 So let's do this standard heat affordability problem. Use the information in the table to calculate the standard heat of the reaction in kilojoules of ammonia reacting with oxygen to yield nitric oxide and gaseous water. So notice they have to give you the fact that it's gaseous. If not, then you won't know what state to put it. A step in the Oswald process for the commercial production of nitric acid. So the first thing you're going to have to do is write the reaction equation. So in this case, it's ammonia gas goes to nitric oxide gas plus H2O gas. And we're looking for, so what would be the next step? Balance it. So this one's not very easy to balance. Looks like it's 4, 5, 4, 6. Yeah? Yeah. You guys got that? Did you get it good though? Yeah, I don't know. Yeah, it's good. And then what do you want to do? Well, you've got to apply the Hess's law, a portion of the heat of formation, right? So what was that reaction, do you guys remember? Are the equation that we're going to be using eventually? Help me out whenever you guys want. Very good, guys. Oh yeah, OK, but you got it both. Yeah, all right. So remember, it's products minus reactants, OK? And remember this sign, summation sign, OK? So now we've got to go back and look at the table, OK? And look for these various compounds or whatever is up here, OK? Not to give them anything away. So gaseous ammonia, what's the heat of formation? You guys see it? Negative 46.1. And we're just going to say the heats of formation of all of these are in kilojoules. I'm sorry, kilojoules per mole. OK, so negative 46.1, what's the heat of formation for oxygen? Zero, right? Good job. Yeah, no, I'm proud of you. That's awesome, OK? So you guys remember, why is it zero? Because it's in its standard state. It's by itself in its standard state. NO, 91.3. This is a positive. And H2O gaseous here. Negative 2.1. Negative, I'm sorry, I can't see, 2.41.8. OK, so now I like to do these brackets and whatnot. So 4 times negative 46.1 kilojoules per mole plus 5 times 0 kilojoules per mole. I'm going to do the products minus the reactants. I was testing you guys. 4 times 91.3 kilojoules per mole plus 6 times negative 2.41.8 kilojoules per mole. 46.1 kilojoules per mole, 5 times 0 kilojoules per mole. OK, I got the same answer I did. I got negative 901.0 kilojoules. So that's the delta H of this reaction. And you can use the heat of formations to do this. Is everybody OK with this type of problem? Any questions about this one? Let's just box this and call it a tick.