2026 AP Chemistry FRQ Question 7

Added: 2026-05-14

[00:00:01]Hey guys, welcome to the final question we're finally here of the 2026 AP Chemistry question number seven.

[00:00:11]Let's get started.

[00:00:13]Sodium metal reacts with oxygen gas according to equation one. Standard enthalpies of formation are provided in table one. Part A Based on the information given, calculate the the value of delta HF for sodium oxide shown that work leads to your answer. Really?

[00:00:35]Uh Are they serious?

[00:00:39]Wait, what?

[00:00:41]Okay. Well, sorry to uh All right, so here we go. Look look take a look. So what does this say? This means per that, right? But what do you see here? You see that, right? So essentially what it means is that um I mean this one's really Look look it's too simple. It's just what? It's um every mole of reaction is actually two moles of sodium um oxide. So there we go. So it's it's essentially stoichiometry, isn't it?

[00:01:14]Right?

[00:01:15]So uh let's solve this um 7A right? What do we do? How do we solve it?

[00:01:23]All right, so to solve this you got negative uh what is that? 828 Is that 828? -828 kJ per mole reaction, right?

[00:01:32]Negative 828 kJ per a mole of the reaction and then stoichiometry says in one mole of the chemical reaction in the equation, you end up with you produce two moles of your sodium oxide.

[00:01:55]All right. That's it.

[00:01:58]All right. Uh because they wanted for sodium oxide. So therefore, what happens? This gives you -414 kJ per mole.

[00:02:09]So, the answer would be -414 kJ per mole. Uh and just do that, you're done.

[00:02:16]Let's go to B.

[00:02:18]Based on an 18.4 g sample of sodium reacts with 12.8 g of O2 according to equation one.

[00:02:25]Calculate the total amount of heat in kilojoules released during this process.

[00:02:28]Uh I can't believe that this is just a simple stoichiometry problem.

[00:02:33]Right? Um All right. Well, let's solve it.

[00:02:36]Stop complaining about how easy it is, but just solve it. 7B, okay?

[00:02:41]All right. So, let's go to solve this.

[00:02:43]We need our equation, so that's uh four sodiums uh and an oxygen yielding two sodium oxide.

[00:02:56]All right. So, it says 18.4 g sample of sodium. So, it's and 12.8 g.

[00:03:07]All right. So, what do we do? We actually have to calculate the moles, right? So, for sodium it's 18.4 g and I don't remember. It's been a year since I taught AP. So, therefore, I don't know if it's 23 or 22.9988.

[00:03:29]So, I'm going to go take a quick peek and yes, 22.99. So, 22.99 g per mole is sodium.

[00:03:39]Right? That gives me uh let's see. Calculator.

[00:03:47]Is 18.4 / 12.99 gives you uh am I doing this right? 18.4 divided by 22.99 is 0.800 so three sig fig 0.

[00:04:08]800 moles of our sodium Let's go for our oxygen for O2. O2 we have 12.8 grams of O2 and then each O2 is 16 or 15.9994 is 16. So it's going to be 32.00 grams for 1 mole.

[00:04:32]What does that come out to? That's 12.8 / 32 that gives me 0.4 00 moles of O2. So who's the who's the limiting? Let's think about this. You need four times as much sodium as you need um oxygen, right? So if you have 0.4 moles of oxygen, four times that is 1.6 moles of sodium that you need. We ain't got that. So what do we know? This is our limit limiting and this is our excess, right?

[00:05:08]You dude if you can't do limiting and reagent problem I well I I don't know why you were taking this test, right? So I'm not going to even explain that, right? So let's go. So that's the limiting. So this is the excess. So sodium is going to dictate how much we're going to create.

[00:05:22]All right, so they want the amount of heat in kilojoules, right? So what do we know about this? Well, look heat of the reaction is what? They have heat of the reaction is um -828 kilojoules, right? So what does that mean? That means it's it this is endo or exothermic guys.

[00:05:43]Right?

[00:05:44]This is right? It's negative. It's exothermic, right? Which means that what? It's released. So, you could actually write it like this as part of the whole equation, right?

[00:05:54]So, if you write it like this, then what happens is that now it seems to make a little bit more sense, right? I mean, unless you, you know, you're really good for for that's fine.

[00:06:03]So, now what we're going to do is we're just going to solve the problem. Sodium is the limiting reagent, so it's going to be 0.800 moles of sodium, right? And then what do we do?

[00:06:15]We know that it's four moles of sodium per one mole of a chemical reaction, for one equation. That's the reaction, right? The for the entire equation, you get four. That's what this is trying to say.

[00:06:32]And then what else we know? We know that the delta H value is -828 kJ per one mole of the reaction, right?

[00:06:45]So, hey, we've got a color, let's use it.

[00:06:49]Hello.

[00:06:50]So, we got this cancels that, this cancels that, leaves us with kilojoules, and that should be our correct answer.

[00:06:57]So, let's do it. And if you punch it into the calculator, 0.8 / 4 * -828, I get -166 kilojoules.

[00:07:13]All right. So, calculate total amount of heat. So, um total heat heat is 166 kilojoules.

[00:07:27]There you go. 7B's done.

[00:07:30]Oh, boy. We're very last question.

[00:07:34]We're going to celebrate after this, right? Lattice energy can be enthalpy.

[00:07:38]Sorry, lattice enthalpy can be defined as the energy required to separate an ionic crystal into gaseous ions.

[00:07:45]Uh rubidium oxide and sodium oxide have similar crystal structures and their lattice energy enthalpies are given in table two. Using Coulomb's law, explain why the lattice enthalpy of rubidium is smaller than that of sodium oxide. And man, like this is this Coulomb's law has been kind of weak in this test, right? I mean, they asked a lot of it, but it's been very weak. So, what's the difference between these two?

[00:08:11]Rubidium is here.

[00:08:14]Sodium is here.

[00:08:15]What do we know? Rubidium is larger, right? So, we know that radius of rubidium is larger than radius of Na, right? So, what? Hey, that's Coulomb's law, right? So, since this Coulomb's law is what? Um F is uh Q1Q2 over R squared, right? Think about that.

[00:08:40]Right? Rubidium having a larger radius means it's dividing by a square of a larger number, which means that your force is going to be smaller. Which means that the the force between rubidium and oxygen is going to be much smaller, right? Which is Right?

[00:08:58]Well done. I just explained it, all right? So, let's just put it in words, wrap it up, and then finish this up, right?

[00:09:05]So, that's seven's uh part seven.

[00:09:10]Question B.

[00:09:12]Right? Really? That's only B?

[00:09:15]Wait. No, hold on.

[00:09:17]Uh that's uh C, right?

[00:09:20]Yeah, I noticed on the uh on question six, I kept writing question five, right? So, I don't know if you caught that.

[00:09:27]All right. So, hello. Can I please Okay.

[00:09:31]All right. So, how do we say this?

[00:09:33]Well, we can say, right? Um it says, "Why the lattice energy" You can say, "Um rubidium has much larger radii, right?

[00:09:51]than that of sodium.

[00:09:57]Okay?

[00:09:59]Since um F is proportional to Q1 Q2 over R squared according to Coulomb law um rubidium oxide right?

[00:10:26]will show a smaller Coulomb big force between uh Rb and O.

[00:10:49]Thus translating to to a smaller lattice energy.

[00:11:05]And we are done.

[00:11:10]That's it, guys. So, that's it. I thank you for staying with me if you actually saw through all the problems. Also, like I said, this is done without any answer key, so if if you do find something that is wrong with it, feel free to call me out.

[00:11:29]Although, I think I got everything right.

[00:11:32]Um and hey, if uh uh College Board answer uh post their answers later, um I don't know, a couple of weeks, right? And if you see something wrong, feel free to comment in the comments.

[00:11:44]Um thanks for watching and um I guess see you next year.