AP Chemistry 2026 Free Response Question 4 - SOLVED!

Added: 2026-05-10

[00:00:00]Hi there, my name is Jeremy Kug 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.

[00:00:13]And this is my walkthrough for free response question number four, which is a short FRQ worth points. If you like what you see or you learned 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 year, I'd be very honored if you'd recommend my videos to next year's AP Chem students. And just for full disclosure, this answer key is my best prediction for the point breakdown. Full scoring guides get released by College Board later in the summer. And as always, other answers that are chemically correct are also acceptable.

[00:00:49]Now, here's my walkthrough for question four. Question four tells us that white phosphorus P4 gas can decompose into P2 gas at elevated temperatures. A balanced chemical equation, KP expression and Lewis diagrams for the chemical species involved in the reaction are given. So we have a a balanced equation and the KP expression. We also have a couple of Lewis diagrams. Part A says the average bond length in P4 molecules is 221 pometers whereas the average bond length in P2 molecules is 189 pometers. Explain why the average bond length in P4 is greater than the average bond length in P2. Well, if we notice P4 has only single bonds. Now, those of you who looked at the 2025 walkthroughs might notice that this P4 structure looks very familiar. In fact, we saw that on last year's exam as well.

[00:01:44]Well, P4 only has single bonds, and those single bonds have weaker attractions between the atoms than these much stronger triple bonds that we see in P2. And so, that's why the bonds in P4 are quite a bit longer than those in P2. So, if you said something like that, then you can give yourself one point.

[00:02:05]Moving on to part B, it says, "A sample of P4 solid is placed in a sealed rigid container and heated to 1,600 Kelvin, at which point the initial partial pressure of P4 gas is 470 atmospheres. Some of the P4 gas decomposes into P2 gas as the system reaches equilibrium. At equilibrium, the partial pressure of P2 gas is determined to be 630 atmospheres.

[00:02:32]Calculate the value of the equilibrium constant KP for the reaction at 1600 Kelvin. Show the work that leads to your answer. And so for this reaction, we're probably going to need to draw out an ice box here. So what I will do is put in the initial uh pressure here. We're given the partial pressure of P4 as 470 atmospheres. And the problem doesn't tell us that we have any P2 in there at the beginning. So it's safe to call that zero atmospheres. The problem also tells us that at equilibrium the partial pressure of P2 is 630 atmospheres. So I can put that in here. Now with this information I have enough information to solve the puzzle here. So when I uh plug in the change row for P2, it looks like it's gone up by about 630 atmospheres.

[00:03:26]And that means that the P4 had to go down but by half as much because this is a 2:1 ratio. So that means that the P4 went down by about 315 atmospheres. And so my equilibrium partial pressure of P4 is about.155 atmospheres.

[00:03:45]And so uh here we have our equilibrium pressure. Now we need to plug these into the equilibrium constant expression given to us here. uh the looks like the P2 we uh were given as 630 that has to be squared because there's a a coefficient of two there and in the denominator we have 0.155 atmospheres.

[00:04:08]So whenever we calculate this we find that 630^2AR divided by.155 gets us about 2.56.

[00:04:19]And so I would estimate that this is probably going to be a two-pointer.

[00:04:24]You'll probably get one point for calculating the 0.155 atmospheres and then we get one point for the 2.56 as your KP value. I forgot to put the check mark up here, but I believe that will probably be a two-pointer on part B.

[00:04:41]Now, part C, it says the reaction is thermodynamically favorable under standard conditions only at temperatures above 1,500 Kelvin. A student claims that the reaction must be endothermic because it is favorable only at high temperatures. Do you agree or disagree and justify your answer using delta S and delta G? Well, you might look at this and wonder how does delta S come into play? Well, we can determine what delta S is for this process because notice that the reaction goes from one molecule of gas P4 to two molecules of P2 gas. So that tells us that entropy is increasing in this process. And so delta S is positive. And the question tells us that this is thermodynamically favorable only at relatively high temperatures.

[00:05:30]You know those above 1500 Kelvin. So if the reaction is thermodynamically favorable at high temperatures, that means that both delta S and delta H have to be positive. And so yes, this reaction does have a positive value for delta H, which means that the reaction is endothermic. And so we should agree with the students claim. So if you said that and gave a reasonable explanation, give yourself one point. So that was question four worth a total of four points. So that's it. I hope this was useful for you as you reflect on your exam performance or get ready for a future 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. Thanks so much for watching. I hope to see you soon.