at an example for this, a case when K is greater than one./nLet's consider K greater than one, and let's look at a reaction for which that is true. Here is one. We have 2NO2 goes to N2O4./nK is 6.84 at 298 Kelvin. You have a delta G knot of minus 4.76 kilojoules per mole./nLet's look at a case where we start with just reactants. We have one bar of the NO2 gas and have no products. Let's consider then what is going to happen and calculate how this goes to equilibrium./nFor this, in this case, we actually have the K is 6.84 and we have the delta G knot minus 4.76 kilojoules per mole./nAgain, if we look at concentration, in this case we are talking about gases so we have partial pressures versus time, we are starting with just reactant, one bar of reactant./nAnd so the reactant will start up high and go lower as it reacts, our reactants./nThere will be no products when we start, but it will be created. And, eventually, when you reach equilibrium it will level off./nIn the beginning, with no products, Q is less than K./nBecause there are more products at equilibrium than there are now, which is there are none now, there are definitely going to be more at equilibrium./nWe have a negative delta G, and we will go in the forward direction of the reaction toward products, which is shown here. Reactants decrease, products form until you reach equilibrium. Now let's calculate what the concentration of these are going to be at equilibrium./nLet's calculate the equilibrium concentrations./nAnd our reaction, again, 2NO2 going to N2O4. Initially, we are going to have 1.000 bar and no product at all./nAs the reaction goes there will be a change, you will be making product, so you will be making product plus x, some amount of product will be formed./nAnd how much reactant will be lost during this change? Minus 2x./nAnd the 2 comes from the stoichiometry of the reaction. Then, at equilibrium, we would have 1.000 minus 2x and x. Now we can calculate K from this to find out what the concentrations of these gases are at equilibrium./nOr, find the concentrations from K, I should say./nK was 6.84 at this particular temperature. And so K is going to be equal to the products over the reactants. The partial pressure of N2O4 over the partial pressure of NO2 squared./nAgain, the stoichiometry of the reaction./nAnd then if we plug in these numbers, or plug in the equation x over 1.000 minus 2x squared./nAnd now you can take this and solve for x. Here, if you solve for x, it comes out at 0.381 bar. And so this is going to be your amount of product./nThat is the amount of N2, the partial pressure for the N2O4./nAnd we can go back and write that up here./nAnd then we can solve for the other one, 1.000 minus 0.381, sorry, 2 times 0.381, and that is going to give you 0.238 bar./nAnd that is going to be your partial pressure of your reactant at equilibrium, 0.238 bar. You see that when you have a K that is greater than one, you have more product than reactant at equilibrium./nAnd you can figure out that exact amount if you know how much you started with and you know K./nAnother thing to point out here is that in this case there is a relationship where you had a delta G knot that was negative 4.76 and that also there was this relationship between K and delta G knot./nIf you rearranged this equation to solve for K, you can see that if K is large then delta G knot is going to be negative and large./nIf you have a larger K, you are going to have a larger negative delta G. There is this relationship between delta G knot and K, and that can be derived from this equation./nYou can also go back and forth between these as well./nWhat that is going to translate into for a K larger than one, more products than reactants at equilibrium./nA couple more things that you should know about equilibrium expressions to be able to work all the problems./nAnd that is that if you want to add up two reactions and form a third reaction, you can also use the information about the equilibrium constants for each individual reaction to get an equilibrium constant for the summed reaction./nIf you add things together you multiply then the equilibrium constants./nCan I add these equations as such and get the third reaction on the bottom? What do I have to do? I have to multiply the second reaction by two./nI need to multiply that reaction by two. And then K3 is going to equal what in terms of K1 and K2?/nK1 times? Yeah, K1 times K2 times K2 or K1 times K2 squared. And then you can get K3 if you know those individual ones./nAnd let me just prove that that is true to you. We will give a little more practice writing out Ks and Qs./nK1 times K2 squared is going to give you K3. I am going to do products over reactants./nThe product is the partial pressure, PCL3 squared, and the reactant for the first one is going to be the partial pressure of P squared times the partial pressure of CL2 cubed./nAnd then K2 squared will have the partial pressure of PCL5 squared over the partial pressure of PCL3 squared, partial pressure of PCL2 squared./nAnd then, just as we did to get these equations to add up, you needed to cancel this one and cancel this one, those will cancel out here as well, and you will be left with an expression --/n-- like this./nAnd if we go back and double check that that is going to be correct then if we have --/nWait. This one is going to be Cl5, sorry./nActually, that is not canceling out appropriately./nIt is right./nOK, good. I am just having a problem canceling everything out. You should be able to add those together. All right. In the next couple of minutes we are going to talk about what happens if we take this system and then we apply stress to it./nWe have our conditions, we reach equilibrium and then a stress is applied. This is the principle of Le Châteier. And this is that a system in equilibrium that is subjected to a stress tends to react to minimize the stress./nAnd I find this a difficult concept for people who go to school at MIT and work at MIT./nAnd the reason why is that in this environment, when there is heavy stress applied, often the behavior is such to add more stress rather than minimize the stress./nYou have to think about this in a way as sort of other systems at other universities maybe down the street might work where a stress is applied and the reaction is, oh, let's minimize the stress. Let's make less stress./nThat is what Le Châteier's principle is about here./nIt turns out to be really useful because chemical reactions, unlike people at MIT, do tend to react in predictable ways to stress. And the chemical reactions react in such a way as to minimize the stress./nAnd this works almost all the time. You can predict the direction of the reaction based on the stress applied if you think about minimizing that stress./nLet's look at some examples. We can look at the simplest case which is adding or removing reactants or products./nWe will look at that, too. We looked at this reaction before. And when you talk about chemical equilibrium, this is the reaction you study, and we will be thinking about this on Friday as well./nHere, again, you start with hydrogen and nitrogen and you're making ammonia./nYou start with just the reactants, no product, and they react and then they level off when equilibrium is reached. What happens then if we're at equilibrium, someone comes along and adds more H2? Which direction will the reaction go? What do you guess? You want to minimize the stress./nMinimize the amount of H2./nIt will go to our products. You will use up more N2 and you will make more product. And they will react and then will come to equilibrium again. And the net composition will have the same ratio once you reach equilibrium./nWhat if you added more product? What is going to happen? It will go in the reverse direction./nAnd so you will end up making more reactants, both H2 and N2, until you reach a new equilibrium again./nYou have the same ratios as you did before. We can think about this. You tend to minimize the stress of the reaction. You add more reactant, you use it up and make more product. You add more product, you use it up and make more reactants./nThe reaction will shift toward product, in this case, when we are adding more hydrogen./nAnd we can think about this in terms of Q and K again. If we think about this and we add more reactant then momentarily Q will drop below K. And so the reaction, when that happens, you will have a negative delta G./nRecall this is our predictive equation that we saw earlier. We can think about the ratio of Q to K in terms of the sign of delta G./nSo with Q less than K, negative delta G, we make more product. This makes sense./nYou came up with the right answer without thinking about this equation, I think, but we can also rationalize it from this equation. And then we saw this before, adding more reactant, making more product, shifting it toward products to the right./nWe also considered adding more product./nAgain, you can rationalize in terms of Q and K and delta G. As you add more product, Q is momentarily larger than K, there is more products over reactants, Q is bigger, so that is going to mean a positive delta G which means that it is spontaneous in the reverse direction so you make more reactants./nYou can think about this in terms of the equation, or you can think about it in terms of minimizing stress./nAgain, we saw this example. We added more product. We made more reactants. To compensate, you add more product, it removes the product./nFinally, what would happen in this case if we remove product? What happens? Right, you would make more product. Again, you can think about this in terms of Le Châteier, you remove product./nTo compensate for that stress you make more product./nOr, you can think about it in terms of Q and K. If you remove product then you are going to have a Q that is less than K because you just pulled product out, so now you have less product than you did at equilibrium./nAnd with Q less than K, you have a negative delta G. And so your system will shift with a negative delta G to the right. It will be spontaneous in the forward direction./nSo, Le Châteier's principle or this equation are really good at predicting stress./nAnd we will continue on Friday talking about Le Châteier and move on to think about other forms of shifting the reaction based on stresses that you add to the system.
Tags // Chemical Equilibrium
Added: April 16, 2009, 11:05 pm
Runtime: 2575.27 | Views: 22239 | Comments:0

Chemical Science - Free Energy of Formation - Lecture 18
Principles of Chemical Science/n * Email this page/nVideo Lectures - Lecture 18/nTopics covered: /nFree Energy of Formation ΔGof/nInstructor: /nProf. Sylvia Ceyer/nTranscript - Lecture 18/nWe also saw, or we tried to understand what the free energy actually was./nAnd we said that when you have, for example, an exothermic reaction that not all of that energy released is going to be released in a form that can do useful work./nSome of that energy that is released, some of that delta H, for example, goes into, or can go into the vibrations and rotations of the product molecules that energy of which is not able to do useful work./nAnd so the amount of energy that we can actually use to do work is this free energy, is delta G right here./nSo it can be something less than delta H. We also saw a condition, which was the oxidation of glucose, where delta G was actually greater than delta H. And that is the case where we have some of that internal energy locked up in the reactants which now is released in the form of this useful work delta G./nAnd I think we had just started to talk about this term delta S, which is the change in the entropy for the reaction./nAnd we talked about how to calculate it. We calculate it from the absolute entropies of the reactants in the products. Delta S for a reaction then is the absolute entropy of each one of a product./nTimes the appropriate stoichiometric number summed over all the products./nAnd then minus the absolute entropy of each one of the reactants times that appropriate stoichiometric number summed over all the reactants. And it is the difference between the two, delta S for the chemical reaction./nAnd S is an absolute entropy that is a result of the third law of thermodynamics, which I mentioned, I think, last time./nYou will talk about it a lot in detail in 5.60. Let's look at an example here at the change in entropy./nBecause I want to show you that you can have a change in entropy that is large enough to make an endothermic reaction spontaneous. You can have what we call an entropy-driven reaction. For example, the melting of ice./nTo melt ice, to go from the solid form of water to the liquid form, that is an endothermic reaction, 6.95 kilojoules per mole./nHowever, going from the solid form of water to the liquid form there is a great increase there in the entropy, in the disorder. And it is that increase in the disorder that allows that reaction to be spontaneous./nWe can calculate the entropy change for that reaction./nThat entropy change then is the absolute entry of the product, which is liquid water, minus the absolute entropy of the reactant, which is solid water. And you can see that delta S here is positive./nIs there a question I can help you out with? Did you have a question? Can I help you out?/nOK. That delta S is positive. We have increased the disorder of the system. And it is that positive delta S then that gives us a delta G that is negative, meaning our reaction in this direction from solid to the liquid is a spontaneous reaction./nDelta G here is delta H, 6.95 minus the temperature at which the reaction is proceeding times that change in the entropy./nAnd, in this case here, delta G now is minus 1.57 kilojoules. That reaction is spontaneous, even though it is endothermic./nAnd it is that increase in the entropy that is driving this reaction here. Now, I want you also to notice something right here. And that is that the change in the entropy usually is in units of joules per degree Kelvin mole./nDelta H, just about always, is in units of kilojoules per mole./nRemember when you're working with delta H, bells ought to go off in your head. Remember that you probably have to do a unit conversation here from joules to kilojoules. This catches a lot of students on an exam, so I am trying to give you some fair warning about this./nNow, what I want to talk about is this free energy of formation, delta G sub f knot./nThe Gibbs free energy of formation is analogous to the enthalpy of formation. This is what we looked at the last time./nThe Gibbs free energy of formation, delta G sub f knot for R standard state, which is one pressure, is the free energy formation of one mole of a compound from its elements in their most stable form at the standard state./nIt is analogous to the heat of formation, only it's the free energy of formation./nIt is tabulated for you just like the enthalpy of formation is. You can look up the free energy of formation for every compound that is known./nHowever, unlike the enthalpy of formation here, the free energy of formation you can also calculate. You can calculate from this expression, delta G equal delta H minus T delta S./nIf you've got a reaction for which you have defined the free energy of formation, so if you know the free energy of formation of a molecule and you know the delta S for the reaction that defines that enthalpy of formation then you can calculate the free energy of formation for every molecule, which is different from the enthalpy of formation./nThat you do have to look up, but you can actually calculate the free energy of formation given that you know the enthalpy of formation and given that you know delta S for the reaction that defines that enthalpy of formation./nQuestions on that? OK. But why is this free energy of formation so important? Well, that's what we have got to look at right now./nHere is a reaction that produces one mole of carbon dioxide. The delta G for this reaction is minus 394 kilojoules per mole./nThe delta G for this reaction is defined as the free energy of formation for CO2. Why? Because this reaction, as written, produces one mole of CO2 from the elements that compose CO2./nBut the elements in their most stable form./nThe elements are carbon. And the most stable form of carbon is graphite. And the element is oxygen. NO2 is the most stable form of oxygen at one bar pressure and room temperature which we are working with here./nThat is the free energy of formation of CO2./nAnd this is very important because that free energy of formation of a molecule defines the molecule's stability relative to decomposition to its elements./nFor example, the forward direction here, this delta G is negative. It is spontaneous in the forward direction. When delta G, a formation, is negative for a molecule, what we say is that the molecule is thermodynamically stable relative to its elements./nBecause the formation of that molecule, that delta G is negative./nThe reaction as spontaneous is written in the forward direction. It is not spontaneous in the reverse direction. The reverse direction, this delta G would be positive./nSo because the delta G for this reaction is negative, we say that CO2 is thermodynamically stable relative to decomposition to its elements, carbon and oxygen./nThis is important. The sign of the free energy of formation for a molecule is important. Had the free energy of formation of CO2 been positive, we would have said that was thermodynamically unstable relative to decomposition to its elements./nBecause this forward reaction, delta G would have been positive, but the reverse reaction would have been negative./nReverse reaction, there would be a spontaneous tendency for CO2 to decompose. But this is an example where CO2 is stable relative to decomposition to its elements./nHere is an example of a molecule where it is thermodynamically unstable relative to decomposition to its elements, benzene./nI write a reaction here that is the formation of one mole of benzene from the elements in the most stable form. The free energy of formation of a mole of benzene here is positive, 124 kilojoules per mole./nIt is positive./nWhat does that mean? Well, it means that the reverse reaction delta G is negative. It means that benzene has a thermodynamic tendency to decompose into its elements. It is unstable relative to decomposition into its elements because it has a positive delta G./nIt is the reverse reaction that is spontaneous./nBut even though a reverse reaction may be spontaneous, it can also be really pretty slow. When was the last time you saw a pint of benzene decompose to graphite and hydrogen? Not recently./nAnd so even though you have a thermodynamic tendency to decompose, it doesn't mean that the rate is going to be fast enough for you to see that in any reasonable amount of time./nThere is a thermodynamic effect and there is a kinetic effect, and you are going to talk about the kinetic effect in much more detail with Professor Drennan a few weeks from now./nBut, at the moment right here, we've got two different names to talk about, thermodynamic stability and kinetic stability, so to speak. And that is the following./nWe call a molecule stable or unstable./nAnd when we use those terms, stable and unstable, we are referring to the delta G of formation. We are referring to the thermodynamic tendency of the molecule to decompose. So benzene here is thermodynamically unstable with respect to decomposition to its elements./nHowever, benzene is what we call nonlabile./nThese terms, labile and nonlabile, refer to the rate with which that thermodynamic tendency is realized. Because the rate of decomposition is so slow such that you never see it, we say benzene is nonlabile./nNonlabile means it is not going to decompose./nNot because it doesn't have the thermodynamic tendency to, but because the rate is just too slow. However, if benzene's rate for decomposition was very fast we would call it a labile molecule./nBut it is not. It is nonlabile, but it is unstable. Unstable refers to thermodynamic tendency. Labile and nonlabile refers to the kinetic, the rate at which that thermodynamic tendency is realized./nWell, like the entropy of formation, delta H sub f, delta G sub f, the free energy of formation, well, that can also be zero./nFor example, for the elements hydrogen, oxygen, chlorine, xenon, all those gases, free energy of formation is zero./nJust the same way in which we described it last time for the enthalpy of formation./nFor the elements bromine here and mercury in the liquid form, delta G sub f, free energy of formation is equal to zero. For the elements carbon in the form of graphite, sodium iron, iodine, all solids, their free energy of formation is also equal to zero./nBecause those are the elements in their most stable form at R standard state./nHowever, look at this. The free energy of formation of bromine in the gas phase, that is not equal to zero. It is not equal to zero because bromine in the gas phase is not the most stable form of bromine at R standard state./nLiquid brome is./nDelta G, a formation of bromine in the gas phase, is not zero. Likewise, here the free energy of formation of diamond, that is not zero, because diamond is not the most stable form of the element carbon at one bar pressure at room temperature./nGraphite is. It is important to look at the phase of the elements that you are dealing with./nDo you want to calculate delta G for a reaction? Well, just like calculating delta H for the reaction, you can use the free energy of formation./nYou need to know the free energy of formation for every one of the products, multiply that by the appropriate stoichiometric number and sum that all up./nAnd then you do the same here for the reactants./nYou need to know the free energy of formation for every single one of the reactants, multiplied by the stoichiometric number, sum that all up and take the difference, and you will have the free energy for that chemical reaction./nNotice this is products minus reactants./nJust like calculating enthalpies from the heats of formation, that was products minus reactants. However, when we calculated enthalpies from bond enthalpies that was reactants minus products./nThat is something you do have to know. Likewise, you can also calculate delta G for a reaction now from knowing delta H for that reaction and also knowing delta S for that reaction./nYou've got your choice, to use the free energies of formation that you can look up or from knowing the enthalpy for a reaction and knowing the entropy for that reaction you can calculate delta G./nYou've got your choice with delta G here, depending on what information you're given or know./nNow what we want to talk about is our ability or inability to control the spontaneity of some chemical reaction to control it by virtue of adjusting the temperature./nLet's take this example here. This is sodium bicarbonate, better known as baking soda./nThis is what you put into the dough of some kind of baked goods that you want to make, muffins or cakes. You put it in there in order to lighten the batter./nAnd, of course, the way this works is that the sodium bicarbonate decomposes. It decomposes to form CO2 and water./nAnd when it decomposes it does so in the oven and the CO2 and the water then expand, they evaporate, but in that dough they kind of make bubbles before they totally evaporate./nAnd around those gas bubbles the dough kind of hardens a little bit and eventually the CO2 and water are driven off./nBut what it leaves behind is a very porous structure in the dough. It leaves you something that you can actually put your teeth into./nHave you ever tried putting your teeth into some cake where somebody left out the sodium bicarbonate? It is an interesting experience. I have done it. Important here. But this is a reaction that is very endothermic, plus 136 kilojoules per mole./nIt is, however, a reaction where there is an increase in the entropy a lot here, increasing the disorder. And let's calculate delta G for this reaction. Well, delta G for this reaction at room temperature is plus 36 kilojoules per mole./nIt is non-spontaneous./nYou know what? That is good because you don't want that reaction to start when the dough batter is still sitting on the kitchen counter because at that point you are not going to be able to harden the dough in any way./nBut let's try to make this reaction now spontaneous by increasing the temperature. Let's make this term here negative, more in absolute value./nWe want to make this term larger than delta H so that we can have a negative delta G./nWell, we can do that by raising the temperature to a baking temperature 350 Fahrenheit, which is about 450 degrees Kelvin. Now, when we calculate delta G at 450 degree Kelvin, what we find is that delta G is now negative./nAnd now we've got a spontaneous reaction./nNow when you get up to this baking temperature this reaction proceeds readily in the forward direction. We have adjusted the spontaneity of this particular reaction by increasing the temperature by making this second term, the T times delta S, larger in absolute magnitude than delta H./nWe have made that reaction spontaneous./nThere are reactions for which we can control the spontaneity by adjusting the temperature. And let's take a look at that. Well, as you know here, delta G is the linear function of the temperature./nLet me just draw a dependence. And, actually, this is for the decomposition of sodium bicarbonate. Let me just draw delta G as a function of the temperature./nThe slope of this line here is minus delta S for that reaction./nYou can see that. This is going to be the slope. I am plotting it versus T. The intercept is delta H. It is the enthalpy change for this reaction. And now let me just draw a zero here, a dotted line./nI drew that zero because I want you to see that at some temperature right here, that I am going to label T star, that delta G is equal to zero./nWhat that means is that at this temperature the sign of delta G is changing. You can see that for temperatures less than T star the value of delta G knot here is greater than zero, meaning the reaction is non-spontaneous./nFor temperatures greater than T star right in here, you see that delta G knot is less than zero./nIt means we have a spontaneous reaction. We have controlled the spontaneity of this reaction by the use of temperature. Let's calculate the value of the temperature at which the spontaneity of this reaction changes./nLet's do that./nTo do that we are going to set delta G knot here equal to zero and then we are just going to solve for T star. We are going to solve for that value of T star that for this delta H and that delta S will make delta G knot equal to zero./nLet's do that. Well, then T star is delta H over delta S./nIf I plug in for the decomposition of sodium bicarbonate, the delta H and the delta S, then T star is 406 degrees Kelvin. So your sodium bicarbonate is not going to decompose until you get to this temperature./nAt least for this reaction it appears we can control the spontaneity of this chemical reaction./nAnd that is the case. In general, if you have a reaction that is endothermic, that is delta H is greater than zero, and if that reaction increases in entropy, delta S is greater than zero, that reaction will be spontaneous for temperatures greater than some temperature T star./nYou can see that by the signs of the reactions here./nIf delta H up here is positive and delta S is positive then you've got to increase this second term, T delta S, the absolute magnitude of it enough so that it is larger than delta H./nAnd you are subtracting the two, well, that is when you are going to get the delta G negative. For large enough temperatures the delta G will be negative./nBut suppose we have a reaction that is exothermic, delta H is less than zero, and a reaction that decreases the entropy, delta S is less than zero./nI plotted that now as a function of the temperature. And you can see that the slope of this line has changed. We now have this positive slope./nWell, in this particular case you are going to have a spontaneous reaction whenever the temperature is less than some temperature T star./nBecause right here that is our T star. And so for temperatures less than that delta G is negative. For temperatures greater than that delta G is positive. Again, you can see that./nIf delta H is less than zero this delta H will be negative./nAnd delta S here is going to be negative. We are going to have a negative times a negative. That is a positive. And so since delta H is negative, T is going to have to be small. This is going to be a positive term./nT is going to have to be small so that an absolute value it is not larger than delta H in order to get a negative delta G./nHowever, suppose we have a reaction that is exothermic, delta H less than zero, and delta S is greater than zero./nWe are increasing the entropy here. Well, this is the best of all worlds. When you have an exothermic reaction that increases the entropy, this reaction is spontaneous at all temperatures. Delta G is going to be negative for all temperatures./nBecause delta H is negative./nAnd then you have minus T delta S. Well, delta S is positive. So you have a negative number plus a negative number, or a negative number minus a positive number, delta G is always negative. Spontaneous at every temperature for those reactions./nYou cannot control the spontaneity of the reaction. And then, finally, you can have a situation where you have an endothermic reaction, delta H greater than zero./nAnd the worst of all case, delta S less than zero./nYou are decreasing the entropy. You are making the system more ordered. In that case, you have a reaction that is never spontaneous. Again, you can see it from the sign. If delta H is positive and delta S here is negative, a negative times a negative is a positive, so we have a positive number plus a positive number, delta G is positive at all temperatures./nSo with temperature there we also cannot adjust the spontaneity./nThis is very important to understand. Now, what we have been talking about so far is delta Gs that have been delta G knots./nWe have been talking about delta Gs, free energies of formation in our standard state of one bar pressure./nWhat that means, for delta G knot, is that the partial pressures of your reactants and your products, they are all one bar. Delta G knot for a reaction is the free energy change for that reaction when there is one atmosphere of each, or one bar of each of the products and the reactants./nLet's take this famous reaction, argon plus boron going to carbon plus deuterium, A plus B, C to D. This delta G knot here is a delta G for a reaction under the conditions of one bar partial pressure for D, one bar partial pressure for C, one bar partial pressure for B, one bar partial pressure for A./nThat is what that means./nBut, you know what, that's not the situation you usually have. Say, for example, you are going to start this reaction. You start the reaction with one bar of A and one bar of B./nBut, of course, you don't have any C or D yet because you haven't run the reaction./nWhat we've got to be able to do is to calculate delta G for conditions which aren't at the standard state, which aren't at one bar D, one bar C, one bar B, one bar A. We've got to be able to do that./nAnd the way you do that is represented here by this expression./nWhere this expression came from is again a subject of great interesting in 5.60. You will see where this expression comes from when you take 5.60, but right now we are going to use it. This is going to allow us to calculate delta G from knowing the standard state delta G knot./nIt is going to allow us to calculate delta G for the reactants and products at any given pressure./nYou see this is delta G equal delta G knot plus RTLN of the following. This has the partial pressure of product C divided by a reference pressure raised to the appropriate stoichiometric number./nAnd that is multiplied by the partial pressure of the product D divided by a reference pressured raised to the appropriate stoichiometric number./nAnd that is all over the partial pressure of A divided by a reference pressure raised to that appropriate stoichiometric number times the partial pressure of the other reactant over some reference pressure raised to the appropriate stoichiometric number./nThat ratio is something that we call a reaction quotient./nWe call it Q. It is the ratio of instantaneous partial pressures. We might be running a reaction, and at any time the reaction may not be complete. At any time we could stop the reaction./nWe can figure out or measure what the partial pressures of reactants and products are./nAnd we can calculate what this Q is knowing what those are from a measurement. That is what the reaction quotient is. Now, our reference pressure is going to be one bar. What I am going to do, and what you can do in this reaction quotient, is that I am going to put PRF, the reference pressure as ones in here./nAnd so I am not going to explicitly write it out./nAnd that is just fine, as long as you put the pressures then here in units of bar. So this is the reaction quotient, the form of it that we are going to use, as long as you put the pressures here in units of bar./nWe've got to remember this because I am now going to use this in a very interesting way./nThis is going to give me delta G at any arbitrary pressures of the reactants and the products. How am I going to use this? I am going to start to talk about the equilibrium constant for the reaction./nI've got my famous A plus B going to C plus D reaction./nAnd you already know that all reactions have some kind of equilibrium. There is a forward reaction and there is a reverse reaction. And we know that, as we said, delta G is less is zero./nIf delta G were less than zero for this reaction as written then it is the forward reaction that is spontaneous./nIf delta G is greater than zero for the reaction as written then it is the reverse reaction that is spontaneous./nAnd now here comes the really important point, and that is if delta G is equal to zero for that reaction as written then we are at chemical equilibrium./nWhen delta G is equal to zero then we are at equilibrium./nNotice this is not delta G knot equal zero. This is delta G equals zero. That is what defines chemical equilibrium, when that delta G that I just showed you is equal to zero./nHere is that expression again that I showed you./nDelta G is delta G knot plus RTLN of this reaction quotient. What I am saying is that we are at equilibrium when this delta G here, this one is equal to zero. Not that one. This one. If that is the case --/nIf you are at equilibrium then I can take this equation here, and what I am going to do is move delta G knot to this side, and the result is delta G knot is equal to minus RTLN of this reaction quotient./nSee where I am going? Delta G was equal to zero. That let me then solve for delta G knot is equal to minus RTLN of this reaction quotient./nNow, under this condition, when delta G is equal to zero, this reaction quotient Q has a special name./nAnd that special name is a thermodynamic equilibrium constant./nWhen we are at equilibrium, this reaction quotient is that equilibrium constant K. This defines the equilibrium constant. What the equilibrium constant tells us is about the relative proportions here of the products to the reactants./nIf the equilibrium is large, we've got a lot of products./nThe numerator has got to be large relative to the reactants. If the equilibrium constant is small, well, then we've got few products relative to the reactants./nThe quotient, Q, that reaction quotient is equal to the equilibrium constant when delta G is equal to zero. Not delta G not./nAnd so that is where this expression comes from. Delta G knot is equal to minus RT times the log of the equilibrium constant./nIf you know delta G knot for a reaction, you can figure out the equilibrium constant. If you know the equilibrium constant by knowing, say, the partial pressures present of the products and the reactants at equilibrium, you can work backwards and get delta G knot./nNow what we are going to do is rearrange these equations here a little bit./nWe are going to rearrange them in a form so that it will be easy to tell whether or not you're at chemical equilibrium./nWe are going to use this reaction quotient, Q, as a measure of whether or not we are at chemical equilibrium./nWe are going to compare Q to K. When Q is equal to K, we are at equilibrium. Remember what Q is. Q is the ratio of the instantaneous partial pressures of the products to the reactants at any time during the reaction./nK is the ration of the equilibrium partial pressures of the products to the reactants./nNow we are going to work on a formalism that is going to allow us to easily compare Q to K so that we can tell whether or not we're at chemical equilibrium. Let's do that. Here is the expression that I wrote earlier that allowed me to calculate delta G at any arbitrary pressures for the products in the reactants./nHere is the expression that I got when delta G was zero when we were at equilibrium./nAnd then delta G is equal to minus RTLNFK. What I am going to do right now is I am going to take this and substitute it in. Let's do that. I am going to substitute this into this more general form./nWhen I do that, I am going to get minus RTLNFK plus RTLNFQ./nThis is just a simple substitution. But now what do I have? I have the difference between two logs. I have a minus LNK here plus an LNQ here. The difference between logs is the log of that quotient of the arguments of those logs./nSo I have RT times LN of Q over K./nThis is the difference in the logs. That is then the log of the quotient of the arguments of those logs. This is going to be great here because look at what we are going to be able to do./nRemember what Q is, the ratio of instantaneous partial pressures./nAnd you know what K is. The ratio of partial pressure is at equilibrium. Bottom line here is the following. If Q is less than K, what that means is we don't have enough products formed compared to what we should have at equilibrium./nThat is, when Q is less than K, we haven't got enough products compared to what we need for chemical equilibrium./nIf Q is less than K, well, then up here in this expression for delta G we have the log of a number smaller than one. The log of number smaller than one is going to be negative./nIf Q is less than K then our delta G is going to be negative./nAnd notice this is delta G, not delta G knot. If our delta G is negative then the reaction is going to be spontaneous in that forward direction. On the other hand, if we've got a situation where Q is greater than K./nThat is where we have more products than what equilibrium says we should have./nWell, in that case, Q over K is going to be a number greater than one. And the log of a number greater than one is always going to be positive. And so our delta G for the reaction as written is going to be greater than zero./nAnd so the forward reaction is not spontaneous, but the reverse reaction is spontaneous./nAnd what is going to happen is that the chemical reaction will proceed in the reverse direction. It will use up all of those extra products that we had for whatever reason to try to attain equilibrium./nHere is another way to look at it./nI am going to plot here Q versus the time of a reaction. And we know when Q is equal to K we are at chemical equilibrium. Suppose we start in a condition some time equal zero where Q is less than K./nIf Q is less than K what that says is that we don't have enough products compared to what we need for chemical equilibrium./nWhen Q is less than K the reaction is going to proceed in the forward direction./nIt is going to make the products so that we can attain equilibrium, so we can make enough products so that we get Q is equal to K. The reaction will proceed in the forward direction./nIf, however, we start in a situation where Q is greater than K that means we've got too many products compared to what our chemical equilibrium says we should have./nAnd so what is going to happen is that the reverse reaction is going to proceed./nIt is going to use up those products and form more reactants so that it can attain the equilibrium value for the partial pressures of the products and the reactants./nQ is going to be our measure of whether or not a reaction is at chemical equilibrium. Very important. Questions?/nR is the gas constant. It is something that will be given to you, you can look up./nOther questions? OK. See you later this week.
Tags // Free Energy
Added: April 16, 2009, 11:11 pm
Runtime: 2867.33 | Views: 17279 | Comments:0

Video lecture: acid-base equilibrium
Acid-base equilibrium is of particular importance in biological systems because the pH at which a reaction occurs is very important. This video presents a lecture on acid-base equilibrium. Edited by Ashraf
Tags // Acid Base Equilibrium
Added: April 16, 2009, 11:15 pm
Runtime: 3004.80 | Views: 18516 | Comments:0