Lecture 17. Final Exam Review
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17
00:00
ChemistryStreptomycinPressureHuman body temperatureFireHuman body temperatureOctane ratingDeformation (mechanics)Electronic cigarettePressureConcentrateSingulettzustandReaction rate constantChemical reactionAmount of substanceCryogenicsGleichgewichtskonstanteWursthülleSodiumCollectingClaus processProcess (computing)MedroxyprogesteroneAlkoholische GärungPilot experimentEconomic sectorComputer animationLecture/Conference
07:44
PressureHuman body temperatureStreptomycinChemical reactionInsulin shock therapyAqueous solutionPhase (waves)Activity (UML)FermiumElfSingulettzustandWursthülleStratotypEthanolChemical reactionAmount of substanceConcentratePeriodateStickstoffatomToll-like receptorPressureHydrogenEssenz <Lebensmittel>Activity (UML)WalkingFireMolar volumeDrainage divideGasKetaminPhosphorusComputer animationLecture/Conference
15:28
PressureActivity (UML)Phase (waves)Aqueous solutionVitalismChemical reactionChemistryCaliforniumRiver deltaChemical reactionEndotherme ReaktionStuffingWaterSetzen <Verfahrenstechnik>Memory-EffektLactitolVerdampfungswärmeConcentrateCalculus (medicine)WursthülleGenePhysical chemistryIceBreed standardBiomolecular structureSilverCoast ProvinceMaskierung <Chemie>River sourceComputer animation
23:25
Breed standardGibbs free energyChemical reactionAageFireStreptomycinLymphangioleiomyomatosisPan (magazine)ScarPressureMolar volumeConjugated linoleic acidAdder stoneTanningSunscreenGesundheitsstörungSpontaneous combustionSilver iodideBET theorySolutionHamMarlMetabolic pathwayFermiumInsulin shock therapyInternational Nonproprietary NameWine tasting descriptorsLactoseImmunohistochemistryThreonineLamb and muttonTinLithiumdeuteridInflammationTin foilFixed action patternColumbia RecordsStress (mechanics)Human body temperatureZeitverschiebungRecreationMaltVancomycin11 (number)Angular milHydrogen chlorideMixtureGene expressionTetrahydrothiopheneTetrahydrocannabinolInfant mortalityMeatSoilChlorbenzolIodineIodideHydrogenDecompositionVitalismAltbierDipol <1,3->CurryAtomic layer depositionRiver deltaGemstoneElectronegativityAgeingChemical reactionCadmium sulfideWaterMemory-EffektSunscreenOctane ratingChemical reactorWine tasting descriptorsGoldRearrangement reactionConstitutive equationSandSense DistrictSetzen <Verfahrenstechnik>Calculus (medicine)GesundheitsstörungKatalaseBreed standardStorage tankWursthülleHuman body temperatureHalogenationReactivity (chemistry)ZunderbeständigkeitScreening (medicine)Meat analogueGolgi apparatusMineralPermacultureElectron donorHydrogenPressureRadioactive decayConcentrateMixtureTiermodellDiamondCoalComposite materialActivation energyTumorImperial Chemical Industries PLCVerdampfungswärmePotenz <Homöopathie>Metabolic pathwaySystemic therapyActive sitePeriodateGibbs free energyChlorineOptical coherence tomographyStuffingCholesterolPaste (rheology)Hope, ArkansasWalkingMolar volumeSea levelSteelElectronic cigaretteData conversionDeathGasDrop (liquid)LactitolAmount of substanceHydrogen chlorideGleichgewichtskonstanteBearing (mechanical)Pascal (unit)Man pageIsland
Transcript: English(auto-generated)
00:08
Okay, so this is it. You know, last night I was thinking that as you get older, time
00:20
just goes fast. I mean, these lectures seem to just drag on and on. I don't know how you feel. But the time, I mean, I can't believe it's been ten weeks now. Anyway. So this is it. So we will, after this, we'll see each other on Thursday morning at 8 o'clock,
00:43
right here. Of course, the people that are in this room right now are the ones that usually show up. The empty seats are the ones that don't always show up. My suggestion folks is that you show up a little early, okay? The final will be about twice as long, okay,
01:07
or should take you about twice as long to do. And in the past, we've had, you know, about 70 minutes for the exam, so maybe it won't be quite twice as long. But I don't want to have you people stringing in here until 810, 815 in the morning. I want to
01:23
pass the exam out at eight. So that doesn't mean you show up at eight. That means you should be here, be in your seats and ready to take the exam and have us passing it out at eight. Otherwise, it's a problem, okay? All right. Well, folks, we're on the home
01:42
stretch here. Just a little bit about Le Chatelier. I've had all kinds of questions. I even had a question last night. This one person tends to be a little bit on the aggressive side. But he came up to me and he said, so do you anticipate that the Thursday
02:04
final exam will be pretty much like the Friday final exam? And I said, well, I said, you know, I learned my lesson on the first exam. I need to make them, you know, different, but I'm not that creative. I can't make it so that there's, I mean, there's going to be
02:22
a Clausius-Clapeyron equation or question on there. And whether you got to take something to the E or you just plug and chug or you solve for T, there's going to be a Clausius-Clapeyron on the night exam and one on the morning exam. So he said, well, would it be possible then for you to get, to e-mail me the exam, the Thursday exam, so I can practice?
02:45
Dang, I tell you. I mean, I'm just shocked that, I mean, he's not even going to try and cheat, you know? He's not going to even ask you. Anyway, so anyway, we will be grading your exams
03:03
on Thursday. We will hopefully post, I'll do what I did before, send out an e-mail that has your scores on the, that evening sometime on Thursday night. We will not, though, post the key
03:23
because of the next exam, okay? So the key will be posted sometime probably Saturday because, once again, we won't start grading the, the other exam goes from 7 o'clock at night until 9 o'clock at night on Friday night. I can't believe that. Anyway, okay. So I have tried in
03:49
this 10 weeks, sometimes I point out this is a good question, okay? This would be a good exam question. I'm hoping, folks, that not only have you learned something about maybe how to predict
04:07
a possible question from my class, but that it will work for you later, all right? Here's an example, okay? I want you to think of how could Don ask a multiple choice question
04:23
from this slide, okay? So Le Chatelier, I love Le Chatelier, okay? So if you change the concentration, can you shift equilibrium and can you change the equilibrium constant, okay? So for this one, if you change the concentration of something, I don't care,
04:46
products or reactants, does that shift equilibrium? Does it change the rate constant? Does it change the rate constant? No. The rate constant, we are trying to maintain that at all
05:03
costs, okay? So we expect it to stay the same. If I say the rate constant is .1, then it's .1 no matter what you dump into it, okay? And so if you pull something out, you dump more in, it re-equilibrates so that you're at .1, all right? So that's
05:21
called a rate constant for a reason, okay? And so you do not change the rate constant, all right, so the equilibrium constant by changing the number of moles or the concentration. How about pressure? I pushed the button too soon, okay? Does it change the equilibrium
05:45
constant? Okay. Volume. And volume and pressure, I'm a little bit confused at this, I mean because I see it as if you double the volume, then you've cut the pressure in half,
06:02
okay? Or if you cut the volume in half, you've increased the pressure, so those to me should be the same, okay? So I didn't even ask you, okay? How about temperature? Is it shift equilibrium? Sure. We have a reaction where we have A plus B goes to C plus D,
06:20
and if there's some heat given off, if we add heat to one side, it makes things happen, okay? Or it makes things slow down. So absolutely, changing the temperature can shift the equilibrium. Does it change the equilibrium constant? Yes, it does. That's the one thing, folks. That is the one thing that changes equilibrium
06:41
constants, is the temperature. So I have to specifically say to you, you know, at this temperature, and I think what you'll find is that in all the examples that we've done so far, it'll say at, I mean some wacky temperature, maybe 1280 degrees, but it's at a certain temperature, the K changes. Now, we know that. We've seen plots, okay? We've seen
07:03
figures where it was like on the Y axis, it was K, and on this axis was temperature, okay? And if you have something, something that's exothermic, gives off energy, as you warm up things, the K at low temperature, once again, this is low temperature here and
07:22
high temperature there, low K, high K, starts here, and it kind of does this, okay? Because as you warm things up, more heat as a product, you've already got heat, so it slows the K, so temperature absolutely can affect K. The catalyst, we're going to talk about catalyst at the end of the lecture,
07:43
it does not shift the equilibrium, okay? And it does not change the equilibrium constant, okay? So, this is something that we will get into in, I don't know, an hour or so. So, folks, this is a little bit of a hint, there are at least 80 people missing, okay? I can see
08:06
if you want to say to turn off your cell phones because there are so many empty seats, okay? So, I want you to look at this. Look at this right here. How many, when I ask a multiple choice question, how many options do I usually give you?
08:25
I like five, okay? Let's say five. So, if I were to ask a question on multiple choice for this, what can I ask? Could I say, when I change
08:41
X equilibrium changes, and I would say concentration, boom, boom, boom, boom, and all of a sudden, which one does not change? Can you see this is a nice question? There's one answer that's different from the other four. How about over here? So, I'm telling you right now that the chances of these two being on the exam,
09:01
at least one of them, is about 90%, okay? It will just say, I don't know how I'd say it, but it will be on there. All right, I want you to do this for me right now. Calculate K for the reaction of ethanol liquid,
09:23
I didn't know how to make an equilibrium sign, I couldn't do that, so this is supposed to, I can even copy and paste, sorry. This is supposed to be two arrows going different ways. Goes to ethanol gas, and the concentration, or the pressure of this gas is .33333 atmospheres.
09:44
I want to know what is K? You've got a liquid on one side, and a gas on the other.
10:01
Where's that guy that always knows the answers? What's the answer, chief? Right there, he says. That's the K, right? K is equal to, you know, products, the pressure of the products over the pressure of the reactants, but this, okay, this has a value of
10:24
not of one, okay? So it's the pressure here over the reactants is equal to K. So in this case, K is equal to .333. No units are necessary, okay? Everybody understand that? Okay, now I want you to calculate the K for this reaction,
10:48
ethanol liquid, ethanol gas, to 253 torr. Who's got an answer? What is it,
11:13
he says it's .333. What is with that? Why isn't it 253? So do me this, for those of
11:30
you who are a little bit confused, there is 760 torr in an atmosphere. So divide that number 253 by 760, okay? Divide that number that I gave you by 760.
11:49
If I've done the math correctly, you get .333. So the problem, folks, is that while I told you at one point that units didn't matter and they would cancel, if you only have,
12:07
if you have an odd number of units, it doesn't work that way, okay? If I've got, you know, H2 on top, certain pressure, and N2 on the bottom, then the units cancel. If I've got, I don't care, five torr, five torr of hydrogen is the product,
12:26
and two torr of nitrogen was the reactants, it's five over two, and guess what? The K is two and a half, okay? And if I converted all that to atmospheres, I'd still get two and a half. The problem is, is that when you just have
12:42
just one thing, okay, there's nothing to divide it by, those units don't cancel, and then the units do matter, okay? So the units do matter, so you can't have, this is the same reaction, you can't have two different K's with the same thing, this is .333 torr, and there's pascals, and there's all kinds of things that you can use,
13:07
okay? So we have to have some normalization, and that is this, okay? I suggest that you either commit this to memory or try and understand the concept.
13:22
For aqueous, it's one molar, okay? One molar are the parameters to use here, this is the reference state. The activity here is the number of moles over one molar, or sorry, the molarity of whatever it is divided by one molar, okay?
13:46
For aqueous, it's the atmosphere partial pressure, the partial pressure is in atmospheres. Now folks, an atmosphere is the same as a bar, okay? But I don't use bars, and I don't use pascals,
14:01
okay? I use atmospheres and I use torr. So it's very possible you'll get a question that says something in torr, and it's something that's, or you're going to get in atmospheres, okay? If I give it to you in atmospheres, then you're good to go. So once again, when you set up an equation for K, you must use atmospheres unless you have the
14:27
same number of moles on top and bottom, okay? And what I mean, folks, is that if you've got A plus B goes to C plus D, and A and B is equal to C and D in terms of the coefficient,
14:41
then everything cancels out. But gases are this. For liquids and solids, folks, it's a one. Okay? So on that last thing, we could say products would be in the partial pressure. We had a .333, okay? The liquid, so it's going to be K is equal to .333
15:06
over, and we had a liquid, over one. If you want to keep track of all your P's and Q's there, that's what you would do. Okay? So it's very important for you to understand that there are times you can use torr, and there are times you cannot use torr.
15:22
There are times you can use millimolar, and there's times you can't use millimolar. All right. So we're going to talk more about delta G today. So that other stuff was more K equilibrium and all. Delta G is equal to, the overall delta G is equal to the delta G of the
15:43
products minus the delta G of reactants. Okay? For everything except for one thing, it's been products minus reactants for 10 weeks. Okay? And that's pretty much how it's going to stay. Products over reactants. So the reactants, this is an energy diagram, folks. If this is energy this way, I'm not sure what's this way, maybe time.
16:04
The reactants have an energy here. If they actually do this, and the products are lower in energy, what that means, folks, is the products are lower in energy, but right here is a big plus heat. Okay? If this is actually lower than this, then it's this, the energy
16:24
had to go somewhere. Okay? The energy is going to be products plus heat. All right? So an exothermic reaction, it looks like this. Endothermic would be from here to here. Okay? You have to give these guys some heat to get it to be up there.
16:41
Certainly not spontaneous. And we know that if delta G is less than zero, okay, if you do a calculation delta is less than zero, that Q is less than K, and it's going to head that direction. All right? If delta G is equal to zero, we're at equilibrium. Okay?
17:00
We're at equilibrium, and Q equals K. And if delta G is positive, it is not spontaneous, and Q is actually bigger than K. So you've got to go from the reactant side, the product side, to the reactant side. So we're going to go into this a lot, but I would suggest
17:23
you just write it down. The delta G is equal to delta G naught plus RT log Q. So that Q has come back. Okay? The Q that I made it sound like early on was not all that important. Turns out to have reared its ugly head again. So it's a good thing that I didn't trivialize it too much earlier. So delta G is equal to delta G naught plus
17:45
RT log Q. We're going to be going over those, several of those, so don't panic with this. And at equilibrium, when delta G is equal to zero, then delta G naught plus RT log K is equal to zero. Okay, do this. Think about it. What are delta G and K at
18:22
equilibrium at 25 degrees C for this? Well, last night I said, well, it's obviously it's this. This. But it's wrong. Okay. I went through this big explanation of if I
18:40
had just water sitting here, is it boiling? Look at this. Does that look like it's boiling to you? That looks like boiling water to me. So I said, oh, well, it's obviously at 25 degrees it isn't boiling. Okay? And therefore, water would rather be liquid than gas at room temperature so that obviously delta G must be equal to greater than zero. Turns out the key word right here is equilibrium, folks. Equilibrium
19:05
delta G equals zero. Do not forget that. So don't read these problems on the exam so fast like I did last night that you screw up. Okay? That's the key. If something is said at equilibrium, folks, it could cut out a whole lot of work that you need
19:21
to do. Look at that. Okay. So we did a, we've done several of these, folks, where it is one reaction that leads to another reaction that leads to another reaction.
19:41
So it's important for you to understand, we did this, I don't know, I can't even remember what chapter enthalpy was. Chapter six, maybe? Chapter seven? I don't know. It doesn't matter. So I would like you to do this. Just humor me. Okay? What is K sub 1 in this, the way it's written right here? A going to B.
20:04
Okay? What is K sub 1? I hope you're going to say that it's going to be the concentration of B over A is equal to K sub 1. Write that down. Please, just write it down. I have a reason for asking you to do this. So here we have A
20:25
going to B is K plus 1. So now I want you to write down what is K sub 2?
20:42
So K sub 2 is B goes to C, and that is K sub 2, and that should be concentration of, K sub 2 is equal to the concentration of C over the concentration of B. And how about this one? From these first two folks,
21:23
you should be able to figure out what this is. Okay? So if you, if you So now just write K1 is equal to B over A, and K2 is equal to C over B. Okay? Multiply those two together. I don't mean K1, K2. I mean what they
21:45
were equal to. Okay? So K1 is equal to B over A, okay? And K2 is equal to C over B. So you know this. You wrote that, all of you probably
22:03
wrote that K sub 3 is equal to C over A, correct? But what I'm saying is that you can show yourself that that's the case, because to get there, from A to C, you knew that B over A was equal to K1, and C over B was equal to K2. So look at this.
22:20
For this first one, K1, B over A. B over A. That's K1, K2 is C over B. That's K2. And then look at these guys cancel. They cancel. So you're just left with C over A. All right? So you can sort of work that way. So in this case, to find K sub 3,
22:44
you just multiply K1 times K2 is equal to K3. For the enthalpies of those reactions, you add them. Okay? So H sub 3 is equal to H1, delta H1 plus delta H2. So folks, this encompasses quite a few
23:04
things. Okay? It encompasses the equilibrium part. This part is to sort of refresh your memory on what we did in chapter 6 or 7, whatever it was. So don't forget how to do this type of thing. This is like a little bit of, not like a Hess's law, but it's a, I love Hess's law. So once again, it's just
23:22
sort of playing with these things. All right. So delta G naught of reaction is equal to the delta H naught of reaction
23:40
minus T delta S naught of reaction. Okay? This is equal to, once you dislike the previous, delta G formation minus of products minus reactants. And I wish I had been able to write something up here, but this, if you remember, folks,
24:02
that delta G of reaction is equal to minus RT log K. Okay? And all this is is a rearrangement of that. Okay? In fact, let's do this right now. Write this exact thing. Okay? And now we're going to do something in reverse.
24:23
We're going to do something in reverse of what we've been doing. So you write that, and then I want you to take the log, the natural log of both sides. The natural log of both sides. That's LN. I suggest you write this down. I mean,
24:44
maybe you've got these kind of brains that allow you to visualize these things. That's wonderful. Just you can't visualize on the exam. I need to see it. So, make sure that you're good going from visualization to putting it on the paper. So, what is, if we take the log of both
25:01
sides, what do we get? What do we get? Log K is equal to delta G over RT. That's right. Okay? So what you're saying is that you just do this, you just take the log of this, makes this a log K, take the log of this,
25:21
you just end up with this. Okay? So you only have to memorize one thing, folks, okay? I tend to memorize that delta G is equal to minus RT log K. But you can play with it any way you want. So, folks, if I give you a K, and I say calculate delta G naught of reaction, that's
25:43
how you do it. Okay? So look at this right here. R is a constant. T is not a constant. Delta G naught is not a constant, and this isn't a constant. So I can, if I give you two or three things here, you can figure out different things. Okay? So once again, this is a nice way I, you know, whether you want to
26:03
memorize that or not, as long as you memorize that delta G naught of reaction is equal to minus RT log K, they both work. Okay, we know this, okay? That if delta G is less than zero, then the reaction is going
26:20
to proceed from reactants to products. If delta G is greater than zero, it's going to come back this direction. Okay? What happens if we increase the temperature in a case where delta G is less than zero? Think about it, okay? Folks, I want you to,
26:42
I've had students come to me and they say, you I can't visualize this, or I can visualize this, but I can't mathematically solve this. There are sometimes when you can do it both ways, okay, when you can visualize it and you say, okay, I can understand this, and then there are other times that you can actually prove it to yourself mathematically, all right? Some of you have
27:01
stronger math skills than others, and I just want you to think about this. If you've got something like this, delta G is negative. What usually does that mean? Not always, but it usually means that there's some heat generated, okay? Remember, heat
27:21
given off, negative delta H. Delta G is equal to delta H minus T delta S, so if you've got something that is favorable, okay, and it's spontaneous, then there are oftentimes, you have heat that's given off, okay? So as you increase the temperature for something like this, is that
27:42
going to make the reaction go faster or slower? I heard faster, but it's slower, okay? K decreases, okay? And if you remember, we had a plot of delta G versus temperature, and on something that
28:01
was exothermic, it looked like this, okay? The higher the temperature, the delta G decreased, all right? Or it was, there's another with K like this, so K was here and temperature was here and it went like this, other ones it went like this. So if you have, folks, if you have heat that is generated, you need to think of this
28:22
as adding heat right there, okay? If you add heat right there, and then you add heat, is that going to force things in this direction or this direction? It goes back this way, okay? If I add, all of you I think understand, if I add, if I double the amount of D, if I double the
28:41
amount of D, does the reaction tend, if it's at equilibrium, at equilibrium, I add more D, which way does the reaction go? Okay. So if, folks, what you don't see, because we don't usually put it in there, is heat. Heat is over here, and it really should be figured in on this equilibrium constant, okay? So if we have heat,
29:03
and now we're at equilibrium, and I add more heat, what happens? Okay? So folks, if you increase the temperature, K tends to decrease when you have a delta G that is smaller than zero, and it'll increase here. Okay? You decrease the temperature, then these things just flip. Boom,
29:22
boom. So once again, I want you to, I've showed you figures, I've showed you plots, I want you to try and understand this. This is just a conceptual thing. There's no math involved here, aside from the fact that you know that the rate, the K equilibrium for this is, you know, C's, whatever the coefficient is, concentration
29:43
of C times the concentration of D divided by the concentration of B and the concentration of A. So if you have something else here, if you add something here, it tends to come back this way. If you add something on this side, it goes that way. So, delta G is not as
30:02
simple, at least it's not simple for my brain, and I don't understand why I struggle with this, but I thought I understood delta G, and now I'm not sure I do. So, the delta G of a reaction, okay, and once again, this is
30:21
the molar, is products minus reactants. Okay? So I've written this, okay, that the value of delta G at a particular stage of the reaction, okay, once again, we've had plots where it was delta G and time, and you remember how I had
30:42
one where I said, okay, this is all the reactant side starting here, and the products, pure products there, pure reactants here, we add stuff in, and this is like the reaction, the progress of the reaction or whatever on this, and this was delta G. And we had a delta G that kind of came down like this,
31:01
and right here, it hit zero. This is where equilibrium was. Okay, so folks, on this screen right now, if this is where equilibrium is, on this scale, okay, there's equilibrium, do you think we have—are we favoring the reactant side or the product side? Reactant.
31:21
Doesn't mean we have product. We have product, right, because this is about, I don't know, a third or a fourth of the way there. So we have some products, but we have more reactants. How about now? I think we're not quite half. How about now? We're starting to favor the products more, more, more, more, more. This is
31:41
one where K, folks, is like 10,000. Okay? K is huge, and that delta G came down like this, and only right there had a minimum of zero. Okay? So equilibrium was right there. Okay? So the bigger the K, the further this way we are. This model of the K, I mean
32:04
if K is 0.00001, we're right here. That delta G went and had a minimum somewhere right here or up there. It doesn't matter where the minimum was. It's just that the minimum happened on this side. So delta G is changing during
32:20
the reaction, and that's what we need to be able to calculate. So the value of delta G at a particular stage of the reaction, somewhere in here, okay, is the difference in the molar Gibbs free energy. Molar means the molar and it's at standard pressure and concentration. Of the products and reactions at the partial
32:43
pressure or concentrations that they have at that stage. Not what you find in a book. Okay? What you find in a book is you have this and you have this, and you look it up, and you say delta G is equal to minus 147 kilojoules per mole. Okay? But
33:01
we're not at equilibrium. We're not at standard. Okay? Not everything works at one molar or one atmosphere. So what this delta G tells us is where we are at that stage, and once again, it's weighted by these coefficients. So delta G of reaction is equal to this is what you find in a
33:23
book. Okay? Reaction is this, plus RT log Q. Q, folks, when Q is equal to K, we're at equilibrium. Right? If Q is smaller than K, we go this direction. If Q is bigger
33:43
than K, we go in this direction. Okay? So Q tells us something, and so that the delta G, this is the, this is like what you really get. Okay? You remember how we talked about delta H? Okay? Delta H is the heat of reaction. The problem is that
34:00
that does not really tell you what, whether it's even spontaneous or not. Okay? Delta H of a negative value means it's exothermic. So you've got a hundred kilojoules per mole. But that didn't tell us how much we got out. We had to invoke entropy, and entropy is
34:23
very dependent on temperature. So then we came up with delta G, which is the Gibbs free energy, which is delta H minus T delta S. So now, in order to tell whether something's spontaneous, you have both the enthalpy and the entropy. Okay? So that was delta G. Now what we're finding is that this
34:41
new delta G is the one that tells us sort of where we are in this reaction. And this, the Q part, okay, I think this is like a little bit, the delta GR is a little bit like a correction factor. The further you are from standard conditions, the bigger or smaller this is. Okay? So look at this
35:00
up here. If we're at standard conditions, this is zero. Okay? As you move further away from standard conditions, this starts to become a little bit more and more important. So if you're way off of standard conditions, then this becomes significant.
35:23
Okay? We're going to do a few problems. Hopefully, when you walk out of here, you won't be any more confused than when you walked in. And Charlie put this together for me. I'm going to put this on the class website. I'm not going to go over it right now, but it, I think it's a nice little review, because I want to get right to this question.
35:41
Let's do this question. This is 10.2 from your book, calculating the Gibbs free energy of reaction from the reaction quotient. Okay? The standard Gibbs free energy of reaction for two SO2 gas plus an O2 gas going to two SO3 gases is
36:01
delta G of reaction naught is equal to 141.7 kilojoules per mole at 25.00 degrees C. What is the Gibbs free energy of reaction when the partial pressure of each gas, this, this, and this is 100 bar, which is 100 atmospheres, folks, okay? 100
36:21
atmospheres, okay? Let's answer that question first. So what is the delta G of reaction, okay? What is the Gibbs free energy of reaction when we have this? Now you're given the standard free energy. So remember, it was
36:42
delta G of reaction is equal to this plus RT log Q. It's a new reaction, so I'll give it to you. If I had just copied and pasted this, I'd have that reaction here. So don't forget, delta GR is
37:00
equal to delta GR naught, which is right there, plus RT log Q. Oh, by the way, there's a mistake
37:34
here in your book. This should say PSO2. SO2 is right there.
37:41
That should be the PSO, because that's the one that's squared. That's on the reactant side. So you've got oxygen, which has a coefficient of one. This is PO2. This should be PSO2. It's your book, not me, this time, okay? And I can't change it. Okay, so Q, folks, that
38:22
part that is sort of the correction factor is 0.01, all right? 0.01. Don't forget, it was delta GR is equal to delta GR naught, which means it's standard, plus RT log Q. Okay?
38:47
Q is that. Okay? Log of that. Anything—what is the log of 1? Everybody, take your calculators, hit enter 1 or whatever, and then hit log of
39:00
it, and ln of 1 is equal to 0. Does that mean anything bigger than 1? 1.1, 1.2, 625 gives you a positive value coming out of that log. Anything less than 1, .99, .90, 0.01 gives you a negative number. Okay? So
39:26
you've got a negative number, maybe not a very big number, but it's a negative number then times RT. Okay? So what is the answer? Okay? Well,
39:42
here's the first part. This is what you get under standard conditions, right there. Delta GR naught is right there. That was given to you. Okay? And then what they did was they multiplied R times T, and they get this. Log of this, that's a small number. Okay, I wish there was one more step in here. I think this value is something like this times
40:01
this is, I think, minus 12 kilojoules. Add them together, and you get this. How many got that? How many cheated by looking at the notes? Okay. So, folks, this is a—yes,
40:21
you cheated? Oh. Yes. Thought you were just a slow answerer. Yes? Which R? Okay, so you look at the units. Delta G is always in joules. Okay? It's always in joules. Enthalpy is always in joules. And so, the R that you
40:41
would use for this is the R that has joules in it, which is the 8.31. Okay? 8.31 joules per degree mole. So, you have that in there, you multiply by T, since you had per degree, the Ts get rid of each other, and you're just left with kilojoules per mole. So,
41:01
your answer—and it has to be—look at this. You've got kilojoules per mole here. You have to have the same units for this to give you this. Now, don't forget, that is in joules. The R is in joules. So, you've got to divide by a thousand if you're going to add to this. What you do not want to do is to say, okay, this is— that this part is 2479 joules
41:23
is what it is. You can't add 279 joules to this and come up with 2,000 and something. You've got to convert this to kilojoules. And then this is the answer. So, folks, what this says—yes.
41:44
So, I sent everybody an email this morning. This morning, and I said, no graphing calculators. We're going to check at the door. I know, I'm not happy about this, but I got some emails from people telling me that they overheard people who said they're going to cheat.
42:01
That they're going to put calculations—they're going to put equations and stuff in their calculators. That's what I heard. I told the class last night they could use graphing calculators, and now I'm saying no. Go to Target, go to the student store, buy a $10 scientific calculator, which still can do a whole lot more than— now I know how to use them. I'm sorry, folks.
42:22
I wish it weren't this way, but think of it this way. If people are actually going to program equations and things into their calculators, they're going to get a better score than they deserve. And since this class is graded on a curve, that means that their cheating is going to maybe put them above you.
42:40
And I do not want you to suffer. So there's the cost-benefit analysis here. So you've got—you've got to spend money $10, okay, which I prefer you not have to spend, and you have to get used to a new calculator. I'm sorry. I don't know.
43:00
What—what would you like to do? No calculator. There's a concept. So, no graphing calculators. Sorry, none. Okay, so look at the difference. At standard conditions,
43:20
this would be 141— negative 141 kilojoules. But under the conditions that we were invoking on this, it went up even greater, okay? It decreased even more. So this is what we've got, okay? So, to try and convince you of this, before we had 100 atmospheres for everything, okay?
43:45
Do it with one atmosphere for everything. One atmosphere. Do it. I'll go back to the previous page. All products and reactions are at one atmosphere.
44:08
Who's got the answer? Yes, you. Negative 141. Why? Because, folks, if Q is one times one over one times one
44:29
or whatever, it's one. Log of one is zero. So this whole thing falls out, and so the delta G of reaction is equal to delta G naught of reaction. Okay, try one and a half atmospheres.
44:42
One and a half atmospheres for everything. Try it. So one and a half atmospheres— oops, where'd they go?
45:03
So it's going to be one and a half squared times one and a half squared times one and a half. So I think your answer is going to be one over one and a half. The Q will be one over one and a half. Because I got a square there and a square there, so those just cancel. So it's just going to be one over 1.5.
45:23
What is the log of one over 1.5? What is the log of one over 1.5? How many don't understand why I'm asking that question?
45:41
Yes. No, no, no, no. I already said— Okay. So, once you put things into this standard state,
46:02
okay, there are no units. It's all divided by one atm. Yes, you're right. We have to kind of do that. Otherwise, you're right. You end up with, you know, unless you have the exact same number of moles of gas, then you end up with the unit in the log. So it's everything. It's going to be 100 atmospheres
46:21
divided by one atmosphere. Gives you 100. That's why there's just hundreds here and no units. Yes. I don't care what you give me. But don't forget, they've got to be the same. Okay, if you want to give it to me in joules, then you've got to multiply by 1,000.
46:42
That 141. So that's a big number. Okay, so what is Q? What is the log of 1 over 1.5? Negative four something, huh? So what is that then? What if we then plug that in,
47:00
what do we get for the RT log Q part? So, right there, what do we get for this? Negative one. So negative one is what we get if we have a half of an, one and a half atmospheres.
47:21
Okay? One and a half atmospheres is pretty darn close to standard. And therefore the correction, this whole thing is negative one. Whereas at 100 atmospheres for each thing, it was a negative 13 or something like that, 12. Okay? So once again, this is the correction factor.
47:41
This is the correction factor that says, the further we are away from standard state, the bigger this thing, the bigger or smaller, depending, this is going to be. Okay? So hopefully you understand when and how to use the delta G thing. Do this one.
48:00
Predicting the value of K from the standard Gibbs free energy of reaction. At 25 degrees C, the standard Gibbs free energy of reaction for half hydrogen gas plus a half iodine gas goes to a whole iodine HI is one plus 1.70 kilojoules per mole.
48:20
Calculate the equilibrium constant for this reaction.
49:09
So you've got temperature and you've got G. So think back to an equation that I gave you where you could calculate K.
49:39
So log K is equal to minus delta G R over,
49:43
or delta G of reaction over RT. So you've got this, you know what this is, and you know what that is. So you just plug this in, 1.7 times 10 to the 3, now we're in joule, we've converted down to joules instead of kilojoules, okay, because then this is in joules.
50:04
So then we just have this in joules, this in joules, this in K, and you come up with the log K is equal to minus 0.685. Yes? We're going to get there.
50:23
It's a good question. Okay, so what they did was they just went ahead and solved, they remembered the equation, okay, that they sort of rearranged it because I always remembered it as delta G is equal to minus RT log K.
50:41
But it doesn't matter. Remember it however you want, and just plug stuff in, and then you just solve for X. All right? So in this case, they did it where log K is equal to this. Okay, now, I want just what K is, not what the log of K is, so I just take this side to the E power, and this side to the E power.
51:01
There's a button on your calculator for that, and you get this. E to the log X is equal to X, so K is equal to .5.
51:29
How many got that? Yes? This? No. These are just to try and show you, you know,
51:41
the relationship between K and whatever. No, you will not have to draw that. I will not ask you a question about that. So how many got that, .5?
52:27
Don't forget this. Okay, do this one. Suppose that in an equilibrium mixture of HCl, Cl2, and H2, the partial pressure of H2 is 4.2 millipascals.
52:41
Once again, I'm not going to give you millipascals. I'm going to give you atmospheres, or tor. For Cl2, it is 8.3. What is the partial pressure of HCl? At 500 degrees K, given that the equilibrium constant, so you're given the equilibrium constant is 4 times 10 of the 18.
53:01
And here's the reaction. H2 plus HCl goes to 2HCl, right there. That's your equation right there. So you want to calculate the equilibrium composition. So what we're asking for is what is the partial pressure of this? Okay?
53:22
Bless you. Okay. So, you got a K, you got these little wacky things, and I'm just going to tell you right now that 4.2 millipascals is equal to 4.2 times 10 to the minus 8 atmospheres.
53:42
So this is 4.2 times 10 to the minus 8 atmospheres, and this is 8.3 times 10 to the minus 8 atmospheres. It's a conversion you're not going to have to do on the exam.
54:47
Okay. So what we know is that K is equal to the pressure in atmospheres squared over the pressure of hydrogen times the pressure of Cl, okay?
55:01
So we can rearrange this and say we want to know what is the pressure of HCl. So we just take the HCl, we move these guys to the other side, and we just say that K is now equal to the pressure of hydrogen times the pressure of Cl2, okay?
55:21
And they do too many steps for me, but because HCl is squared here then, then they take the square root of that whole mess. So here you got the pressure of HCl is equal to, this is the value you are given for the K, this is the pressure of the hydrogen,
55:40
this is the pressure of the chlorine, and because we're looking for this, this is the square, we take the square root of that, and it's equal to about 130, 120. So K is equal to about, I think I got 118. Yes? Pressure has to be ATM.
56:04
Yeah. ATM is a bar, okay? However, if, yeah, let's leave it like that. Yes? Loud.
56:24
Add a 2? Why would you do that? What do you mean a 2?
56:40
Okay, okay. He's asking if instead you had done, instead of a half and a half of the coefficient, you could have just gone 1, 1, and 2. I'm asking you, what is it? Do you think it matters? Do you get the same answer? Okay, then you can do it.
57:03
So, can we do this problem, folks? Okay, let's do this. This is an easy one. All we're doing is we're trying to figure out Q. Unfortunately, we're doing Q. Don't be fooled, folks.
57:21
Look at this. The reaction, whatever the reaction is, oh, never mind, there's no trick. I thought there was a liquid or a solid in there, but I guess it's not. Oh, no, iodine gas, okay. So here, folks, they give you a K. They can't, you can't answer this problem if you don't know what the K is, okay?
57:42
Because the question is, which way is the reaction going to go? So you figure out Q. If Q is bigger than K, then it goes, if Q is bigger than K, it goes this direction. If Q is smaller than K, it goes that direction. So you just plug everything in. K is 46.
58:00
What does that tell you, folks? What does that tell you? If I gave you this and I said draw, draw the delta G versus temperature. Delta G is here, temperature is there. It would go way over here. 46 is a pretty good sized number. So it would minimize, it would reach equilibrium somewhere over in this area. If I said 4600, it would be even further over.
58:23
.46 equilibrium would be reached right about here.
59:06
So in this case, folks, we end up with a Q of 1. Q of 1, K of 46. Which way is the reaction going to go? Toward the products, that's right. So, folks, I'm not trying to confuse you,
59:21
but on a question like this, where you have the same number of moles, you've got two moles of gas on top, and two moles of gas on the bottom, if I had said that all gases are at a pressure of 55 torr,
59:43
55 torr, okay? Instead of .55, guess what? You've got torr squared, you've got torr times torr, everything cancels. It's still going to be 55 squared over 55 and 55, you still get a Q of 1.
01:00:00
one okay so you don't in some cases you don't have to do the conversion you just have to do the conversion when there are different number of moles on top and bottom okay so the reaction is going to proceed toward the products
01:00:20
something quick about catalysts for a catalyst, folks if we have AB here and we're going to CD the distance between here and here okay that is how much energy you get off alright so you got to climb this hill, this activation energy
01:00:41
and then you go down there okay so about this much energy is given off going from AB to C plus D what a catalyst does it just makes the path easier the difference between here and here is the exact same okay
01:01:00
it's it's a little bit like if i said uh... uh... let me see that i want you to go from here to baba island okay I'm here to baba island
01:01:20
but the path of A plus B goes to C plus D that I'm asking half of you to take is you go to baba island from this room uh... but you have to climb over saddleback mountain so your path is from here down to there over the mountain and then to baba island
01:01:44
that's this or there's the one where you leave this building and you walk to the beach that's this one now the amount of energy the difference in potential energy between you being here and in newport beach doesn't matter whether you went to
01:02:01
big bear mountain or kilimanjaro the catalyst provides a different pathway and this believe it or not this little hill right here is really what determines how fast something goes if this is a really a tall one then we're going to
01:02:20
stay more where A and B is okay and not get to C and D even though even though you can see going from here to here makes thermodynamic sense but if this hill is too big you just don't do it a catalyst allows a different path so in engineering folks using a catalyst for this
01:02:41
can sometimes save you millions or billions of dollars or make you millions of billions because you've actually been able to to come up with a way to turn sand into gold okay or coal into diamonds coal to diamonds folks this goes way up there if you can come up with a different way to make diamonds
01:03:03
good for you so it doesn't change the k nor does it shift the equilibrium in the system it just allows a different path
01:03:22
I said all this man I'm good okay so this is it I think Josette was on a trip this yeah she's not in this picture anyway
01:03:41
uh... so folks I hope you learned something I will see you all on Thursday I guess a week a week from today but a little earlier study hard