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Lec. 12. Free Energy & Review Problems

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Lec. 12. Free Energy & Review Problems
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UCI Chem 1B General Chemistry (Spring 2012) Lec 12. General Chemistry -- Free Energy and Review Problems -- Instructor: A.J. Shaka. Ph.D. Description: UCI Chem 1B is the second quarter of General Chemistry and covers the following topics: properties of gases, liquids, solids; changes of state; properties of solutions; stoichiometry; thermochemistry; and thermodynamics.
BetäubungsmittelGibbs-EnergieEntfestigungVorlesung/Konferenz
Chemische ForschungChemische ReaktionAmmoniumverbindungenNitrateChemische ReaktionSingle electron transferAlkoholische LösungAmmoniumnitratChemische ForschungPropylthiouracil <6-Propyl-2-thiouracil>Elektronische ZigaretteBukett <Wein>KochsalzWasserComputeranimation
KohleChemische ReaktionIonenpumpeKonzentratElektronische ZigaretteVerbrennung <Medizin>Bukett <Wein>Vorlesung/Konferenz
Hydroxyoxonorvalin <5-Hydroxy-4-oxonorvalin>Agricultural University of AthensChemischer ProzessComputeranimation
Toll-like-RezeptorenAlkoholgehaltTrimethylaminFaserplatteFülle <Speise>Vorlesung/Konferenz
Chemische ForschungChemische ReaktionKörpertemperaturDruckbelastungAmmoniumnitratAlterungChemische ReaktionWasserfallGibbs-EnergieFunktionelle GruppeDruckabhängigkeitKörpertemperaturWursthülleGesundheitsstörungAdditionsverbindungenInitiator <Chemie>BildungsentropieOrganische ChemieVorlesung/KonferenzComputeranimation
MineralLeicheWasserfallPhthiseStoffwechselwegVorlesung/Konferenz
MilChemischer ProzessBildungsentropieSelbstzündungBildungsentropieGesundheitsstörungWursthülleChemischer ProzessAlterungSystemische Therapie <Pharmakologie>MolekülVakuumverpackungPedosphäreDruckabhängigkeitLinkerFärbenIsopentylnitritAlkoholische LösungSelbstzündungQuerprofilDiamantähnlicher KohlenstoffNanopartikelComputeranimation
MolekülFunkeQuerprofilStoffwechselVorlesung/Konferenz
KörpertemperaturErholungBildungsentropieChemische ForschungGesundheitsstörungSystemische Therapie <Pharmakologie>Kettenlänge <Makromolekül>MolekülbibliothekChemische ReaktionGesteinsbildungMeeresströmungKörpertemperaturDruckabhängigkeitVerdampfungswärmeExothermieWeibliche ToteBildungsentropieComputeranimation
Chemische ForschungBildungsentropieDruckabhängigkeitKörpertemperaturChemischer ProzessVerdampfungswärmeSelbstzündungAmmoniumverbindungenSystemische Therapie <Pharmakologie>DeltaFließgrenzeGletscherzungeBildungsentropiePharmazieMineralGesundheitsstörungAlterungAlkoholische LösungAmmoniumnitratVorlesung/KonferenzComputeranimation
OctaneHydroxyethylcellulosenGalactoseÜbergangsmetallChemische ForschungAtomsondeWasserstoffWasserBeta-FaltblattDeltaSetzen <Verfahrenstechnik>BrennbarkeitVerdampfungswärmeWasserCobaltoxideVan-der-Waals-KraftWasserstoffWassertropfenLinkerMolekulargewichtsbestimmungÜbergangszustandChemische ReaktionOctaneTeilentrahmte MilchExplosionTankChemische ForschungTopizitätKörpertemperaturReaktionsgleichungFettBukett <Wein>Elektronische ZigaretteHalbedelsteinGezeitenstromAlterungMassendichteDruckabhängigkeitChemieingenieurinVorlesung/KonferenzComputeranimation
VulkanisationSchlag <Landwirtschaft>WasserstoffDruckabhängigkeitFülle <Speise>Klinisches ExperimentFarbenindustrieVorlesung/Konferenz
MassendichteChemische ForschungAtomsondeCalcineurinMethanFülle <Speise>CobaltoxideWasserstoffSetzen <Verfahrenstechnik>BrennbarkeitKohlenstoff-14MethanisierungMassendichteKohlenstofffaserTransportErdgasHalbedelsteinPropylthiouracil <6-Propyl-2-thiouracil>AlterungKonzentratEthanReaktionsgleichungKörpergewichtBenzinChemisches ElementGesundheitsstörungKohlendioxidAcetylenWasserVerdampfungswärmeVerbrennung <Medizin>PigmentrußBukett <Wein>BiodieselStöchiometrieIonenbeweglichkeitComputeranimation
EthanMethanChemische ReaktionWasserstoffGasflascheSetzen <Verfahrenstechnik>EthanBenzinPropionaldehydWasserMolekülHalbedelsteinKohlendioxidVerdampfungswärmeVorlesung/KonferenzComputeranimation
AcetylenChemische ReaktionAtomsondeEthanMethanAcetylenMethanisierungMolwärmeBrennbarkeitFlammeVerdampfungswärmeUltraschallschweißenFackelWasserExplosionWasserstoffKohlenstofffaserKörpertemperaturHalbedelsteinElektronische ZigaretteVorlesung/KonferenzComputeranimation
Chemische ForschungWasserstoffChemische ReaktionBiskalcitratumWasserstoffKohlendioxidIsooctanEthanKohlenstofffaserWasserVorlesung/KonferenzComputeranimation
MischenDruckbelastungAlkoholische LösungBoyle-Mariotte-GesetzWasserAtomsondeBruchverhaltenAlkoholische LösungMolekülMischenBoyle-Mariotte-GesetzStoffmengenanteilThermoformenVorlesung/KonferenzComputeranimation
ThylakoidChemische ForschungDruckbelastungAlkoholische LösungBoyle-Mariotte-GesetzTitrationBiskalcitratumStoffmengenanteilBoyle-Mariotte-GesetzAlkoholische LösungDruckabhängigkeitVerbundwerkstoffThermoformenWasserstoffbrückenbindungElektronische ZigaretteVorlesung/KonferenzComputeranimation
Chemische ForschungAtomsondeElektronische ZigaretteBoyle-Mariotte-GesetzAlkoholische LösungWursthülleVorlesung/KonferenzComputeranimation
Chemische ForschungAlkoholische LösungZinnerzDruckbelastungGesundheitsstörungWasserLöslichkeitAtomsondeWasserstoffbrückenbindungBukett <Wein>CobaltoxideKohlendioxidReaktionsmechanismusLöslichkeitLagerungMolekülWasserGesundheitsstörungBoyle-Mariotte-GesetzDisposition <Medizin>WasserstoffIonenbindungDruckabhängigkeitVorlesung/KonferenzComputeranimation
KaugummiChemische ForschungDruckbelastungWasserAtomsondeKonzentratAlkoholische LösungDruckabhängigkeitDurchflussWasserMolvolumenKohlendioxidQuellgebietKohlensäureVorlesung/Konferenz
PhthiseCarbonateMeerOxidschichtHydrogencarbonateKohlenstoff-14WasserQuellgebietMeerKohlendioxidSäureCarbonateKonzentratSchelfeisOrganische ChemieVorlesung/Konferenz
Chemische ForschungMeerMeerwasserMeerwasserFremdstoffMeerEnergiearmes LebensmittelKohlendioxidVorlesung/KonferenzComputeranimation
Chemische ForschungKochsalzlösungMeerFremdstoffKettenlänge <Makromolekül>Vorlesung/KonferenzComputeranimation
LinkerVan-der-Waals-KraftVancomycinDruckbelastungChemische ForschungAtomsondeGesundheitsstörungMilValinReaktionsgleichungContainment <Gentechnologie>StoffmengePasteDistickstoffDruckabhängigkeitSeitenketteBukett <Wein>BlitzschlagsyndromGeysirMelatoninMannoseElektronische ZigaretteVorlesung/Konferenz
Boyle-Mariotte-GesetzHeliumAtomsondeMethanHeliumChemischer ProzessErdgasHelium IIZellfusionBruchverhaltenElektronische ZigaretteRingbrennkammerMethanisierungVorlesung/KonferenzComputeranimation
Chemische ForschungMethanHeliumAtomsondeSterblichkeitRifampicinMischenHeliumChemischer ProzessOktanzahlErdgasGolgi-ApparatWildbachContainment <Gentechnologie>Chemische FormelMethanisierungZellfusionMischenMolekülElektrolytische DissoziationBukett <Wein>MolekulargewichtsbestimmungAtomPotenz <Homöopathie>Vorlesung/KonferenzComputeranimation
WasserVorlesung/Konferenz
Transkript: Englisch(automatisch erzeugt)
We're ahead of schedule slightly.
We're going to do a little bit on free energy today. That part will not be on the exam, however. And then we shall review.
We'll throw out problems and we'll look at them and we'll do them. The other thing you should review is any problem on the last exam because if that same problem comes back and you flub it again, it looks bad, especially if I told you
to look at it, and there I am Monday in the afternoon and three people come by and we work out the problems. 297 don't. So all I can assume is you know what you're doing.
Make sure you know how to do all the problems on the prior exam and fast, okay, not slowly, fast.
Put a stopwatch on it, pretend you're going to flunk out if you don't pass, get that problem right. Don't look at the answer. Ready, go. See how you do. Okay, free energy, some review problems.
And I guess a cautionary tale, so I hope this is the right set.
Most reactions that are favorable, in other words, that give mostly products are exothermic.
They also produce heat, by far the majority. And for a while, it was thought that the reaction being exothermic was a requirement
for it being favorable. In other words, that's how you could tell. If it were exothermic, it would be favorable. If it weren't, then it wouldn't be favorable. But that's not true because if you dissolve ammonium nitrate
in water, the action goes, the ammonium nitrate dissolves, and that's a chemical reaction, and the solution gets cold.
That means it's endothermic, it's absorbing heat. And there are lots of other examples like that. This one's pretty striking because it's quite a big positive number.
And therefore, this discovery forced people to consider, as a chemist, you want to make something, how do you tell if the reaction you've written is going to be favorable? If it isn't, if it produces 10 to the minus 6%
of product, then that's not going to be a going concern in terms of making something. It turns out that the reaction you write doesn't need to necessarily be favorable on its own as long as you're willing to do work.
If you're willing to do work, then you can make it favorable. For example, supposing I have a reaction that produces 10 to the minus 6% of product at equilibrium. If I can take the product out, if I can pump it off
so that the product falls below the equilibrium value, then the reactants make more. And if I keep pumping the products off, the reactants keep making more because the concentration of product is always too low. And so I do a ton of work, and I get my product eventually. But it's a ton of work.
Where did the work come from? The work came from running another reaction that is favorable, like burning coal, which makes electricity that runs the pump. And when you consider the coal burning and then your reaction going and you add them up, it's favorable, and that's why you can do it in the end.
But there's sometimes a reaction hidden in the background that we don't necessarily see, okay? All right, someone got caught cheating on Sapling.
I won't go into details. So I thought we'd have a lesson in honesty since on Sunday you may have seen somebody lost their job. Some people exaggerate accomplishments.
I was the senior manager. I was the senior accountant. I have two degrees. I got my degree a year sooner than I did. I have a higher GPA than is actually true.
Scott Thompson resigned as CEO of Yahoo on Sunday. That's kind of a, would have been a nice job to keep.
He claimed to have a BS degree in computer science. He had a BS degree in accounting. And on his CV or his resume, he put that he had a BS in accounting and computer science, which may sound a little bit better.
But in fact, he did not. He had no such degree, and in the days of public records, that thing is pretty easy to find out pretty quickly. There isn't much cover anymore. People can find out a lot of stuff.
Oh, he and five board members who didn't do the right fact checking are gone. They all resigned. Oftentimes, people will remember you
by either the best thing you ever did or the worst thing you ever did. And everything in between will be forgotten. Make sure they remember you by the best thing you ever did and not, remember the guy who resigned as CEO of Yahoo?
That's who you become. That's how people remember who you are. Hi, I'm Scott Thompson. Ah, the guy who resigned in disgrace. I remember that. I think T.S. Eliot summed this up pretty well in one
of his poems, the awful daring of a moment's surrender which an age of prudence can never retract. Okay, let's get back to our ammonium nitrate.
It's the change, tennis worked out, it's the change in free energy, not just the heat, but the change in free energy that dictates whether a reaction's going
to be favorable in terms of giving more products than reactants at equilibrium or whether the reaction quote will go at constant temperature and pressure. And G named after Gibbs is a state function
and it's partly the heat, that was partly right and then there's an additional term minus T.S. and S is a little bit more abstract. It's also a state function and S, the entropy, has to do
with the amount of disorder, whether something is random or whether it is organized. Organized things have very low entropy, disorganized things have very high entropy.
And if we get the G function, the Gibbs function for both products and reactants, then we can write the change, which is always the final minus the initial. So, in this case, it's G of products minus G of reactants.
And when G is negative, the reaction is favorable. G rolls downhill, just like things fall in a gravitational field.
Chemical reactions or phase transitions start somewhere and if they can fall down, and that means there has to be a path for them to fall down, sometimes they can be like a boulder jammed in a cliff. They're going to fall down, but not yet, that's us.
We're unstable, that's why we die. And you're only at equilibrium when you're dead and gone and the bacteria have finished munching on your corpse and there's nothing left except minerals. Then you're at equilibrium.
The reason we can stay and maintain what we've got is that we consume energy. If we don't eat, then we die. Because we're actually expending a lot of energy to do all the housekeeping to keep everything in order.
Why do things happen at all? And the answer is, and I hope you remember the lecture on solutions, we had some spheres on the right
and some spheres on the left and we just counted. If they just move around at random and we just look once in a while, what's the chance that they're going to be all on one side? The answer is the chance is small.
And when you have as many molecules as you have in this room, while in principle, they could all suddenly rush to the wall and then we all die of lack of air, we don't fear that because that's so unlikely, it's so unlikely
that we discount it. In fact, if we start with a canister of air and the whole room's a vacuum and we open it up after a while, the pressure is equalized everywhere in the room because the molecules don't have any brains and they just go wherever they go and therefore this measure of disorder has
to do with probability. It's much more likely to be dead than alive. That's why you end up dead, unfortunately.
The metric is that if we have an isolated system, so we don't exchange mass, we don't exchange energy, we just have a closed system. We don't look in, there's no light in there. And we let it sit for a very long time and it comes to equilibrium, its entropy will have increased.
It will have just randomly moved to the most likely state and once it reaches the most likely state, it'll just fiddle around there. There'll be some fluctuations. If it's 6 and 6, it might be 5 and 7, but it's only very rarely going to be 1 and 11, for example.
And when we have a lot of particles, the chance that it's ever going to do that is so small that we couldn't wait long enough for it to happen. It would be many, many, many times longer than the age
of the whole universe and we can't wait that long for the air to rush over to the other side of the room. And the universe itself is an isolated, I'll put the D on, system. And so for any spontaneous process in the universe, the entropy of the universe must increase.
That means when you flush the toilet, when you brush your teeth, when you eat something, whatever you do, the entropy of the universe will increase. And the question is, how can we make that more useful because we aren't interested in the whole universe,
we're interested in the system. Or we might be interested in our own health and welfare, in which case I consider my physical being to be the system and the other things around to be the surroundings. And in fact, that's mentally how people carve up the universe,
me and everything else. But everybody's doing that. Oh, if we divide artificially, we make a mental division into some part what we're interested in called the system and the other part called the surroundings, then the entropy
of the system and the surroundings for any spontaneous process is going to be positive entropy if I add them up. And that means that we have to disorder the surroundings to order the system. And when you think about it, that's what humans do big time.
We wreck everything else so that we can have it our way. You eat things, you break down big molecules, you take in energy, you chuck out small molecules as waste,
you produce a ton of heat, you're on fire, dogs are even hotter. The hotter you are, the quicker you go. Dogs are about a degree warmer than us and they fizzle out pretty quickly.
Some other things that have slow metabolism like tortoises can last quite a long time, but they don't do very much. So they sit around, pull their legs in and so forth. And so there's sort of a constant area under the curve.
You can be tall like a spike like that, like a cicada or you can be long and last a long time like a tortoise. The entropy is really a measure of how much extra chaos we're creating. And the chaos depends on the current state.
If I'm at a rock concert and it's loud and people are cheering and clapping so there's a high noise level, then even if I scream, I don't add much to the noise level.
But if I'm in a library or in a church, and it's dead quiet and I start screaming and tearing on, it's very, very noticeable. It's a big intrusion, it's a big change. And likewise, if we have a cold system and we add heat
to it, we disorder it a lot. But if we have a hot system and we add heat to it, we don't disorder it very much at all. And therefore, to keep track of the entropy, we have to take a ratio of the heat that we add because heat disorders things.
It gets things to move around more randomly. And the temperature that we're already at is a measure of how disordered it already is. And so we take a ratio. And to make it a good quantity, we have to add the heat reversibly.
So we add the heat as carefully as possible. And then we measure how much disorder it created. And then we'll have a measure of the entropy. For an exothermic reaction then, the reason why exothermic reactions are favorable is
because the heat they produce disorders the surroundings. That's why they go. They're creating a ton of disorder in the surroundings. And so at constant temperature and pressure T,
if we add some heat to the surroundings, then we know that that's delta H at constant temperature and pressure. And the heat had to come from somewhere, so it's minus delta H of the system over T. That's the change
in entropy for the surroundings. Well, to figure out the total sum, we have to not only add the heat to the surroundings, but we have to see if the system itself disordered.
In ammonium nitrate, heat came in from the surroundings, so the surroundings became more ordered. But because the ammonium nitrate dissolved and could just move around and had many more possibilities than the solution, the ammonium nitrate itself became very disordered, and it became more disordered
than the surroundings became ordered, and so it goes. That's how we keep track. Well, it's always the entropy of the universe, and we could use delta S of the surroundings plus delta S of the system, but we don't like talking about the surroundings because usually we don't measure them.
We don't know that much about them. And so we replace delta S of the surroundings with minus delta H of the system over T, and now we're good to go because this thing here only has state functions and temperature, and that measures whether it's going
to be likely to go. If I multiply both sides of this through by minus delta T, I get minus T delta S of the universe is equal to delta H minus T delta S for the system.
Delta S of the universe has to be greater than zero for any spontaneous process, and that means that minus T delta S has to be less than zero because T is always positive. This is absolute temperature, of course, in Kelvin. And therefore, if this has to be less than zero,
since this is equal to this, this thing has to be less than zero, but this thing is so useful, we call it delta G, and so we say delta G has to be less than zero for a process to be spontaneous at constant temperature and pressure.
And that's why we want to know delta G up front because we want to know what kind of yield we can expect if we're a good chemist and we do everything perfectly, what are we going to get? And then because we aren't perfect, we always get less than that. So just like the reversible work sets a limit
on what we can possibly get, so delta G sets a limit on what kind of yield we can get for any kind of spontaneous process at constant temperature and pressure.
Okay, we'll leave delta G there. We'll come back to delta G next, a week from Thursday. We'll come back to delta G, and we'll continue on it, and we'll see how we can use it to predict the ratio of products to reactants and other things like that.
For now, though, let's press on and do some review problems. And I've tried to make these a little bit topical. They aren't just problems, but these are stories like Aesop's table.
They have a moral at the end. Sometimes it's good to work through a problem just to see how something would work. So if Governor Schwarzenegger says we're all going to be driving hydrogen cars in 10 years, you can say, no,
we won't be, and that's that. Let's have a look. Suppose we want to transition to a hydrogen economy.
Scientists are working on this, chemists, chemical engineers, and so forth, but let's have a look at what we'd have to be able to do. Let's just figure out the volume of hydrogen gas at 1,000 PSI.
That's pretty high pressure, maybe not unbelievably high, but pretty high, 1,000 PSI and 25 Celsius. And remember, if you're going to have a tank at 1,000 PSI, and you park your car in a parking lot out in Indio,
you have to make sure that it doesn't explode. Usually that's going to mean a containment vessel which typically is heavy. The more mass you lug around in your car, the worse the efficiency is.
Okay? Now I've looked up an approximate value for the enthalpy of combustion of octane. Octane is 2.66 kilograms per gallon, and delta H
of combustion is minus 5,500 kilojoules per mole. So it's very, very exothermic. That's why we use it. Okay, let's go through and see what we can do. Let's start with octane. We've got 25 gallons in the tank.
It's 2.66 kilograms per gallon. The gallons go away. I look up the molar mass of octane, and I could do moles per gram, but I've decided to do kilomoles per kilogram. Same thing.
Now the kilograms are gone, and I've got kilomoles. So it turns out in the tank I've got 7,596.3 kilomoles or 7.6 times 10 to the 6 moles of fuel.
Now that I know how much, how many moles I have, and I know I get 5,500 big fat kilojoules of exothermicity per mole, I can figure out that I get 4.2 times 10 to the 10 kilojoules.
That's how much energy manifested as heat that we're putting into the tank. 40 billion kilojoules.
Now let's do hydrogen. Well, for hydrogen, we need to decide on the phase of the water. What I'm going to assume,
but it won't make too much difference, is I'm going to assume that the water is produced as water vapor, as gas, because I don't think in the piston where we're going to get the energy that it's going to, that we can condense the water to water drops
and then milk out extra heat and use that for something. I think that heat just goes out the back. We get water vapor coming out, and then it condenses. If the tailpipe is cold, initially you may see drips of water as the water condenses, or if it's a cold day.
So I'm going to assume that the water is water vapor. Most books always assume it's liquid. But anyway, it won't matter too much. We look up the enthalpy of formation of water vapor at 25 Celsius, and we find that the delta H of formation,
it's exothermic, is minus 241.8 kilojoules per mole. And that's what we're going to get. That's the only product, in fact, is H2 plus a half O2 equals H2O. So this enthalpy of formation is actually the enthalpy
of combustion, because it's burning with oxygen. And therefore, we need to get 40 billion kilojoules. And then per mole of H2O, we get minus 241.8 kilojoules.
And so we can figure out by the balanced chemical reaction that one mole of H2 gives one mole of H2O. So everything goes away, and we have moles of H2 left. 1.73 times 10 to the 7 moles of H2.
It seems kind of big, but it doesn't, offhand to me, this didn't seem so bad. Seems, yeah, seems like a lot. The problem is that the gas has low energy density, even at 1,000 PSI, and so the volume that we're going
to need to lug around is going to be uncomfortably large. Let's figure out how large it's going to be. Well, we aren't given any van der Waals constants on hydrogen or anything like that. Therefore, all we can do is assume hydrogen's an ideal gas.
Good enough for the back of the envelope here. So without any further assumptions, we just use the ideal gas equation, and we say, the temperature, we'll assume that the temperature's 25 Celsius, and, of course, whenever we see that, we say, no, straight to Kelvin,
then back to Celsius or Fahrenheit, whatever the public's more comfortable with. V is equal to nRT upon P. We know how many moles. We know R. We know T, and we know 1,000 PSI.
Ooh, PSI, pounds per square inch. What's that? Well, we have to convert to atmospheres because this is atmospheres. If we write in all the units and we don't convert it, we have atmospheres over PSI times liters in the answer,
and that's a clue that we aren't done. We convert 14.7 PSI per atmosphere, and we find we need 62 million liters of hydrogen. Question is, how big is that?
A liter is not so big, but millions of them. It's getting to be large, and if you work it out, this requires a cubic tank that's about as tall
as me on each side. That's what you need. In other words, it's almost as big as a whole smart car. Just the fuel tank. By the way, is the smart car a smart choice?
Only if you don't value your life. The smart car doesn't have very good mileage either compared to how tiny it looks, and if you have a collision with another car that Mercedes Benz makes,
like just a regular sedan, the smart car is annihilated, and if you figure out the acceleration of the crash test dummy, you're dead, and the other car just kind of hears this crumpling sound like paper tearing and keeps driving.
It doesn't slow down at all. The smart car is called a bus, where you get 50 people on and you're getting the same mileage as the smart car, and you're completely safe under most circumstances. Okay, the 25-gallon gas tank is only 94.6 liters,
and what this means is that even if you can get around all the problems with refueling and stuff, who's going to be driving a car with this huge thing on the back? Well, okay, you make the pressure higher, you say.
Make it 3,000 PSI, but there's a limit to that before you need really, really heavy cylinders, and they're heavy to carry, and that wrecks your mileage compared to a lighter car.
So my conclusion is that the hydrogen economy is going to face a lot of practical problems before it's going to contribute to mobile transportation, so I wouldn't hold my breath. And keep in mind that although we figured it out with heat, it's really delta G that we need to figure,
because if we figure delta G, we can tell how much work we can get. We'll get to that later. But heat is reasonably good. It's close. All right, let's try another one. Well, hydrogen may not work.
Let's try some other clean-burning fuels, some things that don't produce particulates and soot and stuff that gasoline tends to do and diesel does big time, and let's see if we can make it go out of any of those.
So let's compare then the energy density of methane, natural gas, ethane, C2H6, acetylene, C2H2, and hydrogen, H2, for mobile transportation.
The delta H values of formation for these guys, are respectively minus 17.9 kilojoules per mole for methane, minus 20 kilojoules per mole for ethane, and plus 54.2 kilojoules per mole for acetylene.
Okay. So we'll start with those, and let's just go through them one by one and see how it pans out. We have to do two things. We have to look up the delta H of formation of the products,
and on an exam, of course, you would be, yes, of course, on an exam, of course, you would be given a table of the delta H of formation of these things. Don't memorize numbers that you can look up.
We want delta H of combustion. We have to balance it, and then we have to use Hess's law to figure out what it comes to, and the delta H
of formation values for CO2 are minus 393 kilojoules per mole, and for water vapor, I'm going to assume they're all making water vapor, so at least it's a fair playing field,
is minus 241.8 kilojoules per mole. We balance it, and the way we balance it is we start with CH4, and we write plus O2. And for complete combustion, which we don't always get in a vehicle, we get 100% CO2 and H2O.
Those are the products. If you see the word, if you see the phrase, complete combustion, that means CO2 plus H2O. If the fuel has nitrogen, I think I told you, you have to be careful. You can get different things dependent on the condition.
I have one carbon atom, so that means I have to have one CO2, because CO2 is the only thing with carbon. I've got four hydrogens, so I have to have two H2Os, because this is the only thing with hydrogen, and then I just figure out how many oxygens I need to make sure oxygen atoms aren't disappearing, and there I go.
And now I can figure out delta H of combustion. It's delta H of formation of the product minus the reactants. Delta H of formation of CO2 is minus 393.5. Plus there's two moles of this, so I have to make two
of them, plus two times minus 241.8 minus, this is an element. Delta H of formation is zero. This is minus 17.9, and if I total this up,
I find for methane I get minus 859.2 kilojoules per mole. That's a lot better than hydrogen, and that's why you see some cars marked CNG on the bumper already, because they're running on compressed natural gas.
And the reason why it's better than hydrogen is that unfortunately most of the oomph comes from forming CO2. But CO2 is the villain currently, and if we're burning any kind of carbon fuel and forming CO2
and letting the gas just go up, then we're going to be increasing the concentration of CO2, yes. Well, we have to know that. Well, we have to know the complete combustion means that the products are CO2 and H2O.
That's what we have to know. We have to know complete combustion really means CO2 and H2O, and then we have to balance it, and then we just use the numbers and the stoichiometric coefficients.
Okay, let's go to, let's see, I think the next one's ethane, yes. For ethane, we have two carbons. Therefore, we get two moles of CO2. We have six hydrogens, so we get three moles of H2O,
and then we just have to balance the oxygens so that we end up with seven on each side. And so that's seven-halves moles of O2. The computer that controls how the air intake and fuel intake goes into the cylinder has
to know what fuel you're burning. Or it doesn't know what this coefficient is, and therefore, if you take a car with a computer with fuel injection, and you just put in some other fuel which has different stoichiometry, your car won't work right.
It'll be awful. It'll either bump and chunk around, or it'll produce, smell terrible, or it'll knock. You'll get all kinds of problems. So you have to use the fuel that they're assuming you're going to use. And here, we do better precisely, unfortunately,
because we're forming two moles of CO2, and CO2's the 800-pound gorilla because it's got that huge minus 393.5 in its favor. It's very favorable to form it. It's a very stable molecule. We take 2 times minus 393.5 plus 3 times the water,
and then we subtract the delta H of formation of ethane, and delta H of formation of O2 is 0, and we get minus 1492.4 kilojoules per mole. Therefore, ethane is much better than methane,
if you had to choose, because for a mole of gas, it takes the same volume, and you can go a lot farther with ethane. And propane will be even better, and then you go all the way down, and you end up using gasoline again.
Now, ethane is superior to methane in terms of the heat released. We get almost twice as much, not quite, as with CH4. Now, let's look at acetylene.
For acetylene, we've got two carbons again, but only two hydrogens, so we only get one mole of H2O, and again, we do the O2 at the end. It doesn't really matter what it is on an exam, because this isn't going to contribute to the delta H of combustion. So if you're in a hurry, you don't even need
to balance the O2, because it's not going to change the answer. For this one, then, we take, again, 2 times 393.5 plus minus 241.8 minus, and then this is a positive number, so this now gives us extra oomph.
We get minus 1083 kilojoules per mole, so acetylene is close to ethane, not quite as good, but acetylene makes a much hotter flame, and the reason why acetylene makes a much hotter flame is
because it only makes one mole of H2O, and H2O has quite a high heat capacity. Therefore, if I can burn something and not make as much H2O, then the heat that's released can heat up the rest of the products to a much higher temperature,
and it's for that reason that we use acetylene and a torch to do welding and things like that that require high temperature, and you won't see anybody using ethane in a torch. In fact, it's called an oxyacetylene torch for precisely that reason.
Acetylene is quite an explosion hazard, however. Acetylene, if you look at it wrong, it says, did you say something, and it blows up. And so you have to take that into account, too, the propensity for something to go wrong
as well has to figure in. And then finally, if you go to hydrogen, we get this dinky little minus 241, and it's just not competitive.
So sadly, while hydrogen is the best environmentally, because it only produces water and doesn't produce any CO2 that's a greenhouse gas, because it doesn't produce any carbon dioxide, it's wimpy,
because it's precisely producing the carbon dioxide that gives you the energy. So you're kind of in a jam. If you had to choose of these three, you'd choose ethane as the best energy density, but it's still much, much worse than isooctane.
Okay. Let's do a vapor pressure problem. Let's go back to mixtures. Remember, we've got mixtures of two things, and they both contribute molecules to the vapor,
and the more volatile one will have more in the vapor. Oh, we've got two substances. They're pure, pure compounds, A and B, and we've measured the total vapor pressure.
We've just measured the pressure. We haven't measured how much of each one's there, but we've taken a certain number of moles, one mole total, and it's either all A, all B, or something in between.
And based on the data, just from the vapor pressure, we want to ask the following question. Do A and B form an ideal solution? We have to remember what an ideal solution is. We remember an ideal gas from the ideal gas equation of state. For an ideal solution, we have a different measure.
Now remember, we have these mole fractions, chi A and chi B, and we start with 100% of B, and the vapor pressure is 150 torr, and then we make 20,
80, and it's 160.6, 40, 60, 169.9, 60, 40, 173.9, 80, 20, 172.6, and then 100%, 170.
And we want to know, do A and B form an ideal solution? Well, an ideal solution is one that follows Raoult's Law at all compositions. That's the definition.
If it doesn't follow Raoult's Law at all compositions, it's not an ideal solution. And Raoult's Law says that the vapor pressure is the mole fraction times the vapor pressure of the pure liquid.
That's the partial pressure of each. So if I add them up, the total pressure P, which is the vapor pressure of A in the solution plus the vapor pressure of B in the solution,
it has to be a straight line. Therefore, it has to go, since the endpoints are 150 and 170, it has to be a straight line in between. Otherwise, it's not following Raoult's Law, and therefore it's not an ideal solution.
And remember, there are positive deviations from Raoult's Law where the vapor pressure is higher than you would predict, and there are negative deviations where the vapor pressure is lower. That means that A and B are kind of sticking to each other.
They're sort of sticking, and so they aren't bothering to go up into the vapor because they're hanging together. If it's a positive deviation, it could be that one of them likes to form hydrogen bonds and pushes the other guy out. Just get out of here, and so the vapor pressure is higher than you would have predicted.
Therefore, if we make this plot, we're given the data, and all we have to do is make this plot and see if it's straight. And if we make the plot, we see that it's not. It's not straight, and therefore this is not an ideal
solution, end of story, because it's quite curved. And in fact, in this case, it's making a positive deviation from Raoult's Law because the vapor pressure is higher
than we would have predicted from a straight line. And that means that A and B, quote, dislike each other. They prefer to be separate so they kind of push each other out of the solution, and therefore it's not ideal.
Okay, another one. Let's do Henry's Law. The solubility of CO2 gas in pure water is known to be 34 millimolar at one atmosphere pressure of CO2.
Let's ask, what is the solubility of CO2 in water under atmospheric conditions? And let's assume that the partial pressure of CO2 in the atmosphere is 380 ppm.
I think it's higher than that now, but let's assume the partial pressure is 3.8 times 10 to the minus 4 atmospheres. That's the partial pressure of CO2.
By the way, if you're running and you're getting tired, there could be two reasons why you're getting tired. Well, there could be three reasons. The first reason is you're out of shape. That's most likely. The second reason is you aren't getting enough oxygen.
So you should really concentrate on breathing in like crazy. And the third reason is you aren't exhausting enough CO2. So you should breathe out like crazy. In fact, it's usually the third reason why you get tired.
So if you're running and you're running and you're coming onto the tape and there's somebody on your shoulder and you're winding it into the red and then suddenly a big bear jumps on your back and it's very unpleasant, that's because the pH
of your blood is changing. And when the pH of your blood changes, your brain says, that'll be enough of that. Let's ease off. That's the signal. And then the other part of your brain says, no way. Because I want to win.
But that's quite a battle of wills because you're fighting against something that's kind of an automatic mechanism and you have to override it in order to keep going. And there are limits to what you can do. Anyway, let's assume that CO2 follows Henry's law.
We know that CO2 can't make hydrogen bonds. CO2 is a non-polar molecule. Water is a polar molecule that can form hydrogen bonds. There's no way they're going to form an ideal solution. But we still know that if we don't have much of the solute,
that it's still a straight line initially before it goes haywire and curve. But that the straight line has a different slope than the actual vapor pressure of the pure material. And that's Henry's law, which we usually write
as the concentration in solution is proportional to the partial pressure of the gas. And this proportionality constant here is something different. So it's just some constant K. So the concentration in moles per liter is equal
to some constant times the pressure, the partial pressure of the gas in question above the liquid. Let's have a look then. At one atmosphere pressure, the concentration's .034 molar. And from that, we can get the Henry's law constant
because we can say it's .034 molar at one atmosphere. So Henry's law constant, K, is .034 molar per atmosphere. Now, if we're going to go up to really high pressures, we have to be careful because it won't follow Henry's law because if I keep squeezing and squeezing
and squeezing and pushing CO2 in and disrupting the water, just making it go in, then it may not behave the same way. But we're going to be below this because the partial pressure of CO2 in the atmosphere is very tiny. So it will follow Henry's law.
And we can use this constant, K. And then we just insert the other pressure. We say the concentration and solution in moles per liter is equal to the Henry's constant, which we figured out, times whatever the pressure is of CO2 in the atmosphere, which we said was 3.8 times 10 to the minus 4.
And so we end up with 12.9 micromolar as the concentration of CO2. We have to be a little bit careful here for the following reason.
We're assuming that CO2 doesn't react with water. But CO2 does react with water. Over time, we get carbonic acid, H2CO3, that's part of the source of your misery when you're sprinting toward the tape, is producing too much
of that and making your blood acidic. And unfortunately, I think it's going to be the source of some misery going forward because ocean acidification results from dissolved CO2 reacting
with carbonate, which is in the ocean. And unfortunately, carbonate is what all the marine organisms use, like coral and anything with a shell. Carbonate is what they use to make their exoskeleton.
As we increase the CO2 partial pressure in the atmosphere, we increase the concentration of CO2 in the ocean, but now we're talking about a separate problem that's nothing to do with the planet warming up. This is a separate problem.
And if we consume then all the carbonate ions in the ocean by adding a lot of CO2, millions of tons, then there's less around for these little guys to make their skeletons out of.
And here's what the measurements show. Here is the CO2 stops at 2,000. I should get some more data. It's not only going up, not only is it going up faster than linearly, but it seems to be going something like this
as everybody in China decides to drive an SUV. If you look at the seawater partial pressure of CO2, that's going up.
And if you look at the seawater pH, the pH is going down. And that means that the ocean is becoming more acidic, and that means that these guys are dissolving. When they dissolve, they die. When they die, there's no food for anybody on the planet
because most of our food comes from the ocean one way or another. If you simulate what happens to these pteropods, these guys are food for everything
from whales to everything else. And if you put them in the pH that we predict we're going to have in 2100 in saline, then their shell dissolves. They won't be able to survive this.
They will die. When they die at the bottom of the food chain, everything above them dies too. And then there's nothing there. So seafood may become scarce, and we may become scarce too
because that's nature's way. If you swing things way out of balance, you can do that for a while, and then suddenly, the pendulum swings back the other way. And the only way it can balance out is if we go extinct. Then everything will be hunky-dory again.
and it'll be paradise. Millions of other kinds of animals and we'll be gone. Like the dinosaurs. Okay, let's do another one. This is a different one. Back to the good old days.
Boy, what would a lazy professor do? You're sitting there and you're thinking, golly, should I pick up a new problem or should I just give them the Van der Waals gas? Because I told them to do it.
Cut, paste. That's one problem. We've got a mole of a Van der Waals gas. The constants are 5.46 liters squared atmosphere per mole squared for A, the attraction part.
And .05 liters per mole is B, which is the repulsion part. Something to do with the volume of the liquid. And we want to study it at half an atmosphere and 100 degrees C. And we want to know what volume it'll occupy because we're going to do this in a lab and we have
to get a container to do it. Well, we have to remember the Van der Waals equation of state. That's not too hard. You've seen that many times. And then we have to ask which variable we're solving for.
I think I've told you that if you're solving for A, that could be a variation. The volume's this, what's A? You've got to know B as well. Or what's P? That's all easy because you can isolate them easily.
With V, the problem is you can isolate V but you get a cubic equation and it takes forever to solve it and it's very prone to error. With N, N is a little bit trickier because N's on both sides. Never guess when the thing is on both sides.
At least get everything to one side. Because if both sides change when you guess, first this is too low and this is too high, and then you guess another one, this is too high, this is too low, it gets very confusing. Better to have the target be fixed so that you can hit it.
So V is on the left-hand side only and we start with some guess and in the absence of any better knowledge, the guess I choose is always what the ideal gas would predict. I start with NRT over P, I know all these guys.
I plug them in and then I just guess values of V until I hit the target. Let's try it. I always set up a table to make sure I'm not going around in circles.
I'm going to set up a table. I'm going to have V and then I'm just calling it F of V but all it is is the left-hand side. I put in .5 atmospheres for P. I put in 5.46. The number of moles is one but if it isn't, be careful because it's N squared over V squared.
B and then the number of moles is one, minus .05 there and I just plug in and the first, the NRT, the value I want to get, I put in T 100 Celsius and so forth, I get 15.31 liter atmospheres.
My first guess is NRT over P. P is half an atmosphere so it's 30.62, just twice this. So I guess 30.62. I shove it in here and here and I multiply it out and I get 15.463, 15.31.
It's too big. Now how too big is it? Well, deliberately make a mistake and then we will see how to correct the mistake. Here's what I'm going to do. I'm going to say that this is about 17 units too big or so
and so I'm going to subtract the same from this. I'm going to subtract the difference between these, saying I want to make this smaller so this comes out smaller
and if I, well, it's actually more like 15 I guess, sorry, 15. So if I subtract .15 from this, my next guess is 30.47 and then I see that it didn't come down to 15.31,
it only came down about half to 15.39 and then I say, why is that? The answer is because the pressure is half an atmosphere and it's basically P times V is the big guy and then the rest of the guys are small and since P is half an atmosphere, I have to go down twice as much as I thought.
So I still don't have it, it's still too big. But now I say, okay, it went down, I'm still 8 units too big so I want to come down 16 units, .16 from this because it's .08 too big
and I see that when I lower it, it comes down by half as much as what I change. And so my next quote guess seems very smart. I come down to 30.31 and I bingo, I hit 15.31 right away and then I'm done, okay?
Usually you can get it with three guesses and that's why guessing, although it sounds kind of like it's a joke, it's not. It's actually extremely powerful and very fast.
And of course, you've checked the answer because you've actually calculated it so you know it's right and you know you haven't hit a bad button because you see the trend. It's very robust. Okay, here's a challenge.
You set up a problem. I say set up this problem. But leave N unspecified and you tell me what N is, okay?
You know N should be 1 mole. Leave it unspecified. Guess the N from the ideal gas and then guess until you get it, okay? Because you can guess for any variable, not just V. Okay, let's do one more.
This one is definitely worth paying attention to. Suppose helium gas occurs at 7% by volume in methane, natural gas from fracking. What would be the percentage after one stage of effusion through a tiny orifice into a large evacuated chamber?
What would be the percentage of helium after two such processes in succession? So we're trying to purify helium and get it away from the natural gas.
And as you'll recall, the rate of effusion because if fusion is a random process, your chance of going through the hole depends mostly on how fast you're going. Because if you've got a big container and a small hole
and you're a little molecule or atom rattling around, if you're going ding, ding, ding, ding, ding, ding, ding, ding, ding, ding, ding, ding, basically you find your way out pretty quickly. If you're a big guy and you're moving around slowly, you're getting hit by all these flies and you're an elephant, it takes you a very long time
to find your way out through the hole because it takes you a very long time to go anywhere. And that's why the ratio goes like the square root of the masses because that's how the most probable speed, the mean speed, and the root mean square speed, they all have the inverse square root of the mass and then 3RT, 2RT, whatever on the top.
And T is the same for both of them if they're in the same container. So that goes away. So the rate of effusion, R1 over R2 is M2 over M1 to the 1 half power or the square root of M2 over M1. In round figures, methane is 16 grams per mole and helium is 4.
And therefore, the rate of helium compared to the rate of methane is the molar mass of methane, 16, divided by the molar mass of helium, 4, square root of 16
over 4, square root of 4 is 2. It's a little different from that, but that'll do. That's the rate, but that's not the enrichment because the enrichment depends on the rate and then how much of each guy you started with.
If we let the rate of effusion of methane be 1 relatively, and then that of helium is 2, it's only really the ratio that matters, then we can figure out the percent of helium. The percent of helium is how much helium got out.
It's how much we started with times 2, the rate at which it gets out, divided by the total, which is the rate at which methane gets out, which is .93, it's 93% methane, times 1 plus .07 times 2 because that's how much helium.
And I work this out and this number comes to be .131 or 13.1%. That's the enrichment after one stage of effusion. And then if I do another stage of effusion, I start with the 13.1 and I just do it again.
And I'll let you work out if you do it a lot of times, there's a slick way to work out the formula without going bum, bum, bum one after another. But anyway, for now let's just do the next one. We now start with 13.1% helium.
It effuses out again through the second hole twice as fast. And then we start with 86.9% methane because that's what's left. It comes out at rate 1 plus .131 again times 2. And this works out to be .232.
So now after two stages we're up to 23.2% enrichment. And as we get higher and higher enrichment, it gets harder and harder, unfortunately, to make progress. So for example, if the mixture's already 99% helium
and then we effuse it through the orifice, we enrich it to 99.5. And that means to get to 99.9 we still have to do a bunch more stages. And if we want 99.99, it's a bunch more.
And if we want it really, really pure, we're going to get really tired of doing this. And that's why if you take any material and you specify that its purity is really high, it's ever so expensive. No matter what it is.
Even pure water is relatively expensive. If you say it can't have anything in it, that's a special price. Okay? So you only specify the purity that you actually need, not higher. Okay. We'll leave it there and we'll continue on these kinds of problems on Thursday.
And we'll just go through them one by one. Yeah?