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### Spontaneity, Entropy, & Free Energy, Part V

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 0:56
• Equations: ∆G of Reaction, ∆G°, and K
• Example 1: Question
• Example 1: Part A
• Example 1: Part B
• Example 2
• Example 3
• lnK = (- ∆H° ÷ R) ( 1 ÷ T) + ( ∆S° ÷ R)
• Maximum Work

### Transcription: Spontaneity, Entropy, & Free Energy, Part V

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

Today, we are going to close out our discussion of spontaneity, entropy, and free energy.0005

I know we spent a fair amount of time on this, and in the earlier part of the lessons, we probably didn't get a chance to do too many actual examples, but that is not a problem--it's really very important that we actually get a sense of what is going on.0012

So, if there is any theme that I hope has sort of carried forward through all of these lessons, it's that if we understand what is happening, nine times out of ten the math should be able to fall out by just being able to see: this is happening, this is happening--putting pieces of the puzzle together.0026

Let's go ahead and continue our discussion of free energy and RT, ΔS, and the relationships that exist among them, and we will close out with a final discussion of the maximum work that is obtainable from a spontaneous process.0041

OK, so earlier we discussed what we are going to continue discussing--the equation--one of the primary equations that we have been dealing with, which is that the ΔG of the reaction (in other words, a reaction that is not necessarily taking place under standard conditions) is equal to the standard ΔG (25 degrees Celsius, 1 atmosphere, things like that...the standard conditions), plus RT ln(Q).0055

Q is the reaction quotient; it was the same as any equilibrium expression, except the values in the concentration terms (or the pressure terms, for gaseous reactions) are the ones at any given moment.0090

So, if we start a reaction, at any given moment we can go ahead and put those values into Q, multiply by the RT, find the ΔG, and that will give us the ΔG of the reaction.0105

That will tell us which way the reaction is going to go, based on those circumstances.0118

And remember, we said that a reaction is going to go toward bringing the ΔG to 0.0123

So, a negative ΔG of reaction means it's going to go to the right, as written (or, from your perspective, that way); a positive ΔG means it's going to be spontaneous in reverse as written.0132

That the means the reaction is going to go to the left.0143

A ΔG=0 (ΔG of reaction equal to 0)--that means that the reaction is at equilibrium, and that is what it is going to seek out.0145

A reaction seeks equilibrium, not necessarily completion; so completion is something entirely different.0154

OK, so let's go ahead and keep going with this.0160

At equilibrium, as we said, the ΔG of the reaction is equal to 0; so at equilibrium, let's set this left side equal to 0, except I'm going to write this over to the right and set it equal to 0.0165

So, we have ΔG, standard free energy, is equal to -RT (oh, wait, I'm sorry: let me go ahead and actually write out the equation, so that we can do the algebra): ΔG + RT ln(Q), which, under equilibrium conditions, becomes K.0180

That is the whole idea: the Q is for any time during the process, but when we have actually reached equilibrium, we are using the equilibrium constant, so ln(K) (I'm going to put eq down at the bottom...well, no; you know what, I'll just leave it as ln(K)) is equal to 0, because the ΔG of the reaction is 0; it's at equilibrium.0199

And now, we solve for ΔG=RT ln(K).0222

This relationship right here is profoundly important; this establishes a relationship between the free energy of reaction and the equilibrium constant.0229

Let's say it again: this establishes a relationship between the free energy of the reaction and the equilibrium constant of that reaction--profoundly important.0240

If we wanted to actually solve for K, we can keep doing some of the math; so it becomes like this--so we get: ln(K) is equal to -ΔG/RT, and the logarithm...the inverse function is the exponential, so we exponentiate both sides.0250

That cancels that, and what we end up with is that the equilibrium constant is equal to (oh, let's not get so crazy here)...0270

So, let me flip things around: the equilibrium constant of a given reaction is equal to e, raised to the power of -ΔG/RT.0282

And again, we want to make sure we watch our units: temperature is in Kelvin; R is the 8.3145 Joules per mole-Kelvin, or Joules per Kelvin (that is the unit itself)...so it's in Joules--that is what is important.0293

And the ΔG should be in Joules, whereas most thermodynamic data that is at the end of books, or in tables...the ΔG and the ΔH are in kilojoules, remember?0312

So, when working with thermodynamics, we have to make sure the Joules and kilojoules match; otherwise, the numbers will be completely off.0320

OK, so there you go: these two expressions are profoundly, profoundly, profoundly important.0326

This is pretty much it as far as thermodynamics; I mean, for all practical purposes, the two important equations that you should really take away from this are: ΔG=ΔH-TΔS (that is a profoundly important one) and the ΔG of reaction is equal to standard ΔG, plus RT ln(Q).0338

These two equations pretty much cover most of what you are going to be doing with thermodynamics.0362

If you are dealing with equilibrium situations, well, you just set this equal to 0, solve, and change this to K.0368

But this is the standard equation; these are the two that, if you are going to memorize anything, memorize these.0374

But again, when you are taking the AP exam, these equations are going to be available for you; so you just need to know what they mean.0380

There is a relationship between the free energy, the enthalpy, the entropy, and the temperature.0387

There is also a relationship between the free energy of a reaction in non-standard conditions, the free energy, and the reaction quotient (which is expressed as a relationship between the standard free energy and the equilibrium constant)--very, very important.0393

OK, well, let's do a problem.0410

OK, Example 1: Consider the reaction: nitrogen gas, plus 3 hydrogen gas (three moles of hydrogen gas) goes to 2 moles of ammonia gas (so I'll put g, so that we know that we are talking about all gaseous states here; it's going to be important).0414

Consider that at 25 degrees Celsius, which is 298 Kelvin.0443

OK, we want you to predict the direction in which the reaction will move (why don't I use up more room here) in order to reach equilibrium under the following conditions.0448

We are going to present two different situations, two scenarios, and you are going to decide which way the equilibrium is actually going to shift under those conditions.0481

Under the following conditions: A is going to be: The partial pressure of NH3 (if you haven't noticed yet, which I'm sure you have, chemistry...as most physical sciences--well, at least the hard physical sciences...is very notationally intensive) is 1.2 atmospheres.0491

The partial pressure of N2 equals 1.5 atmospheres; and the partial pressure of H2 is equal to 0.01 atmospheres.0516

What that means is: we are going to have a flask, and we are going to pump in 1.2 atmospheres worth of ammonia; we are going to pump in 1.5 atmospheres of nitrogen and only .01 atmospheres of hydrogen gas.0529

Well, the system is going to move until it finds an equilibrium position; and that equilibrium position is defined by the equilibrium constant.0544

Well, we are going to use our thermodynamics to decide which way we are going to move.0552

OK, and I'll just write down Scenario B, and then we'll do that after we do A, obviously.0559

p of NH3 = 1.2 atmospheres; p of N2 equals 1.5 atmospheres (the same), but the hydrogen concentration is 1.5 atmospheres (so very, very different; so we want to see what actually is going to happen--if it's going to make a difference).0567

OK, A: well, let's see...we are going to use, of course, our ΔG of a reaction is equal to the standard ΔG + RT ln(Q).0589

Now, Q, because this is at any given moment; we don't know if it's at equilibrium or not; that is what the ΔG of reaction is going to tell us.0604

We need to find this thing--we know R; we know T; we know ln (Q)--we'll do that one in just a moment.0612

Let me just go ahead and write the expression for Q: it's going to be, again, products over reactants; so it's going to be the partial pressure of NH3, squared, divided by the partial pressure of nitrogen, raised to the power of 1, times the partial pressure of hydrogen, raised to the power of 3.0621

The 2 and the 3--these are the stoichiometric coefficients, remember?--they show up, and when we are dealing with gas, gas, gas, we can just write it like this.0640

We don't have to write concentration, because gas pressure is a variation of concentration; it's just the flip side of concentration, moles per liter.0648

Remember, if you rearrange the PV=nRT, gas pressure and concentration are actually the same thing.0656

OK, so ΔG...well, ΔG is going to equal 2 times the free energy of formation of this, minus 3 times the free energy of formation of that, minus 1 times the free energy of formation of this.0663

Products minus reactants, multiplied by their stoichiometric coefficients (the number of moles)--using a table of thermodynamic data, we end up with 2 times (you know what, let me just write it out)...0680

2 moles, times -17 kilojoules per mole, minus 1 moles times 0 kilojoules per mole (right?--H2 and N2--they are elements; the free energy of formation of the elements is 0), plus 3 moles times 0 kilojoules per mole; so this is gone; what we are left with is a ΔG which is equal to -34 kilojoules (because mole and mole cancel, so we are left with -34 kilojoules).0697

OK, so now, let's go ahead and put our value in here, and put in the values--these values--into the Q.0737

We get: ΔG of the reaction is equal to -34,000 Joules (so again, we have to work in Joules, because R is in Joules per Kelvin), plus 8.3145, times the temperature, which is 25 degrees Celsius, which is 298 Kelvin, times the natural logarithm of K.0748

The partial pressure of NH3 squared (that is...let me do this in red...right here), 1.2, squared, divided by the partial pressure of N2, which is 1.50, times the partial pressure of hydrogen, which is 0.01, cubed: when we do all of this math (I'll write out both of the numbers, at least): -33,000, plus 34,180...which ends up being positive 1,180 Joules, or 1.18 kilojoules.0781

This is positive; so, since the ΔG of the reaction is positive--it's greater than 0--that means the reaction is spontaneous to the left.0860

This implies...so, the ΔG (well, let's actually move on to the next page here) is positive; this implies that the reaction will move to the left.0873

That means it is spontaneous to the left: ΔG--when it's positive, it's spontaneous to the left.0896

That means that N2 + 3 H2 goes to 2 NH3; when you put all of those in a flask, the reaction is going to move this way.0903

In other words, ammonia is going to decompose; nitrogen is going to form, and hydrogen is going to form.0917

That is what it means, "moving to the left"; it's going to move this way to reach equilibrium.0923

Now, let's do part B.0929

Part B: well, the same thing; we are going to use the same equation, ΔG of the reaction is equal to the standard ΔG, plus RT ln(Q), except this time, Q is going to be a little bit different.0931

When we put these numbers in, it equals -33,000, plus 8.3145, times 298; this part is the same, but now, the concentrations are different.0945

Now, we have 1.2 squared, over 1.5, 1.5 cubed; now, what we get is -33,000, minus 3,115, for a total of -36,115 Joules, or -36.1 kilojoules.0966

Ki--that's nice; kilojoules.0991

Now, ΔG of the reaction is negative; a negative ΔG means that it is spontaneous as written.0995

Negative: that means the reaction will proceed to the right until it reaches equilibrium, at which point it will stop, because that is equilibrium.1007

There you go: ΔG of reaction equals ΔG standard, plus RT ln(Q).1035

That is it: nothing fancy--a basic, straight thermodynamic equation; good.1043

Let's do another example (let me go back to blue here).1051

OK, let's see: so iron rusts as follows: the reaction is: 4 molecules of iron, solid metal (you know what, let me write this a little more clearly here...these stray lines again...), plus 3 molecules of oxygen gas, form 2 molecules of Fe2O3.1061

Well, Fe2O3 is not really a molecule, per se; it's an ionic compound, so it's sort of an extended crystal, if you will.1103

Now, find (well, I'll say--wow, these lines; I don't know what it is about these things--they are just all over the place; OK)...using a thermo table (in other words, the table of thermodynamics data), find the K--find the equilibrium constant for this reaction.1114

Find the K at 25 degrees Celsius.1148

OK, well, we know how to do that: they want us to find K; well, we know the reaction; we know the relationship is ΔG=-RT ln(K) (right?--we're at equilibrium).1152

We just solve that equation by setting the ΔG of reaction equal to 0; so let's go ahead and use this version of it: K is equal to e-ΔG/RT.1168

OK, so let's find our ΔG.1183

Well, our standard free energy change equals products minus reactants, so you're going to end up with 2 times -740, minus 4 times 0, plus 3 times 0.1186

Oxygen gas is an element--the free energy is 0; iron metal--anything that is elemental--anything that is in element form--free energy is 0.1203

So, that doesn't count; so the ΔG ends up being -1480 kilojoules (right?--2 moles, times -740 kilojoules per mole--that is -1480 kilojoules).1212

So now, we write: K is equal to e to the -1480000 (we have to work in Joules, because R is in Joules per Kelvin); R is 8.3145; and temperature is 298...I'm sorry; this is minus; so minus, minus.1229

That minus sign is not part of the ΔG; so minus...ΔG is negative...so it's minus, minus.1260

There we go; so we end up with (when we do the math) e to the 597.1267

Now, what can you tell me about e to the 597--is that a big number?1275

Yes, that is a massive number; it is absolutely massive.1279

That means that...when you see an equilibrium constant that is this big, that means that, once this thing reaches equilibrium--this reaction reaches equilibrium--it is so far to the right that, literally, there is absolutely no reactant left.1286

There is no iron, and there is no oxygen gas, left.1304

It is so far to the right--so, basically, this is what we would call completion.1308

Now, this is still equilibrium: the fact of the matter is, there are always going to be a few atoms of iron left--and certainly, oxygen you are never going to run out of, because oxygen is in the atmosphere.1313

The limiting reactant in this particular reaction is going to be the amount of metal that you have rusting.1326

Well, it's true--it is going to reach equilibrium--but see how high the equilibrium constant is: that means that the products are so much, and the reactants are absolutely 0.1332

This is virtually infinite--in fact, this is infinite.1343

So, it is equilibrium, but a high equilibrium constant pretty much tells you you are talking about something that is completion.1345

There is no reactant left; it is all product.1354

OK, we would call this thermodynamically favorable.1358

You will often hear this term: "Is this reaction thermodynamically favorable or unfavorable?"--yes, as written, this is thermodynamically favorable at 25 degrees Celsius.1364

The equilibrium constant is very, very high, and clearly, the ΔG is very, very, very negative.1375

Minus one million, four hundred and eighty thousand Joules--that is extraordinary: thermodynamically favorable.1382

So, in some sense, as you see, the thermal energy and the equilibrium constant are sort of almost the same thing.1398

They tell you what is going on; but there is a relationship between them, and that is it--they are not the same thing; I'm sorry.1403

OK, so let's do another example here.1413

Example 3: Using ΔG=ΔH-TΔS and ΔG=-RT ln(K), show that, for an endothermic reaction, a system already at equilibrium will shift to the right when the temperature is increased.1418

OK, so let's make sure we understand this: we want to use these equations, and we want to demonstrate quantitatively, with an equation, that when we have an endothermic reaction that is already at equilibrium, that if we raise the temperature, then it's going to push the equilibrium to the right, meaning it's going to keep going (no, I'm sorry--shift to the right--yes, that it will actually shift and push the equation to the right)--meaning more product will form.1498

We want to demonstrate this with an equation.1525

Well, we already know this from Le Chatelier; so, you remember, if you have some A + B going to C + D, an endothermic process is one that absorbs heat; and when we think about heat and Le Chatelier's Principle, we just think about it as another reactant.1528

If we add heat to the system, well, Le Chatelier's Principle says that the system is going to react by trying to compensate for what it is that I have done.1546

If I add heat to the system, it's going to try to deplete that heat; the only way to use up this heat is to pull the reaction that way.1554

By pulling the reaction that way, heat is going to be used up, so it pushes it to the right.1562

Well, again, heat is just another product--so, like anything else--if I add A to the system, it's going to push it right; if I add B...well, if I add heat, it's going to push it to the right.1567

So, we know this qualitatively--we know that it's going to move to the right; now, we want to demonstrate--we want to show that the work that we have done in thermodynamics confirms what we have done earlier with discussions of equilibrium.1577

That is what we want to do; and if the equations fit, if the equations actually confirm this, that is really, really good evidence that we are on the right track--that these are valid equations.1589

OK, so let's do the math.1602

Well, we know that ΔH...we just wrote down ΔG=ΔH-TΔS; and we also have that ΔG equals -RT ln(K); so just set them equal to each other.1605

So, Δ...I'm going to write, actually, this way: I'm going to write -RT ln(K) is equal to ΔH-TΔS (right?--because they are both equal to ΔG, so they are equal to each other; it's the transitive property of mathematics).1620

Well, let me divide by -RT; so, I get ln(K) is equal to ΔH over (well, actually, any time you divide by a negative, you should really just go ahead and put the negative on top there, in the numerator) RT, and then plus ΔS/R.1640

OK, so this is the equation that we have; so now, let's see what happens for an endothermic process.1663

Well, endo- means that the ΔH is positive, right?1669

OK, if the ΔH is positive, that means this term here is going to be negative.1679

ln(K)--I don't need to exponentiate this; as K rises, ln(K) rises; so as ln(K) rises, K rises; I can leave it as logarithm--it's not a problem.1689

Logarithm is just an accounting tool...it's just a way of making math easier--it doesn't actually mean something different.1700

So again, ΔH is positive for an endothermic reaction; that means this whole term is going to be negative (right?--negative of a positive number).1706

Well, now, what happens...so, that means that the equilibrium constant depends on this term and this term: this makes it more positive; this makes it more negative.1715

Well, what happens as I raise the temperature on this term?1726

Watch: as I raise--as this temperature increases, as this number goes up, this whole term becomes smaller.1729

Because it becomes smaller, this whole thing becomes bigger: it has less of an effect on the total sum.1739

Let me say that again: "Endothermic" means that the ΔH is positive; if ΔH is positive, that means this whole term is negative; well, if this whole term is negative, as we raise the temperature for a system at equilibrium, this whole negative number gets smaller negative.1749

This doesn't change; so, if I have...for example, if this were 100, and this were -50, the sum would be 50.1769

Well, if this were 100, as temperature rises...this is in the denominator; if this is in the denominator, that means this fraction becomes tinier and tinier.1780

Let's say now it becomes -10; well, now, 100-10 is 90; so we have gone from 50 to 90; we have gone up.1788

The K actually increases; the logarithm increases--the K increases.1797

When K increases, that means it is pushing the reaction to the right.1805

That is the whole idea behind the equilibrium constant: as the equilibrium constant rises, that means the reaction is going to the right; as an equilibrium constant declines, that means it is going to the left.1810

When the equilibrium constant is what it is supposed to be, that is measured, that means the system is at equilibrium.1821

I hope that made sense; just follow the math--that is all that is going on here.1829

OK, and the same thing if you want to go ahead and go through the process for an exothermic reaction--but I'll let you think about that one.1834

Let me go ahead and erase these and make some more room.1842

Well, actually, you know what, I think I'll put those numbers back; I'm sorry--yes, let's go ahead and put those numbers back.1851

Let's say, for example, if this were 100 and this were -50 to start, we are looking at 50 for ln(K).1856

But, as the temperature rises, this gets smaller and smaller, so this is going to go to, let's say, -40; this doesn't change; this goes to 60; this is going to go to -30 as the temperature rises; this temperature is in the denominator; this is 100; this is going to be 70; and so on.1869

The ln(K), is going to rise; that means the K is rising; as the K rises, that means it's going to move to the right.1885

OK, now let's rewrite this equation here: so this was the end of our example; now we are going to do something to this equation.1895

ln(K) is equal to...I'm going to pull out a...-ΔH/R, times 1/T, plus ΔS/R.1905

I pulled out...I separated this into 1/T and a little coefficient here.1919

Well, for a given reaction, under certain conditions, ΔH/R, ΔH, and ΔS are fixed.1924

So, notice the form that this thing takes: it looks like: y=m, times x, +b.1931

It's a linear equation; so, when you plot 1/T, one over the temperature, on the x-axis...so let's say I ran a series of experiments--one of them is 0 degrees, one at 10, one at 20, one at 30, one at 40, one at 50, one at 60...all those temperatures, and for each of those temperatures, let's say, I measured an equilibrium constant, I would have a temperature and a K; a temperature and a K; a temperature and a K.1939

If I did--if I plotted the 1 over the temperature on the x-axis, and plotted the ln of the K on the y-axis (so let's make sure my x's and y's don't actually look alike here--"y-axis," and I'll make this a clear "x-axis"), then you will get a straight line.1976

It's fantastic--it's pretty incredible, isn't it?2010

And we can do this--we can measure equilibrium constants according to temperature--not a problem.2014

When we do a series of experiments at different temperatures and different equilibrium constants, we actually get a straight line.2019

Well, the slope of that line that we get gives us a way of finding the ΔH for that reaction.2024

The y-intercept for that line gives us a way of finding the entropy change for that reaction.2033

This is extraordinary--this is absolutely extraordinary--this is fantastic!2039

Really, it's truly amazing that you have two thermodynamic pieces of data, simply from temperature and an equilibrium constant.2044

You wouldn't think that that is the case, but there it is.2052

So, let's write out: the slope equals -ΔH/R, and the y-intercept equals ΔS/R.2056

That is actually pretty extraordinary.2076

Now, you are probably saying to yourself, "Well, wait a minute: if we do different temperatures, don't ΔH and ΔS actually depend on temperature?"2078

They do; however, within a narrow temperature range, these are actually very, very, very good approximations.2086

And again, we are not talking about 500 degrees; we are talking about 40, 50, 60, maybe 100 degrees; actually, the equations work out beautifully.2093

So, plot 1/T on the x-axis; plot the logarithm of the equilibrium constant on the y-axis; you will get a straight line.2103

The slope of that straight line is equal to that--it gives you a way of finding ΔH; and the y-intercept is this, ΔS/R; it gives you a way of finding the entropy change for that reaction.2113

That is pretty extraordinary; and, once you have ΔH and ΔS at a given temperature, you can find ΔG, because ΔG equals ΔH-TΔS.2124

You see how everything is sort of--the circle is closing in on itself in thermodynamics.2134

There is a lot going on with thermodynamics, but the equations that actually govern thermodynamic behavior are very, very few, and what seems to be a lot going on--it's not that there is a lot going on; we just need to wrap our minds around new concepts.2139

That is the thing--that is all that is going on here.2154

OK, so let's close off this discussion with a discussion of a concept: the idea of maximum work.2156

I'm just going to go ahead and write the equation: Workmax is equal to ΔG.2166

So, in other words, when I have a given process that has a given amount of free energy, the maximum amount of work that I can extract from that process is equal to the amount of free energy for that process.2173

Let me write that out: The maximum work obtainable from a process at constant T and P is equal to the free energy.2189

Let's write this in a different way: Now, under certain conditions (conditions that you don't necessarily need to concern yourself at this point--maybe later on in your careers), ΔG (I should say ΔG, not ΔG standard) is the amount of energy free to do useful work.2222

Unlike heat--heat is not very...you can't really use heat; it's just sort of a very disordered form of energy.2260

But free energy of a process actually can be used to do work for us; that is the whole idea.2268

That is what a battery does--do work.2275

OK, under certain conditions, ΔG is the amount of energy that is free to do useful work for a spontaneous process.2280

Obviously, if a process is not spontaneous, that means we have to do work to make it happen; so, this is for a spontaneous process.2291

What is a spontaneous process?--it's one where the ΔG is negative.2302

OK, now, if the process is not spontaneous, well, what ΔG tells you--ΔG is the minimum work we have to do (which we don't want to do) to make it happen.2307

Well, that is the whole idea: we want to deal with spontaneous processes, because, if a spontaneous process as written has some free energy available to it, we want to be able to exploit that process--we want to find a way to extract, to have the process actually take place, so that we can take the free energy and use it to do useful work.2342

If a process isn't spontaneous as written...so, for example, if I have a certain reaction that I really, really like, and I would really like to do, because it would make our life easier...well, if thermodynamically it's unfavorable--if you have a positive ΔG--well, you are not going to get anything out of it.2362

In order to have that process work, you are going to actually have to do work to make it happen.2379

Well, what is the point of us doing work?--the whole idea is to be able to extract work from a process that happens spontaneously!2384

This is why people look for catalysts: we know that there are a whole bunch of processes that are thermodynamically favorable--the beginning is here; the ending is here--lots of free energy change.2391

We want to be able to extract that free energy, but...so, thermodynamically, it's favorable...but remember that region in between, the kinetics?2402

It doesn't happen quickly: if we can find a catalyst to lower that activation energy so that we can make this process happen quickly, we can actually use that energy to do work.2409

But for a non-spontaneous process where the reactants are here and the products are here, the ΔG is positive...why bother looking for a catalyst?--because a catalyst isn't going to do anything.2420

The reaction is not going to happen spontaneously.2431

If we want to make it happen, we are actually going to have to push it uphill; that is a waste of energy.2433

So, this is the value of doing a thermodynamic analysis: we don't want to be chasing after a process if the process is not going to give us anything.2439

The history of science is replete with things like this: so, we want to understand what is going on underneath, so that we can make decisions about future things.2450

We don't want to end up trying to roll a rock up a hill, thinking that rolling a rock up a hill is going to get us something, because it's not.2459

It is just going to be a waste of energy.2466

OK, so final comments: in any real process (which, of course, means everything in life), the work obtained is always, always, always (oh, it might be nice if I actually spelled that properly) less than Wmax, because some of the energy (most of the energy, in fact) is always lost as heat.2470

In other words, the entropy of the universe, because it has to increase--a spontaneous process means ΔS of the universe is greater than 0--that means the entropy of the universe has to increase.2532

If the entropy of the universe didn't increase--if it stayed that way--then yes, we could obtain all of the maximum work.2546

But, because that is the case--because energy is lost as heat to the universe--the entropy of the universe increases.2553

That is what that means: entropy of the universe must increase.2563

That is actually the second law of thermodynamics.2576

OK, so in any real process, the work obtained is always less than the maximum work, because some of the energy is lost as heat.2579

So, if we were to take, for example, the gas that you burn in your cars: it actually releases a lot of energy.2588

Well, all that free energy is available to do work: the problem is, in the process of actually extracting it to do work, much of that energy--most of it--is actually lost as heat.2598

Even though we can, let's say, get 500 units of energy from a process, it is more practical--what we really end up with, if we are lucky, is maybe 100 to 150 units; that other 350 to 400 units is actually lost as heat.2613

There is nothing we can do about it.2630

So, this was put really, really beautifully by a thermodynamicist once, and he said it like this.2632

He said, "The first law of thermodynamics says that the best you can do is break even"--in other words, in conservation of energy, you can't get more out than you put in.2639

"The second law says you can't break even."2650

That is the whole idea; so there is a certain amount of energy that is available; spontaneous process--we can use that free energy to do work.2653

But, in the process of doing work, some of that energy is lost as heat; the other energy--only a fraction of that--is what actually ends up doing the work that we want it to do.2664

The heat that is lost--that is what increases the entropy of the universe.2673

That is why it's a spontaneous process.2678

Hopefully it's starting to sort of make sense, and some of the pieces are starting to come together.2680

OK, well, that pretty much finishes our discussion of thermodynamics.2686

In our next lesson, we are going to start discussing electrochemistry.2689

Thank you for joining us here at Educator.com.2694

See you next time; goodbye.2695