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Lecture Comments (11)

2 answers

Last reply by: shashikanth sothuku
Fri Sep 5, 2014 4:54 AM

Post by shashikanth sothuku on September 4, 2014

Hello professor,

in the example you told, @110d centigrade  the delta s (sys + surr) is positive. that means that delta s (univ) is positive. but in general delta s (univ) in this real world is negative. how can i figure this out?

1 answer

Last reply by: Professor Hovasapian
Thu Mar 13, 2014 8:49 PM

Post by Christian Fischer on March 5, 2014

Hi Raffi, Nice lecture. I have a quick question

at 19 minutes you mention that the sign of Delta S_surroundings depends on heat flow: So if a system is Exothermix delta S_sur > 0, but does that also Always imply that the Delta S_system is always less than zero since heat is leaving the system?

2 answers

Last reply by: Professor Hovasapian
Thu Apr 25, 2013 1:30 AM

Post by Sdiq Al-atroushi on April 25, 2013

Hello professor,

DO you have any lectures on mechanisms (elementary steps, reaction intermediates, Rate determining step) and enzyme kinetics?

Thanks

1 answer

Last reply by: Professor Hovasapian
Sun Apr 21, 2013 4:12 AM

Post by Antie Chen on April 21, 2013

In the equation (delta)G=(delta)H-T(delta)S
the (delta)S is the entropy of system, but why it's system? why it isn't surroundings or universe?

Spontaneity, Entropy, & Free Energy, Part II

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Spontaneity, Entropy, Free Energy 1:30
    • ∆S of Universe = ∆S of System + ∆S of Surrounding
    • Convention
    • Examining a System
    • Thermodynamic Property: Sign of ∆S
    • Thermodynamic Property: Magnitude of ∆S
    • Deriving Equation: ∆S of Surrounding = -∆H / T
    • Example 1
    • Free Energy Equations

Transcription: Spontaneity, Entropy, & Free Energy, Part II

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

Today, we are going to continue our discussion of spontaneity, entropy, and free energy.0004

Last lesson, we spent a fair amount of time just discussing entropy, discussing what things are; we started off with a review of general thermodynamic principles.0009

But, it was entropy that we were concerned with, and we closed off by discussing the notion of spontaneity--that a spontaneous process is when the entropy of the universe is greater than 0.0019

If it happens to be less than 0, then it is spontaneous in the other direction.0031

So, let's just jump in, continue our discussion of entropy, and we're going to try to do it in the context of something that is familiar to us--something that we experience every day: the melting of ice.0038

We have an intuitive sense of what is going on when ice melts, or when water freezes; so we are going to use that to sort of wrap our minds around this notion of entropy, and how the entropy of the universe is broken up, and made up of two part--the entropy of the system and the entropy of the surroundings--and what that means.0051

And then, towards the end, we are going to introduce this thing called free energy.0070

Free energy is a way of discussing entropy; it's an accounting tool, if you will--it's a thermodynamicist's way of accounting for entropy without dealing with it directly.0074

So, let's just jump right on in and see what we can do.0086

Remember, we said that if the ΔS of the universe is bigger than 0, then we have a spontaneous process.0090

In other words, it will happen without our interference; it will just naturally tend toward that direction, as written ("as written," meaning if you wrote a chemical reaction, and it turns out that the entropy increases, then it will happen).0104

OK, so now, let's rewrite our equation: so ΔS of the universe is equal to ΔS of the system, plus ΔS of the surroundings.0119

OK, so each term--the system and the surroundings--the change in entropy of each of those, because they are separate entities--contributes to the tendency for a given process (I'll write "reaction") to proceed spontaneously.0134

So there we go: whenever I'm considering a particular process, a particular reaction, I have two things going on: I have the change in entropy of the system itself, which is the reaction itself; but I also have the change in entropy of the surroundings--the environment in which the reaction actually takes place.0180

Both of those entropies contribute to the ΔS universe; it's the ΔS universe that we are interested in.0200

So, we have to account for both of these things.0207

So, before we actually do the example, let me just write a convention that we are going to use; and this convention makes absolutely perfect sense--in fact, I probably don't have to write it, but I do want to be as complete as possible.0211

It is something that you know, just from your experience.0228

Adding energy to a system to a system (or adding energy to anything)--adding energy to something (I'll just say "something"--I won't say "system," because we speak about system and surroundings; you can actually add energy to the surroundings also, right?--an exothermic process is heat leaving a system, going to the surroundings, so you are adding energy to the surroundings) raises its entropy, because it causes (let's see...) an increase in the random motion of the molecules (or atoms, as the case may be--I'll put "molecules and atoms").0231

So, remember what we said when we defined what entropy is: entropy--think of entropy as chaos.0296

You know that, if I add energy to a system (like heat it up), the molecules--the atoms--they start to move faster; they are bouncing around more vigorously.0301

Well, that is more chaotic; it is not just sitting there sort of doing its thing--now, you have sort of supercharged the system, and now it's more chaotic.0312

So, adding energy to a system increases the entropy; that is just something I wanted to throw out there, because it's going to be an issue when we discuss what we discuss next.0322

But it's natural to think about that; I mean, you know that, just naturally speaking.0330

OK, so let's examine a system that we are familiar with (let me go back to...you know what...that is fine; let me go back to blue); so, let's examine a system we are familiar with, or "with which we are familiar," for those of you who prefer correct grammar.0336

OK, H2O liquid going to H2O gas--liquid water turning into water vapor: OK, we know that at lower temperatures (and by "lower temperatures," I mean temperature greater than 0, less than 100--somewhere above the freezing point, but below the boiling point--that is what I mean by "lower"), it is spontaneous to the left.0363

In other words, when the temperature is below 100 degrees and above 0 degrees, water vapor spontaneously turns into liquid water.0400

That is the whole idea; it is below its boiling point--below its boiling point, it wants to be liquid water.0410

Therefore, at those temperatures between 0 and 100, if there is any water vapor around, it will more than likely condense--well, not "more than likely"--it does; it naturally condenses; it's spontaneous to the left.0415

That is what it means, "to the left"--it means it goes from gas to liquid--"spontaneous to the left."0430

Now, at T above 100 degrees Celsius (which is boiling point), now it is spontaneous to the right.0438

In other words, liquid goes to water vapor; it's boiling: to the right.0449

In other words, if I took a beaker of water, and if I all of a sudden stuck it in an oven that is at 200 degrees Celsius--well, the water is going to be gone in about 5 minutes, because liquid water will spontaneously turn into water vapor.0458

That is the whole idea: I haven't done anything--I have stuck it into a set of circumstances where the temperature is high, and now, spontaneously, the water will evaporate.0471

OK, it's spontaneous to the right.0480

All right, so (now I'll do this in red) clearly, temperature has something has something to do (I'll actually write it out--temp) with whether ΔS of the universe is (I won't say "greater than 0"--I'll say) positive or negative.0484

So, at the lower temperatures, it's spontaneous to the left; in other words, as written, H2O liquid to H2O gas--it's spontaneous this way, which means that the entropy is actually going to be less than 0.0533

In other words, it's not spontaneous as written; perhaps it would be better if I eliminate this one.0551

As written, at lower temperatures, it is spontaneous this way--which means that the change in entropy is going to be less than 0.0556

As written--"spontaneous as written"--means that ΔS is greater than 0; that is for temperatures above 100.0566

This is how I have written it; OK--I hope that makes sense.0575

OK, so clearly, temperature has something to do with whether the ΔS of the universe is positive or negative; the question is how.0581

OK, how is this possible?0589

Well, we know that the ΔS of the universe is equal to the ΔS of the system, plus the ΔS of the surroundings.0593

OK, since it is temperature-dependent (this is temperature-dependent; we know that it is from our experience), that means either this is temperature-dependent or that is temperature-dependent.0608

One of these is--maybe both; well, let's take a look.0625

ΔS of the system (let's deal with this one first) is equal to the entropy of the product, minus the entropy of the reactants (right?--that is what ΔS is for a reaction).0630

Recall, our reaction is: H2O liquid to H2O gas; that is what is happening.0652

So, in this reaction, this is the product; this is the reactant.0663

Well, what do you think has a higher entropy--a mole of water gas or a mole of water liquid?0668

Which is more disordered--which is more random--which is more chaotic?0676

Well, you know this already: a gas that is bouncing around, all crazy, is more chaotic than the liquid, which is sitting there, all nice and peaceful.0681

One mole of liquid water at 25 degrees Celsius is about 18 milliliters (18 grams, 18 milliliters--1 gram per milliliter, so the volume it occupies is 18 milliliters).0692

At 100 degrees Celsius, one mole of water occupies a volume of about 31 liters; 18 milliliters versus 31 liters--there are a lot more possibilities for all of the water molecules to be occupying places all over those 31 milliliters than contained here.0709

This is ordered; this is disordered; therefore, the S of the products, the entropy of the products, is higher than the entropy of the reactants.0730

Therefore, ΔS of the system is positive; it is greater than 0, because the product entropy is bigger than the reactant entropy--entropy of water vapor is bigger than the entropy of liquid water.0744

Therefore, the ΔS of the system is positive; it is greater than 0.0767

OK, here is a statement: In fact, in general, ΔS of a system is normally a fixed quantity.0771

The system in one state and the system in another state (in other words, the products and the reactants)--they have sort of a fixed entropic value--a fixed entropy value.0800

There is not much that can be said about it; it's a fixed quantity.0810

OK, we want to discuss (let me rewrite this, actually--"it's a fixed quantity"...)--whenever you are dealing with the entropy of a given system, we are always going to be talking about some sort of a reaction: in this case, this is the reaction.0815

The change in entropy of the system is going to be a fixed quantity.0833

If I look at a table of thermodynamic data (the ones that I used for ΔHs, for enthalpies), there is also a column for entropy values--not ΔS values, entropy values.0837

I can find the entropy of all the products, subtract the entropy of the reactants, and I get my change in entropy of my system.0847

So, I can always calculate the change in entropy, if I happen to have a nice, complete table of thermodynamic data.0855

But, it is usually a fixed quantity: this quantity, as it turns out, is not temperature-dependent--it is the next quantity.0862

Now, I won't exactly--we won't--go into why this is the case, but this is the one that is going to be temperature-dependent.0871

OK, so we won't discuss why here; you may discuss it in your subsequent courses, in an actual formal course in thermodynamics, where you spend the whole quarter or semester in thermodynamics; you may discuss it there.0877

But know that ΔS of the surroundings (in other words, this one) is what is variable, and depends on the flow of heat at a given temperature.0898

As it turns out, it's the ΔS of the surroundings that is the one that is temperature-dependent.0942

It depends on heat flow--how heat is flowing into and out of the system (or, in this case, into and out of the surroundings).0948

OK, now we said before that a thermodynamic quantity (thermodynamic property) has two components: a sign and a magnitude.0957

It has two components: it has a sign, which indicates direction (for example, enthalpy--if enthalpy is negative, that means heat is leaving the system; if enthalpy is positive, that means it's coming into the system; so a sign tells us which direction this property is flowing), and 2) it has a magnitude.0979

Magnitude doesn't need to be explained; it is a magnitude.1008

OK, now, the sign of ΔS (surroundings) depends on the direction of heat flow.1013

Here is how it depends: exothermic, from the system's point of view--an exothermic reaction releases heat from the system into the surroundings.1040

The ΔS of the surroundings is greater than 0; that makes sense--here is why.1060

If heat is leaving (so let's say here is my boundary; this is my system; this is my surroundings)--an exothermic process: heat is leaving the system and going into the surroundings; well, we said that when we put energy into something, it increases the entropy of that something.1067

Well, heat is leaving the system (exothermic) and going into the surroundings; energy is going into the surroundings; the entropy goes up.1088

The entropy is greater than 0; there is nothing new here.1095

Endothermic: an endothermic process--the ΔS of the surroundings is less than 0.1100

OK, an endothermic process means heat is coming into the system.1109

Heat is leaving the surroundings; if you drop the energy of the surroundings, you drop its entropy; entropy is less than 0.1114

That takes care of the sign.1123

The magnitude: the magnitude of ΔS of your surroundings--that is what depends on temperature.1126

It depends on temperature, and here is the definition: the change in entropy of the surroundings is equal to negative ΔH, divided by absolute temperature (temperature in Kelvin).1140

ΔH is the change in enthalpy of the system, OK?1153

Any time you don't see a subscript, we are talking about the system; any time we talk about surroundings, we will say "surroundings."1161

But, if you don't see a subscript (let me put that in blue--if you don't see a subscript) on a thermodynamic property, it's the system's point of view.1169

If you don't see a subscript, it is the system's point of view.1183

So, the magnitude of the change in entropy of the surroundings is equal to the negative of the enthalpy, divided by the temperature.1200

Well, an exothermic process--the enthalpy is negative; negative, negative makes it positive; that is what this says.1210

If the enthalpy is positive (an endothermic process), negative means that it is less than 0.1218

OK, so let's go back to our H2O: so we have H2O, liquid, going to H2O gas.1224

The ΔS of the system, we said, is positive; and the ΔS of the surroundings (so let me write the ΔH for this--the ΔH for this is positive 44 kilojoules)--well, ΔS of 44 kilojoules--negative 44 over T (we are not worried about T just yet)--this is negative.1237

Now, you see: this is equal to ΔS of the universe; when the ΔS of the universe is greater than 0, the process is spontaneous as written.1263

When the ΔS of the universe is less than 0, the process is spontaneous in the reverse direction.1275

One of these--this is equal to this plus that; there is a positive term, and there is a negative term; depending on what these are...well, ΔS of the surroundings is negative; not only that, it is also temperature-dependent.1282

It is equal to -ΔH over temperature.1299

We see that, at lower temperatures, if this number is low, this quantity here is large.1305

Because it is large, it tends to dominate this sum; therefore, even though this is positive, the magnitude of this will be bigger than this magnitude, and you will end up with a negative entropy at lower...which means it's going to be spontaneous in that direction.1319

However, as you raise the temperature and raise the temperature and raise the temperature, once you get to 100 degrees Celsius or higher, once this temperature rises, notice, this is an inverse relationship; the temperature is in the denominator.1339

As the denominator gets bigger (the temperature rises), the magnitude gets smaller; therefore, the effect of the negative sign diminishes.1353

Therefore, it stays positive, and now it becomes spontaneous in this direction.1363

I hope that makes sense--let me say it again.1369

The entropy of the universe of this particular reaction consists of the surroundings and the system.1373

The system, in this particular case, is positive, because the products minus the reactants gives you a more chaotic system (less chaotic system: the change--products minus reactants--is positive).1380

The ΔS of the surroundings is equal to -ΔH /T; ΔH is positive; therefore, this value is negative; but the temperature is in the denominator.1393

As the temperature rises (as the denominator gets bigger), this whole fraction gets smaller; therefore, the effect of the negative sign diminishes.1403

It comes to a point where this completely dominates, and you end up with a ΔS which is greater than 0, positive, which means make it spontaneous as written.1412

That is why, above 100 degrees Celsius, liquid turns into water vapor gas.1420

Below 100 degrees Celsius, water vapor gas turns into liquid.1426

It's pretty extraordinary.1431

OK, let me actually draw it out: if we mark this as 0, well, at 25 degrees (you know what, let me make this a little bit smaller here, because I want a little bit more room)...so let's just say: At 25 degrees Celsius, this is lower temperature (between 0...).1433

If this is our 0, well, our ΔS of the system goes here, and then the ΔS of the surroundings is bigger; so we end up with a negative value.1454

This plus this is the ΔS of the universe; we end up negative, so it's spontaneous in the reverse direction: gas goes to liquid (or liquid doesn't go to gas--however you want to look at it).1472

At, let's say, 110 degrees Celsius (we start at 0), well, the ΔS of the system is going to be the same; and then, the ΔS of the surroundings is going to take us here; we're going to end up positive, and it's going to be spontaneous as written, from left to right--liquid water will become water vapor.1485

That is what is happening here; so, what is important is: ΔS surrounding equals -ΔH over T; this is the first of our great thermodynamic equations.1510

The change in entropy of the surroundings is equal to the negative of the enthalpy of the system, divided by the absolute temperature at which the particular reaction is being run.1526

That is it--a profoundly important equation.1537

Heat, divided by temperature; heat divided by temperature; Joules over Kelvin, Joules over Kelvin; that is entropy.1540

OK, let's do an example.1551

Example 1: OK, antimony is found sometimes as a sulfide ore.1556

In other words, it's antimony sulfide, as opposed to just pure antimony; we have to find a way to pull out the antimony--separate it from the sulfur.1579

OK, and it is Sb2S3.1588

Now, pure antimony is recovered by reducing the antimony ion in this ore to pure antimony metal, with iron metal (we use iron to reduce it).1593

Iron gets oxidized, and it becomes iron sulfide; well, here is the reaction.1620

Sb2S3, plus 3 moles of iron atoms, produces 2 moles of pure antimony metal, plus iron (2) sulfide, FeS.1628

Let's erase some of these random lines here.1644

The (oh, this is driving me crazy) ΔH for this is -125 kilojoules.1650

This is an exothermic reaction; as written, it releases heat.1661

The system is exothermic; OK.1667

So, we want you to calculate (just a little bit of practice, getting used to using the equation) ΔS of the surroundings at 25 degrees Celsius.1671

Well, OK; we know what ΔS of surroundings is: ΔS of our surroundings is equal to -ΔH, divided by T, is equal to negative, negative, 125 kilojoules, divided by 25 degrees Celsius.1685

Absolute temperature--meaning in Kelvin: so 298 K; we end up with 0.419 kilojoules per Kelvin, or (the more standard unit of entropy is Joules per Kelvin) 419 Joules per Kelvin.1706

There you go: so, that is what we wanted to sort of get to--our first real equation dealing with entropy that we were actually able to quantify in terms of something that we have dealt with before.1732

So, hopefully, the discussion and the process that we actually got to made sense; but this is our first primary equation.1753

OK, so now, we are going to introduce something called the Gibbs free energy, or free energy.1761

Let's just go ahead and start that discussion; we'll do some definitions, and then we'll leave it for now, and then we'll continue on with an example in the next lesson.1770

So, for a system, define free energy.1781

OK, free energy is symbolized by a G; it equal H minus TS; the free energy of a system is equal to the enthalpy of that system, minus the product of the temperature of that system times the entropy of that system.1800

Well, again, we are interested in Δs; so, at a constant temperature, we have ΔG=ΔH-TΔS.1820

I cannot, cannot, cannot overemphasize the importance of this equation.1845

The free energy of the system, the change in the free energy in the system (and we will talk about what we mean in just a minute) is equal to the change in enthalpy of the system, minus the temperature at which the thing is taking place times the change in entropy.1851

OK, now let's do a little bit of mathematical manipulation; so let me rewrite it: I will write: ΔG=ΔH-TΔS.1865

Well, we said that, at constant temperature and pressure, the ΔS of the surroundings is equal to -ΔH/T, right?1879

Let's move this around: I have -TΔSsurr=ΔH; let me take this, which is ΔH and stick it in there; I get ΔG is equal to -T, times ΔS of the surroundings.1909

-T times ΔS (and remember what we said--no subscript--it means the system--so I'll just go ahead and write "system"); that means ΔG is equal to -T, times ΔS of the surroundings, minus ΔS of the system; now let me divide by negative T; I get -ΔG/T (it doesn't matter where I put the negative sign; I always prefer to put it on the numerator, just to be consistent) equals ΔS surroundings (I'm sorry, plus, because I pulled the negative out).1934

Well, what is ΔSsurr, plus the ΔS of the system?--it's the ΔS of the universe.1971

ΔS of the universe--let's take a look at what we have just done here.1978

I have defined this thing called free energy; in a little bit of mathematical manipulation, I was able to come up with an expression that involves temperature for the ΔS of the universe.1985

We said that the ΔS of the universe--if it's greater than 0, we have a spontaneous process; the only way that it is greater than 0 is if ΔG is negative (because negative, negative makes it positive).2001

Let me say that again: "ΔS of the universe is greater than 0" implies that the process is spontaneous as written.2018

In order for ΔS of the universe to be 0, that means -ΔG/T has to be greater than 0, right?2032

That means -ΔG has to be greater than 0; that means ΔG has to be less than 0.2042

So now, we have a way of discussing spontaneity under conditions of constant temperature and pressure (which is most chemistry--it takes place under constant temperature and pressure).2054

Now I don't have to worry about entropy; instead of talking about entropy, I can just talk about free energy.2068

Free energy is a form of entropy; the universe--ΔS of the universe--I didn't have to break it up; so now, I have free energy, temperature, ΔS of the universe.2073

So now...if you remember, from the thermodynamic tables at the back of your book, they have three columns: ΔH (enthalpy) in kilojoules per mole; ΔG (which is also kilojoules per mole), and S (entropy, which is Joules per mole-Kelvin).2087

So now, if I want to talk about whether something is spontaneous, like a reaction, I don't have to worry about its entropy.2104

I can if I want to--it's not a problem--or I can just talk about free energy.2111

And again, the important equation is...well, actually, I'll get to it in just a second.2116

Let me just finish this part up--let me write it down--so: Under conditions of constant temperature and pressure, which is what matters, which is...actually, I should say which are...the conditions of interest to chemists (not just chemists, though), ΔS of the universe is equal to -ΔG/T.2122

ΔS universe positive implies that ΔG is negative, implies spontaneous.2192

There you go; and let me write the equation: ΔG=ΔH-TΔS.2209

I'm going to make a little bit of a broad statement here: it's the most important thermodynamic equation for chemists.2220

Certainly, you can argue this--but it's pretty important.2225

"The most important thermodynamic equation for chemists"; OK.2229

So, for a reaction (rxn), the enthalpy, the entropy, and temperature all contribute to whether the reaction is spontaneous as written.2241

That what this is all about--"is spontaneous as written."2269

The enthalpy (which I can get from the table of data); the temperature at which I am running the particular reaction; the ΔS (which I can get from a table of thermodynamic data)--if I take the ΔH, minus T (absolute, Kelvin) ΔS, and I get a ΔG; if that ΔG is less than 0, the reaction is spontaneous as written.2279

If the reaction's ΔG is positive, that means it's spontaneous in the other direction as written.2302

If the ΔG equals 0, that means the system is at equilibrium.2309

If the ΔG is at 0, that means the system is at equilibrium--very, very profoundly important.2314

OK, I'm going to go ahead and put little circles on top here.2320

If you remember what those circles mean, they mean "at standard temperature and pressure, and standard conditions" (meaning 1 atmosphere, 1 mole per liter for solutions, 25 degrees Celsius, things like that).2325

These are the values that are written in the table of thermodynamic data; so I think, let's go ahead and...it's nice to just go ahead and put these in.2338

So again, this is a profoundly, profoundly, profoundly important equation.2347

If you don't take anything away from your study of chemistry, at least take this away from it, because it will come in handy, and you will impress a lot of people later on in your graduate studies if you can actually draw out this equation and know what it's about.2351

It says that a spontaneous process depends on three things at constant temperature and pressure.2365

It depends on the (well, "under conditions of constant temperature and pressure"--mostly it's under conditions of constant pressure) enthalpy, the entropy, and the conditions under which you are actually running that particular process.2370

OK, thank you for joining us here at Educator.com for a further discussion of spontaneity, entropy, and free energy.2384

We will see you next time, and we will start working some problems in thermodynamics.2392

Take care; goodbye.2395