 So, it's great that we can use enthalpies to talk about chemical reactions. The heat for a constant pressure reaction is the enthalpy of that reaction. The bad news, however, is that we have a nearly infinite variety of chemical reactions to consider. We've got many thousands of different chemical compounds, millions of different chemical compounds. Every conceivable combination of any of those compounds with any other might produce a different reaction. So we have an enormous number of chemical reactions. And each one of which has its own enthalpy of reaction. And that's far too many for us to actually perform every reaction and measure the heat of the reaction, the enthalpy of the reaction for each of those reactions. So we need a way of understanding enthalpies of chemical reactions without having to perform all the reactions themselves individually. The good news is we can use Hess's law and he's a formation to do that. So we'll use this example. This reaction as an example, let's say we have methanol, CH3OH, and we'll talk about combustion of methanol. So liquid methanol burning in oxygen gas to produce carbon dioxide, gas, and water, vapor. So if I balance that reaction, I'm going to need two methanol which will produce two CO2s and four oxygens, the four hydrogens times two is eight, so I need four waters. And total of eight oxygens on the left means I need four O2s on the right. So there's a balanced chemical reaction for the combustion of methanol in oxygen to produce CO2 and H2O. And our question is what is the enthalpy of this reaction? What is delta H of this chemical reaction? Without performing the actual reaction itself. So the way we can think about this is using Hess's law to break down this reaction into other ones that we might know the answers to. And a pretty common trick is to take each one of the compounds involved in this reaction, both the reactants and the products, and decompose the reactants into elements in their standard states. So methanol composed of carbons and hydrogens and oxygens in this particular structure. Let's say I break that down to form carbon and hydrogen and oxygen as elements in their standard states. So that's going to be hydrogen gas and oxygen gas. And the standard state of carbon is graphite. So two methanol will produce two atoms of graphite, six, eight hydrogens. That'll be four molecules of H2. And one molecule of O2. So that's what I'll call reaction number one. Breaking down methanol into elements in their standard states. And once I've got these elements in their standard states, I can also, actually, I've misbalanced my chemical reaction, I think. I forgot about these two oxygens. So that leaves only three O2s that I need in the balanced chemical reaction. Just to double check, two oxygens there, six oxygens here. That's up to eight. Four and four makes eight. So now it's balanced correctly. Reaction number two, I'll take these oxygens, break them down into elements in their standard state. But O2 is already a pure element in a standard state. So let me write the O2s as O2 and O2 and O2. So in this reaction, I've broken down these three oxygens into these three oxygens. So I haven't really done anything in reaction two. I've just moved these oxygens down here. But now that I've got all elements in their standard states, I can start reassembling them, building up these molecules out of these pieces. So if I want to make, let's start with the CO2s. If I want to make some CO2s, two CO2s are going to require me to take these two carbons and two of my oxygens. So that's why I broke the oxygen down into pieces, two of the oxygens. And reassemble those into CO2s. So that's step number three. Step number four is the pieces that I have left over. The four hydrogens and the two oxygen molecules combine those to make four water molecules. So that's reaction number four. So my arrows have gotten a little bit crazy, but reactants break them down into elements, take those elements and reassemble them into products. If my reaction is actually successfully balanced, then all the pieces I create by breaking down the reactants can be used to reassemble the products. Now the good news is, because we know Hess's law, we know that the delta H for this reaction, the original reaction we wrote down, doesn't matter what path we take to get there, starting from these reactants to get to these products. We can either do the reaction directly as it's written, or we can take this more circuitous route, steps number one and two to break down the elements, steps number three and four to build them back up. So we can say that delta H for the reaction we're actually interested in is the enthalpies for steps number one, two, three and four summed together. But each one of these steps, one, two, three and four, can be related to enthalpies of formation by design. That's why we wrote things down into elements in their standard states. Let's start with step number three. Step number three was combining two graphites and two oxygen molecules to form two CO2s. So I'm forming two molecules of CO2 out of elements in their standard states. By definition, that's the heat of formation of CO2 twice, because I'm forming two molecules of CO2. Likewise for step number four, that's forming four molecules of water out of elements, H2 and O2, in their standard states. So step number four is four times the enthalpy of formation of H2O. Steps one and two are almost as simple. They require one extra step in thought, however. Step number one involves taking two methanols and breaking it down into carbons and hydrogens and oxygen elements in their standard states. So that's the reverse of a formation reaction. It's a breaking elements down instead of forming elements. So step number one is negative heat of formation of methanol. Because I'm breaking down two molecules of methanol, it's negative twice the heat of formation of methanol. And the second step, we're breaking down the oxygens into elements in their standard states. So I can call that three times the heat of formation of O2 gas. So we've been able to write each of these steps one, two, three, four, in terms of heats of formation. And the total enthalpy of the reaction, if I move back over to this side, the enthalpy of the reaction, if I just combine those results, is going to be twice the molar enthalpy of formation of CO2, the first product. And four times the molar enthalpy of formation of H2O, minus twice the molar enthalpy of formation of methanol, minus three times the molar enthalpy of formation of oxygen gas. And we can see now that this is always going to work in the same way. If I have any reactants turning into any products, when I break them down into elements in their standard states, that's going to be the reverse of some enthalpies of formation. When I reassemble them into products, that's going to be some positive enthalpies of formation. So in general, it's always going to be the case that the enthalpy of a reaction is going to be the sum of a bunch of coefficients, this 2, 4, negative 2, negative 3. Those came exactly from the coefficients in this reaction. Two moles of CO2, four moles of H2O, those are my products. Two moles of methanol, three moles of O2 in the reactants, those show up with negative signs. So we can write that as the sum of a bunch of heats of formation of individual compounds. So I'm going to sum over all the reactants, all the products, in my chemical reaction. And then the number I multiply each of those heats of formation by, we call that a stoichiometric coefficient. So that's just the coefficient in the balanced chemical reaction. With the extra caveat that the reactants get negative numbers for their stoichiometric coefficients, the products get positive numbers for their stoichiometric coefficients. And that ensures that we recognize that we're breaking down the reactants and the negative sign we're forming the products with a positive sign. So regardless of what the chemical reaction is, we don't actually have to go to the trouble of drawing this chart writing out the individual reactions. It's always going to be true that the enthalpy of the reaction is the sum of the heats of formation of the products with their stoichiometric coefficients minus the sum of the heats of formation of the reactants with their own stoichiometric coefficients giving us that minus sign. So that's the origin of this rule that you may already know. Empty B of a chemical reaction is heats of formation of products minus heats of formation of products. This is a slightly more general way of writing that. A couple of comments that we'll make about doing this process with Hess's law. Number one, we didn't mention it numerically. We haven't plugged numbers into this equation yet. That'll wait until the next video. But if I were to think about what number I plug in for the heat of formation of oxygen gas, this reaction number two is not actually a reaction that does anything. 302s got turned into 302s. And that's because the reactant here was already an element in its standard state. So whether that's a forward heat of formation of elements or a reverse, doesn't matter. That reaction doesn't do anything. So it's always going to be true that the heat of formation is 0 for anything that's already a pure element in its standard state. So that's important to recognize. The other thing that's useful to point out at this point is that all we need now in order to calculate the enthalpy of reaction for any reaction we can think of, any reaction we can write down, whether anybody's actually performed that experiment in a lab yet or not, is just the heats of formation. So there's still a lot of different compounds that have their own individual heats of formation, enthalpies of formation. And there's long tables containing all those heats of formation. But the good news is once we've tabulated the enthalpy of formation for methanol and for CO2 and H2O, we don't even need to tabulate it for oxygen because we know that one's going to be 0. We now have the building blocks to construct the enthalpies of reaction for any chemical reaction that involves those reactants and products. So tabulating the enthalpies of reaction for a large list of compounds is much easier than tabulating the enthalpies of reaction for a nearly infinite list of possible chemical reactions. So now that we know how to use these enthalpies of formation, we can look up calculated or measured enthalpies of formation in a table and use those to calculate enthalpies of reaction for just about anything we want.