For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

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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

### AP Chemistry Online Prep Course

### 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 NH _{3} (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 N _{2} equals 1.5 atmospheres; and the partial pressure of H_{2} 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 NH _{3} = 1.2 atmospheres; p of N_{2} 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 NH _{3}, 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?--H _{2} and N_{2}--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 NH _{3} squared (that is...let me do this in red...right here), 1.2, squared, divided by the partial pressure of N_{2}, 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 N _{2} + 3 H_{2} goes to 2 NH_{3}; 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 Fe _{2}O_{3}.*1061

*Well, Fe _{2}O_{3} 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: Work _{max} 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 W _{max}, 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

1 answer

Last reply by: Professor Hovasapian

Mon Apr 27, 2015 2:54 AM

Post by Jinbin Chen on April 14, 2015

Hi, Mr. Raffi!

I am currently preparing for a national chemistry exam, and on one practice exam, there is a problem that asks to find the equilibrium constant for the dimerization of NO2 (g) to form N2O4 (g) after finding out the delta G of this reaction.

We have to use the equation Delta G = -RTln(K), but when we find out the value of K, is this value Kp or Kc? And since we have to use molarity to calculate Kc, what is the unit we have to use to calculate Kp (atm, torr, or something else)?

1 answer

Last reply by: Professor Hovasapian

Wed Apr 8, 2015 3:44 AM

Post by Lyngage Tan on April 8, 2015

indeed wonderful lectures professor. ill be watching your series on physical chemistry as well. this is really helping me alot. i hope educator can also make a series on Analytical Chemistry which you can discuss more on gravimetric analysis and back titration problems.

1 answer

Last reply by: Professor Hovasapian

Sat Mar 28, 2015 10:29 PM

Post by Sherwin Yu on March 15, 2015

Professor Raffi,

We learned in the reaction equilibrium chapter, if we start with just nitrogen and hydrogen, the reaction goes forward and when the ammonia has accumulated to certain amount, the reverse reaction takes place, when the forward and reverse reaction rates are the same, the reaction reaches equilibrium. But according to delta G calculation, if a forward reaction at a given moment is negative, then the reverse reaction must have positive delta G. How can reaction occur simultaneously from both sides before it reaches equilibrium?

Thanks a lot for answering this question for me.

1 answer

Last reply by: Professor Hovasapian

Tue Jul 23, 2013 5:52 AM

Post by Alicia Morris on July 22, 2013

Hello Raffi,

At 13:30 for Example 1, should standard delta G have been -34,000 instead of -33,000? On the line right above, we calculated -34,000 and I wanted to make sure I'm not missing a step. Thank you!

1 answer

Last reply by: Professor Hovasapian

Mon Apr 22, 2013 6:40 PM

Post by Antie Chen on April 22, 2013

Hello Raffi, a small question about the equation lnK=-(delta)standard heat/R*T+(delta)entropy/R

I can understand the infect of temperature in the -(delta)standard heat/R*T,

however when the temperature increase the entropy won't increase?

1 answer

Last reply by: ahmed alzeory

Mon Dec 3, 2012 9:37 PM

Post by Professor Hovasapian on December 3, 2012

Hi Ahmed,

I hope you're well.

I'm hoping you'll call me Raffi -- the Prof or Mr. is too formal for my taste.

Ahmed, as it turns out, I copied one of the numbers incorrectly -- the 33,000 J for Standard Delta G. -- so your calculation is correct, MINE is incorrect. For some strange reason I used 33,000 instead of the 34,000 J. The final Delta G for the Rxn is still positive, but it is YOUR number ( in Joules, not kilojoules) that is the right one, NOT 1180 J.

Thank you for bringing this up -- it is very much appreciated when the students using these videos find errors. I am notorious for silly errors like this....

Thank you, again. Take good care, and best wishes for a happy year.

Raffi

0 answers

Post by ahmed alzeory on December 2, 2012

hello Mr Raffi ( i dont know weather to call you prof or mr sorry)

for example 1 part A i got 129 kj i dont know y i keep getting that answer