 So if calorimetry is a useful way to determine the enthalpy change for a reaction, we can ask ourselves, can we also do calorimetry at constant volume rather than just at constant pressure as we've seen previously? Because after all, we know that energy and enthalpy are very analogous. Analogous, if we do something at constant volume, we learn something about the energy in the same way that we learn something about the enthalpy when we do it at constant pressure. So let's suppose we're interested in a different type of chemical reaction, and this one will be useful for describing measurements at constant volume. So let's say we have 10 grams of sucrose. So we measure out 10 grams of sugar, and we want to burn that sugar. We want to have that sugar undergo combustion. So the sucrose combusts according to solid sucrose. If I combine it with oxygen, it will burn, and it will give me CO2 and H2O. So that's what happens when a substance combusts. The carbons turn into carbon dioxide, the hydrogens turn into water, and it may take some oxygen to make that happen. If we balance that reaction, 12 carbons are going to form 12 CO2s, the 22 hydrogens are going to form 11 waters, and then if I have 24 plus 11 is 35 oxygens on this side, 11 of which came from the sucrose that leaves 12, sorry, 24 divided by 2 is 12 molecules worth of oxygen that is going to take to balance that reaction. So we might be interested in what is the enthalpy change when I burn a mole of sucrose, enthalpy change for that reaction. I can call that a combustion, an enthalpy of combustion. So the question is what is the enthalpy of combustion when I burn a mole of sucrose, and I can do that in an experiment where I'll say I measure out 10 grams of sucrose, and I cause this reaction to happen, and I want to use a calorimeter to say when that reaction happens and heat is given off, it's going to raise the temperature of the surroundings. What do I surround that system with in order to measure the temperature? I can't just do it as simply as dropping the sugar into a coffee cup full of water because I can't make this reaction happen underwater. The sucrose will dissolve, but then I won't be able to burn it in oxygen. So for that type of process, what we often use, so here's a little illustration of my experiment. So there's a bad cartoon of a pile of sugar sitting in a container of some sort. What I'm going to do, so I can cause this reaction to happen by applying some spark to this sugar. So if I light it on fire, essentially, that will initiate this combustion reaction. I can do that even in a closed container by running some wires into the container. So if I apply sufficiently large voltage to this sample, I can cause a spark, I can initiate this combustion reaction, I can burn the sugar inside this container. And I've done it in a closed container so that if I surround the whole thing with water, so here's a container that I've surrounded with water, in fact I can put a lid on the entire container. And then both the volume of this sample that contains the sucrose and the volume of this container that contains the water, those are both kept at constant volume. So I do this typically in a steel container that's not going to expand or contract very much when this reaction takes place. And I need to have it be made of a fairly sturdy material because sometimes the thing that I combust inside this chamber will burn nice and gently. It may release several moles of gas as sucrose does or even worse it might explode. So the name that we give to this sort of experiment or this sort of instrument is called a bomb calorimeter because perhaps what I'm doing inside this container is actually causing something to detonate or explode or perhaps I'm doing something milder like just combusting it or burning it. So inside this bomb calorimeter we're at constant volume conditions and that's necessitated by the fact that we can't do it open in the atmosphere because then the CO2 and the H2O that were produced would leave into the atmosphere and we can't measure the temperature. We can't measure the temperature of the surroundings if we let things escape from the surroundings too easily. So when I'm doing this process at constant volume then it turns out we'll be needing to use this expression. So let me go ahead and finish setting up this example. If I take 10 grams of sucrose and I combust it in a bomb calorimeter from previous experiments we need to know, we need to have measured the heat capacity of the calorimeter itself. So remember in our previous example where we did constant pressure calorimetry all I gave you was the heat capacity of the water that surrounded the dissolution reaction that took place because that's essentially all the surroundings were. In this case I've got a metal container, I've got wires, I've got the water, I've got the outer casing. The thing that raises in temperature in a minute I'm going to tell you how much the temperature increased when I burned sucrose in oxygen. The thing that increased in temperature was not just the water but also the container and the wires and all the other apparatus. So the heat capacity is not the heat capacity of water, it's going to be the heat capacity of the entire apparatus. So from doing experiments you can measure, in fact those are the wrong units, not kilojoules per mole, you can measure how much energy, how much heat it takes to raise the temperature of the device filled with water by some amount. And in this case let's say that we've done that experiment and we've measured the heat capacity of this bomb calorimeter to be 38.5 kilojoules per Kelvin. When I apply 38.5 kilojoules of heat to the calorimeter it raises the entire thing by one Kelvin in temperature. So when I burn some sucrose, and let's see the other piece of information I'm going to have to give you is the molar mass of sucrose, 342 grams per mole. So this is some number of moles of sucrose that we'll calculate in a moment. When I burn that number of moles of sucrose in oxygen generating these products that raises the temperature of the calorimeter by a very carefully measured 4.27 Kelvin. So let's say maybe it started at 298 Kelvin and it increased in temperature to 302 Kelvin and some change. So that's what we measure in the laboratory. The question now is what is the molar enthalpy of combustion of sucrose that we can measure from that calorimetry experiment? So we can start by saying we know the amount of heat given off by the chemical reaction that raised the temperature of the calorimeter. That's the heat capacity times the change in temperature. It takes 38.5 kilojoules to raise the temperature by one Kelvin. We actually raised the temperature by 4.27 Kelvin. So a total of 164 kilojoules were given off when we burned that 10 grams of sucrose. The next step is connecting that heat to some thermodynamic property. But fortunately or unfortunately we've done that process at constant volume in this bomb calorimeter. So this heat that was given off is the delta U of the process. So we can calculate relatively easily the internal energy of combustion. For 10 grams of sucrose it was 164 kilojoules. For a mole of sucrose it would be 164 kilojoules divided by however many moles of sucrose we burned. So I can convert the 10 grams to moles. So 164 divided by this quantity 10 over 342.3. So that gives us a number. The internal energy of combustion, the molar internal energy of combustion when I burned sucrose works out to be 5,630 kilojoules per mole. When I burn a mole of sucrose that's the change in internal energy. And in fact the sign of that reaction we have to be careful with this. The sign of U because this reaction increased the temperature of the water. So clearly the sign of Q is positive when we're considering water to be the system. Heat went into the water and increased its temperature. On the other hand from the point of view of this chemical reaction heat was released in that chemical reaction. That same heat was transferred to the water. So the sign of this delta U of combustion that is a negative number. So depending on your point of view either the if you're the water then heat flowed into you the system. If you are the chemical reaction then heat flowed out. So that's how we keep track of the signs. The next question though is how do we get from a delta U to a delta H? So it's relatively easy to measure the delta U for this process because we did it at a constant volume. If we're more interested in the enthalpy than in the energy how do we connect those two together? And so if we recall our definition enthalpy is U plus PV. So a change in enthalpy must come from a change in U and a change in P times V. If the process this combustion process this is mostly composed of gases. I've got a mole of sucrose but quite a few more moles of oxygen and CO2 and H2O involved in that process. So the P and the V that we're talking about here any PV work done in the process involves those gases. So as an ideal gas PV is equal to NRT and R is not changing so we can certainly not worry about what the change in the value of R. T is changing. T changes from about 300 Kelvin to about 300 Kelvin. So in other words it doesn't change by very much. It changed by 4 Kelvin but that's a relatively small fraction of the total temperature. So I can say approximately if I'm willing to be a little bit approximate delta NRT I'm not going to be too far wrong if I just say well the temperature is 298 or 302 or whichever one of these temperatures I want to use. And really the big change comes from the change in the number of moles because considering the gases in this process I started with 12 moles of oxygen and by burning the sugar I converted those to 12 moles of CO2 and 11 moles of H2O. If they're all behaving as ideal gases I've converted 12 moles to 23 moles of gas. So in our case I'm converting 11, I'm gaining the change in the number of moles is plus 11. I've gained 11 moles of gas every time I do the reaction once. So this unit may look a little confusing. What I want to say here is that the change in the number of moles of gas so plus 11 that's 11 moles of gas per mole of times that the reaction proceeds. So when this reaction goes once I gain 11 molecules of gas, if it goes a mole of times I gain 11 moles of gas. So unit wise that's what we need to do if I use my delta U of 5630 kilojoules per mole add that to 11 times the gas constant in joules per mole kelvin multiplied by the temperature. Notice that I've got kilojoules per mole added to joules per mole after I cancel the Kelvins in this expression. So I also need to include unit conversion to convert those joules into kilojoules. So when I do that the units will end up in kilojoules per mole. And you can see after we divide by 1000 so 11 times 8 times 300 kelvin is a pretty large number but after dividing by 1000 joules that only works out to be about 30 kilojoules per mole. So we end up with let's see this was a negative 5630 kilojoules per mole after I add this roughly 30 kilojoules per mole I end up with about negative 5600 kilojoules per mole as the enthalpy of combustion. So notice what we've done here in doing a constant volume calorimetry experiment we can relatively easily calculate the delta U of the process. We have to do a little bit of extra work and this equation is very useful for converting delta H of a reaction into a delta U of a reaction and vice versa. So we can do a little bit of extra work to convert that delta U for the process into the delta H that we may be more interested in. So that's an example of doing calorimetry at constant volume. The next thing we'll do since we've started to label some of our enthalpy and energy changes with these subscripts like a delta H of combustion turns out there's a lot of different interesting types of enthalpies that we can label so we'll list a few of those in the next lecture.