AP Chemistry Spontaneity, Entropy, & Free Energy, Part III

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to start on Part 3 of our discussion of spontaneity, entropy, and free energy.0004

Last time, we left off--we introduced this very, very important equation, which was...well, we introduced the notion of free energy, if you remember: so, ΔG=ΔH-TΔS.0010

Now, up until this point, we had been talking about entropy, and how the entropy of the universe (which consists of the entropy of the system, plus the entropy of the surroundings)--when that is greater than 0, that defines a spontaneous process, and how, if it is less than 0, then it's spontaneous in reverse, as written for a given process.0032

If it is equal to 0, that change in entropy of the universe, then the system is at equilibrium--nothing is going to happen.0055

Well, we did a little manipulating; we introduced this thing called free energy, and as it turns out, in order for something to be spontaneous, the free energy should be negative.0062

So, just to rewrite here--you can certainly go back to the previous lessons, but just a quick review: for ΔG less than 0 (which is negative), that implies a spontaneous process.0073

And, if it's positive--it's greater than 0--that means it's spontaneous in reverse.0084

A ΔG, free energy--if it's equal to 0, that means we are at equilibrium.0089

So, there is a relationship among the enthalpy of a given system, the entropy of a given system, and the temperature at which this particular reaction is taking place.0095

ΔG is dependent on these three factors: the enthalpy (or the heat, ΔH), the temperature, and the ΔS for a given reaction.0106

This is our primary equation, at least for this particular lesson--a very, very important equation.0116

What we are going to do is: now, we are actually going to start to get into the quantitative aspects: we spent a lot of time discussing entropy, discussing free energy, trying to wrap our mind around this notion; all of that is very, very valuable, but now, we're going to start actually doing some problems.0124

We are going to examine the relationship here by examining a process that you actually are very familiar with--the melting of ice, or the freezing of ice.0140

Because you are familiar with it, you will understand what the numbers mean, and it will make this equation make more sense in the context of your experience.0150

Let's just go ahead and start.0159

Example 1: We want to examine the melting of ice at 3 different temperatures--the melting of ice at -20 degrees Celsius, 0 degrees Celsius, and positive 20 degrees Celsius.0162

I have a tendency to put a positive sign on things that are positive, as opposed to just sort of leaving them out there.0182

The process that we are interested in is...so, as written: H₂O, solid, going to H₂O, liquid--in other words, the melting of ice.0186

OK, let's just go ahead, and we're going to set this up as a table, so that we can compare all of the numbers.0203

This might end up actually taking...well, you know what, I'll try to do this all on one page; hopefully I can do it here; let's see.0208

Let me move this one up a little bit, so that I'll have a little bit more room.0216

Our reaction is going to be (let me write my reaction in red): H₂O, solid (ice), going to H₂O, liquid (this is a 2; I know my 2's never look like 2's); now, let me go back to blue.0224

We have -20 degrees Celsius--that is going to be one column; we are going to have 0 degrees Celsius--that is another column; and we are going to have 20 degrees Celsius--that is going to be our third column.0238

OK, so the first thing we want to write down here is going to be, of course, the temperature, which is in Kelvin.0251

And then, we are going to write out the ΔH, which is going to be in kilojoules per mole; and then we are going to write the ΔS; OK, and I'll put a little circle at the top right--that means standard conditions (25 degrees Celsius, one atmosphere pressure for gases, one mole per liter concentration for aqueous solutions...just general standard values; these are the values that you find at the end in those thermodynamic tables).0257

OK, ΔS, which is going to be in Joules per mole-Kelvin: notice, of course, one is in kilojoules; one is in Joules; it is very, very important.0289

So, when we work this equation, we need to either convert this to kilojoules, or (more often than not) we convert the ΔH to Joules--because our units have to match.0298

OK, our next one is going to be ΔS of the surroundings; this ΔS right here--this first ΔS--remember, we said if it doesn't have a subscript, we are talking about the system--that is just a default position.0309

But, this one, ΔS surroundings--that is the one that was equal to the -ΔH/T.0324

So, ΔS of the universe--that is the sum of the previous two terms; we have the TΔS; and we have the ΔG.0331

So again, this is the equation that we are going to use, and we are just going to fill this table again.0343

OK, so the temperature--this is going to be 253 degrees; I'm actually going to work my way down.0348

253 degrees: the ΔH--well, what I do is: I look up in my thermodynamic tables: the ΔH of a reaction is the sum of the ΔH's of the products, minus the sum of the ΔH's of formation of the reactants.0356

I just look them up: H₂O, liquid, minus H₂O, solid; and I end up with 6.03x10³, 6.03x10³; actually, you know what, this is not in kilojoules per mole; I'm sorry, I already changed it to Joules per mole.0372

So, again, when you are working with ΔH and ΔS, you should probably work in Joules; you can do kilojoules, but it's probably best to work in Joules.0396

OK, the ΔS: the ΔS of the system--you do it the same way: what you do is: you look up in the thermodynamic tables for the entropy of liquid water, and then you look up the entropy--the standard entropy--of solid water (ice).0405

You take the products minus the reactants; when you do that, you get 22.1.0421

So again, these are all just state functions (the entropy, the free energy, the ΔH), and you have thermodynamic tables, so you can use them to find the Δs.0426

OK, the ΔS of the surroundings: this is the one that is interesting; remember, we said that the ΔS of the surroundings (let me write it down here in red--remember that equation, ΔS of the surroundings) equals ΔH, divided by the temperature, Kelvin?0436

Because, the ΔS of the surroundings depends on heat flow, and heat is ΔH.0455

So, if I take this number, divided by this number, negative sign, I'm going to end up with -23.8.0459

OK, ΔS of the universe is equal to the ΔS of the surroundings, plus the ΔS of the system.0477

When I add these two, I end up with -1.73; notice, the ΔS is negative--a ΔS that is negative--ΔS of the universe being negative--means that it is spontaneous in the reverse direction.0482

Let's do the TΔS; so we'll multiply the T times the ΔS of the system right here; and we end up with 5591.3; and then, when we calculate the ΔG (which is ΔH-TΔS), we end up with 439.0498

And again, this is in Joules per mole.0525

So notice: ΔG is positive; ΔS universe is negative.0530

One is just a different way of looking at the other; that is why we have this equation.0535

When ΔG is negative, that means it's spontaneous; when ΔS is positive (of the universe), that means it's spontaneous.0542

439 is not negative--it's positive; that means it's spontaneous in the reverse direction; but you know that already--at negative 20 degrees Celsius, your experience tells you that ice doesn't melt into liquid water; but at -20 degrees Celsius, liquid water turns into ice.0549

As written--I wrote it as solid to liquid--at -20 degrees Celsius, this ΔG and this ΔS of the universe say that it goes in that direction; it's spontaneous in reverse.0567

Now, let's do the 20 degrees Celsius first, before I go to the 0.0580

The temperature is 293; the ΔH (again, that is the same; that doesn't change)--that is 6.03x10³; the ΔS is 22.1 (that is still the same).0584

Now, when I do the ΔS of the surroundings (which is this equation, -ΔH/T), now T is a lot higher.0599

So, what I end up with is (not -23.8, but): I end up with -20.6; and when I add this and that together (this and that together--that is what gives me this), I end up with positive 1.5.0609

+1.5: that is my ΔS of the universe.0625

That means it is spontaneous as written.0629

I do my TΔS; I end up with 6475.3; and then, I end up with a -445 for a standard free energy change.0631

This is negative; ΔS of the universe is positive; that means it's spontaneous as written.0645

You know from your experience that at 20 degrees Celsius, above the freezing point of water, solid ice turns into liquid water without you doing anything; it just melts at 20 degrees Celsius.0651

It is spontaneous--no interruption from anything else; at that temperature, that is what happens.0662

Now, we will fill in the 0 column: 0 is 273 Kelvin; this is the same; this is going to be 6.03x10³; this is 22.1; this one is going to be -22.1.0667

When I add those together, I get ΔS is equal to 0; when I take TΔS, I end up with (guess what?) 6.03x10³.0686

ΔH-TΔS, that minus that: I get 0.0698

So, 0: free energy is 0; the system is at equilibrium.0707

That means water and ice--liquid water and solid ice--are at equilibrium; there is no net change.0711

Back and forth: ice is melting; water is turning into ice; ice is melting; water is turning into ice; it's a cyclic process--there is no net change.0718

So, as you see, in this particular case, what you know about ice is corroborated by what we have been talking about--I should say, what we have been talking about in terms of entropy and free energy is corroborated by your experience.0726

At -20 degrees Celsius, as written, water will freeze; that is what this says--the free energy is positive; the ΔS of the universe is negative.0742

At 20 degrees Celsius, yes, solid ice will turn into liquid water; it's spontaneous as written--I wrote it from left to right, ice to water.0753

-445: well, for chemists--again, for chemists--we have this thing...we have this equation where we can calculate ΔH's; we can calculate ΔS's based on thermodynamics, and find the ΔG; for chemists, it's free energy that actually works, because most of the time, under chemical conditions, you are actually working with constant temperature and pressure.0765

This is why this whole notion of free energy comes into effect.0792

You are probably wondering, "Well, wait a minute: this ΔS actually tells us the same thing"--you are right.0797

It does, but free energy, as it turns out, ends up being something that we can actually measure, better than we can entropy; it's easier to deal with.0801

That is the reason why we actually use free energy, as opposed to working with entropy itself--the entropy of the universe.0812

This is just the free energy of the system--that is the nice thing about this: this is talking about the system.0820

This ΔS has to be positive for a spontaneous process; it has to be ΔS of the entire universe, so I have to take the entropy of the universe, the entropy of the surroundings and the entropy of the system, into consideration--whereas, with this ΔG, I only have to worry about the system.0827

That is what is nice about it.0844

OK, so what can we take away from all this?--so all these things (namely temperature, enthalpy, and entropy) contribute to spontaneity.0845

But, temperature has the capacity to shift the direction of spontaneity.0872

You just saw that...well, let me see; actually, should I say...yes, that is fine; so, in the example that we just did, you saw that, at higher temperatures, one term dominates; at lower temperatures, the other term dominates.0899

So, it controls which direction a reaction is going to be spontaneous in, at any given moment.0925

ΔH and ΔS for a particular reaction--they are going to be fixed; you can't do anything about that, but what you can change is the temperature.0932

And by changing the temperature, you can actually take a non-spontaneous process and make it spontaneous.0941

Or, by lowering the temperature, you can take something that is very, very spontaneous and maybe ease it up a little bit.0946

Temperature is your control device, in this case.0952

OK, let's do another example: so we have Example 2: At what temperature is the following process spontaneous?0957

OK, the process that we are concerned about is: CH₃OH (which is liquid methanol)--at what temperature will it actually become gaseous methanol?0981

OK, we want to know: At what temperature is the following process spontaneous?--the process is liquid methanol turning into gaseous methanol.0997

The boiling point--that is what they are asking for.1006

"Spontaneous process"--meaning, if you don't do anything to it, at what temperature will this happen without any interference, just naturally?1009

OK, well, let's calculate what we are going to use: we are going to use ΔG=ΔH-TΔS.1016

OK, this is what we are looking for now: in this case, we are looking for temperature.1028

Well, what does "spontaneous" mean?--"spontaneous" means that ΔG is going to be less than 0; so we want ΔG (which is equal to ΔH-TΔS) to be less than 0.1033

In other words, we want ΔH to be less than TΔS (and I'm going to...well, I'll just leave it as is); so T--we want the temperature to be greater than ΔH/ΔS.1045

Well, we can calculate ΔH and ΔS from thermodynamic data in the back; so let's do it.1063

Let's see: ΔH, when I calculate it (oh, that's fine; I'll do it down here), equals the products minus the reactants; what you get is -201, minus a -239, equals 38 kilojoules per mole.1070

OK, so this is an endothermic process; you knew that anyway.1098

A liquid going to a gas--it needs heat, so this is an endothermic process.1101

The ΔS...when you look up the entropies in the back, you end up with 240, minus 127, equals 113; this is Joules per mole-Kelvin.1106

OK, this is kilojoules; this is Joules; so now, when you put these numbers into here, into here, and into here, we want T greater than ΔH/ΔS; ΔH is in kilojoules per mole; this is in Joules per mole-Kelvin; I need to convert this to Joules (that is the only thing you have to watch out for).1122

Temperature has to be greater than...it's going to be 38,000 Joules per mole, over 113 Joules per mole-Kelvin.1146

Joules per mole cancels Joules per mole, and you end up with a temperature greater than 336 Kelvin (or 63 degrees Celsius).1158

So, as written, this process--if the temperature is greater than 63 degrees Celsius, this will happen.1172

At 63 degrees Celsius, liquid methanol will start to become gaseous methanol--just automatically.1180

And, in fact, it will keep going that way until all of the liquid methanol is gone.1185

At temperatures below 63 degrees Celsius, it's the other way around: gaseous methanol will tend to condense and form liquid methanol.1189

At 63 degrees Celsius exactly, there will be an equilibrium between liquid methanol and gaseous methanol.1198

There will be no more gas methanol formed; there will be no more liquid actually formed; it will be in a dynamic equilibrium.1206

Some will evaporate; some will condense; it will go back and forth, but there will be no net change.1214

So, our magic number is 336 Kelvin, 63 degrees Celsius.1219

And again, we are just using our basic equation, ΔG=ΔH-TΔS.1225

And, in this case, we just solved for T, the key idea being "spontaneous": spontaneous is what allowed me to use this greater than/less than symbol.1230

OK, let's see: so again, watch your units: in your thermodynamic tables, ΔH's are in kilojoules per mole.1249

I'll talk about this again next lesson, but it can't be overemphasized: ΔH's are in kilojoules per mole; ΔG's are in kilojoules per mole; S's are in Joules per mole-Kelvin.1266

Let me write that down: n.b. means nota bene--it means make sure you get this one: ΔH is in kilojoules per mole.1280

ΔG is in kilojoules per mole.1297

S--in thermodynamic data, it isn't ΔS; in thermodynamic data, it's actually just the entropy of that species.1303

The entropy is Joules per mole per Kelvin.1312

OK, now, let's shift gears just a little bit: now, we said earlier that ΔS of the surroundings is determined by heat flow.1320

That is where the equation came from: ΔS of the surroundings is equal to -ΔH, over temperature; it's determined by the heat flowing in and out of the system, or into or out of the surroundings.1347

Well, what about ΔS of the system?--well, ΔS of the system--at least in reactions where reactants and products are gaseous--(let me write this a little better) is determined by the number of gas particles.1359

The idea is this: the more particles means more entropy--more chaos--more disorder.1420

More disorder, more chaos, higher entropy: that is it--that is all it means--but you know this already; I mean, that is the whole idea.1432

When we talk about entropy, we are talking about general chaos; if you have 500 gas particles versus 5 gas particles, well, clearly, the 500 gas particles is a more chaotic system: it has a higher entropy.1442

That is it; there is nothing here that is not intuitive.1455

Now, let's tie this together with a reaction, to show what this actually means.1458

Let's take the reaction of nitrogen gas, plus hydrogen gas, to form ammonia gas.1464

OK, notice: on this side, the reactant side, we have four particles of gas.1473

We have one molecule of nitrogen, three molecules of hydrogen; that is four particles of gas.1478

On the right side, we have two particles of ammonia gas; the identity doesn't matter--what matters is the number of particles.1485

We have two particles: the entropy of the products is less than the entropy of the reactants.1492

OK, the system, in going from this to this, has become less disordered; it has become more ordered; it has become less chaotic.1502

It just depends on which word you want to use: order, disorder...it just depends on how you think about it.1516

We went from 4 gas particles to 2 gas particles, so our entropy is actually less.1521

Now, that means, since the entropy of the products is less, then the entropy of the reactants--the ΔS--is going to be less than 0.1527

You know that already--not just quantitatively; you know that because you are going from something that is highly disordered (or more disordered than this), so the change in entropy is actually negative.1539

That is why, if you take the entropy of this side, it's going to be a certain number (let's say 50), and at this side, maybe 150.1554

It is always products minus reactants: 50-150 is a negative 100.1562

Again, when dealing with reactions that have gas particles as reactants and/or products, the entropy of the system is determined by the number of gas particles, on the one side or on the other, of the reaction.1567

OK, now, if we actually confirm this by looking up, in a data table, the S of this, the S of this, the S of this, and doing entropy of products minus entropy of reactants, we actually end up getting: ΔS is equal to -199 Joules per mole-Kelvin.1586

So, it is actually confirmed when we do the calculation.1609

OK, so now, let's do a final example here, Example 3: What are ΔH and ΔG for the above reaction (for the N₂ + 3 H₂ going to 2 NH₃)?1613

OK, well, here is our equation; well, we have the ΔS already--that is good; we need that: ΔS equals -199 Joules per mole-Kelvin.1641

When we do the ΔH, we are going to look up the ΔH of NH₃ gas, minus the ΔH of this and this, and if you remember, the ΔH's of formation for elements (hydrogen gas and nitrogen gas) is 0, so it's only that number.1656

You end up with -92 kilojoules...actually, this is Joules per Kelvin; we have already accounted for the moles.1672

We have already accounted for the moles because we are taking 2 times the entropy of this, minus 1 times the entropy of that, plus 3 times the entropy of this.1688

We have already canceled the moles, so it's just Joules per Kelvin; my apologies.1697

And, this is just kilojoules, not kilojoules per mole.1701

OK, so now, we just plug it into the equation, making sure to match our units; we get ΔG=ΔH-TΔS.1703

It equals -92,000 Joules, -298 Kelvin, times -199 Joules per Kelvin.1721

Kelvin cancels Kelvin; our final answer is going to be in Joules; and what we end up with is -32,700 Joules, or -32.7 kilojoules.1736

So, ΔG is -32.7 kilojoules; this is at 25 degrees Celsius, 298 Kelvin.1754

So, at 25 degrees Celsius, this reaction (under normal conditions--1 atmosphere pressure) is going to be spontaneous.1762

Now again, spontaneous doesn't mean fast; it doesn't mean this actual reaction will actually go; it might be something else.1773

This is a thermodynamic discussion: thermodynamics only has to do with energy at the beginning and energy at the end, or energy here and energy there; it just tells us what the relative energies are of products and reactants.1779

It is: if I left this reaction alone, eventually, it will happen.1792

It doesn't tell me how fast it will happen; that belongs to the domain of kinetics.1796

OK, thank you for joining us here at Educator.com to sort of start to round out the discussion of spontaneity, entropy, and free energy.1802

We'll see you next time; goodbye.1809