AP Chemistry Cram Session 2026

追加: 2026-05-05

[00:00:00]Hi there everybody. My name is Jeremy Kug and this is the place for all things AP Chemistry. As I record this video, we are days away from the May 5th, 2026 AP Chemistry exam. So, the purpose of this video is to do some last minute cramming. We're going to go through some of the high points of all nine units, the entire AP Chemistry course, and hit the things that uh I think will probably pop up on the AP exam to help you get ready to ace that exam here in just a few days. By the way, if you're new to my channel, welcome. I have all kinds of playlists here to help you get you ready for the AP Chemistry exam, honors chemistry as well, if you happen to be in that course. Uh, also, uh, don't forget that the guided notes for the presentation that you're about to see here are available on the ultimate review packet for AP Chemistry. There's a link for that in the description down below. If you haven't gotten the AP Chemistry Ultimate review packet yet, it's a great resource. It has review videos for all nine units. It has full study guides. It has practice questions in the AP exam simulator to make it look just like you'll see on the big exam day. So, uh, check that out if you haven't done so already. Let's jump right into our cram session and we're going to start with some unit one material here and we're going to look at mass percent. Now in order to find the mass percent of any compound you first of all want to find out what is the molecular mass or the formula mass of that compound. So in the case of NF3 just as an example nitrogen triloride you'd have to find the molecular mass of that. So nitrogen has an atomic mass of about 14.01 florine is about 19.00 and we have three of those. So that's 57 total. When you add those together, you find that the total molecular mass of NF3 is about 71.01 atomic mass units. Now, to find the mass percent of each element in that compound, you need to take the individual masses for each of those elements and divide it by the total. So, for example, nitrogen would be 14.01 divided by 71.01.

[00:02:15]When you calculate that, you find it's about 19.73%.

[00:02:20]You do the same thing for florine. Its total is 57 and you divide that by the 71.01 and find that florine's uh percentage in that compound is going to be about 80.27%.

[00:02:34]So this is how you find the mass percent for a compound. By the way, if we did this correctly, the two individual mass percents should add up to about 100%, which they do in this case. What if we have aluminum oxide? Well, it's the same process. We have the atomic mass for aluminum, which is about 26.98. And there are two of those. So, that's 53.96 total for the aluminum. We have three oxygens at 16 a piece. So, that's about 48. And so the total formula mass for this compound is about 101.96 when you add those together. So to find the percentage by mass of each element, we take the value for each element and divide it by the total. So for aluminum that's 53.96 divided by the 101.96 total. So that's about 52.92%.

[00:03:29]We do the same thing for oxygen. 48 divided by that total and it's about 47.08%.

[00:03:36]And once again, if we did this correctly, the two percentages should add up to 100 and they indeed do. So that's how we find mass percent for any compound that you could be uh given on the exam. Now let's take a look at another concept here and this is the idea that if you have uh compounds that are made of elements in the same family or the same column or group of the periodic table they should have very similar chemical properties. So just as an example of that, if we have potassium oxide and lithium oxide, you'd expect those two compounds to be very similar chemically because, well, first of all, they're both oxides. But more importantly, the cations ions in those oxides are both alkali metals. They're both found in group one of the periodic table. So you'd expect K2O and Li2O to be very similar chemically. If we had magnesium oxide, MGO, well, Mg is in group two, so it probably would not be as similar chemically to these two compounds right here. Now, let's move on and take a look at mass spectrometry.

[00:04:43]So, at some point on the exam, you're probably going to see a graph that looks like this, where we have a pure element analyzed through the mass spectrometry process. Now each peak that we see here represents an isotope in that pure element. So we can say that the three peaks represent the fact that this element has three isotopes. Now the higher the peak, the more abundant that isotope is. So we can say that the isotope that has a mass number of 88 is by far the most abundant. In fact, about 83 or 84% of all the atoms of that element seem to have that mass. And then 86 and 87 are the mass numbers that are much less abundant. Now, we can use this diagram, this mass spectrometry diagram here to estimate the average atomic mass of this element. So, what we do is we take the mass numbers for each of these.

[00:05:40]So the first one is 86 and we multiply that by the decimal equivalent of its percentage abundance. So it's about 10%.

[00:05:47]So that would be10 and so that's 8.6.

[00:05:51]The isotope that has a mass number of 87 is about 7% abundance. So we multiply those across. And then the 88 mass number isotope is about 83% abundance.

[00:06:03]We multiply those. Once we get those products, we just add those together and we find that our estimate for the average atomic mass of this element is about 87.73.

[00:06:14]Now, this is just an estimate, but we can still take that number and match it up to the periodic table and see that this is very very close to the atomic mass for strontium SR. So we can confidently say that this mass spec diagram here is very likely representing a sample of strontium. So be able to analyze a mass spectrometry diagram like we see here. Understand how many isotopes it has, the significance of those heights and be able to make that mathematical estimate of which element we happen to be looking at. Now let's move on to electron configurations. At some point you'll see something about electron configurations on the exam.

[00:06:59]You'll be asked to write one or analyze one. Let's just do some practice here.

[00:07:03]Let's look at sulfur. So we can see where sulfur is on the periodic table here. And just like we always do, we start at the beginning of the periodic table and work our way through the sections, the sub levels of the periodic table. So it goes 1 s2 and then 2 s2 2 p6 because there's six boxes there and then 3 s2 and notice that sulfur is in the 3p section. So it's 1 2 3 four boxes into that. So it's 3 p4. So that's how you'd write the ground state electron configuration for sulfur. Now what if we have the sulfur ion which has a -2 charge? Well, well, if it has a -2 charge, that means that it's gained two electrons. So, we're going to have to add two electrons here to make this 3p6 at the end of its electron configuration. So, that's the electron configuration for the S2 negative ion.

[00:08:04]Let's try another one. Let's try magnesium. So, we can see where magnesium is on the table. And just like we had before, we start at the beginning and work our way through. So, it's going to start with 1 s2 and then 2 s2 and then 2 p6 and finally magnesium is in the 3s section. It's the second element in so it's 3 s2. So that's magnesium.

[00:08:26]Now what if we have the magnesium cation Mg2+?

[00:08:31]Well, remember that the positive2 charge implies that that atom has lost two electrons. So if it's lost two electrons, it's going to lose those electrons from the veilent shell, that outermost energy level. So that 3s2 is basically going to go away. So it's going to be down to 1 s2 2 s2 2 p6. Now you might notice that in both of these cases where we have ions, those two ions are iso electronic with noble gases. In the case of uh sulfide, it seems to be very similar to the argon atom, isn't in its electron configuration. And the magnesium cation is very similar to the neon electron configuration. And that's how most of these ions are. Most ions, annions, cations, ions, they're going to have that noble gas configuration. That's very very stable for ions most of the time. Now, let's take a look at the florine atom. We can see where that is. And once again, we start at the beginning. It's 1 s2 and then 2 s2 and then 2 p5 because we're in the 2p section. One, two, three, four, five boxes into that. Now, what if we have the F+ ion? Now, you have to be careful here because we're not used to seeing florine with a positive charge.

[00:09:49]It normally has a negative charge, doesn't it? Well, remember that a positive charge means that it has lost one electron. So we just have to take one electron away uh from the outermost suble here. So that's going to be 2p4.

[00:10:04]So 1 s2 2 s2 2 p4 for the f+ ion which is not very common but they could ask you about that on the exam. Now what about one that's a bit longer here? How about arsenic? So we see where arsenic is is on the table. And once again we start at the beginning. So, it's 1 S2, 2 S2, 2 P6, 3 S2, 3 P6. Then we go back back around to 4s2.

[00:10:35]3d has 10 boxes. So, it's 3d10 and then 4 p3 because arsenic is the third element in that 4p section. So, just so you know, this is how you'll be writing electron configurations. If at some point you learned how to do some exceptions to the electron configuration, I know that there are some elements that do not follow the alphab principle, they're not going to ask you about those on the AP exam. There are some weird ones like chromium or copper. They're not going to ask you about those on the AP exam.

[00:11:07]They're going to keep it pretty simple as far as electron configurations go.

[00:11:10]Now, let's take a look at some periodic trends. Probably one of the most important ones is atomic radius. You need to remember that the atomic radius is generally going to be largest or greatest at the uh left and the bottom of the periodic table. So those will have the largest atoms. And the reason for that is that atoms at the bottom of the table have more electron shells. And so as atoms add electron shells, well those outermost electrons become farther and farther away from the nucleus. And so atoms with more electron shells are going to have a greater radius than atoms that have fewer electron shells.

[00:11:51]And what about the left versus the right? Well, atoms toward the left of the periodic table are going to have what we call a less effective nuclear charge. What that means is they have fewer protons to pull in their energy levels. And so atoms on the the left side have fewer protons. And so those atoms are quite a bit larger than atoms on the right side of the periodic table.

[00:12:16]Atoms on the right side have more protons and so they can pull in those energy levels much more tightly. So that's the trend for atomic radius. Now what about ionic radius? It's kind of similar, but notice that it's a little different as well. Here I have a pictorial representation of that. Here we have the ionic radi of several ions here. And what I want you to notice is that generally speaking, notice that the positive ions or those cations ions are going to normally be smaller than the negative ions or the annions. And so if we look at just some examples here, if we look at the the diagram, you'll see that nitride, oxide, fluoride, as well as the sodium, magnesium, and aluminum ions all have the same number of electrons. So if you start looking at the numbers there, you'll see that all six of those ions have exactly 10 electrons. So we say that they are iso electronic. They have the same electron configurations. So why is it that one of them is much larger than the others? Why is it that nitrite is so much larger than aluminum if they have the same number of electrons? Well, it comes down to the number of protons. When you have iso electronic species like this, the atom that has more protons is going to be the smallest. So in the case of aluminum, if you look at your periodic table there, aluminum has a whopping 13 protons to pull in those 10 electrons.

[00:13:51]While nitrogen only has seven protons to pull in the same 10 electrons. So nitrogen well its size is going to be much larger. If it's a tugof-war remember that seven versus 10 well the 10 is going to win. So those electrons are able to spread out and you know become much uh farther apart there in that atom. On the other hand in aluminum imagine that tugofwar we have 13 protons versus 10 electrons. Well the protons are going to win and they're going to be able to pull in those uh electrons much more tightly. So the one that has more protons is the smallest and the one that has the fewest protons is the largest in the case of those ionic radi. Now another periodic trend that's quite important is first ionization energy.

[00:14:39]Generally speaking, it's going to be the greatest toward the right and the top of the periodic table. So atoms like helium and neon and florine, these are going to have very high first ionization energies. By the way, first ionization energy just means the energy that's required to remove that very last electron, that outermost electron in the last suble from an atom. Uh the reason for this is that atoms at the top of the periodic table have fewer electron shells and so that last shell is closer to the nucleus and so it has a stronger attraction to its outermost electron. So in an extreme case, think of helium.

[00:15:21]Helium has two protons and and two electrons. Well, those two electrons are really really close to the helium nucleus. So, it's almost impossible to remove an electron from helium. Well, on the other hand, think of something that, you know, has more uh energy levels or more electron shells like, you know, let's say krypton. Now, krypton is a noble gas. It's going to have a a fairly high first ionization energy, but it does have more electron shells. So that last electron shell is just much farther away from the nucleus than it would be in helium. So that last electron is so far away, it's a lot easier to basically pluck that electron away and remove it.

[00:16:04]So generally speaking, at the top of the table, you'll have a higher first ionization energy. Toward the bottom, it's going to be lower. Now toward the the right side of the table, it's going to have a greater effective nuclear charge. So, it's kind of like we said earlier, it's going to have more protons to attract those electrons. And so, since it has more protons to attract its its electron shells, it's going to require more energy to remove that last electron. So, that's why the noble gases have a very high first ionization energy, while the alkalion metals have a very low first ionization energy. It's very easy to remove those because you know the uh the lack of protons compared to the number of electron shells makes it very easy to remove that last electron. Now let's move on to something else beyond periodic trends and this is photoeleron spectroscopy. So at some point on the exam you're going to have to interpret a photoeleron spectroscopy diagram. So in this case here I want you to notice that we have five peaks. Now every peak represents a suble in this uh atom. And the way that you write these is you know from left to right it goes 1 s and then you have 2 s and then 2 p and then 3s and then 3 p. So this should look familiar to you. This is the same order that we write the sub levels for an electron configuration. So if you can write an electron configuration, you can do a photoeleron spectroscopy diagram.

[00:17:48]Now the heights of these peaks represent the relative number of electrons in that suble. So you're used to doing electron configurations, I imagine. So notice that the 1s, the 2s, and the 3s peaks are all the same height. Well, you know that S sub levels generally have two electrons in them. So, all of those have two. So, that's why I'll put a two for those. Now, for the 2p peak, notice that it's three times taller than the 2s peak or any of those S peaks. So, it must have three times as many electrons. So, that's why the 2p is 2p6.

[00:18:26]Now, notice that the 3p peak is 2 and a half times taller than the s peaks.

[00:18:32]Well, that means that that 3p suble must have 2 and a half times as many electrons. So 2 and a half * 2 is five.

[00:18:40]And so we can say that that's 3 p5. Now if you know your electron configurations or you just happen to have a periodic table with you, which you will on the exam, you can match this up and see that this pes diagram matches up to chlorine.

[00:18:55]So this evidently represents chlorine.

[00:18:58]Now I want you to notice that if we have this other PEES diagram, we can you know write out those peaks again. We have the 1s, 2s, 2p, 3s and 3 p. But notice our our 1s are will all be two. Our 2 p is still 2p6.

[00:19:18]We have a 3s2 and the 3p this time has the same height as the 2p peak. So it must be 3 p6. So this must be argon. I want you to notice that the peaks on the argon are shifted a little bit to the left, especially those uh inner peaks.

[00:19:38]And so the 1s and 2s and 2p, those are uh farther toward the left. And there's a reason for that. There's a general rule that the more protons an atom has, the greater the binding energy of its core electrons will be. And so since argon has more protons than chlorine, you'd expect the binding energy of those inner electrons to be a little bit higher than it would be for chlorine. So that's why your inner uh peaks here are shifted a little bit to the left on argon compared to chlorine. So just be aware of that and understand how to interpret a pees diagram. Now moving on to some unit 2 content. Let's take a look at how to draw Lewis electron diagrams. You've probably had some practice with this, but let's just get a little bit more practice. In the case of SF2, sulfur dioxide, we always write the central atom, of course, in the middle.

[00:20:37]So, sulfur's in the middle. We have our two florines. I strongly recommend that you draw these by starting with the outermost electrons and outermost atoms and work your way toward the middle.

[00:20:47]That's a good way to avoid mistakes. So according to the periodic table, florine has seven veence electrons. So I'll draw seven dots around that florine there on the left and then I'll draw seven dots around the florine over here on the right. So we have that. Now sulfur has six veence electrons according to the periodic table. So I'll draw those in.

[00:21:09]We have the six here. And remember the goal is for everything to have eight if it's possible. And looks like everything here actually does have eight. So whenever we draw the overall structure, we see that there are two shared pairs here that will create a pair of single bonds here and we have a couple of lone pairs. So this is going to be our overall leis electron dot diagram for SF2. We have the two single bonds and we have the two lone pairs or the two unshared pairs on the central atom. Now how about CO2?

[00:21:47]Now, if you're not sure what the central atom is, it's usually the first atom given to you. If it's not obvious, the AP exam will usually draw it out for you in the appropriate order. So, we'll have something like this. Like always, you want to start with the outside and work your way toward the inside. So, chlorine has seven veence electrons. So, I have seven on the one here on the left. I have seven veence electrons on the chlorine on the right. And I have six veence electrons from the oxygen because that's in group 16. Now carbon is in group 14. So it has four dots. I'll put the four dots for the carbon. And so we have the right number of dots. But notice that not everything has eight. So I'm going to have to move some dots around to make that carbon in the middle have eight. So I'm going to take a pair of dots from that oxygen on top there.

[00:22:40]Just kind of drag that down to the bottom like this. And now everything has eight. So it looks like we have a couple of shared pairs between the oxygen and the carbon to make a double bond. And then we have single bonds between the carbon and each of those chlorine. So this is going to be our overall leis electron dot diagram. We have one double bond and we have two single bonds in this in the structure of this molecule.

[00:23:09]Now let's go a bit further here and talk about the types of bonds and the hybridization that we're going to have for these. So in the case of SF2, remember that every single bond is a sigma bond. Every double bond is composed of one sigma and one pi bond and every triple bond is composed of one sigma and two pi bonds. So in SF2 we have two single bonds. So that means we have two sigma bonds here. To determine the hybridization, we have to take the number of electron domains and pair that up to a hybridization value. Anything that has two electron domains will be sp hybridized. If a central atom has three electron domains, it's sp2 hybridized.

[00:24:00]And if it has four electron domains, it's sp3 hybridized. So in this particular structure here we have 1 2 3 four electron domains two bonding regions here plus the two lone pairs. So 2 plus two is four. So we have four uh electron domains or a steerup number of four as it's sometimes called. So that's an sp3 hybridization. Now let's try this one here. We have this molecule and if we're asked to count up the total types of bonds here. Well we have a single bond. So that's a sigma. We have a triple bond. So that's a sigma and two pi. So when you add that up, that's a total of two sigma and two pi bonds. And as far as the hybridization, we have two electron domains. We have two bonding regions here. We have the the single and the triple. No lone pairs here. So two electron domain. So that would correspond to sp hybridization. Now, what about a case where we have multiple central atoms like we have here? Well, we add up the sigma and pi bonds just the same way as we always do. Notice that we have a total of 1 2 3 4 5 six single bonds. So, those are all going to be sigas. We have a double bond here.

[00:25:20]So, that's a sigma and a pi. So, when you add that up, that's going to be a total of seven sigma bonds and one pi bond. Now there are three central atoms here. That means all three of those will have their own hybridization.

[00:25:34]Notice that on the first atom here, the the carbon atom, we have four electron domain. So that's going to be sp3.

[00:25:42]On the second carbon atom, that has three electron domains. So that's sp2.

[00:25:48]And on the oxygen atom, we have four electron domains. We have two bonding regions plus two lone pairs. That's four electron domain. So that's an sp3 hybridization. So understand that every central atom will have its own hybridization. Now also be aware that we can take these same structures and talk about the molecular geometries and bond angles. By the way, you have to know those. You have to basically memorize those for the exam. So back to that SF2 molecule, it has two sigma bonds or or two bonding regions and uh two lone pairs. So two and two is going to be a bent structure and that's 104.5°.

[00:26:33]Now there's a reason that most chemists when they have this molecule, they actually draw it like this. That actually represents the fact that this does have a bent structure. If you draw it the other way, it looks like it might be a linear structure and it's not.

[00:26:46]That's not an incorrect way to draw it, but it is more correct to draw it like this, where the structure is actually bent. Now, for our next structure here, we have three bonding regions with zero lone pairs. So, that's going to be trigonal planer, and that's a bond angle of 120 degrees. So, you need to know that. And notice that on the next structure, we have three central atoms.

[00:27:11]So, there are going to be three molecular geometries and three bond angles. So the first carbon atom there has four bonding regions with zero lone pairs. So four and zero is tetrahedral with a 109.5 degree bond angle. The second carbon atom has three bonding regions with zero lone pairs. So three and zero is trional planer 120° once again. And then on our oxygen atom we have two bonding regions with two lone pairs. So two and two is bent 104.5 degrees. Now just so you know on the AP exam they're probably going to be more lenient on those bond angles than your teacher might be. So instead of saying 104.5° it's okay to just u assume that those are going to be the same as your tetrahedral bond angles of 109.5°.

[00:28:07]So those are the bond angles you have to know. 109.5 degrees 120 degrees and if it's linear that's going to be 180 degrees. One other thing that you have to know about these structures is is it polar or non-polar? Does it have a dipole moment or not? Well, in the case of SF4 we can see that there is a lopsidedness to this here. There is a region an unbalanced region of negative charge. So this would be a polar molecule or in other words as we sometimes say it has a dipole moment.

[00:28:41]Now whenever you have a molecule that is a polar structure it's going to have both London dispersion forces and dipole dipole forces. So if it asks you the intermolecular forces you want to list both of those. All molecules have London dispersion forces. Polar molecules also have dipole dipole forces. Now, in the second molecular structure, we see that there is a lopsidedness to this as well.

[00:29:07]We have these lone pairs here that aren't really being canceled out by anything. So, the N2H4 molecule is polar as well. And notice that it has London dispersion forces, but it also has hydrogen bonding. Whenever you see a molecule that has the NH or an O bond or an FH bond on it, it's also going to have hydrogen bonding. If you want to say it has dipole dipole forces, I guess that's okay, too. You're not required to say that though because hydrogen bonding can be sometimes described as a special type of dipole dipole force, but sometimes we can say it's kind of its own thing as well. So, we say it has LDF, London dispersion forces, and hydrogen bonding. How about this last structure down here? Well, notice that there are no unshared electron pairs.

[00:29:56]Uh, basically everything cancels out.

[00:29:59]There are there's no lopsidedness to this and so we can say that this molecule is a nonpolar structure. So it does not have a dipole moment and since it's a non-polar molecule the only intermolecular force it's going to have will be London dispersion forces. If it's a non-polar structure it only has London dispersion forces. As you can see we're kind of slowly moving into unit three here talking about intermolecular forces and uh such things. Let's take a look at how we can predict some relative melting points as we kind of backtrack a little bit here into unit two. If you're asked to compare the melting points of CA2 and CAS, the first thing you want to do is look at the charges of the ions. So in the case of CA2, we have a plus2 and a minus1 as we can predict from the periodic table. On the other hand, CAS is a +2 minus2. Kulom's law tells us that the higher the magnitude of the charge, the stronger the particle attraction is going to be.

[00:31:07]So we can say that a plus 2 minus2 is going to have a stronger kulumbic attraction than a plus 2 minus one. So that's why calcium sulfide has a much higher melting point than calcium chloride. We can look these numbers up and see that indeed that's the case.

[00:31:25]Calcium sulfide has a melting point of well over 2500 degrees Celsius and calcium chloride is less than 800° C.

[00:31:35]Now, what if you have a case where it's a tie? Like in these three cases here, we have sodium broomemide, potassium broomemide, and rubidium bromide. If you look at the charges of the ions there in those compounds, well, they're all plus one minus one. So, uh, you can't really use the ionic charges there to predict the relative melting point. So, if it's a tie, you wouldn't look at the ionic size. And we can look at the periodic table here and see that first of all, they're all broomemides. So, it comes down to the alkalion metal ionic size.

[00:32:09]So, notice that sodium has the smallest ionic size of these three cations ions.

[00:32:16]Potassium is is a larger ion and rubidium is the largest of these three.

[00:32:21]So remember Kulm's law tells us that the larger the ion, the weaker the attraction is going to be and the smaller the ion, the stronger the attraction is going to be. So that's why we'd say that sodium broomemide has the highest melting point of these three.

[00:32:38]Potassium bromide is in the middle and rubidium broomemide would be the lowest.

[00:32:42]And we can look these up and notice that that is indeed what happens here. So the larger the ion is, the weaker the attraction will be and the lower the melting point is probably going to be as well. Now, what if we're asked to compare the melting points of sodium fluoride and potassium chloride? Well, we can look at the ionic charges and see that in both of those cases, we have a + one minus1 charge. So the charge doesn't help us. we have to go back to the ion sizes and compare those. So when we compare sodium and fluoride, we see that that is a smaller ionic pair compared to potassium and chloride which are both larger. So that tells us that the colombic attractions will be stronger between sodium and fluoride than they are for potassium and chloride. So, as you translate that to a melting point, that tells us that potassium chloride should have a lower melting point and sodium fluoride should have a higher melting point. And if you look that up, you'll find that that is indeed what happened. So, be able to use charges and ionic sizes to compare relative melting points of these ionic compounds. Now, as we talk about ionic compounds, we know that a lot of those, in fact, probably most ionic compounds are at least a little bit soluble in water. And the way this works is if you take, you know, sodium fluoride, which we just talked about, and dissolve that into water.

[00:34:05]Well, there are forces between the polar molecule in the solvent, which is usually going to be water, and the ions in the ionic compound. So in the case of sodium fluoride, we know that the positive pole of water is going to surround the fluoride annions and literally drag those out into solution like this. That's how that works. And then for the sodium ions, well in that case, the negative pole of water, which is the oxygen side, is going to surround the sodium ions and once again literally drag those out into solution. And that's how an ionic compound dissolves in water. And the force that we see here that's causing this to take place is called an ion dipole force. It's a force literally between the ion and the dipole of uh that solvent, which in this case, of course, is going to be water. Now, let's move on to another unit three concept here, which is gas laws. So, if you're working on unit three, make sure that you know how to work with the ideal gas law. I can almost guarantee there's going to be an ideal gas law problem somewhere in that exam. So in this problem, it says a gas mixture is collected in a 250 L rigid container at a temperature of 325.0 Kelvin. If the pressure inside the container is measured to be 1.072 atmospheres, how many moles of gas are in the mixture? So once again, we have a pressure, we have a volume, we have a temperature. It's asking about moles.

[00:35:42]That's PV equals NRT. So, we're just going to plug and chug into that equation. The pressure is 1.072 atmospheres. The volume is given to us as 250 L. We're trying to find N, the number of moles. So, that's our unknown.

[00:35:59]The value for R is 0.08206 liter atmospheres per mole Kelvin.

[00:36:05]uh equation and that constant will be given to you on the equation sheet just so you know you you do not have to memorize that the temperature in Kelvin is 325.0 so now we just have to do some algebra here and we find that the N is 05 moles. So that's a pretty straightforward ideal gas law problem.

[00:36:28]But what if we have this question? What if we go a step further and say that if it is determined that 0 660 moles of the mixture is nitrogen.

[00:36:38]The other gas would compose how many moles? So notice the question is asking us about a gas mixture. So we have two gases here and the way that you do that is just to subtract. If we have a total number of moles of 10005 well to find the other gas we subtract the moles of nitrogen. So you take 0.15 moles minus 0660 moles and find that the other gas is 0345 moles. Now with that being said, let's take this a step further yet. If it is determined that this other gas has a mass of 697 g, what is the molar mass of this gas? Now to determine the molar mass of a gas, it's very helpful to remember that the units for molar mass are grams per mole. That literally means grams divided by moles. So the problem tells us that the mass of this gas is 697 g. We just said in the last part of the problem that the number of moles of gas here is 0345 moles. So all you have to do on your calculator is divide 697 divided by 0.0345.

[00:37:56]And you'll find that the molar mass is about 20.2 g per mole. And if you match this up to the periodic table, you can see that this is most likely going to be neon gas. And so you can actually use the ideal gas law and concepts of gases here to actually determine the identity of an unknown gas.

[00:38:18]Now something else from unit 3 that you are going to see on the AP exam is spectrophotometry.

[00:38:24]So if you have never had a chance to do a spectrophotometry lab uh that's okay you can still do well on this question.

[00:38:31]The way spectrophotometry is done is you have this graph which is this is called a calibration curve that we're creating here. sometimes called a best fit line.

[00:38:41]And the concentration of whatever substance we're trying to analyze is going to be on the x axis and the absorbance that is read off of the instrument is going to be on the y ais.

[00:38:54]So the first thing that you have to do is you run a blank which means that it's going to have a concentration of zero and at concentration zero the absorbance should also be zero. So we're going to plot a point there basically at the origin of our graph. And then we're going to run several known concentrations and record their absorbances. And so that's what you do throughout the lab. We have different known concentrations and we plot those on the graph and then we draw a best fit line. So you can eyeball that or you know draw that. They'll normally have that drawn for you on the exam on the AP exam. Now what you'll do next is you'll take the absorbance of the unknown and match it up to the line to find its concentration. So for example, if we take the unknown and its absorbance is about 34, well we just match that up to the line as you can see here and then we can drop down to the x axis and see that the concentration seems to be somewhere around 0.16.

[00:39:57]So you can figure out the concentration of an unknown using its absorbance. Now sometimes on the exam they'll ask you some questions about outliers and this requires you to understand the beer Lambert law the absorbance equals epsilon uh time BC and I do have a video for that but just briefly we can say that if you have an outlier like this where the absorbance is lower than the rest of the line that tells us that the concentration is less than we're expecting it to be. So, if you have an outlier that's below the rest of the line, that probably tells us that it was contaminated with water or possibly contaminated with some other substance uh that got in there, you know, perhaps of a lower concentration. Now, if you have an outlier that's too high, well, there are a couple reasons for that.

[00:40:48]It's possible that it got contaminated with a solution that had a a higher concentration than it was supposed to have. So, that could cause it to be too high. Also, uh, if you touch the sides of the test tube or the sides of that cuette, your fingerprints will sometimes cause less light to pass through the sample than should pass through and that will cause the absorbance to be lower than it should be. So, there are some reasons why you could have outliers and so be aware of that as you work spectral photometry questions on the exam. Now, as we move on to unit four, an important part of unit four is working with net ionic equations. You may be given a type of equation. It might be a redux reaction, a precipitation reaction, possibly even acid base. And you might be asked to answer some questions about it. So, for example, if we have a piece of magnesium metal that's added to a solution of zinc 2 nitrate, it's important to realize that the magnesium is going to react with the zinc. Usually metals react with metal ions in redux reactions whereas non-metals react with non-metal ions. So in this case we have the metal which is magnesium and we have the metal ion which is zinc 2+ and so those are going to react with each other. That's going to leave nitrate as the spectator ion. Remember spectator ions are there but they just really aren't doing anything in the course of the reaction. So in the case of this reaction, magnesium the metal is going to be oxidized. Now you need to know that magnesium has a plus2 charge when it becomes an ion. We can predict that from the periodic table. And then the ion form of the metal is going to be reduced down into its metallic form. So that's just going to become plain old metallic zinc. So we have to balance these half reactions. And we do that by adding two electrons to the product side for the magnesium half reaction. And for the zinc, we have to add two electrons to the reactant side of that. And once we have those, we can add these two half reactions together and get the overall balanced equation as we have here. And of course, you should be able to see that since magnesium's charge is going up, that is an oxidation. Since the the zinc charge is going down from plus two down to zero, that is a reduction. Or if you prefer, we're losing electrons in the case of magnesium. So that's oxidation. And in the case of zinc, it's gaining electrons. So that is reduction.

[00:43:25]So there are a couple different ways to talk about oxidation and reduction.

[00:43:30]Sometimes you'll have just some random equations that you have to write equations for or random reactions. In in this case, magnesium metal is burned in air. So, magnesium of course is Mg. And anytime something is being burned, it's being reacted with oxygen O2. So, when you put those together, that gives us MGO. And yes, you need to know that it's Mg and not Mg2 because Mg has a plus2 charge and oxide has a negative -2 charge. You know, those cancel out.

[00:44:02]Whenever you write an equation, you are normally expected to balance it. So when you balance that, you balance the oxygens and then balance the magnesiums.

[00:44:11]And that's our balanced equation. Also be aware of how to write the equations for combustion reactions. Like in this case, a sample of ethylene gas C2H4 is burned in air. So ethylene is C2H4. And by the way, you're not expected to have those uh organic compounds memorized. If they give you something like that ethylene, they'll tell you what the formula is. But you are expected to know that when something is burned in air, it is reacted with oxygen. You're also expected to know that for the complete combustion of any hydrocarbon, the products are always carbon dioxide and water. You do have to know that. And you are expected to be able to balance this equation. So you balance the carbons by putting a two in front of CO2. You balance the hydrogens by putting a two in front of water. And then you balance the oxygen. I see two oxygens on the left side. I see six oxygen on the right side. So to balance that, I need a three in front of O2. So now I have a balanced equation.

[00:45:19]As we take a look at this one here, this is a precipitation reaction. Solutions of calcium nitrate and sodium phosphate are mixed in a test tube. So calcium nitrate is a mixture of calcium ions and nitrate ions. Of course, sodium phosphate would be a mixture of sodium ions and phosphate ions. And just so you know, they will most likely tell you the formulas of these ions. If you forget that phosphate is P43 negative, that's okay. They'll most likely tell you that on the exam. You are expected to know, however, that when these two ions try to swap partners, the sodium and the nitrate, well, that's soluble. So, those are going to be spectator ions. And so, the pair of ions that will form something will be calcium and phosphate.

[00:46:06]So, when this reaction takes place, it's going to be the calcium and the phosphate that are going to get together to make the precipitate. Like we said, sodium and nitrate are spectator ions.

[00:46:16]They're there, but they're just not going to do anything. basically they're going to stay in solution because you know all nitrates and all sodium compounds are soluble. So the product is going to be calcium phosphate Ca3 P42.

[00:46:31]And so when you write the overall balanced equation uh the net ionic equation that is it's going to look something like this. You are expected to balance this. So we have three calciums and two phosphates and that is our net ionic equation for this process. Now as we move on with unit four, let's try some stoeometry here.

[00:46:52]Let's say we have this balanced equation and we have this problem. It says in a chemical analysis, excess silver ions are added to a solution containing chloride ions. After the reaction, the total amount of AgCl solid is weighed and found to have a mass of 0.445 g. How many grams of chloride ions were present in the original sample? So what we have to do here is to take the 0.445 grams of Agl that are mentioned in the problem here and figure out how many grams of chloride or Cl minus we had originally.

[00:47:32]And we're going to use that balanced equation and stochometry to do this.

[00:47:36]Remember when we have a stochometry problem we have that three-step process that I showed you how to do earlier in my course. So step one is convert to moles. Step two is the mole ratio and step three is convert to your final unit. So step one is to convert to moles. So in the denominator we have to put grams. In the numerator we have to put moles. And if we consult the periodic table we can see that there are 143.32 gram in one mole of AgCl. They might give you that. You might have to add it up on your own. Well grams are out now.

[00:48:11]So we're in moles of AgCl. Our second step is to uh use the mole ratio. So the mole ratio AgCl has to go on the bottom and chloride Cl minus goes on the top.

[00:48:23]And we can consult the balanced equation and see that since both of these have coefficients of one, it's going to be a one one mole ratio. So AgCl is out.

[00:48:33]We're now in moles of chloride, but we want to be in grams of chloride. So that's what the third step is for, to convert to that final unit of grams. So moles have to go on the bottom, grams on the top. And we can consult the periodic table and see that there are 35.45 g in one mole of chloride Cl negative.

[00:48:54]So moles are out. And now we can do our math. We take445 divided by 143.32 times 35.45 and we find that the amount is.110 grams of chloride. So that's how you'd solve a simple stochometry problem like this. Let's say we have a solution stochometry problem because sometimes they'll ask about that on the exam as well. Here we have the balanced equation and this time the problem says a student adds 125 ml of.1 molar silver nitrate to a solution of excess sodium carbonate. Assuming the reaction goes to completion, how many moles of silver carbonate should be produced? Well, the first thing we have to ask ourselves is how many moles of silver ions were used? We can look at the question up here. It says it was 125 milliliters of.1 molar silver nitrate.

[00:49:57]So we know that to find the number of moles in solution we take the marity and times it by the volume in liters. So we just have to take the 0.125 lers and multiply that by 0.1 molar of our silver ion. So we get 0125 moles. So that's the first part of this.

[00:50:18]We have 0125 moles of Ag+. Now that's all very good. But the question is asking us how many moles of Ag2 CO3 can be produced. So we're actually trying to convert to moles of Ag2 CO3. Well, now we can do our mole ratio. We're starting with moles here. We're trying to get to moles. So we can use our mole ratio. And in the mole ratio, we have to put Ag+ on the bottom this time. And Ag2 CO3 goes on the top. We can consult the balanced equation and see that we have a coefficient of one in front of silver carbonate and a coefficient of two in front of Ag+. So we can cancel Ag+ top and bottom. And we're basically going to take 0.0125 moles and divide that by two and we find the answer is 6.25* 10 -3rd moles of silver carbonate. Now let's take a look at unit five and go on to kinetics. Now, a lot of students have trouble with kinetics. It is fairly heavy on the mathematics, but if you've had practice on this, I don't think you'll have much trouble with the kinetics problems. Here we have a very typical problem with an initial rate of reaction shown here. It's measured at a specific temperature, 350 Kelvin, and we have three trials where we have initial concentrations of both reactants and the initial rate of reaction given to us.

[00:51:45]Now, how do we find the order of the reaction with respect to A2? Well, in order to do that, we have to find two trials where A2 is the only reactant that's changing. And so, I would recommend using trials one and two because notice that D2 is held constant whereas A2 is being doubled from trial one to trial two. Now A2 is doubling but what's happening to the rate from trial one to trial two? Well notice it's also basically doubling isn't it? So if we have the concentration doubling and the rate doubling well the power that makes that a true statement is a one. So this is first order. So we say that A2 is first order. Now how do we do the order with respect to D2? Well, we have to do the same thing except this time we have to find two trials where D2 is the only reactant that's changing concentration.

[00:52:46]And for this one, I would recommend trials one and three because notice that from trial one to trial three, A2 stays constant while D2 is doubling. Now, notice that D2 doubles. What happens to the rate from trial one to trial 3?

[00:53:04]Well, notice that in that case, it quadruples. It goes from 1.12 to 4.47.

[00:53:09]That's a quadrupling, isn't it? So, when the concentration doubles and the rate quadruples, what power makes that a true statement? Well, it would be the power of two. So, we'd say that D2 is going to be second order. So, we have A2 is first order, D2 is second order. And so we can say that the overall order is third order. Remember the overall order is just the sum of all the individual orders in the equation. So this is a third order overall. Now let's take this a step further here and let's write the rate law for the reaction. So it has to be in a very specific format. It's always rate equals K times the concentration of the first reactant raised to its power times the next concentration of its reactant raised to whatever order it is. So in this case A2 is first order. So that's the power of one. D2 is second order. So that's a power of two. So write it like this.

[00:54:13]Rate equals K * A2 * D2.

[00:54:19]Now, students will often make mistakes on these. A very common mistake is students will forget the K. Okay? If you don't have the K in there, it's wrong.

[00:54:30]You have to have the rate constant.

[00:54:32]That's part of the equation. Also, some students will forget the rate equals part of this. If you forget the rate equals and just write K A2 * D2, it's wrong because a rate law is an equation.

[00:54:46]You have to have the whole thing there.

[00:54:47]So rate equals K * A2 * D2 ^2. So if you can do the orders, writing the rate law should be pretty easy, but don't get tripped up on those common mistakes. Now let's determine the rate constant for the reaction at this temperature.

[00:55:03]Essentially, this is just a plugandchug.

[00:55:05]You take any one of those three trials and plug it into the rate law. So we have the rate law that we just figured out. Now, we're going to plug in some numbers from there. I'll just use trial one. The rate in trial one is 1.12 * 10 -3rd marity per second. We're solving for K. So that's our unknown.

[00:55:25]The A2 is 04 molar. And then the D2 concentration is 03 molar and we do have to square that because that's second order. Now determining the numerical value of K is pretty easy. This is just simple algebra. It's about 31 if we have the proper number of sigfigs here. The hard part is getting the units. Anytime they ask you for the rate constant, they do expect you to get the right units.

[00:55:55]And this can be a little tricky. In fact, a lot of students make mistakes on determining the units. Here's how I do it. Notice that on the right side of this equation, we have marity time marity squared. So that's marity cubed.

[00:56:11]So I'm going to multiply the right side by marity to the -3rd to get rid of that. Now as you know in algebra when you do something to the right side you have to do the same thing to the other side. So I'm also going to multiply the the left side by marity to the -3rd. And now I can figure out my units because I have marity to the -3rd times marity which is marity to the2 time reciprocal seconds. And so those are my units. It's going to be a marity to the -2* seconds to the minus1 or whatever equivalent you have to that. If you want to put it all in the denominator and make your powers all positive, that that's fine as well as long as you're going to use the right units. Something else to look out for, make sure that you use the right time unit. Sometimes we assume that we're always using seconds.

[00:57:04]Sometimes they will trip you up and put minutes. they'll say marity per minute or like marity per hour or something like that. Well, you have to match the time unit to whatever time unit they use in the problem. So, if it's minutes, use minutes. In this case, it's seconds. So, just be careful there on determining the rate constant and the rate law. Now, let's go a step further here and try to determine an order graphically. So, here we have a balanced equation. We have 2 HI yields H2 plus I2 and we're given some graphs. Now the key here is to focus on the one that's a straight line.

[00:57:45]If we're asked to determine the order of the reaction with respect to HI, like I said, focus on the straight line. Only one of the graphs, exactly one, is going to be a straight line. So that would be this graph right here. One over HI versus time. Now, you have to know this, okay? They're not going to tell you this on the exam or on the equation sheet, but you have to know that if it's the natural log versus time graph that's a straight line, it's first order. You have to know that. If it's the concentration versus time graph that's a straight line, it's zeroth order. You have to know that. If it's the one over concentration versus time graph, that's a straight line. It's second order and you have to know that. So in this case, notice it's the one over the concentration versus time. That's a straight line. So that means this is a second order process. And what if it asks you to write the rate law? Well, it's rate equals K times the reactant HI raised to the second power. Like I said before, don't leave out the K. That's a common mistake. And don't leave out the rate equals part. That's a common mistake as well. Now, how could you determine the rate constant? Well, the rate constant is found by looking at that straight line graph. It's always going to be the absolute value of the slope of the line. So, whichever graph is the straight line, just calculate the slope of that. And the absolute value of that slope is the rate constant. In this case, since it's second order and the slope is is positive anyway, you don't even have to say absolute value of the slope. is just the slope of that line.

[00:59:33]Now let's take a look at something else with u mechanisms here in kinetics and let's say we have a mechanism. So we have two steps here and this uh mechanism is given to us for a reaction.

[00:59:45]They may ask us to identify the reaction intermediate.

[00:59:50]Well, remember that a reaction intermediate is a particle or a molecule that is produced in an early step that is consumed in a later step. So notice that in this particular equation, BF2 is produced in step one and then it's consumed in step two and BF2 never actually appears in the overall balanced equation. So, our reaction intermediate is going to be BF2.

[01:00:19]Now, if we're asked to write the overall balanced equation, well, we would add those two steps together. And notice that we've canceled out BF2. So, that's out of the picture now. And when you add up what's left, we have BF3 plus 22 yields B plus O2 F2 plus O2F. If they give us some extra information and tell us what the slow step is and what the fast step is, we can also determine the rate determining step. Remember the slow step is always the rate determining step. You know the slow step slows the whole thing down. Doesn't matter how fast the fast step is, but the slow step slows the whole thing down. So that's why we say that the first step is the rate determining step because it is the slow step. Now if we know that we can also write the rate law for this reaction. The rate law for the reaction is basically the same as the rate law for the slow step. So we would write it as rate equals K times the concentration of the first reactant in that slow step times the next reactant in that slow step whatever that may be. So in this case it's rate equals K time BF3 concentration times O2 concentration. So all you have to do is focus on the slow step to write the rate law for a reaction like this. As we move on to unit six, which is about thermochemistry, we have a very typical problem involving transfer of heat.

[01:01:52]Let's take a look at this question. It says, "A student adds 0.652 gram of warm copper metal to a 93.54 g sample of water. After stirring, the temperature of the copper metal drops from 425.5 degrees C down to 23.2 degrees C. The specific heat capacity of copper metal is 385 JW per g° C. And the specific heat capacity for water is 4.18 JW per g° C. Assume that no heat is lost to the surroundings. And the first part of the question says, how many jewels of heat were lost by the copper? So for this problem, we're going to use Q= MC delta T. So we're going to plug that in and we're going to solve for Q, which is jewels. So Q is our unknown. M is the mass of the copper. So if we look back in the problem, it says that the mass is 652 g. We're going to plug that in for the mass. Now C is the specific heat capacity of the copper and the problem tells us that that is 385 jewels per gram degrees Celsius. So that gets plugged in for C. And for delta T, notice that the copper drops from 425.5° C all the way down to 23.2° C. So if you subtract that, you find that this is a loss of 402.3°C.

[01:03:22]So that's why I'm putting 402.3° C in for delta T because the temperature is dropping. So now if we multiply these by each other, we find that the Q is 101 jewels. Now that negative sign right there implies that the heat was lost.

[01:03:40]And that makes sense because the temperature went down. So the answer is 101 jewels of heat were lost in part A.

[01:03:47]Now part B says how many jewels of heat are gained by the water. Now we have to focus on the last sentence in the question that says assume that no heat is lost to the surrounding. So we're assuming that however much heat was lost by the copper is equal to the heat that's gained by the water. So if 101 jewels were lost by the copper, that means 101 jewels had to have been gained by the water. We're assuming that there's a direct transfer of heat from the copper to the water. Now, part C says, what was the original temperature of the water? Well, once again, we're going to use Q= MC delta T for this once again, but this time we're looking at this problem from the point of view of the water, not from the copper. So, Q is going to be 101 jewels because we know that the water is going to pick up that 101 jewels. Now, M is the mass of the water. The problem says that's 93.54 gram of water. So that's plugged in for M. Now the C, the specific heat of water is 4.18 Jew per g° C. We're solving for delta T in this problem. So we can use algebra and find that delta T is.26° C. Now that's the change in temperature of the water. But that's not what the question's asking. It says, what was the original temperature of the water? Well, we know that the water temperature had to go up and we know that the final temperature of the water is equal to the final temperature of the copper. Now, the problem does not come out and say that, but we need to understand that in terms of something called thermal equilibrium. If you take something that has a high temperature that's very hot and you drop it into something that's at a lower temperature like water in this case well the temperature of the copper the hot material is going to drop and the temperature of the colder material the water is going to rise until they equilibrate. The two materials will have the same final temperature. So the problem says the final temperature of the copper is 23.2° that means the final temperature of the water is also 23.2°. So if it had gained 26°, what was it before that? Well, we have to take 23.2° and subtract.26° and find that it's about 22.9° C. So that's the original temperature of the water. So that's how you would solve a problem like this involving Q= MC delta T. Now, another diagram that you're very likely to see on the AP exam is a heating curve that looks like this.

[01:06:30]So, I want you to notice that I have the different regions of this labeled.

[01:06:34]Remember that as heat is added, temperature is going to go up if a phase is not changing. However, if a phase is changing, the temperature stays constant. And we can see this as we look at the heating curve. Notice that as we start out, if the material is a solid, we add heat and the temperature goes up.

[01:06:55]And that makes sense. It would steadily go up as you add heat to it. However, once that solid reaches the melting point, if you keep adding heat, as the product is melting or as the material is melting, the temperature does not change. It stays constant over the course of the melting process. Once all of the material is melted and it's a liquid. Now, when we add heat again, well, the temperature continues to rise until it hits the boiling point. Now, once that liquid hits the boiling point, if you add heat to it, the temperature is not going to go up. That heat is invested into the boiling process. So, it causes it to boil. And then once it's all boiled away and the material is completely a gas, well, then the temperature starts to rise again. So we can basically generalize this and say that as we have a solid or a liquid or a gas and we add heat to it, the temperature rises because the average kinetic energy of the particles is increasing. Once you're at the melting point or the boiling point though and you're adding heat, the kinetic energy on average of those particles stays the same. It's the potential energy that is increasing whenever we uh change those phases. And by the way, this curve is called a heating curve. We could take the curve and put it in reverse where we're cooling it down, that would be called a cooling curve. And instead of having bo melting and boiling, you'd have condensation and freezing. Uh but the concept is essentially the same. It looks the same except it's in reverse.

[01:08:29]Now, let's take a look at another concept from unit six. And this one says the delta H of the following reaction is equal to 91.4 KJ per mole. Determine the heat of formation of sodium chloride solid. So this is where we have to use the equation the delta H of a reaction is equal to the sum of all the heats of formation of the products minus the sum of all the heats of formation of the reactants. And so we're going to use the data here to determine the heat of formation of the NaCCl. Now we do need some information. And they're going to have to give this to you on the exam, but they would tell you that the heat of formation of Na3 is -365.4 kJ per mole. And since there are two moles of that, we have to multiply it by two. Now, we don't know what the enthalpy of formation of NAC is. We're going to call that X because we're going to solve for that. Now, there are two moles of this since there's a coefficient of two in front. So, that's basically 2X. The enthaly of formation of oxygen gas is zero as it is for all elements in their most natural state at standard conditions. We do have three moles of that but of course 0* 3 is still zero. Now notice that the sum of the reactants is well it's just -730.8 k. The sum of the products is 2x. 2x plus 0 is just 2x. And the way this works is delta H equals the sum of the products minus the sum of the reactants.

[01:10:07]So the problem does tell us what the delta H of the reaction is. Up in the problem, it says that the delta H is 91.4 K. So we're going to fill that in.

[01:10:19]The value for the sum of the products is 2x. So we fill that in. And the value for the reactants total is -730.8 8 K. Now be careful with your signs here because as you know two negatives make a positive. So this is saying 911.4 K= 2X plus 730.8.

[01:10:43]That's a common mistake. Students will mess up the signs and they'll end up getting the wrong answer. So be very careful here. So we're going to subtract 730.8 8 from both sides and we get 2x = -822.2.

[01:11:02]Now we can divide both sides by 2 and we find that the answer is -411.1 kJ per mole. So that means x which was the enthalpy or the heat of formation of NaCCl is -411.1 kJ per mole. So that's how you can use this equation or this this concept here to solve for the delta H of a reaction or the enthalpy of formation of any one of those uh substances in the reaction if they give you the others. Now as we move on to unit seven and talk about equilibrium for a bit, let's start with writing equilibrium constant expressions. As a reminder, this is products over reactants raised to the power of the coefficients. Remember that we only include gases and aquous solutions. Liquids and solids don't really have a concentration that can change and so those liquids and solids are going to be omitted from any equilibrium constant expression. So let's say we have this reaction right here and we're asked to write the KC expression for that. So remember KC is in terms of concentration. That's what that C stands for. So KC is going to be equal to the concentration of the aluminum 3+ ions all over the concentration of the silver plus ions quantity cubed. And so that's how we'd write that. Couple of common mistakes.

[01:12:33]Students will sometimes forget the KC equals. Okay, if you leave off the KC equals, it's not right. It has to be an expression. It has to be in the form of an equation. Otherwise, it's not correct. Some students will forget the three. They'll forget the power. And of course, some students will put the solids in there. Remember, solids are not included in the equilibrium constant expression. Also, since this is in terms of concentration, marity, you have to use the brackets. The brackets around the substances imply marity and so you have to have the brackets there otherwise it's not correct. Now how about this equation right here? We have 2 N2O gas yields 2 N O gas plus N2 gas.

[01:13:20]This time let's write the KP expression for this one. So KP is in terms of pressure. So, KP equals the partial pressure of N O quantity squared times the partial pressure of N2 all over the partial pressure of N2O quantity squared. So, it's the same idea.

[01:13:42]Products over reactants raised to the power of the coefficients except this time it has to be in terms of partial pressures. So we have to have it in that P with the little sub uh N O or N2 or N2O in this case. Common mistakes. Once again, some students will forget the KP equals. If you forget that, it's wrong.

[01:14:05]Also, since this is in terms of pressure, you cannot use the brackets.

[01:14:09]If you use the brackets, it's wrong. If it's KP, it has to be in terms of pressure. So, put the P in there and the appropriate power. So be very careful as you're writing those equilibrium constant expressions. It's not hard, but you have to do it a very specific way, otherwise you don't get credit for it.

[01:14:28]Now, let's do a fairly fundamental equilibrium problem. Here we have a problem where it says that at a certain temperature, the reaction below occurs and the chemist investigates the process by filling a container with Cl2O gas to a pressure of 1.50 atmospheres. After the mixture retains equilibrium, the pressure of Cl2O has dropped to 0.10 atmospheres. Calculate the equilibrium constant KP at this temperature. So for equilibrium problems like this where we're talking about an initial value and maybe a final value as well, I would strongly recommend solving the problem using the icebox method. IC stands for initial change and equilibrium. So as we organize our data, it's going to make it a lot easier to calculate and determine what we have at the end and then of course calculate our constant. Now the problem tells us that we start out by filling the container with Cl2O to a pressure of 150 atmosphere. So that means that the initial pressure of Cl2O is 150 atmospheres. The problem doesn't say anything about adding chlorine or oxygen to this. So, it's safe to assume that the initial pressures of those two substances will be zero. So, I'll fill in zeros for those. Now, the problem does tell us that once we're at equilibrium, the pressure of Cl2 is 0.10 atmosphere. So, I'm going to plug that in here. 010 atmospheres in the equilibrium position for Cl2O.

[01:16:06]Now, with this data, I can solve the puzzle and figure out the equilibrium pressures of everything else. So, it looks like Cl2O dropped by 1.40 atmosphere. So, I'm going to fill that in here. -1.40 in the change row. Now, for Cl2, since it's a product, it's going to go in the opposite direction.

[01:16:30]Reactants went down, products have to go up. And since it's a 2:2 ratio, it's going to go up by the same magnitude as the Cl2O went down. So it's going to go up by 1.40 atmospheres. Now the oxygen's also going to go up, but notice it's a 2:1 ratio. So that means it's going to go up by half as much as the chlorine went up. So it's going to go up by.70 atmospheres. Now that means our equilibrium pressures are going to be 1.40 atmospheres for chlorine and.70 atmospheres for oxygen. Now this is all very good but this is not what the question's asking. It asks us to calculate the equilibrium constant KP at this condition. So we're going to have to write the expression for KP. Once again that's products over reactants raised to the power of the coefficient.

[01:17:28]So, KP equals the partial pressure of chlorine squared times the partial pressure of oxygen all over the partial pressure of Cl2O squared. And now I can plug those numbers I got from the ice box into that expression. And when I do that and you use my calculator, I find that the value for KP is 1.4* 4 * 10 the 2 or if you didn't use scientific notation about 140. So that is the answer for KP. Now I want you to notice something about the value of KP. I want you to notice that the value of this equilibrium constant is relatively large. It's not the largest equilibrium constant we've ever seen, but it is significantly greater than one. So, anytime we have a relatively large value for K, that means that we're going to have a lot of products and we're not going to have very much reactant. And if you look at the equilibrium pressures, you'll see that that's exactly what happened. We had a whole lot more products than we have reactants at equilibrium. So, once again, that implies a large equilibrium constant. If it had been the opposite, a very small equilibrium constant, much less than one, you know, like one to the negative fourth or something like that, that means you'll have a lot of reactants left, not a whole lot of products. And so, be aware that the magnitude of an equilibrium constant means something as to the relative amounts of reactants and products that you'll have at equilibrium. Let's take a look at this next question here. It says at the same temperature as the previous example, a mixture of three gases was added to a closed container. The partial pressure of Cl2O gas was 26 atmospheres.

[01:19:16]The partial pressure of Cl2 gas was 1.02 atmospheres and the partial pressure of oxygen was 1.10 atmosphere. Part A asks, what is the total pressure in the container? Well, remember that the total pressure is equal to the sum of all the individual partial pressures. So, all we have to do is add those three partial pressures together. So, 26 atmospheres plus 1.02 atmospheres plus 1.10 atmospheres equal 2.38 atmospheres total pressure in the container. Now, part B says, as the reaction proceeds, which gases will increase in pressure and which gases will decrease in pressure?

[01:20:03]Now, notice that the mixture that we're given here in the problem is not necessarily at equilibrium. We're going to have to use Q versus K and decide in which direction the reaction is going to proceed. So, we have to write the expression for Q. And just so you know, it's written exactly as we'd write the expression for K, except we can't call it K because it's not at equilibrium. So it's the same as it was before. Products over reactants raised to the power of the coefficients. This time we're going to plug in those partial pressures given to us in the problem. So for chlorine, that's 1.02 atmospheres, which will be squared. For oxygen gas, that's 1.10 atmospheres. And for Cl2O gas that's 26 atmospheres which will be squared. So when you solve for Q on your calculator you'll find that the reaction quotient is equal to 17 approximately. Now this is where we have to compare Q versus K.

[01:21:05]So you might remember in the last example we calculated K to be about 140 1.4 * 10 2. So Q is much less than K.

[01:21:16]Now, what does that mean? Well, if you've forgotten, you know, Q is less than K. So, the reaction proceeds toward the products. Now, if you forget that, because that's easy to do, use the Pac-Man rule. Okay? Take that little less than sign and make it into a Pac-Man like this. And so, that tells us that since the Pac-Man is eating toward the right, the reaction is going to proceed toward the right. And so that means that we're going to have more chlorine and more oxygen. You know, we're going to have more products. Those will increase. And the reactant, the Cl2O is going to decrease. And so this is a Q versus K question. So you have to plug the numbers into the Q expression.

[01:22:00]Compare Q with K. And if it helps, use the Pac-Man rule. Since the Pac-Man is eating toward the right, that means the products are going to uh increase or be favored in this mixture. Now, let's take a look at another uh part of equilibrium, which is Lhatier's principle. And this question says, if the reaction below is at equilibrium, how would each of the following changes subsequently affect the partial pressure of oxygen in the reaction vessel? So the first change is we remove some SO3. So remember if we remove a reactant the reaction is going to proceed so that it replenishes whatever has been removed whatever substance has been removed. So if SO3 is removed well the reaction is going to try to replenish that SO3. And in order to do that, it's going to have to react some SO2 and O2 to make that happen. So the O2 is going to decrease.

[01:23:03]And so that's the answer for part A. Now part B says, what if some oxygen is added? If we add some oxygen or anything in fact for that matter, the reaction is going to proceed so that it tries to deplete some of the material that was just added. So if we add O2, well that O2 is going to be used up uh to produce some product. So the O2 should go down once again. So your partial pressure of oxygen will decrease. How about if we add some helium gas? Well, don't forget anytime we add an inert gas or some gas that really has nothing to do with this reaction, there's going to be no change to the equilibrium. So that's why it's no change in this case. Part D says, what if the pressure is decreased by increasing the volume? Well, if we increase the volume, the reaction is going to proceed towards the side of the reaction that occupies more space or has more moles of gas. Well, notice that on the reactant side, we have three moles of gas. On the product side, we have two moles of gas. So if we increase the volume, you know, decrease the pressure in this reaction, it's going to proceed toward the reactant side. And so that means that we would expect SO2 and O2 to increase. So the O2 is going to increase. At the same time, the SO3 partial pressure would go down. So you have to count the number of moles of gas and compare them to see how that works.

[01:24:38]Now in part E, what if the temperature of the container is increased? Well, this is where that delta H comes in. I want you to notice that delta H is a negative value for this particular reaction, which implies that the reaction is exofothermic.

[01:24:55]And if the reaction is exothermic, that means heat is a product. That's what exothermic means. Heat is being given off as a product. So if we increase the temperature, well, we're basically adding heat, aren't we? So adding heat is going to force the reaction to proceed in the opposite direction back toward the reactants. And so if we add heat, you're going to expect O2 and SO2 to increase while the SO3 is going to decrease. And so that's how you answer questions involving Lhatier's principle.

[01:25:31]Now, as we move on, let's take a look at some acid base problems here. And let's look at some relationships between hydrronium ions, hydroxide ions, pH, and p. And by the way, these equations are going to be given to you on the equation sheet. So, you don't have to go home and memorize all this stuff. You just have to know how to use it. So one relationship is that the concentration of the hydrronium ions times the concentration of the hydroxide ions will always be equal to KW. And at 25° C, KW equals 1.0 * 10 -14.

[01:26:11]So at 25° C, this relationship holds true. The concentration of hydrronium times the concentration of hydroxide equals 1.0 * 10us14th.

[01:26:25]So if you know the hydrronium ion concentration, you can calculate the hydroxide ion concentration and vice versa as well. pH is equal to the negative log of the hydrronium ion concentration and PO is equal to the negative log of the hydroxide ion concentration. So, let's use these equations here to answer a few questions. Let's say that we have a solution that at 25° C has a pH of 9.33.

[01:26:55]Let's determine its p, its hydrronium ion concentration, and its hydroxide ion concentration. Well, let's do the easy one first. If we know the pH, we can figure out what the p is. One relationship that you might know is that pH plus p equals 14. They're going to give you that equation on the equation sheet by the way. I know it's not on this slide here, but they will tell you that. So to find the p, we just have to take 14 minus 9.33.

[01:27:28]And you'll find that the p is 4.67 for this solution. So let's take a look at the hydrronium ion concentration which is you know the negative antilogue. So we have to take 10 to the negative pH power in this case. So the pH is 9.33. So we just take 10^ the 9.33 and we find that the hydrronium ion concentration here is 4.7 * 10 -10th moles per liter. Now let's do the other one. The hydroxide ion concentration. There are a couple different ways to do this. You could take KW 1* 10us4th and divide it by that number that we just got. Or you could take the negative antilogue of the p. We just said that the p is 4.67. So we could take 10 to the -4.67 power and find that the hydroxide ion concentration is 2.1* 105th. So, two different ways to get the same answer. Now, as we continue talking about uh acids and bases, you have to know what a strong acid is and a strong base by looking at it. You're expected to know those. For AP chemistry, there are six strong acids that you have to know. And these are the six.

[01:28:49]Hydrochloric acid, hydromic, hydrootic, nitric, sulfuric, and perchloric. And the concentration of the acid is equal to the hydrronium ion concentration. So for example, if we have a solution of 0.85 molar nitric acid, the way we find the pH of that is just take the negative log of 0.85. Since strong acids are 100% dissociated, we can assume that the concentration of hydrronium is the same as the concentration of the acid. So the negative log of 085 is about 0.07. So that's the pH of that particular acid solution. Now there are strong bases as well. Your group one and two hydroxides are your strong bases. Now some of those are more common than others. Lithium hydroxide, sodium hydroxide, and potassium hydroxide are probably your most common strong bases. Every now and then they may include calcium hydroxide, strronium hydroxide and berium hydroxide as well. Those are also fairly common.

[01:29:58]For the strong bases, the concentration of the compound is equal to the hydroxide ion concentration. Now, if it's a group two hydroxide, you do have to multiply that by two since there are two hydroxides for each of your u of your components there. So if we have 050 molar sodium hydroxide as an example, if we take negative log of 050, that's going to be our PO. It's a base, so it's p this time. So that's 30. And so if the PO is30, we have to subtract that from 14 to find that the pH is 13.70.

[01:30:40]Now what if it's one of those group two hydroxides like 0.021 molar strontium hydroxide? Well, in that case we have to multiply the concentration by two to find the hydroxide concentration because there's a 2 to1 ratio there. So uh we'd have to take negative log of 042 in order to get the p. So in that case the p would be 1.38 and we subtract from 14 to find the pH.

[01:31:10]That's going to be 12.62.

[01:31:12]So this is how you find the pH of a strong acid and the p and pH as well of a strong base. Now what if it's not a strong acid or a strong base? What if it's a weak acid or a weak base? Well, remember weak acids and weak bases do not dissociate completely. They are partially reacted. And that's why we have to work those problems as equilibrium problems. And so let's say we have an example where we're asked to determine the pH of 95 molar acetic acid. And we have the Ka here of 1.8* 10 - 5th. So I'm going to write the equation for the dissociation of this acid. We have the H C2H302 plus water. And don't forget that since acetic acid is an acid, it's going to donate NH+ to the base, which is water in this case. So the products are going to be hydrronium because it's, you know, water has received that H+. And then after acetic acid has donated the H+, what's left is C2 H3O2 negative. So that's our equation for the dissociation of this acid. Since it is in equilibrium, we're going to draw an ice box here. Initial change in equilibrium. And I'll start plugging in some numbers. Since the concentration of the acid is 0.95 molar, that's the initial concentration of the acetic acid. Water is a pure liquid, so it's not even going to be a part of our discussion here. We're going to leave that column blank. And we essentially don't have any of our products here. So I'm going to fill in zero molar for both of the products. Now this is all we're given. So for the change on acetic acid, it's going to be minus x. And for hydrronium and acetate, it's going to be plus x. So our equilibrium values are n5 minus x x and x. So we're going to take those values and we're going to plug them into the equilibrium constant expression. So we write that as K a equals hydrronium ion concentration times the acetate ion concentration all over the acetic acid concentration.

[01:33:27]Don't forget water is a pure liquid. So it does not factor into this equilibrium constant expression at all. We're going to plug and chug. The Ka is 1.8* 105th.

[01:33:40]Our hydrronium and acetate are both X.

[01:33:43]Those are right out of the ice box there. And our acetic acid concentration at equilibrium is.95 minus x. Now, as you look at this equation, you might see that this looks like it's setting us up for maybe a quadratic equation or something like that. Well, let's try to avoid that. We have a very small value for K a. So, it is safe to ignore that minus x right there, that 5% rule as we sometimes call it. And now we can cross multiply and take the square root. So we do that and we find that X is equal to 4.1 * 10 -3rd molar. So if that's the concentration of hydrronium then the way we find the pH is we take the negative log of that. So the negative log of 4.1 * 10 -3rd is 2.38.

[01:34:37]So that's how we find the pH of this solution here. Now, sometimes you'll be asked for the percent dissociation. The percent dissociation for an acid is just the value of X that you got in your algebra equation there divided by the initial concentration of the acid times 100 of course to turn that to a percent.

[01:35:00]So the X was 4.1 * 10 -3rd and we divide that by the initial concentration of the acid 0.95 and we find that that is about 0.43%.

[01:35:13]So that's a fairly small amount a small percentage of our acetic acid molecules that actually underwent the dissociation process. And just so you know, on the AP chemistry exam, you're never going to be asked to use the quadratic equation.

[01:35:30]Okay? On these equilibrium problems, they're going to be able to work out mathematically or you'll be able to use the 5% rule to solve them. You're never going to have to use the quadratic equation. So, you don't have to worry about that on the AP exam. Sometimes they may ask you about the acidity or the basicity of a salt or an ionic compound. So, how do you determine if a salt is acidic or basic like for example aluminum chloride? Well, there's a little trick that you can do to figure that out. The way that you determine if a salt is an acid or a base is you add it to water. Now, I'm going to write water as HOH just to make the double replacement reaction products a little bit easier here. and we're going to see what the products are. So, aluminum chloride added to water in theory would give us aluminum hydroxide and HCl. Now, just so you know, this reaction does not actually take place. This is just a trick to help us see if the aluminum chloride is acidic or basic. Now, that aluminum hydroxide that is a theoretical product here is a weak base. It's not a group one or two hydroxide. So, it must be a weak base. HCl, however, is on our list of the big six strong acids, isn't it? So, it's a strong acid. So, we have a strong acid with a weak base. So, who's going to win? Well, it's the acid that wins. And so, aluminum chloride is acidic. That's how you do this. Now, how about this? NBr, sodium broomemide. Is that acidic or basic? Well, let's check it out. We're going to theoretically react this with water. HOH. And the two products when you, you know, react those, we have NaOH and HBr. We're just kind of acting like this is a double replacement reaction like you might have learned in first year chemistry. So sodium hydroxide is a strong base. That's on our list of, you know, strong bases, group one hydroxides. And then HBr is a strong acid. That's on our list of the big six, isn't it? So a strong acid, strong base, they cancel each other out and so it's neutral. So sodium bromide is going to be a neutral solution when it's dissolved in water. How about this one?

[01:37:49]CF2.

[01:37:51]Well, same thing. We use that little trick here where we add it to water and our theoretical products would be calcium hydroxide and HF once you, you know, pair those up. So calcium hydroxide is a strong base, isn't it? A group two hydroxide is a strong base. HF on the other hand is a weak acid, isn't it? This is not on our list of the big six strong acids. So HF is a weak acid.

[01:38:18]So which one wins this time? Strong base, weak acid. Well, the base wins, doesn't it? So that means this ionic compound is slightly basic. So, that's a little trick that you can use to determine if a salt or an ionic compound is going to be acidic or basic when it's dissolved into water. Now, let's move on to another very important part of unit 8, which involves acidbased titrations.

[01:38:43]Now, normally in an acidbased titration, our goal is to determine the concentration of an acid or of a base.

[01:38:49]Uh, and so let's say we have a titration and there's a titration curve that goes along with this. And here's the question. And it says if 10 milliliters of the weak acid was used in this titration, determine the acid's concentration if the NaOH was.110 molar. So for this type of problem, I would strongly recommend using the titration equation. M A VA equals MB VB.

[01:39:17]The marity of the acid times the volume of the acid equals the marity of the base times the volume of the base. that we're just going to plug and chug here.

[01:39:26]The marity of the acid is what we're trying to solve for. It says, determine the acid's concentration. So, MA is our unknown. Now, VA is actually given to us in the problem. It says that 10 milliliters of the weak acid was used.

[01:39:41]So, that's our VA. Now, MB is the marity of the base. It tells us that the marity of the base is.1 molar. So, that gets plugged in for M subb. and then the volume of the base.

[01:39:55]Well, the problem doesn't specifically say that, but we can determine that by looking at the titration curve. So, looking at the titration curve, we see that the inflection point is denoted by the letter Y here on this particular titration curve. And if we go down to the bottom of that X axis, we can see that it required about 28 milliliters to get to that point. So we're going to use 28 ml as the volume of the base in this case here. So now we can use algebra and solve for m sub a. We find that m sub a is about 31 molar of that acid. So that's the first part of this problem.

[01:40:40]Now let's go a little bit further here and let's estimate the Ka of the weak acid that was used in this titration.

[01:40:49]Now the way we do that is we focus on the half equivalence point. This is a very important point in the titration and on the AP exam they like to ask about this. So be very aware that at the halfway point or the half equivalence point of the titration pH equals the pKa. So if we reach the equivalence point at 28 milliliters that means that the half equivalence point is half of that or 14 milliliters. So if we look at 14 milliliters here on this uh graph, it seems like that's denoted by point W.

[01:41:26]And if we look at the pH at point W, it's about 4.5.

[01:41:31]So that means that the pKa is 4.5. If we're asked to estimate the Ka, well that's the negative antilogue of 4.5. So in other words, we have to take 10 to the 4.5 power. And when we do that, we find that the Ka is about 3 * 10 -5th.

[01:41:52]And so you can use the titration curve to estimate the pKa and the Ka of the weak acid used in the titration. Now let's take this a step further and let's look at the same titration curve and we're going to estimate the primary reaction component other than water of course present in the titration flask at each of the following points. So I want you to notice that at point W that is the half equivalence point. So at the half equivalence point the concentration of the weak acid is exactly the same as the concentration of the conjugate base.

[01:42:30]And so it's important to realize that they're equal. Okay? That's why pH equals pKa. That has to do with the Henderson Hasselbach equation that we're not going to cover in this video, but it is a part of unit 8 as well. Now, how about point X? Notice that at point X, we have a point in this titration where it's more basic than the halfway point.

[01:42:52]Now, if it's more basic than the halfway point, the pH is higher. So, the pH is more basic. That means that the conjugate base is going to predominate.

[01:43:02]So, we would have more A minus than we'd have HA. So, conjugate base is going to predominate. How about at point Y? Well, at point Y, that is the exact equivalence point of this titration. So, all of your HA is gone. It's all been reacted. So, all you have left is the A negative. So, your conjugate base is the primary component at this particular point and only this particular point actually in the titration because your HA is is all gone. It's it isn't even there anymore. Now, point Z. At point Z, notice that we have overshot the titration. All of our HA is gone. U, in fact, even your A minus, even though some of it's still there, you're continuing to add hydroxide. And so, at that point, the strong base predominates, and the primary uh component here is going to be the O minus. And so, your hydroxide ion is going to be in excess. Now, it doesn't ask this particularly on here, but what if the question were point T? What's going to predominate at point T? Well, notice that at point T, right here, we're at a point that is more acidic, you know, the pH is less. So, it's more acidic than the halfway point. So, if it's more acidic than the halfway point, then the acid is going to predominate.

[01:44:28]And so, at point T, we'd have primarily HA, we'd have primarily the acid. We'd have a little bit of a minus though as well. So that's how to tell what ion or what component is going to predominate at different points of the titration.

[01:44:46]Now let's move into our final unit of the course which is unit nine. And we're going to take a look at thermodynamics, specifically the second law of thermodynamics. And we're going to talk about entropy here for a bit. Entropy can be described in a couple different ways. It can be described as a measure of the total number of possible energy states or microates in a material or a measure of how well dispersed a material is. These are good ways to talk about entropy. Sometimes it helps us to uh to think about it in terms of how much disorder there is as well. For example, if we have a solid, we know that a solid uh is not dispersed much at all. The molecules are in a perfect crystallin structure. There's a lot of order there.

[01:45:31]Very little entropy compared to other states. In a liquid, on the other hand, the molecules are able to move around more. They're more possible energy states, more microates. It's a little bit more dispersed than a solid is. So, a liquid has more entropy than a solid will. Most of the time in an aquous solution, usually we have even more entropy because we have the liquid solvent, but we also have little particles of the solute swimming around in there as well. So normally aquous solutions will be more dispersed than a pure liquid. So they'll normally have even more entropy. A gas on the other hand, well you can see that the particles are really far apart in a gas.

[01:46:12]It's very dispersed, much more so than a solid or a liquid. So gases will normally have the most entropy of these different states. So we need to remember as well that if there's a tie, it comes down to temperature. materials at higher temperatures are going to have more entropy than those at lower temperatures. So, if you have two samples of water and they're both liquid water, well, if one sample has a temperature of 90° C and one has a temperature of 10° C, generally speaking, the one at the higher temperature will have more entropy, more motion, uh more microates and uh you essentially more dispersion. Now, if the temperature is the same and the state is the same, we break that tie by looking at how many molecules we have. A container with more molecules will have more entropy than those with fewer molecules. So, five molecules would have more entropy than two molecules if we're at the same state and the same temperature. So, we can use this information here to predict the signs for delta S for different processes.

[01:47:23]Like for example here we have water liquid changing to water gas. Well liquid has a lower entropy than a gas.

[01:47:32]So we're increasing in entropy. So our delta s should be positive since entropy is going up. What about this one here?

[01:47:40]In this reaction we have uh a mixture of a solid and a gas being converted to all solid. Well a mixture of a solid and a gas is you know fairly distinct there.

[01:47:52]But if it's converted to all solid, we have a drop in entropy. So that delta S sign would be negative. We're going to have less dispersion at the end of the process than we would at the beginning of the process. How about this one? In this reaction, notice that we have all gas molecules. So looking at the state of matter is not going to help us. We're going to have to break that tie by looking at how many molecules we have.

[01:48:19]So in this case we have five gas molecules on the left side 2 + 3 and on the product side we have six 2+ 4. So if it goes from five to six that's an increase in number of particles. So it's an increase in dispersion increase in entropy. So it is a positive delta s.

[01:48:40]How about this last example? N2 gas plus O2 gas yields 2 N O gas. They're all gas molecules, so that doesn't help us. We have to look at the number of particles.

[01:48:52]On the reactant side, we have two. On the product side, we also have two. So, guess what? We really can't tell. We don't know exactly what the sign for delta S is going to be. We can probably predict, however, that the delta S is going to be relatively close to zero for a reaction like this. Now as we talk about entropy and enthalpy, we can say that enthalpy and entropy are the two driving forces for chemical reactions.

[01:49:22]And as we ask, will a reaction take place? Well, that has to do with thermodynamic favorability or as some textbooks call it spontaneity or how spontaneous a reaction is. Now, like I said, the two forces that determine whether or not a reaction will be thermodynamically favored are enthalpy, that's our delta H, and our entropy, our delta S. And generally speaking, the universe likes exothermic reactions. If a reaction is exothermic, that will drive the reaction to take place. Also, the universe likes it when entropy increases. And so if you have a reaction that has a positive value for delta S, that's going to drive that reaction toward favorability. Now, if you ever have a reaction where both of those are the case, it's both exothermic and it increases in entropy. Well, the universe likes both of those things. So that reaction or that process is going to be thermodynamically favored at all temperatures. Now, if it's the opposite, if you have a reaction that's endothermic and it has decreasing entropy, well, the universe doesn't like either of those things. It doesn't like endothermic processes. It doesn't like it when entropy decreases. So, that process would never be favored at any temperature. So, we have those cases.

[01:50:46]But what if only one of those factors is favorable? Well, then it's only thermodynamically favored at certain temperatures. If they're both positive, delta H is positive, delta S is positive, it's going to be favored at high, relatively high temperatures. If they're both negative, delta H is negative, delta S is negative, it'll be favored at relatively low temperatures. Now, how do we measure thermodynamic favorability? Well, there are several ways. One way is the delta G or what we sometimes call the Gibbs free energy is equal to the delta H of a process minus the temperature in Kelvin times the delta S. This is just a plugandchug equation. We have those four factors in there. If you know any of the three, you can solve for the fourth one.

[01:51:37]We could calculate delta G using this equation. Delta G equals R time the temperature in Kelvin times the natural log of the equilibrium constant K. Now whenever you use this equation you have to be careful because the R is in units of jewels. A very common mistake using this equation is students will try to use 0.08206 for R. Don't use that here. This is in jewels. So the value for R when it's in jewels is 8.314.

[01:52:10]temperature of course is is in Kelvin and natural log of that equilibrium constant and you can solve for delta G that way and whatever you do however you solve for delta G and there are several ways to do that as you can see here the delta G is going to be either positive or negative if the delta G is positive the reaction is not going to be thermodynamically favored and if delta G is a negative number then the reaction is going to be thermodynamic ically favored. And so you can, you know, plug into these equations here and depending upon what you're given, you can solve for delta G and determine if it's going to be favored or not. Now, as we move into the last part of unit 9, this is about electrochemistry.

[01:52:56]And most of the time we're working with galvanic cells in electrochemistry. A galvanic cell is for all practical purposes a battery. This is a thermodynamically favored process where a redux reaction is taking place.

[01:53:10]There's a reduction on one side of the cell. There's an oxidation on the other side of the cell. And let's say we have this particular cell right here. And we have aluminum on one side, we have zinc on the other. And I want you to notice that there is a wire that is traveling between those two uh electrodes. And there is a light bulb or some other kind of load or a voltmeter or something here that is connecting them. We also have a salt bridge here in the middle. And the purpose of that salt bridge is to allow ions to move back and forth somewhat freely to balance our charge here. So generally speaking, each half reaction is going to be given to you on the exam or in some kind of a data table as a reduction. That's just how it's done.

[01:54:00]And so here we have the two half reactions and notice that every half reaction has a voltage associated with it. So the way we uh analyze this is we first of all have to determine what's the cathode and what's the anode. Well, we're going to use this equation right here. the EC cell or the total voltage of the cell equals the E of the cathode the reduction potential for the cathode minus the E of the anode the reduction potential for the anode. Now we have to subtract these two values in such a way that we get a positive value each cell the voltage for any galvanic cell has to be positive and so we're going to subtract these two in such a way that we get a positive number. Now the only way to do that is to put the.76 volts in the position of the cathode and the -1.66 volts in the position of the anode. And when we do that we can see that the overall voltage here is.90 volts. And so that's the voltage of this galvanic cell. Now just so you know they're not going to give you this equation on the exam. Okay? You have to know that. You have to memorize it. EC cell equals E cathode minus E anode. You should learn that. That way you can figure out what the EC cell is. Now, whichever one is in the first position, in this case, the.76 volts is the cathode. So that means that zinc is the cathode and whichever one is in the second position represents the anode. So the negative 1.66 or the aluminum is the anode. At the cathode, reduction takes place. At the anode, oxidation takes place. Okay, that's always how it works.

[01:55:52]Reduction takes place at the cathode.

[01:55:55]Oxidation takes place at the anode in every galvanic cell. If you want to use the pneummonic aid red cat and an ox, that may help you to keep that straight as well. Now, as we look at how the electrons flow through the wire, they always go from anode to cathode. So this is the direction in which electrons will flow. They always flow through the wire in a galvanic cell from anode to cathode.

[01:56:21]AC A to C. AC alphabetical order or think of AC like that. It always goes from anode to cathode. And if we want to take this equation and actually write the overall balanced equation, well, we're going to take the half reaction for the cathode and write it as it is.

[01:56:40]So that's the zinc half reaction. The one that represents the anode, we're going to have to flip that because one of these is the oxidation and aluminum was the oxidation. So we have to flip it. Flip an ode. Flip the one with the anode. So aluminum is going to be written as Al yields Al3+ 3 electrons. Now it's time to add these two half reactions together. But notice that if we try to do that, the electrons will not cancel out. We have to make these half reactions add up so that the electrons cancel each other out. So we're going to multiply the top half reaction by three and the second half reaction by two. And so now we have six electrons that will cancel out because we can't have electrons in our overall balanced equation. So our overall balanced redux reaction here is 3 ZN2+ plus 2 Al yields 3 ZN plus 2 Al3+.

[01:57:40]And remember we can say that whenever we have two metallic electrodes in a galvanic cell like this, the mass of the cathode is going to increase. The cat gets fat. If it helps you to remember that the zinc cathode increases in mass, the anode, in this case, the aluminum is going to decrease in mass. Now, if you forget that, you can just look at the ballast equation. Notice that one of the products is zinc. And so, you'd expect that the product is going to increase.

[01:58:11]We're going to have more zinc produced as the reaction proceeds. And aluminum is a reactant, Al, so it's going to get used up. the aluminum anode gets corroded away or slowly eaten away as the galvanic cell proceeds. And so this is how we can analyze galvanic cells in problems like these.

[01:58:32]Well, we've made it to the end of this very long course and this long video and I hope I've reviewed some things that will help you boost your score and get you ready to slay that AP Chemistry exam. I know this was a long video, so if you've made it this far, please make sure you hit that like button and leave a comment down below, too. And let me know if this video was helpful to you.

[01:58:55]And don't forget that it's not too late to get the AP Chemistry Ultimate Review Packet to get all the resources you need to succeed on the big exam coming up.

[01:59:04]Hey, thanks for sticking it out to the end. Now, go out there and show the College Board and show the AP exam who's boss. Keep up the good work. You got this. And I'll see you next time.