Greetings! My name is Dibash Sarkar and I am here to assist you with the first Free Response question on the AP Chemistry Exam 2023. This is a substantial question that accounts for 10 out of the total 46 points on the Free Response section.
As of today, May 4th, 2023, the official answer key has not yet been released. Therefore, I would like to clarify that the answers I provide are solely my interpretation and not representative of the College Board.
While I am a post-graduate in Biotechnology and Chemistry and have been teaching in this field for the past 5 years, I am not immune to making mistakes. However, it is essential to note that the AP readers will accept any answer that is chemically and factually accurate, regardless of whether it aligns with the official answer key or not.
So, with that in mind, here’s question two.
AP Chemistry Exam 2023 Free Response Question 2 Answer
Question 2 (a) Answer
In problem two, we’re working with aluminum chloride here. In this case, it’s a molecular substance.
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And the question in Part A says how many grams of CL gas can be formed from 1.25 moles of aluminum chloride gas? So this is essentially just a stoichiometry problem. So we’re going to start with the 1.25
moles of aluminum chloride, and we have to do a mole ratio. So step one is always convert to moles, and we’re in moles, so we can go right to our mole ratio. Now, so aluminum chloride goes on the bottom, and the question is CL.
So that goes on top, and we can use the balanced equation to put the coefficients in. This is a three to 1 mol ratio and aluminum chloride is out. We’re in moles of CL.
You want to be in grams of CL, so we have to convert to grams. So moles on the bottom, 1 mol on the bottom, and grams of chlorine on the top. That’s 35.45
grams. Just as a clarification, it’s not diatomic in this particular case, so that’s why we don’t double it. So moles are out.
And when you do the math, we get an answer of about 133 grams of chlorine. So that’s your answer for part A.Â
Question 2 (b) Answer
So moving on to the second part of this, we have a Hess’s Law question.
And so on this one, calculate the value of Delta H for this first reaction up here, using reactions two, three and four. So this is a Hess’s Law question. So the first thing I notice is that aluminum chloride gas needs to be on the left side up here.
Well, in reaction two, it’s on the wrong side, has the proper coefficient, it’s just on the wrong side. So I have to flip that reaction right there. So when I flip reaction two, it changes the sign of Delta H.
So it’s going to look like this. It’s positive 583. Now, the next thing I notice is that the CL gas needs to have a three in front of it, needs to be on the right side.
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Well, reaction four shows chlorine CL gas as being on the proper side, but it has a coefficient of only two. So I have to multiply that reaction by three halves. So I’m going to multiply reaction four by three halves.
And when I do that, the 243 gets scaled up to a positive 364.5 kilojoules per mole of reaction. Then I notice that the aluminum needs to have the gas on the right side, and this does as well in reaction three.
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So I’m actually going to leave reaction three alone and leave it just the way it is. I’ll keep it like it is. So that’s positive 326 kilojoules per mole.
And so now I’m ready to add these three reactions together. And I can cancel the three halves chlorine on both sides. I can cancel the aluminum solid on both sides.
So everything adds up the way it’s supposed to for reaction one up here. And when I add these together, I get that Delta H is about positive 1274 kilojoules per mole. So that’s my answer.
Question 2 (c) (i) & (ii) Answer
So on to the next part here, part C. We have a potential energy diagram. Every now and then we see one of these.
What is the bond length? Well, the bond length is going to be whatever has the lowest potential energy. So your lowest potential energy is right here at the 200 pecometer internuclear distance. And so that’s your answer for C One.
Now, C Two talks about a completely different bond, talks about the aluminum chlorine bond, and it says it’s 220 PECO meters. And the bond energy is 425 kilojoules per mole. Draw the potential energy curve.
So be careful. Here how we do this. If it’s at 225 PECO meters, I’m sorry, 220 PECO meters, that means that right around here is where it’s going to have its lowest point.
So somewhere actually 225 is on this side of it. So right around here. And then if it’s at 425 a kilojoules per mole, then it’s going to be right down here between the 400 and 450.
So it’s going to have its lowest point right where those two lines cross. So you can draw it, draw it kind of like this, not drawing very well on the screen, but you get the idea there, that’s where it’s going to have its lowest point, and then it’s going to kind of, kind of go up from there. Okay, so something like that, it needs to have its lowest point right there around the 225.
Or I keep saying that, 220 PECO meters. 220 pecometers. So actually that is right, right there.
I was right the first time, right there at 220 PECO meters and at 425 kilojoules per mol. So right around that spot right there is where it should meet its lowest point. So give yourself a point if you said that.
Question 2 (d) (i) & (ii) Answer
Moving on to Part D, we have some Lewis structures here. And Part D one says we have a trigonal planar geometry. So which diagram can be eliminated based on geometry.
Well, there are a couple of things I would eliminate automatically for diagram two, that one. Diagram two is out for a couple of reasons. If you have three sigma bonds and the unshared pair there, that’s not trigonal planar, that’s trigonal pyramidal.
And so diagram two is out just based on the geometry. I’m going to toss in this as well. Diagram two is also incorrect because it has the wrong total number of electron dots.
If you add these up, three chlorines at seven apiece gets you 21 and aluminum three more that should be 24. You should have 24 dots represented well. Diagram two actually has 26 and so diagram two is wrong on so many levels.
So I hope you crossed out diagram two. So give yourself a point for that. Now on D two it says which is the best representation based on formal charges.
Diagram one and three both are okay as far as the octet rule and as far as total number of electrons go. But the fact is diagram one is much better because if you look at the formal charges in these, they all have a formal charge of zero. The chlorines over here you’ve got seven.
It has seven from the periodic table minus the seven assigned to each of them. So seven minus seven is zero. Aluminum has three valence electrons according to the periodic table and then subtract 1233 minus three is zero.
So that’s the best configuration there. Diagram three, on the other hand, not everything has a formal charge of zero. The aluminum actually has a formal charge of negative one because you’ve got your three from the periodic table minus 1234, that double bond counts twice.
So it’s three minus four is a negative one. So that’s not ideal. And chlorine actually has a positive one, or at least this right hand chlorine has a positive one.
You have the seven from the periodic table minus 123456. So that’s positive one. So yeah, diagram three is not the best choice.
Diagram one is the best. So give yourself a point if you got that one.Â
Question 2 (e) Answer
Now this next one here, we’re going to take a look at the equilibrium constant expression for this reaction and once again it’s just products over reactants raised to the power of the coefficient.
So it’s the partial pressure of the al two CL six all over the partial pressure of the ALCL three squared. So you need to have it in this form, not concentration, has to be with the p’s, has to be with the partial pressures. So one point for that one.
Question 2 (e) Answer
Now we can take a look at the last part here for part F. We have a particle level diagram and it says calculate the value of KP for the reaction if the total pressure in the container is 22.1 atmospheres.
Well, I can look at the little chart here and see that the partial pressure of the aluminum chloride is 710 or at least the mole fraction is seven tenths. There are seven out of the ten. So that’s seven tenths.
Multiply that by the total pressure and I get 15.5 atmospheres is the partial pressure of the ALCL three. The partial pressure of the Al two, CL six is well, three tenths times the total pressure.
So three out of your ten, that’s your mole fraction times the total pressure. 22.1 gets me the 6.63
atmospheres for that one. And then all I have to do now is plug those partial pressures into the equilibrium constant expression that we got up in Part E. And when I plug those numbers in and chug out the answer, I get an answer of about 00:26.
So there we have those answers. And so I believe, and of course I’m just making up some points I would believe that this one is a three pointer. I believe.
I think they would give you one point for the 15.5 atmospheres, one point for the 6.63 atmospheres and then a third point for plugging in to the KP expression correctly and getting an answer there.
I think the other answers would be one point apiece if I added these up correctly. So this is a ten pointer. And that’s question two.
I hope that was relatively easy to follow, and I hope you got as many of those ten points as you can. Check out the Question 3 from the Link Below or everything you need for AP Chemistry. I hope you like it and find it helpful.
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