 So, let's try this one. It says the reaction between hydrogen and oxygen to form water vapor has a heat of reaction or standard enthalpy change, right? Negative 484 kilojoules, so let's write out what that means. So hydrogen, gas plus oxygen, gas, goes to water, gas. So first we've got to balance our equation, right? Two, and then what we'll do is write at the end over here Delta H, whenever it's that's standard conditions. So in this case, it's like the molar quantities. So negative four 84 kilojoules. So does this reaction spontaneous or non-spontaneous? We can ask already. spontaneous, right? It's giving off heat. It says what is the energy in kilojoules for the reaction of 0.5 moles of hydrogen and 0.25 moles of oxygen at atmospheric pressure if the volume change is negative 5.6 liters. Okay, so that's a lot of information. But some things it's telling us are the number of moles. So we got the number of moles of hydrogen which is 0.50 moles hydrogen, and we got the number of moles of oxygen. Why do you think they gave us these two moles? Numbers of moles. What do you think we've got to figure out? We normally have to figure out what we're doing in this situation. Come on guys. I know you guys know this. That's the first thing you'll always have to do. Mole ratio of what and why? But why? The limiting reagent is what you have to do, right? So you have to figure out what the limiting reagent is first. All right, so is there a limiting reagent here? So it says there's 0.5 moles of H2 0.25 moles of O2, right? So let's figure it out. How do we figure it out? Well, we just multiply like this, right? So for every one mole of O2, we should look for two moles of H2, cancel, cancel. So that equals what? 0.50 moles of H2. So in this case, there's no limiting reagent. So now it says what is the atmospheric, so what is the standard atmospheric pressure? So the P external would be 1.0018. And it also gives us the change in volume which is negative 5.6 Okay, so what else do we know? Is this equation based on 0.5 moles of hydrogen? Is this either reaction based on 5.5 moles of hydrogen? How many moles is it based on? Two moles, right? So we're going to have to find the delta H of the reaction. So everybody understand what we're doing? So we've got 0.50 moles of H2 and for every two moles of H2 we've got negative 484 kilojoules of energy being formed. We have 121 kilojoules of energy for the heat of this particular reaction. Is everybody okay with that? So we now have to remember that delta H equals delta E plus P external times delta V So in this case it's the delta H of the reaction that we just figured out. And now all we have to do is isolate the variable and plug in J. So delta E is what we're looking for. So that's going to equal delta H minus P E X T times delta V. Everybody got that? Well, this is delta H of the reaction. Sorry. Does that make sense to everybody? Let's put it over here. So we've got some more room. So delta H of the reaction, where is that? It's here, right? 121, negative 121. So what have we got? 121, yes. So this is the way I like to do these ones. Big bracket like that. P external, right, which is 1.0018 times delta V So it's 18 liters units of energy. It is, but can we add it to kilojoules or subtract it to kilojoules? So what do we got to do to convert it to kilojoules, okay? So in order to do that, you guys remember the conversion factor? Yeah, so 101.3 joules, right? R is 1 liter, ATM. So that'll cancel that, that, that. But this is still in joules and we want it in kilojoules, right? 1,000 joules is 1 kilojoule. That's why I had those big brackets. Cancel, cancel, and now we're in kilojoules, right? Okay, so 1 times negative 5.6 times 101.3 divided by 1,000. So notice, how many kilojoules is that? Did anybody calculate it with me? Not a single person? Yeah, okay, so negative 0.567, right? How, is that a big or small number relative to 121? Tiny, right? So that factor doesn't play a huge role. Remember we were talking about that earlier? How much of a role the PV term actually plays? Very tiny role. So let's write it out here. So we'll have a little bit more room. So negative 121 kilojoules. And then what we're, what are we doing? Subtracting 0, well, we're subtracting, sorry. Subtracting the negative. Negative 0.567, we'll just say kilojoules. Like that, and so we'll be adding to it. So I got this number, negative 120, well, 0.43, right? But when we do our sig figs, it's going to be negative 120 kilojoules, okay? So in this case, that's delta E, how much energy? Notice, the amount of energy between delta H and delta E is very, yeah, the amount of difference is very tiny. Okay, are there any questions on that one? So it's a pretty involved process. That's probably the, one of the harder questions that you can get on this thing, okay?