 So, among all the different flavors or types of enthalpy, reaction enthalpy that we've seen, one of the most important is the enthalpy of combustion. So, combustion of course is a process that's important for a lot of different processes. We use combustion to cook with. We use it to generate a lot of our energy. We use it in many cases to drive our cars. So combustion as a process is very important and we can use thermodynamics to understand the details of combustion at a slightly deeper level. In particular, we can learn using what we know about thermodynamics to understand the temperature at which flames burn when combustion takes place. So to illustrate what I mean by that and why that's an interesting question, you're familiar most likely with burning a number of different types of fuels you might burn. You know, if you cook on a gas stove, the natural gas is mostly methane. If you use a Bunsen burner in a chemistry lab, those most likely are burning propane. If you weld with an acetylene torch, then you're burning acetylene. If you drive a gasoline powered car, you're using a mix of hydrocarbons that might be, include some octane for example. But a range of different hydrocarbons we can burn. All of those, combustion in every one of those cases, we combine those with oxygen. The carbon turns into CO2, the hydrogen turns into steam or water. So combustion is the same chemical process in each of those cases, just the stoichiometry of the reaction works out to be different. And yet the flames, the temperature at which these burn, the temperature of the flames that are generated when these different hydrocarbons combust in a flame turn out to be very different. To understand why that would be true, let's take the simplest of these as an example, just to keep the stoichiometry relatively easy for the first example. If we burn methane, so again, maybe we're cooking on a gas stove and burning natural gas or primarily methane, if we burn methane and oxygen, this will generate CO2 and H2O. If I burn one mole of methane, one carbon becomes one carbon, so I've got one mole of CO2, four hydrogens need two molecules of water, or two moles of water, so I've got a total of two plus two is four oxygens, so that means two O2 molecules on the reactant side of the reaction. So there's my balanced reaction, that's the combustion reaction for methane. The enthalpy of combustion for methane, so the enthalpy change associated with that reaction, enthalpy of combustion for methane is minus 890.3 kilojoules per mole, the number we can get by looking it up in a table or by calculating it from the heats of formation of products minus reactants, or by doing it in the lab and measuring the enthalpy in a calorimetry experiment, we can get that number a number of different ways. That's the enthalpy change for this combustion reaction. That's enough information, that plus one additional assumption that I'll mention in a minute is enough to let us calculate the temperature at which methane will burn. So that assumption, let's say, let me draw a picture of what's going on in, so I'll draw a flame, which I've drawn as a candle, but in any of these examples, if I'm burning methane, I've got some source of methane, my natural gas line on my stove or whatever, the oxygen that's required to burn the methane, that's just in the air. So methane and oxygen both enter this flame and then this chemical reaction takes place. The thing I need to understand about the flame, what I eventually use this flame for, if I'm burning methane on my stove, I'm going to use it to heat a pot of water or whatever it is I'm cooking. So there's some heat that's going to get transferred to something else at some point, that process is relatively slow. The process of getting the heat from the flame to something else takes a significant amount of time. What's going to happen very quickly when this reaction liberates some heat, so this enthalpy change is negative, which means the enthalpy of the products is less than the enthalpy of the reactants. So if I just do this reaction, it has a negative enthalpy change, so that would normally generate heat that gets given off to the environment and that's what heats my pot of water. That process, like I've said, is relatively slow. The first place that heat goes, that energy in the form of enthalpy goes, is to heating up the products of the reaction themselves. So the first thing that gets heated up is the CO2 and the H2O. So in fact, those atoms that make up the molecules of CO2 and H2O, those are the same atoms that started in my methane NO2, so it's the same atoms that were there originally. So in that sense, the heat has not left the system and we can assume, at least under some conditions, that the reaction is fast enough, that the reaction is adiabatic, meaning that the heat in the reaction, no heat was transferred to the surroundings in the reaction. All of the energy stayed within, all of the enthalpy stayed within the system, which means that the enthalpy change of the products, that energy was used to heat up the products and at least initially didn't transfer to anything outside of those individual atoms or molecules. So if we do this process at constant pressure, if it's open to the environment, we know that heat and enthalpy change are the same thing at constant pressure, that's why we define the enthalpy, is to be able to make statements like this one. So if we do this at constant pressure, enthalpy and heat are both equal to zero for this adiabatic process. So the thing that's zero, the enthalpy that's equal to zero, is the full enthalpy change for the reaction, not just the enthalpy of combustion, not just the enthalpy change when the reactants get converted to products, but also the enthalpy change that occurs when these products are heated up. So the way physically to understand what's going on is I generate 890 kilojoules every time I burn a mole of methane and then I spend all 890 kilojoules of that energy in heating up the CO2 and H2O to a much higher temperature than where they started. That's why a flame burns hot is because that energy is used to heat up the products. So in this thermodynamic equation, the total enthalpy is the enthalpy change of the reaction, heat of combustion, plus since I've got a mole of CO2, if my heat capacity is constant, which is a big assumption, we'll come back to that statement in just a second. But as an initial guess, if I say CO2 has some heat capacity, the total enthalpy change of the CO2 is its heat capacity times the change in temperature. So for each mole of CO2 I have, I've got one mole participating in this reaction, I've got heat capacity times delta T. If I've raised the temperature by some amount delta T, I've also got two moles of H2O, so I also need to include twice the heat capacity of water vapor times the same increase in temperature. The whole system, all of the products are going to be heated to the same temperature of this flame. So this delta T and this delta T are the same number, I've raised the temperature of the CO2 and the H2O by the same amount. So if I rearrange this equation, what I'm interested in, I know that enthalpy of combustion, I know I can look up the heat capacity of CO2 and H2O. What I don't know is the amount by which this reaction will have increased the temperature of the products. So I want to solve for delta T, so let's rearrange this equation first and say, move the enthalpy of combustion over the other side, negative enthalpy of combustion is equal to these heat capacities, heat capacity of CO2 and heat capacity of H2O twice and all times delta T. Then if I want to solve for delta T, that's going to be negative enthalpy of combustion divided by my heat capacities. So these are numbers that I can look up and plug in to calculate the temperature change. We're going to do that in just a second. And this would be exactly how to proceed with the calculation if the chemical reaction occurred as I've drawn it, if we're burning methane in oxygen. And of course we are, there's oxygen in the air, that's the oxygen that's used in this combustion reaction. And of course air is composed not just of oxygen, but also of nitrogen. Nitrogen doesn't participate in the chemical reaction, so normally we would avoid thinking about it when talking about the chemistry of this reaction. But once we start talking about the fact that the enthalpy gets transferred to the products and heats them up, it's unavoidable that if oxygen diffuses into this flame, nitrogen is diffusing into the flame as well. So in the immediate environment of the flame, which is at several thousand degrees, I'm not only heating up the CO2 and H2O that I generate by burning the methane, but I'm also going to need to heat up the nitrogen as well. So for a more accurate estimate of the flame temperature in air, we're going to need to include the effects of the nitrogen as well. So remembering that air is 21 parts oxygen, so the O2 to N2 ratio is 21 parts oxygen for every 78 parts nitrogen, and the other 1% is argon and other trace components. So I need to know that ratio because for every two moles of oxygen that I need to diffuse into the flame and burn with the methane, or help burn the methane, I'm going to have, so for every two moles of O2, there's 21 parts of O2 in air. That ratio needs to be the same as 78 parts of N2 in the air, so some number of moles of N2. So roughly, not exactly, roughly four times as much N2 as O2 in the air, if I work this ratio out, it turns out to be that instead of two moles of oxygen, this will work out to be 7.4 moles of nitrogen are the stoichiometric amount to the two moles of oxygen. So if I go back and say the things I need to heat up in this flame include not only one mole of CO2 and two moles of H2O every time I burn a mole of methane, but I'm also going to need to heat up 7.4 moles of N2. So again, in this parentheses, I've got 7.4 times the heat capacity of N2, and also in the denominator of this fraction, I've got 7.4 times the heat capacity of N2. So that's accounting for the fact that I need to heat up the N2 that's a passive participant in this reaction just because it's in the same vicinity as the oxygen. So now, we're in a position where we'll get a reasonable answer if we plug the numbers into this expression. We know that enthalpy of combustion, 890.3 kilojoules per mole, let's pay attention to the units. So that's negative 890 with a negative sign becomes positive in the numerator. Heat capacities of CO2 and H2O and N2, the heat capacities I'm going to need to use there are not the heat capacities at room temperature, but the average heat capacity over the range of temperatures. So this temperature increase is going to be on the order of thousands of degrees. Flames are relatively hot. So we need the average heat capacity over the range from room temperature up to the temperature of the flame. So I'll just let you know what those numbers are. A reasonable number to use for the average heat capacity of CO2 between room temperature and flame temperature is about 55 joules per mole Kelvin. For water, it's about 42, and for nitrogen, it's about 33 joules per mole Kelvin. So a few things to notice about that. As we've seen, heat capacities are larger for these triatomic molecules than for diatomic molecules for reasons related to the Equipartition Theorem. Also, yeah, that's enough information about the heat capacity. So if we plug these numbers in, notice that the units, joules per mole Kelvin and the denominator, I've got kilojoules per mole in the numerator. So to keep the units straight, let's multiply the enthalpy of combustion by 1,000 so that we've got units of joules per mole that will cancel the new units of joule per mole in the denominator. And what we get when we do that math, 890,300 divided by the quantity 55 plus 2 times 42 plus 33. When I do that math with a couple of sig figs, oops, that's not correct. One step ahead of myself. That number works out to be 2,300 Kelvin is the increase in temperature. So what that tells us is when I burn methane, whatever the initial temperature was, the temperature is going to increase by 2,300 Kelvin. So the actual flame temperature, if I start out near room temperature, that's where my 2,600 Kelvin comes from. So start out near room temperature, increase the temperature by 2,300, and I find out that the temperature of this flame is going to be around 2,600 Kelvin. So or 2,300 Celsius roughly if you prefer. So a few comments about that number, that's a reasonable estimate in purely ideal conditions. So the notes I'll make will be this calculation is for having made the assumption that this flame burns adiabatically, the heat generated in this reaction immediately goes to raise the temperature of the products and the nitrogen in the air before any of it gets transferred to other things like the pot that's sitting on the flame and things like that. So it's under relatively ideal adiabatic conditions. If the reaction is not completely adiabatic, then the temperature increase will be less. So if it's non-adiabatic, adiabatic, the flame temperature will decrease. Another thing to notice is that we've done this calculation for burning methane in air, a 21 to 78 mixture of oxygen and nitrogen. The calculation would be different if I were burning in pure oxygen. If you take the effort to burn a flame in pure oxygen rather than oxygen with nitrogen, then the flame temperature will be different and you can just leave out the nitrogen to do that. So in oxygen, I'll point out that the flame temperature will be considerably higher in oxygen because we don't have to go to the effort of the thermodynamic effort of heating up the N2. I'll also point out that with the same type of calculation, we've done the example for methane. We could do it for propane. We could do it for acetylene. The only difference in those cases is the stoichiometry of this reaction works out differently. We combine it with a different number of moles of oxygen, which bring with them a different number of moles of N2, and we generate a stoichiometrically different number of moles of CO2 and H2O. So every molecule, every fuel burns at a different flame temperature because the ratio of carbons to hydrogens is different in that particular fuel. So the stoichiometry of the reaction is a little bit different. And then lastly, I'll remind us that we used a constant, just a single number for the heat capacity, assuming that the heat capacity was a relatively constant number between room temperature and this flame temperature, which is certainly not true, or assuming that this average heat capacity is a reasonably good estimate. In fact, we know that the heat capacity depends on the temperature, sometimes relatively strongly at these high temperatures. So the more accurate way of doing the calculation, if we want more than just a couple of sig figs worth of accuracy out of this flame temperature, would be to treat the dependence of the heat capacity on the temperature. So in other words, instead of saying heat capacity times delta T, we would replace this with something like an integral of the heat capacity over the temperature from some initial to some final temperature. That makes the math of doing the calculation quite a bit more complicated. But of course, it gives you a much better, more accurate estimate for the flame temperature. But this overall procedure, depending on whether you want to burn your fuel in oxygen or in air, what your fuel is, how accurately you want to treat the heat capacities, can give you a pretty good estimate of the temperature at which different fuels will burn.