AP Chemistry 2026 FRQ #1 Answered

追加: 2026-05-12

[00:00:01]All right. So, they released the 2026 APFRQs. So, I'm going to show you guys how to solve the first one here. This one's not too bad. I was actually happy to see this one. So, for part A, the first section, it says, write the ground state electron configuration of the potassium ion. So, when potassium becomes an ion, it loses a single electron. And so, originally it has 19 electrons. So, then we go to the electron configuration of argon. And then, so now we're looking at 18. So then for the potassium ion it configuration is 1 s2 2 s2 2 p6 3 s2 and then 3p6. That 4s1 electron is the one that got ionized and removed. So then this is going to be our configuration for our first question here. You can also do the noble gas configuration but you can't just put the argon noble gas.

[00:00:53]I can't just put argon in brackets.

[00:00:55]That's just that's the cheat code. You guys can't do that. So then you have to pick the noble gas right before it, which would then be neon. And then so you pick up with the 10 electrons from neon. So then it's probably going to be something like this.

[00:01:09]So for the second part, they then ask, well, which has the larger radius when you're comparing the potassium ion and the potassium atom? And then explain your reason using the principles of atomic structure. So then like I was just saying before it becomes an ion that last occupied suble is 4s1. So then that's on the fourth ring compared to now the potassium ion's outermost ring only being on the third energy level. So then that is your justification for the atomic structure. So then we would say that the potassium atom has a larger radius.

[00:01:48]And then my justification would be because it has one more energy level in comparison to our ion here where it's like as soon as it becomes ionized, boom, then you go down to the third energy level.

[00:02:08]For part B, it gives us our data table. And so someone's doing um an enaly reaction here and they're dissolving the potassium chloride and water. You have the given masses for both substances.

[00:02:20]And then you have the initial temperature of the water and then the final temperature of the solution once your potassium chloride is completely dissolved. Part B says, describe how the student can use the temperature readings during the experiment to determine when the dissolution is complete. So when you dissolve a salt, you kind of just look for the salt to no longer be solid, right? But they want you to use the temperature readings to identify when it is completed. And then so we do have a change in temperature here from the initial and the final values. And so you want to look for that constant temperature. So when temperature becomes constant, that is when your dissolution is most likely finished. Because if the salt is completely ionized now and is being attracted to water through ion dipole, the change in those attracting forces is going to result in the change in kinetic energy of the solution. And then that becomes constant. That is then your final temperature of the solution. That is when the dissolution is finished.

[00:03:23]For C part I, it asks us to calculate the magnitude of the thermal energy, the Q value in units of jewels that is transferred during the dissolution. So when the salt dissolves, what does that transfer of energy look like? And they want us to show the work that leads to our answer. And then we're assuming that the solution has a heat capacity of 3.95 JW per g*°.

[00:03:45]So we're going to use our Q= MC delta T equation where now we're calculating Q of the solution and then so we need the mass of the solution which means that we now have to take both of these into account. It can't just be one of the two. So then in our setup here we have the 97.5 plus the 6.80 multiplied by the given heat capacity which is here. So then the 3.95 and then we have our change in temperature. And so let me just make some space here.

[00:04:23]We then have our final temperature minus the initial temperature.

[00:04:28]And then when we do our calculations here, we end up getting -1400.75 jewels. And then so that is the transfer of energy in jewels when you have your salt completely dissolve based on the given mass value and the given heat capacity of the solution.

[00:04:48]For double I it wants us to calculate the value of the molar enaly of the solution. So the total change in energy in k per mole based on the amount of the KCL that dissolved which is also given but given in moles. So the 00912 moles and then show your work that leads to your answer. And so we already have the Q value in jewels, but we are eventually going to have to convert that to kilogjles. So you can do that initially or you can do that at the end. So in order to convert jewels to k, we end up dividing by a,000. So then we have the -1400.75 /,000 and then that converts our number into this lovely 1.4 basically k. So then that's going to be the numerator of our expression because delta H is equal to energy or Q / moles which would then be N. And then so and this is still negative. Don't forget that it's negative. So we have - 1.40 K divided by the mole amount given in the prompt which is 0.912 moles.

[00:06:03]And then we end up getting a number which then is going to be -15.36 K per mole. Even if you end up converting to KJ at the end before you convert your number comes out to be -100 oh sorry -15,359 JW per mole. You divide that number by a,000 boom you end up getting the same thing. So then this is going to be our answer for oh lovely.

[00:06:35]This is going to be our answer for part double I. So then D says the calerimeter is not perfectly insulated which means that you have some transfer of energy from the surroundings to the inside of the calerimeter. And then they want to know would this cause the magnitude of the delta H of the solution that we calculated in our part here to be greater than less than or equal to the accepted value and then justify our answer. And so because there's a loss or an addition of energy into the calerimeter, we can't assume that this Q value that we had is 100% accurate. And then so they're saying hypothetically, how would your actual compare to our experimental one?

[00:07:20]So my prediction is that it's actually going to be less than because of this Q value here, right?

[00:07:29]Like the number of moles is going to be constant. Like the moles is not going to change, but it really comes down to this change in temperature here, which would result in a different Q value that we calculate. And so if it's not perfectly insulated, the Q, sorry, the delta T would be a little bit different. My guess is that the delta t is not going to change as drastically. So that means that overall your delta t will be smaller which will then result in a smaller q value. And then if we use a smaller q value that gives us a smaller delta h. So that is how we came up with the less than. So then that whole justification. less than because the delta t would be smaller due to a loss of energy to the surrounding. ings or it would even be something where because you're decreasing the temperature of the solution, the air in the room is probably warmer than 21.1°C.

[00:08:45]So that warmer temperature of the air would actually increase this final temperature value and then so it's not as small as it should be, which still results in a smaller delta t.

[00:08:58]So I don't know, just be on the safe side. It could be loss or gain of energy here but definitely going to give you a smaller delta t overall.

[00:09:08]So here in this scenario we have equal mass values of our two salts. We have ribidium chloride instead of potassium chloride. However, even though we have the equal mass values, their molar masses are not the same. Where if you divide by this respective molar mass, this gives you fewer moles in comparison to this having a greater number of moles.

[00:09:31]So they tell us that the molar enalpy of these solutions is pretty much the same.

[00:09:36]So if we want to convert our mole amounts into jewels using the molar enalpy, you'll actually get a smaller amount of energy.

[00:09:47]If you have smaller moles or fewer moles compared here, you'll get a greater amount of energy because you have greater moles.

[00:09:55]So then it's not asking about well just the change in energy. It specifically wants the magnitude of the delta T. So then it says well if you use the rubidium chloride instead how would its delta T compared to the delta T when you use the potassium chloride. So like we said fewer moles means that you're going to have a smaller amount of energy. And so if you have a smaller amount of energy that means your deltat T is also going to be smaller. So we would say that the delta T of RBC is smaller than the delta T of KCO because there are fewer moles of RB. VCO which produces or requires less energy.

[00:11:06]So then if that energy value is different, you're going to have a different delta T as a result.

[00:11:16]So for part fi they give us the Ksp of the rubidium salt at 20° C and they want us to write the net ionic equation.

[00:11:27]So for I here we have the unionized salt which is a solid and then we end up with the rubidium ionizing with a plus one charge because it's group one and then chloride ionizing with a negative one charge because it is group 17. And so those would be the respective charges and then that is our net ionic equation.

[00:11:54]Even though this is reversible, I'm pretty sure you could just write it as just going in the forward direction.

[00:12:02]And then you get your separated ions.

[00:12:04]Boom. Because water doesn't really play a role in the net ionic equation here.

[00:12:12]For double I wants us to calculate the molar solubility of the KCl in a saturated solution at the same temperature. And then they want us to show our work. So then we can use the Ksp expression where Ksp is equal to the marity of our ions multiplied by each other and then they give us the Ksp which is the 57 and then we just plug in our variables for both of these. So because it's a 1:1 ratio we end up with 57 being equal to the molar solubility time the solubility. So we get S^ squ. So to undo the square we take the square root. So S is then equal to the square t of 57 which is approximately 7.5. So that is our molar solubility. Oh and thenh don't forget your units 7.5 molar.

[00:13:08]Triple I then wants us to compare the solubility here to a solution that already has one molar of KCl dissolved.

[00:13:17]So then they're asking will the molar solubility of rubidium in the 1 molar KCl solution be greater less than or equal to the molar solubility that we just calculated here in part double I.

[00:13:29]And so there is a common ion between these two salts right the chloride ion.

[00:13:34]So if we think about this we already have one molar of the chloride ion present before this even proceeds in the forward direction. That's our common ion where if a common ion is present, it will inhibit our salt from ionizing and proceeding in the forward direction.

[00:13:52]So if we think about this as well, this term here, it won't just be s, it'll actually be 1.0 minus s or sorry or 1.0 plus s. My bad. Yeah, it'd be positive. And then so this expression now becomes 1 + s. So then if we rearrange this we end up getting a smaller number here which means that if we take the square root of it this number should then be smaller as a result. So then we would say that the solubility is going to be less than due to the presence of a common ion which is the chloride ion.

[00:14:52]This prevents the forward reaction.

[00:15:05]So the solubility will decrease for the RBCL for the rubidium chloride.