AP Chemistry 2026 Free Response Question 2 - 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.

[00:00:12]And this is my walkthrough for free response question number two, which is a long FRQ worth 10 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 guidelines get released by the College Board later in the summer. And as always, other answers that are chemically correct are also acceptable.

[00:00:48]Now, here's my walkthrough for question two. Question two had quite a bit of uh content in here from several different units. We had some vesper, we had some redux, we had some electrochemistry, and we had some kinetics in here as well.

[00:01:02]So, a lot of variety. In question two, it says, "Answer the following questions about the chromate ion CR42 negative and the dromate ion CR272 negative." And part A says the bonds between chromium and oxygen and CR42 negative can be represented as coalent bonds. One possible Lewis diagram for the chromate ion is shown in figure one.

[00:01:26]And we have the structure here. And part one says based on vesper theory predict the molecular geometry of the chromate ion. Well, we see that the chromium is touching four atoms and there are no lone pairs. So four and zero would correspond to a tetrahedral molecular geometry. So if you said that, give yourself one point based on that one.

[00:01:50]Now part two says on figure two, draw a complete leis diagram for a resonance structure of chromate that results in a formal charge of zero on two of the oxygen atoms and a formal charge of negative one on the other two oxygen atoms. Your diagram should contain the same number of veence electrons as in figure one. Now if you back up to figure one and just start counting dots, we know that each bond represents two dots.

[00:02:15]So we actually have a total of 32 dots or 32 veence electrons represented in figure 1. So in figure two we also need to have 32 dots total. And the only way to make that work is to have uh an octet around every one of these oxygen atoms and to have a total of two single bonds and two double bonds. If you add these up, you'll notice that we do indeed have 32 veence electrons total considering that each of the bonds has two veence electrons uh in it. Now the question says it has to be a formal charge of zero on two of the oxygen atoms and negative one on the other two oxygen atoms. And that's exactly what we have here. The oxygen atom right here and down here both have negative -1 formal charges because remember oxygen has six veence electrons according to the periodic table because it's in group 16.

[00:03:09]And then we count up the the dots 1 2 3 4 5 6 and the one bond here counts as as seven. So 6 - 7 is -1 for both of those.

[00:03:19]And then these two oxygens both have formal charges of zero. We have our six original veence electrons from group 16.

[00:03:27]And then we have 1 2 1 2 3 4 and 5 6. So 6 - 6 is 0ero. And both of those have the same uh shape essentially. So we have that. If you were able to draw that, give yourself a point for part A2.

[00:03:44]Now moving on to part B. It says when reacted with a strong acid like nitric acid, chromate can be converted to dromate and water in a reversible reaction. So part one says, write a balanced net ionic equation for the reaction between chromate and a strong acid. Well, we're going to need to write it like this. We have H+, which is our uh our form of acid. Anytime we have a strong acid, we should write it as either H+ or H30+. In this case, it's just more convenient to write it as H+.

[00:04:16]Uh and we have C42 negative. And we do need to have a reversible double-headed arrow here because the question specifically says that this is a reversible reaction. So we need to have a double-headed arrow there. And then the products are dromate CR272 minus and a water molecule here. The state symbols are optional. You don't have to have the aq and the l in parenthesis. We do have to balance this though. And I can balance this by balancing my chromiums with a two right here. and then balancing my hydrogens. I have two in the water and then maybe put a two right there. And now I have a balanced equation. So make sure that you have something like this. Common mistakes. I'm guessing that there will be a lot of students that don't put the double-headed arrow here. They forget that this is a reversible reaction. I also believe that there will be a lot of students who forget to balance it.

[00:05:09]They'll probably forget to put the twos here. If you put hydrronium H30O plus instead of H+, that's fine. uh the only difference is uh you would balance it a little bit differently if you did that but other than that that is your equation. Part two asks is the reaction a redux reaction? Justify your answer based on the oxidation number of chromium. Well, if we take a look at chromium in the dromate ion over here, we can determine its oxidation number.

[00:05:37]If we call chromium x and we have four oxygens at -2 a piece, that's minus8 and the whole ion adds up to minus2. We find that chromium on the reactant side has a plus six oxidation number. On the product side, we can do something similar except this time we have two chromium. So we'll call that 2x. And there are seven oxygens at -2 a piece.

[00:05:59]So that's -14. And the whole thing still adds up to minus2. So we can use some algebra here and find that the chromium is also pos6. So since the oxidation number of chromium is not changing from reactant side to product side we can say that this is not a redux reaction. So one point for that one. So one point for part one writing the equation and balancing it correctly and then one point for calculating your chromium oxidation number and realizing that it is not a redux reaction. So moving on to uh the next part here we have some information that applies to parts C and D and we have an acidic solution where the dromate ion reacts to form C3 according to this equation right here.

[00:06:45]And then we have some of that CR3 that can be electrolyed to plate objects with a thin layer of chromium metal according to this new equation equation two. And part C says, is the reaction represented by equation two thermodynamically favorable or thermodynamically unfavorable under standard conditions?

[00:07:06]Justify your answer with a calculation of delta G and show the work that leads to your answer. Now we can use the equation delta G equals NF to solve for delta G here. The delta G is what we're solving for. So that's our unknown. The n is the number of electrons transferred over the course of this process. In the equation, the half reaction here it says that there are six electrons. So six is going to go in here for our n. Now f is faraday's constant which is 96,485 kms per mole of electrons. And then e is the voltage and it's given to us as negative 1.32 volt. So I'll plug in -1.32 volts in for the E. Now I'm using jewels per kumm here just so we can follow the units and see how this works.

[00:07:58]Whenever we do the calculations here we see that uh kms will cancel and electrons will cancel. So we're left with jewels per mole as our unit and our calculated value for delta G is positive 764,000 Jew per mole. Now since the value for delta G is positive that tells us that the reaction will be thermodynamically unfavorable at these standard conditions. So basically one point for calculating delta G and realizing that it is thermodynamically unfavorable. I'm expecting that there will be some mistakes with the mathematics here. Some students forgetting the signs and maybe getting the wrong sign for delta G. I'm also expecting that there will be some mistakes where some students get a positive value for delta G, but don't realize that that positive means it's unfavorable. They may get that flipped around and think that positive means it's favorable, which is is not the case. But one point for part C. Part D is a stochometry problem. It says, "Calculate the number of grams of chromium that could be plated onto the surface of a steel rod when 15 amps of current is applied for 3,250 seconds. Show the work that leads to your answer." So in this problem here, the first thing you have to do is calculate the charge in kulom. So I equals Q overt. Faraday's law will help you to do this. We're just going to plug and chug here. I is our current, so that's 15 amps. And the time is 3,250 seconds. So that's our T. And we're solving for Q, which is uh the charge in kulms. When you cross multiply, you find that we have about 48,800 kms for that. So give yourself one point if you said that. Now we can take the 48,800 kms and do some stochometry here.

[00:09:53]We're going to take that and convert it to grams of chromium. So way down here at the end, we'll have our grams of chromium. This is just a stochometry.

[00:10:00]Step one is always to convert to moles.

[00:10:03]So that means we have to have kulms on the bottom and moles on top. And we can use Faraday's constant here to realize that one mole of electrons is equivalent to 96,485 km. So kuloms are out top and bottom.

[00:10:18]We're now in moles of electrons. Step two is our mole ratio. And we can look at the balanced equation, equation two, and put electrons on the bottom. and chromium on the top. And we can see that this is a 1:6 mole ratio. One chromium requires six electrons in order to produce that. So electrons are out top and bottom. We're now in moles of chromium, but we want to be in grams of chromium. So that's where we have to do step three, convert to our final unit. So moles will have to go in the denominator, and grams will have to go in the numerator. And that's 52.00 g in one mole of chromium. When you look at that on the periodic table, we can cancel moles top and bottom. And on our calculator, we take 48,800 divided by 96,485 divided by 6 * 52 and we'll get an answer of 4.38 g of chromium. So if you got that, give yourself one point as well. And so uh there are a lot of possible mistakes here, but this is a pretty straightforward stochometry problem.

[00:11:25]Hopefully you got that right. As we move on to part E, we have some kinetics here. It says in a separate experiment, an iron wire is placed directly in a solution of 050 molar dromate with a pH of 1.99. The concentration of dromate remaining is recorded over time and the results are presented in figure 3. So we have this uh graph of the concentration of dromate over time and the question says explain how the data in figure three support the conclusion that the reaction is first order with respect to dromate. Now as you look at this one thing that jumps out to me is that you know every so often the concentration of dromate gets cut pretty much exactly in half. Notice that over every let's say every 20 minute interval the value of dromate decreases by half. For example during the first 20 minutes from 0 minutes to 20 minutes the concentration changes from 0.5 molar down to right at 0.25 molar. That's seems to be the halflife of this 20 minutes. Well guess what? During the second 20 minutes from 20 to 40 the same thing happens. It gets cut in half again from 0.25 25 molar to what seems to be pretty close to you know half of that about 0.125 molar I would say well this is a sign that the reaction is first order that after every u very fixed interval we have the concentration cut in half so if you said something like that or something similar to that uh then give yourself one point I'm guessing that a common mistake here is that there are going to be some students who uh use figure four instead that shows natural log versus time as a straight line. And they'll try to use that as an explanation. Unfortunately, I don't think that's going to be acceptable because the question specifically says, explain how the data in figure three support this conclusion.

[00:13:24]So, you're going to have to do something with figure three and use that to explain why it's first order. So, uh using figure four, I believe, is not going to get you the point here. But moving on to figure four, it says this information applies to parts F and G.

[00:13:38]The student creates a plot of the natural log of dromate versus time for a trial with an initial concentration of 050 molar dromate as shown in figure 4 and we have this straight line. And by the way, the natural log versus time being a straight line does show that this is a first order process. Uh but of course that doesn't answer the previous question. Part F says, calculate the value of the rate constant for the reaction. Report your answer in units of reciprocal minutes and show the work that leads to your answer. Well, we know that whenever we have a straight line graph like this, the rate constant K is just equal to the absolute value of the slope. And so, we have to essentially calculate the slope of this line and take the absolute value of that. So, in order to calculate the slope of this, I'm going to just take a couple of ordered pairs off of this. And so, a couple of good ordered pairs would be 0 comma.70 seems to be a good ordered pair right around there. And then how about the last one? How about 18, -1.3.

[00:14:43]That's another ordered pair. If I plug those into a slope equation, rise over run, I get60 divided by 18 minutes. And I just went ahead and made that positive since it's going to have to be positive anyway. And so my K is about 033 minutes. And so that is the answer for the rate constant. So if you said that, give yourself one point. And by the way, uh you know earlier on the last part I said that the half-life was 20 minutes.

[00:15:15]Well, just a little fun fact here. If you were to take the first order halflife equation, half-life equals 693 divided by K, if you take 693 and divide it by 033 reciprocal minutes, you'll find that the halflife is really close to 20 minutes. And so that is another way to kind of calculate the halflife there. Now moving on to the next part here, part G. It says that the student then runs a second trial of the experiment under identical conditions, but uses an initial concentration of 0.25 molar dromate instead of 050 molar dromate. On figure four, carefully draw the plot of natural log of dromate versus time that would be expected for the second experiment. The point at time zero has already been plotted. It's kind of nice that they've already plotted that initial point for us. It's interesting that we have identical conditions. So that means that the identical conditions include an identical temperature. And we should know that rate constant is temperature dependent. So as long as we're at the same temperature and the same conditions, we should have the same rate constant. So the same rate constant means we should have the same slope. And so essentially, you want to draw a line that is parallel to the other line that just, you know, starts at that other point. So that's what your line should look like that you plotted. Hopefully they'll be fairly generous. I know sometimes it's a little tough to get the slope exactly the same, but hopefully you will get something very close to that. So give yourself a point if you got that one correct. So this was another uh free response question worth 10 points. I hope you did well on this.

[00:16:58]This had several units uh in it and there was a lot going on here. 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 and I hope to see you soon.