 Alright, so we know that we can combine enthalpies of formation to predict the reaction enthalpy for any reaction we are interested in if we know the heats of formation, the enthalpies of formation. However, the tabulated enthalpies of reaction that will usually come across, those are tabulated at a particular temperature, usually 298 Kelvin. So what that means is we can predict the enthalpy of a reaction at 298 Kelvin, but if the enthalpy of formation is different at different temperatures, which it is, then that's the only temperature we're able to predict the enthalpies at using the Hess's law approach that we know right now. So to illustrate how to deal with the problem when we want to know a reaction enthalpy at a temperature other than 298 Kelvin, let's consider a reaction that we're typically not very interested in at room temperature, but we might be much more interested in at elevated temperatures, and that's the water gas shift reaction. So this reaction, not very difficult to balance. If we take water and carbon monoxide, we can essentially transfer one of the oxidants, we can oxidize the carbon monoxide down to carbon dioxide and reduce the hydrogens to elemental hydrogen. So these are all gas phase reactants and products. This is an important industrial reaction in part for the production of hydrogen. This is one way to produce hydrogen gas is by reacting H2O and carbon monoxide. And we, let's say, let's calculate the delta H of this reaction using heat formation as a way to do an example. So the heats of formation for H2O, carbon monoxide, carbon dioxide, and H2. We can look those up in a table of enthalpies of formation. And if we do that, water has an enthalpy of formation of negative 241.8. These are all going to be in kilojoules per mole. Carbon monoxides is also negative 110.5, yes, 5 kilojoules per mole. CO2 is even more stable, 393.5 kilojoules per mole. So those numbers are all negative, indicating that when I form these compounds out of elements in their standard states, the enthalpy decreases. That's what the negative sign tells us. These are more stable than the elements themselves. If I go to look up the enthalpy of formation of H2, gaseous H2, I won't find it in a typical table of reaction enthalpies. And that's because we're expected to know and to remember that the enthalpy of formation for anything that is already an element in its standard state, gaseous hydrogen is an element in its standard state, that by definition has enthalpy of formation of 0. So that's enough information now to calculate the enthalpy of reaction for this water-gas-shift reaction. So we take stoichiometric coefficients are all 1s. Sometimes they're negative 1s for the reactants, positive 1s for the products. But I just take all these elements, enthalpies of formation for the products, CO2, kilojoules per mole, add that to H2. Sorry, I've got the wrong value here for CO2. CO2, the product, has an enthalpy of formation of 393.5, kilojoules per mole, hydrogen, the other product, has an enthalpy of formation of 0. I add to that the enthalpy of formation for H2O, negative 241.8. Being careful with my signs, the enthalpy of formation itself is negative, and the stoichiometric coefficient in front of H2O is a 1, but it's a 1 with a negative sign because it's a reactant. Likewise for the other reactant, carbon monoxide, stoichiometric coefficient is negative 1. Enthalpy of formation for carbon monoxide is negative 110. And if I do the arithmetic and combine those numbers, we find that the enthalpy of this reaction is negative 41.2 kilojoules per mole. As I mentioned, though, this chemical reaction, it's important industrially, but it's not important industrially at 298 Kelvin. And because the numbers we've looked up, the table we looked these up in will tell us that those are all correct at 298 Kelvin. So this result, if we were to do this reaction at 298 Kelvin, that would be the enthalpy of the reaction. Let me write the reaction one more time. So we've calculated the enthalpy of reaction at 298 Kelvin. Let's say we're interested in knowing the enthalpy of this reaction at a more industrially relevant temperature. We'd like to know the enthalpy of reaction at 700 Kelvin, which is a temperature more like it would be performed when we're actually producing hydrogen gas. What we know is a different number. We know the enthalpy of this reaction, 298 Kelvin. So these are the conditions where we know the answer. This enthalpy of reaction we've just calculated, that's negative 41.8, no, 41.2 kilojoules per mole. So how do we find this answer when we know this one already? Hess's law is, again, our friend. It doesn't matter how we get from these reactions to these products. It doesn't matter whether we perform the reaction at 700 Kelvin as written or whether we take some other path to the answer. And another path that we could take would be let's take our water molecules and cool them down from 700 degrees to 298. Take our carbon monoxide and cool them down. Once we've got water and carbon monoxide at 298 Kelvin, we can react them at 298 Kelvin, at least hypothetically, giving off 41.2 kilojoules per mole in enthalpy to produce carbon dioxide and hydrogen at 298 Kelvin. And once we've got products at 298 Kelvin, we can heat both of those back up to 700 Kelvin. And the net result of those five steps, cooling the two reactants, reacting them at 298, heating them back up to 700 Kelvin, the net result is the same products that we were interested in at the temperature we were interested in. So regardless of whether we actually do that process or not, we know what the enthalpy of this reaction at 700 Kelvin is by combining the five steps. So the delta H of the reaction at 700 Kelvin is going to be, let's say, we need to give this one a number, step number five, let's say. Step number five has an enthalpy of reaction that we already know. If I add that to, let's do step three and four first, because that's maybe the slightly easier direction to think about, heating up from 298 Kelvin to 700 Kelvin, that's going to require heat. That's going to require the input of heat. The enthalpy required for that step, remembering that enthalpy is heat capacity times the change in temperature. I can write step number three. The enthalpy is going to be heat capacity of CO2 multiplied by the change in temperature. And that change in temperature is 700 Kelvin minus 298 Kelvin for that heating step. I would need moles times molar heat capacity times the change in temperature, but my stoichiometric coefficient is 1. I'm only heating up one mole of CO2. Likewise for the H2, I'm heating up one mole of H2 by the same amount. For steps number one and two, I'm cooling them down, so I can write those as heat capacity of H2O times, let's keep it the same delta T. So this delta T is also a positive number. So I'll throw in a negative sign reflecting that I'm cooling down rather than heating up. And then for step number two, heat capacity of carbon monoxide multiplied by delta T with the same negative sign. So now as long as we know both the enthalpies of formation that gave us this enthalpy of reaction at 298 Kelvin, we can compute the enthalpy of reaction at any other temperature we want by summing up that enthalpy of reaction that we can obtain with the heat capacities of the reactants and the products multiplied by that change in temperature. So again, the general equation that we can use. So if we want to know enthalpy of a reaction at some temperature that is not 298, we can relate it to the temperature enthalpy of the reaction at some temperature we do know, perhaps 298, as long as we add in heat capacities times temperature for all the different reactants and products multiplied by their individual stoichiometric coefficients. Notice that the numbers I'm multiplying by here, the one, the one, the negative one, the negative one, those are the same as the stoichiometric coefficients, plus one for these two products, negative one for these two reactants, same stoichiometric coefficients we used when we calculate the enthalpy of a chemical reaction. So that would be how we calculate the enthalpy of a chemical reaction at a temperature different than the one we're interested in. Before we go and plug numbers into this equation to find out how much the enthalpy of this reaction might change at 700 Kelvin compared to 298 Kelvin, I'll point out that, first of all, I'll point out that these should be molar, enthalpies, molar heats of, sorry, molar heat capacities. We already know something about molar heat capacities. You might think that the molar heat capacity of gases, these are all gases that behave relatively ideally, carbon dioxide, H2O, CO. If we treat them as ideal gases, you might say, heat capacity, molar heat capacity of an ideal gas, let's just call each one of those five halves R, as we've seen for, at least for the 3D particle in a box. At this point, that would be a reasonable assumption, but it turns out when we actually measure or look up the values of the heat capacities of individual gases, this turns out not to be that good a guess after all. So we were, have an unpleasant surprise if we tried to use what we already know about heat capacities in this equation. So that's our next step is to try to understand a little more about what these heat capacities actually are because they're not always five halves R. So that's what's coming next.