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/nA... | more...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. | less...