 Let's try this Hess's Law type problem. So this one says from the following heats of combustion, it gives you these three reactions reacting with oxygen, so they're combustion reactions. And then it gives you the standard heat of each of the reactions there on the right. And then it wants you to calculate the enthalpy of formation of methanol from its elements. And I've written that equation down here on the bottom for you all. So let's go about figuring out how to do this. So what we need to know is the final reaction equation that we're looking for, and we need to get the things that are on this side of the final reaction equation, on this side of the other reaction equations, the component in the reaction equations. So hopefully you can see something like C graphite is there, right? H2 is there, albeit not in the right coefficient quantities, do you see that? And O2 is there as well. But if we look over here, methanol, CH3OH, is on the reactant side. And in our final reaction, it's on the product side. So that gives us a clue as to what we want to do. So for this reaction here, we're actually going to flip it around, OK? So let's turn it around. I've got my blue pen out, so we can turn it around with a different color. So 2H2O gas, O2 gas, we're here, methanol, liquid, 3 halves, O2 gas, OK? So that doesn't only mess with the reactants and products, but it changes the heat of reaction from, since this was an exothermic, it's going to turn it into an endothermic reaction, OK? So if you want to remember your reaction coordinates, OK, and how the forward reaction and reverse reaction will be the same delta H, but opposite values, OK? So what we're going to have here is going to be the new delta H of reaction. I'll put a little. So it's going to be positive 726 kilojoules per mole, like that. Does that make sense? OK. So I'm going to erase that top reaction so it doesn't mess with us, OK? And so our blue reaction is the one that we're interested in now, at least where those processes that were in that initial reaction are concerned, OK? So now hopefully we see, well, here the coefficient of H2 is 2, but up here it's 1. So eventually we're going to add all these things up and it's going to be this. We're going to add all these things up and it's going to be this. But if we see, we only got 1H here, 1H2 here, right? And we've got 2H2s here. So what are we going to have to do? We're going to have to multiply this equation by 2. So these are just like algebraic equations. So what you do to one thing in the equation, you've got to do to everything, OK? So if we take this and say we're going to multiply this by 2, like that, we've also got to take this over here and multiply it by 2, like that, OK? So what's our new equation going to be? I'm just going to rewrite it here with blue coefficients. So 1 times 2 is 2, 1 half times 2 is just 1, right? And 1 times 2 is 2, OK? So the new delta H of reaction, well, what is it going to be? It's going to be negative 285.8 times 2, right? 285.8 times 2 and I get negative 571.6 kilojoules per month. Are you OK with what we've done so far? OK. So now, well, all we've got to do is add these things up and anything that's on both sides of the reaction, we do what, too? Cancel that, right? OK. So hopefully, so just go through every one of the species that you see on each side of the reaction to see if they're on the other side. And the same number of moles or coefficients, whatever you prefer to think about. So let's start with carbon dioxide. So overall, we've got one carbon dioxide on this side of the equation, right? And we look over here, what do we got? One carbon dioxide. So what's going to happen to it? Cancel. Cancel, right? So cancel, cancel. OK, look at the 2H2O. So we got here. Do we got 2H2O somewhere on the other side? Yes. So those are going to be canceled, like that, OK? So now, C graphite, right? Carbon graphite, do we see that anywhere over here? No. No? So we don't cancel it, OK? So let's do the 2H2s. Do we see them anywhere over here? No. No, so we don't cancel them. So how many O2s do we have? One, two, right? So we have two. What is 3 halves? That's one and a half, right? 1.5. So let's just write that up there, 1.5, so it doesn't confuse us. So can we do that? Is that all right? So let's just say 1.5. So if we have 2Os here and 1.5 over there, what's going to happen? We're going to cancel one, right? And cancel another half there. So we're going to have half and O2 there, OK? And none there, OK? So what do we have here? C graphite, we said there's none over there. H2s, there's none over there. Half O2, is there any more O2s over there? No. And then we add these things up. So we got C graphite plus 2H2Os plus 1.5 O2. Is that what we have down here? Yes. And over here, all we've got is the one methanol liquid, and that's what we have over there, right? So now what we're going to do, so we've got it right. So we know that all we have to do is add up these three numbers, and that'll give us the heat of formation of this molecule, OK? So let's go ahead and do that. So if you want to, label these as like that. So heat of formation is going to be delta H reaction 1, that first one, plus delta H reaction plus delta H, OK? And I'm just saying because I changed that one in, I changed that one, OK? So just plug and chug down, right? So heat of the first reaction is going to be plus 726.4 kilojoules per mole plus a negative 393.5 kilojoules per mole plus a negative 571.6 kilojoules per mole, like that, OK? So now let's just add those numbers up. 726.4 minus 393.5, subtract that, or subtract 7, or 571.6 from that, and I get negative 230.7 kilojoules. So is this an exothermic reaction or an endothermic reaction? Is it giving off heat, or is it taking in heat? What is the negative mean? The heat is leaving, so what is it doing? It's giving off heat, right? So is this an exothermic reaction or an endothermic reaction? Exothermic. Any questions on Hess's law? No.