AP Chemistry 2026 Free Response Question 7 - SOLVED!

Added: 2026-05-10

[00:00:00]Hi there. My name is Jeremy Krug and this is the place for all things AP Chemistry. As I record this video, the free response questions for the 2026 AP Chemistry exam have just been released and this is my walk-through for free response question number seven, which is a short FRQ worth four points. If you like what you see or you learn something from this video, remember to smash that like button and leave a comment down below. If my videos have been helpful to you this school year, I'd be very honored if you'd recommend my videos to next year's AP Chemistry students. And just for full disclosure, this answer key is my best prediction for the point breakdown. Full scoring guidelines get released by College Board later in the summer. And as always, other answers that are chemically correct are also acceptable. Now, here's my walk-through for question number seven.

[00:00:55]Question seven was a thermochemistry question. We have a little bit more thermochemistry than you might have been expecting on the FRQ section here.

[00:01:02]Question seven says that sodium metal reacts with oxygen gas according to equation one. So, we have the equation.

[00:01:09]It is balanced and we have a delta H given for that value as -828 kJ per mole of reaction. And we also have standard enthalpies of formation provided here in the table. And part A says based on the information given, calculate the value of the enthalpy of formation or heat of formation for sodium oxide solid and show the work that leads to your answer. So, in order to do this, we have to use the equation delta H equals the sum of all the enthalpies of formation of the products minus the sum of all the enthalpies of formation of the reactants. So, I'm going to go ahead and just rewrite the equation here and we'll calculate this.

[00:01:49]So, for sodium, it says that the enthalpy of formation is zero. So, we plug that in there and there are four moles of this, so we times it by 4, but of course that's still 0 kJ. Now, for oxygen gas, that's 0 kJ per mole, and we have 1 mole of that, so that's so that's still 0 kJ. Now, for sodium oxide, we don't know what that is, that's the question mark, so we're going to call that X. We do have 2 moles of that, so that is going to be 2X. Now, the sum of all the reactants, 0 + 0 is, well, just 0 kJ, and for the products, it's just 2X kJ. So, we need to use the equation ΔH equals the sum of the products minus the sum of the reactants.

[00:02:32]The ΔH is given to us in the problem is -828 kJ. The products, we just said will be 2X, so I plug that in for products, and the reactants are 0. So, we just have to solve for X here. So, 2X is -828 kJ, and so X is equal to -414 kJ per mole. So, if you got that as your answer, then give yourself one point for that one right there. Moving on to part B, we have the same equation, but this time it says an 18.4 g sample of sodium solid reacts with 12.8 g of oxygen gas according to equation 1. Calculate the total amount of heat in kJ released during this process. Show the work that leads to your answer. So, once again, we're going to uh start with both of these values, the 18.4 g of sodium, and then later I'll do the 12.8 g of oxygen.

[00:03:32]But, basically, we have to convert both of these to kilojoules. This is a stoichiometry problem using heat. Now, just like in any stoichiometry problem, step one is to convert to moles. So, I do that, I put 22.99 g on the bottom, 1 mole on top, and And step two, I'm going to do my combined mole ratio and convert to final unit where I put four moles of sodium on the bottom because there's a four in front of Na right here. And on top we have -828 kJ. So that's because that's our delta H of the reaction. I didn't cancel these out but we could cancel out grams and cancel out sodium and cancel out moles as well if you want to. And we're left with unit of kilojoules. So if you do the math here 18.4 / 22.99 * -828 / 4 gets us a negative 166 kJ which implies that there are 166 kJ released in that potential process. Now let's see what happens with the 12.8 g of oxygen.

[00:04:40]I do the same thing. Step one is convert to moles. So this time it's 1 mole over 32 g.

[00:04:46]And the second step is the combined mole ratio and convert to final unit. So this time it's -828 kJ on top and 1 mole of oxygen on the bottom because there's a one right here understood in front of that O2. So I can cancel out grams and I can cancel out moles. I can cancel out O2 and I find that it's -331 kJ which implies that we have 331 kJ released. So we go with the smaller. This is a limiting reactant problem. So we go with the smaller quantity of heat which would be 166 kJ released. Now I'm not sure exactly how they're going to break the points down here. I'm pretty sure that you'll get one point for getting the 166 kJ released and you might get a second point for maybe the other one as well.

[00:05:38]I'm not exactly sure how they'll break down the points for question seven but I'm pretty sure that part B is going to be worth two points simply because of the quantity of work involved with that.

[00:05:50]Now, let's go into part C here, and this one is a lattice enthalpy question. This is kind of moving back to unit two, and it says lattice enthalpy can be defined as the energy required to separate an ionic crystal into gaseous ions.

[00:06:03]Rb2O and Na2O have similar crystal structures, and their lattice enthalpies are given in table two. So, we have the the table here. Using Coulomb's law, explain why the lattice enthalpy of Rb2O is smaller than that of Na2O.

[00:06:21]Now, remember, whenever we look at lattice enthalpies or melting points or something like that, the first thing you want to look at would be the charges.

[00:06:29]Well, notice that the charges are equal.

[00:06:32]Rb2O is a +1 -2 pair, and sodium oxide is also a +1 -2 pair. So, the charge is not going to help us here. We have to break the tie by looking at the ionic sizes. And if you look at the ionic sizes or ionic radii of Rb+ and Na+, we can see that the ionic radius of the rubidium ion is quite a bit greater than that of the sodium ion, and that's because rubidium just has a couple more uh electron shells. And because of this, the ions in Rb2O are going to be farther apart from each other than the ions in Na2O.

[00:07:10]And this is going to cause the Coulombic attractions between ions in Rb2O to be weaker. Right? Coulomb's law says that the farther apart charged particles are, the weaker their attractive forces are.

[00:07:22]So, since the forces are weaker in Rb2O, we're going to have a lower lattice enthalpy. And if you want to, you could go ahead and say, you know, also has a lower melting point as well, even though the question doesn't really ask about that.

[00:07:35]But yeah, that is why this happens. So, if you said that or something like that and gave a reasonable explanation, give yourself one point. So, that's question seven. We had a lot about thermochemistry with a little bit of uh lattice enthalpy and ionic attraction and Coulomb's law thrown in there for good measure. So, that's it. I hope this was useful for you as you reflect on your exam performance for question seven or you get ready for a future chemistry exam. And if you are getting ready for a future AP Chem exam, remember that my ultimate review packet and ultimate exam slayer have dozens of practice free response questions and nearly a thousand practice multiple choice questions with full explanations to help you slay your AP exam. The link is in the description down below. Thank you so much for watching this video and all of my videos this year. Have a great summer. I hope to see you again soon.