Lecture 22. Midterm 2 Exam Review.
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Transcript: English(auto-generated)
00:07
Okay, so mid-term 2 is coming right up. We're going to review it today. I'm going to tell you in some detail what's going to be on it.
00:26
Now, quiz 6 had an error on problem 2. I know several of you brought this to our attention. So we're going to re-grade all of them. All right, and basically what we're going to do is just credit everybody with problem 2.
00:44
The issue is for a particular reaction, this was supposed to be the reaction Gibbs, the standard reaction Gibbs energy with the superscript of 0 here. You can see that there is no superscript. So in reality, the correct answer would be
01:01
about K equilibrium, we can then conclude nothing. That would have been the correct answer. But we're actually just going to give you credit for the, everyone, give everyone credit for that problem. Okay, so we're going to be revising this histogram. I'm assuming that it's going to shift to the right.
01:23
I guess it would have to. We're going to give everyone credit for problem 2. All right, Steven's going to hold a mid-term 2 review session today, starting at around 5 over in Bren Hall. Okay, he's just going to work problems.
01:42
So if you think that might be useful for you, attend that. I think he might even be planning 2 review sessions. Steven, is that right? Okay. So, yeah, stay tuned to the Facebook page
02:01
and he'll let you know when those are going to be, but the one today is definitely 5. Is that right? Yeah. So almost every video that we've generated has been posted to YouTube. We're only missing the last couple. These videos have to be processed and edited
02:21
and it's very time consuming and so it's impossible to get these things posted immediately after the lecture. It may not be obvious to you, but there's a lot of work that has to be done between the time the raw video is taken on these two machines back here and it's put together in a form where the two video feeds are brought together
02:46
and all of this takes time. And so hopefully we'll have everything up through lecture 20 posted by the end of the day or so. Right now lecture 19 is still missing, but Sean's fixing that up for us and so maybe by the end of the day we'll be able to post that
03:02
if those are interesting to you. How many people care about these videos? Have used them? That many. Alright. So if you want to find these, there's two ways to do that, you go to YouTube, search on my name
03:22
and they all pop up in random order unfortunately, alright, and this list goes all the way down to all, includes all of them that are posted on YouTube or if you go to our lecture page, I try to post them manually and I have some degree of success at doing this,
03:42
but I'm usually behind what's on YouTube, in fact I'm always behind and so if you want the latest, just go directly to YouTube and search on my name and you can find it. Okay, so what's Midterm 2 cover? In principle it covers everything on this slide. All of Chapter 15, Chapter 16, Sections 3, 4 and 5,
04:07
Chapter 17, Sections 1, 2, 4 and 5, Chapter 19, first three sections of that, alright, all of this is thermodynamics. This is the only kinetics that we talked about so far
04:22
at the beginning of Chapter 19. And so as I thought about all of this material and writing this exam from a practical perspective I can't ask questions about all of this unless the questions are just trivial and on these exams we don't want to ask trivial questions,
04:41
we want to ask more probing questions and so what we've decided to do is actually only ask about thermodynamics. There won't be any kinetics on the midterm and in fact problem 1, there's going to be two problems on the exam,
05:03
problem 1 is going to relate to entropy in the Carnot cycle and the lecture that keys on that the best is Lecture 13 but if you look, Lecture 14 is there, Lecture 12 are also impacting this concept, alright, but Lecture 13 is sort of the central lecture relating
05:22
to this concept and problem 2 is going to be about the Gibbs energy of reaction, the equilibrium constant and things related to that and that's the central lecture, there's Lecture 17 it turns out. Okay, so I'm going to recap some highlights from Lectures 13 and 17 in this lecture, try to emphasize a few points
05:42
that maybe weren't emphasized enough the first time around. Alright, now I already made strong statements about what the extra credit problem was going to be and yesterday when I started to work on the exam I started to look through, I Googled and I started to find out what information you would be able to find
06:02
about this problem on the internet and I basically concluded that it would be hard for you to get the right answer on this extra credit problem. I found some really good papers about Le Chatelier's principle and the Haber process
06:21
but these papers are mathematically intense, I'm not showing you the last 8 pages of this paper but it's jammed full of equations and it talks about non-ideality and it talks about concepts that we didn't really talk about in class at all. So, I'd like in the next couple of slides to explain
06:44
to you what I was thinking, right? I'm not going to ask this question on the midterm exam, I'm going to ask a different question but I want to explain it to you. At the end of what I'm explaining here I want you to tell me if this is what you're going to write down.
07:03
So, if it was I'm going to apologize. Alright, here's the ammonia synthesis reaction. Alright, here's its equilibrium constant. If we double, hypothetically, if we double the partial pressure of nitrogen in order
07:23
for this equilibrium constant to stay constant and independent of the values of these partial pressures, this partial pressure of ammonia is going to have to increase by root 2 because the square of that ammonia partial pressure has got to double in order for this K to stay constant, right?
07:42
And so, of course, this conforms exactly to the prediction of Le Chatelier's Principle, right? If we increase N2 we expect to increase the amount of ammonia that's produced, alright? If we increase N2 here we push this equilibrium to the right
08:01
and generate more ammonia. That's exactly what Le Chatelier's Principle predicts. Alright, why is the Bosch process, the Haber Bosch process different from that prediction? Why would we expect it not to adhere to that prediction?
08:21
The key is in the catalyst particle. What I'm trying to represent here is an iron micro-particle on some kind of inert support, right? In the ammonia synthesis reaction in the Haber Bosch process, iron is used as a catalyst. What does the iron do, right? Basically what the iron does is it takes apart
08:43
and generates, so it takes apart the H2, takes apart the N2 and produces activated hydrogen and nitrogen on the iron surface. Now this isn't actually literally correct the way I've drawn this, but all I mean to represent by this is
09:01
that we've created on the catalyst surface activated nitrogen and activated hydrogen. Okay? And you can think about that, one way to think about that is that it pulls the nitrogen apart and produces a nitrogen species that's bonded to the hydrogen. It wouldn't have, this bond here is not meant
09:20
to represent a single bond necessarily. The presence of the iron catalyst, activated nitrogen and hydrogen are produced on the iron surface. Ammonia is formed from these activated species so the reaction is occurring on the catalyst surface. Once you activate the nitrogen and the hydrogen, then they can react to give you ammonia under conditions that are much milder than would exist
09:42
in the absence of the catalyst, right? That's the whole point of the catalyst is to reduce the pressure and reduce the temperature necessary to drive this reaction forward. Alright, now once you understand this picture, once you agree that this picture is qualitatively correct,
10:01
I think you can see that if we increase the partial pressure of nitrogen and we get rid of most of the hydrogen on this surface because there's conservation of catalyst area. In other words, if we saturate the catalyst with nitrogen, that doesn't leave very much room for hydrogen, does it? And so the point is if the partial pressure
10:22
of nitrogen is increased, the part will become saturated with nitrogen at the expense of activated hydrogen. That's the key, right? If I fill up these binding sites with nitrogen and I get rid of most of the hydrogen, is the reaction going to be more efficient at that point or less efficient?
10:41
Well, the first order, this catalyst particle is going to operate at peak efficiency when the mole ratio between hydrogen and nitrogen is 3 to 1 because that's the ratio we need to make ammonia, right? Right, so the zero order picture in your mind should be, we want 3 times as much hydrogen
11:00
on this iron surface as there is nitrogen. Under those conditions, we would naively expect the reaction to be most efficient and generate the most ammonia but that won't happen. If you increase N2, you will saturate these binding sites with nitrogen and you will make less ammonia.
11:21
In this case, the outcome would be the shift of equilibrium further towards the reactant state in defiance of the Le Chatelier's principle. Alright, this is the difference, alright? When you've got the catalyst particle, you have to think about what's on the catalyst surface, alright? If you can't generate a stoichiometric mixture
11:41
of the reactants on the catalyst surface so that they can react efficiently, then you're going to favor the reactants relative to the products. Does that make sense? How many people were about to say that on the midterm? Well, I apologize to you then.
12:06
Okay, I believe you. What we're going to do instead is we're going to pull a problem directly off midterm 1. Not a whole problem because they're pretty long but part of a problem, alright?
12:22
Now, it's not going to be exactly the same so don't memorize a problem. I'm going to change the numbers but it's going to be virtually identical to a problem that was on midterm 1. Now midterm 1, most of you didn't dominate midterm 1
12:41
and so I think this is a great idea because this is going to force everybody who wants to get these extra credit points to go back and look at midterm 1, make sure they understand how to do those problems because if you want these 10 extra credit points, I might make it even 20. I haven't thought about it yet. I'm going to write it today, you're going to have to know how
13:02
to do the problems on midterm 1 and I don't suspect most of you have gone back and gone through and figured what did I do wrong, how could I have gotten that right? Now you have to do that, okay? So I should have given more thought to the Haber-Bosch problem
13:22
for the extra credit but I think if we had assigned that problem it would have just been a disaster for everyone except this young lady right over here. Okay, so what we want to do in what's left of this lecture is zoom through this material
13:43
and hit the high points and zoom through this material and hit the high points, okay? And then it's going to be, you don't have to worry about kinetics on Friday, there isn't going to be any kinetics on the midterm exam, right? You've got to focus on these two topics which I think are really at the crux of the thermodynamics
14:03
that we talked about since midterm 1, right? You can focus all your attention on that, you can work on nothing but those kinds of problems. So problem 1 is going to be all about entropy in the Carnot cycle. We've got two definitions for entropy, we've got a statistical definition for entropy, K log W
14:22
and we've got a thermodynamic definition for entropy. This is what it looks like when we're talking about a reversible process, right? DS is DQ for reversible process divided by T. When we're talking about the Carnot cycle, heat is fluxing in and out of this cycle, all right?
14:43
We're converting heat into work but there's no mass fluxes, okay? So we're not talking about an isolated system, we're not talking about an open system, we're talking about a closed system, right? Closed in the sense that there could be fluxes of heat and energy, all right, but no fluxes of material, right?
15:05
So the Carnot cycle is the most efficient existing cycle capable of converting a given amount of thermal energy into work or vice versa. That's what this little schematic diagram is supposed to indicate. This is a hot source, right, a reservoir,
15:25
this is a cold reservoir, all right, and we're extracting heat between the hot reservoir and the cold reservoir and some of that heat flux gets converted into work, right? That's what we're trying to represent here. So what Carnot showed us is that we can think
15:43
about a process that looks like this one, all right? What are we looking at here? This is a pressure volume plot, this is an isotherm at T1, this is an isotherm at T2, this is the starting point for the Carnot cycle up here at T1, this is the higher temperature and so there are four processes
16:04
that make up every Carnot cycle, right? There's a reversible isothermal expansion, then there's an adiabatic expansion, volume's getting bigger here, getting bigger here, now there's a compression, isothermal compression and an adiabatic compression. Those are the four steps.
16:23
We spend so much time talking about this Carnot cycle because every pressure volume process can be carved up into little tiny Carnot cycles and if you think about this, you know, once you've got this mosaic of Carnot cycles, all right, this downward process
16:44
for this guy is the upward process for this guy and so these lines where you've got adjacent Carnot cycles like this, that line goes away because that cancels with that and this cancels with this and that cancels with that.
17:04
So the borders of these Carnot cycles that are mosaic together all disappear and we're left with a boundary that looks like it would be like a sawtooth but if you imagine that there's 10,000 of these Carnot cycles it starts
17:20
to smooth out and as you increase the number of Carnot cycles you start to match up with this purple border and consequently it's telling you about the efficiency of this random process even though this doesn't look anything like a Carnot cycle, the Carnot predictions apply to it. That's why it's so important.
17:43
Now, normally in the Carnot cycle we're talking about a reversible, reversible processes and we're talking about an ideal gas but it also places a rigid upper limit on the efficiency that you can achieve for a real process with real gases that are not acting reversibly.
18:00
It places a rigid upper limit on the efficiency of those processes. It's important for that reason too even though we're not usually talking about ideal gases, reversible processes. So what do we know for sure about this Carnot cycle? Well, this is an isothermal expansion here.
18:21
The internal energy of an ideal gas only depends on its temperature. The internal energy of an ideal gas only depends on its temperature. It doesn't depend on anything else. Consequently, this is an isotherm so the internal energy of the gas is not changing anywhere along this isotherm.
18:41
It's constant. The internal energy of the gas is constant on the isotherm because the internal energy of an ideal gas only depends on its temperature. Here the temperature is not changing. So if that's true and this is an expansion
19:00
so work is less than zero that means that Q has to be greater than zero because Q plus W has to equal zero. So we know Q is greater than zero here. In other words, we have a flux of heat into the system.
19:22
Then down here, same story, this is the other isotherm at colder temperatures. That's a colder temperature than that. Internal energy of an ideal gas only depends on temperature and the temperature is constant here and so that's not changing. Work is greater than zero so Q has got to be less than zero so there's a flux of heat out of the system.
19:44
These two guys are adiabats so there's no fluxes of heat there at all by definition because their adiabatic Q is zero. That's what we know before we even start thinking too hard about this. The efficiency for a Carnot cycle is defined
20:01
by this equation. Don't try to derive this equation. It's just a definition. It's the work that's performed in response to the temperature gradient divided by the heat that's absorbed by the system. Just the absorbed heat. That's only the heat at the high temperature,
20:20
the heat flux at the high temperature. Okay, so it's, that's the work, that's the heat, work over heat, heat transferred from the hot reservoir to the system. That number doesn't matter at all. All right, we get 25 percent for this process. All right, how efficient is the Carnot cycle?
20:44
We can write the efficiency in terms of the temperatures of those two reservoirs, the hot reservoir and the cold reservoir and we went through and derived this equation laboriously. All right, we can write the work that happens in each one of those four steps.
21:01
This is the isothermal expansion. This is the adiabatic expansion. This is the isothermal compression. This is the adiabatic compression. That is the work that happens in an adiabatic process. All right, how do we know that?
21:21
The internal energy of an ideal gas only depends on its temperature. The internal energy of an ideal gas only depends on its temperature. So, consequently, I can calculate the change in the internal energy and I know Q's not changing because it's an adiabatic process.
21:41
Q is zero, so that means the work's got to be that, right, because that's the change in the internal energy. The change in the internal energy equals work for an adiabatic process. All right, so what's confusing about this is that's the constant volume heat capacity
22:01
and the volume's changing. Isn't that confusing? I think it's confusing. All right, it's still correct. That constant volume heat capacity doesn't depend on the volume. It's volume independent. For an ideal gas, it's 3R over 2, full stop, right?
22:21
The constant volume heat capacity is 3R over 2. It doesn't depend on V. Okay, so now, just looking at this equation, I can see right away that these two terms are going to cancel because they're just going to be, that's just the opposite of that because all I'm changing is the direction
22:41
of this integration, right? Okay, and so I end up with these two terms. And then I can simplify those. These equations look like they can't possibly be right, but I have triple checked them. And I am confident that they are exactly correct.
23:01
All right, so there's this relationship between these two temperatures, T2 and T1. All right, that's what these T2 and T1 refers to. This is V1 and V4. That's V1 and V4. Okay, so this expression applies to any two temperature volume points on an adiabat.
23:25
All right, gamma is just CP over CV. What's CP? What is it numerically for an ideal gas? Five-halves R. All right, so this is five-halves R over three-halves R, right, for an ideal gas.
23:47
Okay, so we can make those substitutions and do some algebra, divide, substitute, my goodness. We get this expression once we've done that, right?
24:02
That's the total work. And the heat transferred from the hot reservoir only now, this is the total work for the whole Carnot cycle. Here's the heat transferred only from the hot reservoir. Of course, there's heat transferred in the cold reservoir as well, but we ignore that when we're calculating the efficiency. And so now we can plug in these numbers into the,
24:21
plug that into that, plug that into that, simplify it. We get this. Now, I'm not showing you this so that we can derive this expression for the efficiency of the Carnot cycle. I'm showing it to you because in the process of deriving this equation, we learn a lot about how to calculate entropy changes, internal energy changes, heat,
24:43
work, we can calculate that for each step of the Carnot cycle. That'd be a good thing to know how to do. What is the entropy change for each of the four steps? The reversible Carnot cycle, this is the expression that we have, the thermodynamic expression
25:01
for entropy is this equation right here. So since the entropy is a state function, we know that we can just add the entropy changes for each one of these four steps, boom, boom, boom, and boom. Not only that, because it's a closed cycle, because the Carnot cycle ends up where it started,
25:20
we know the total entropy change has got to be zero. Right? It has to be. It's a state function, so if I integrate over the whole cycle, no matter what that cycle looks like, as long as I end up where I started, delta S has got to be zero. That's what that means right there. Steps two and three are adiabatic, so Q is zero,
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so these two are easy to calculate. Right? That's an adiabatic process. That's an adiabatic process. Q is zero, so the entropy change is zero. For the other two processes, I can just calculate what the heat flux is from the work. All right? Those are the temperatures, and so this is the total. We already know the total cycle has got to be zero,
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so that means that has got to equal that, right? All right, so in terms of entropy, nothing happens for the adiabatic processes. Right? There's no entropy change for the adiabatic process. It's just a vertical line, here and here.
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Here's the entropy change that we can calculate for the isothermal processes. All right? So in terms of temperature and entropy, we get a box. That's what the Carnot cycle looks like. Okay? Now, the Carnot cycle applies to reversible processes.
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All right? What if the process is irreversible, the efficiency is lower, all right, it is bounded by the efficiency of the reversible process, but it can be equal to or less than that of the reversible process, and we went through and did a derivation of an equation
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that looks just exactly like this equation right here. All right? So this is called the Clausius Inequality. It says if we integrate the entropy changes over any cyclical process that does it work, not every step is reversible. All right? If there's one irreversible process embedded
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in some cyclical series of processes that end up at the same place, all right, you're ending up where you started, all right, this is going to be less than or equal to zero. Yes, that's Clausius. Okay, and so this Clausius Inequality leads directly
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to the second law of thermodynamics, which says that for all processes, reversible and irreversible, DS is greater than or equal to greater than for irreversible processes and equal to for reversible processes, DQ over T. This is the most
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rigorous statement of the second law of thermodynamics. And we derived it by doing this and then talking about this and we talked also about this and noticed that and then we got to this. Okay, so this equation makes predictions
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about three types of processes. There are spontaneous and irreversible processes where that is an inequality, right? DS is going to be greater than DQ over T. For reversible processes, it's equal. For processes that are not spontaneous, it's less. That's why the entropy is important. It tells us about spontaneity.
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Okay, some simple but important examples. Reversible phase transition. For phase transitions, there's a well-defined enthalpy of the phase transition. All right, we can look it up. There are tables of enthalpies of phase transitions
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and the change in the entropy is just that delta H, a change in the enthalpy divided by the temperature at which the phase transition is occurring. We're talking here about thermal phase transitions, right, phase transitions that occur as we change the temperature, like melting, like boiling. If we plot the entropy as a function of temperature,
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what this equation tells us is that there are stepwise changes in the entropy when we melt, when we boil, right? There are stepwise changes in the entropy that occur. Increases in the case of these phase transitions that we're talking about right now, right? There's more entropy in the liquid than in the solid.
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More entropy in the gas than in the liquid. That's what these vertical lines are telling us, right? There's a stepwise increase in the entropy as we undergo these phase transitions. In terms of heating or cooling, we can deduce that the change in the entropy is related to the heat capacity
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and we can write this expression for the change in the entropy where this C depends on whether the process that we're talking about occurs at constant volume or constant pressure. If it's a constant pressure heating, we're using C sub P. If it's a constant volume heating, we're using C sub V
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and we use the C that's appropriate for the constraints that we're imposing on the system while we do the heating. Okay, and so if we plot, this is the change in the entropy divided by NR, but essentially it's proportional to the change in the entropy as a function of the temperature. This is what the entropy looks like for a gas as a function
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of temperature, has this curved shape. It's increasing and we can talk about changes in volume. DQ is minus D. So for pressure volume work, we know this is delta S. So consequently, we can relate pressure and volume to Q. That allows us
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to write this expression from our normal general expression for the entropy and then by substituting from the ideal gas equation, we can equate the change in entropy with this expression right here. All right, log V final over V initial.
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That tells us what's happening to the entropy as a function of volume and we can also, by isolating P, derive this expression for the change in the pressure. Notice here there's a minus sign, here there's not. Entropy's going down as we increase the pressure. It's going up as we increase the volume.
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Okay, so if we want to calculate the entropy for any kind of change, we can now do that. All right, we can calculate it as a function of temperature using this expression right here or as a function of volume using this expression right here or as a function of pressure using this expression right
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over here, don't forget in the case of pressure only there's a minus sign. Right? Final over initial, final over initial, final over initial, temperature, volume, pressure. Okay, so these are pretty simple expressions. We can even derive them if we have to.
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Okay, because this is a statement to, so if two things happen, calculate the entropy change when a gas is heated and expanded, when a gas is pressurized and cooled, right?
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Any combination of things, all right, the nice thing about the entropy is it's a state function so you can calculate one thing at a time. You know, you can get from point A to point B any way you want to and you should get the right answer if you're calculating each segment correctly. All right, where a segment could be a change in temperature
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or change in volume or change in pressure. Calculate any one of those things in any order sequentially and if you do it correctly, you should get the right delta S for the composite process that might involve several different variables. All right, that's the beauty of this. S is a state function.
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Because S unlike you, you can't, and then, yeah. Calculate the entropy change when argon gas is, so this is straight out of lecture 13. All right, calculate the entropy change with argon gas at 25 degrees and 1 atm in a container volume 100 cubic centimeters is compressed to 5 cubic centimeters
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while simultaneously being heated. So here we're changing the volume. You may assume reversible processes and CV and CP are constant. All right, the total change in entropy is equal to the change in entropy for each one of these two processes done sequentially. It shouldn't matter which order you do them in either.
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Process 1, process 2, process 1 is isothermal compression, process 2 is isovolumetric heating. Okay, and we worked this out so in the case, so just looking at the signs, in this particular case, it was not possible to predict definitely whether the answer would be, whether delta S would be positive or negative
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for the process because one term is negative and the other term is positive. All right, so in this case, you can't really predict if the answer is going to be positive or negative before you do the calculation. All right, if they were both positive, you could say, yes, that's definitely positive. So, process 1, a compression to constant temperature,
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that's this guy. We're only using this expression here to calculate N. Remember, we pointed out in this particular case that these two Rs do not cancel here. We're calculating, we're using the Rs to calculate N so you don't want to cancel the Rs.
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Units work out, plug in the numbers, this is the delta S that we get for the first process. All right, negative, we predicted it would be, good. All right, seems to indicate that we could have done it right. After this compression, we're at 25C and 5 cubic centimeters, process 2 is heating
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at constant volume to 45 degrees C. All right, that's this equation right here. We're going to use CV equals 3R over 2. All right, we're going to assume this is an ideal, this argon gas is behaving ideally, so we need to multiply by the number of moles, that's what we're doing right here.
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Plug in all the numbers, this is the delta S that we get. We predicted it would be positive. It's a pretty small number, but it is positive, small and positive. All right, and so then we just add these two delta S's together and we should get for the composite process, that's the total delta S. We can say
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that the process is not spontaneous based on the sign of the delta S that we calculated. Okay, so that's the whirlwind tour through problem 1. All right, any questions on problem 1?
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You'll be calculating delta S, saying something about a Carnot cycle, heat, work, internal energy, these are concepts that were many of them also on midterm 1. Right, really no escaping that.
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Okay, so problem 2 is going to be about the Gibbs energy of reaction. How do individual reactant product species contribute to G, the Gibbs energy? All right, we've looked at this curve endlessly. What can we deduce about the equilibrium state?
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We look at the mixing of two isomers of hydrogen. We did some algebra. We derived something called the chemical potential. We said the chemical potential is the Gibbs energy that we can attribute to a single chemical species.
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All right, it's the Gibbs energy that we can assign to the, it's the contribution of a single chemical species to the Gibbs energy. That's what that equation means even though there's a partial derivative here with respect to the number of moles. You look at that, it's just puzzling. All right, the way I think about this partial derivative, as you already know, is that I think about it as a scale.
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This is the total G of the system and the partial derivative of the Gibbs energy with respect to the moles of the blue stuff you get by just thinking about removing a little bit of the blue stuff making a measurement of the total Gibbs energy and the difference before and after you've removed that little bit of blue stuff.
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That's the partial derivative of the Gibbs energy with respect to the blue stuff while you hold the green stuff and the orange stuff constant. This may seem childish, but without this picture, it really doesn't make intuitive sense to me. But with this picture, I understand it perfectly.
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Okay, so about this chemical potential, basically everything that's true for the Gibbs energy is also true for the chemical potential. I don't think there's any exceptions to that except that the chemical potential is always referring to a single.
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It's the Gibbs energy that we assign to a single chemical species. So the units are energy per mole. Pressure dependence is the same as G. Temperature dependence is the same as G. Depends on the concentration is something that we hadn't talked about, right? What is the dependence of the chemical potential on the concentration of that species?
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We mentioned that there's this thing called the activity. We said the activity has two parts to it. There's an activity coefficient in the concentration. In this equation right here, you can see very clearly how the chemical potential relates to the concentration, right? It's directly related to the log of the concentration.
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And in the dilute limit, right, if A is very dilute, this gamma is equal to 1 and we can see very clearly what the relationship is between concentration and the chemical potential. This activity coefficient in the dilute limit is equal to 1. Otherwise, it's going to be less than 1.
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There's very few exceptions to that. But we haven't said hardly anything about the activity coefficient. You're not going to have to answer questions about activity effects on midterm 2, all right? But you should understand that this is the relationship between chemical potential and the activity.
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That's where we get the concentration dependence is from that. Okay, so we already understand that the Gibbs energy is minimized upon the approach to equilibrium. This issue of the bowing of the Gibbs energy is an important issue to us.
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It's important to understand qualitatively where that comes from. Because if we didn't have that bowing effect, right, if there was just a straight line that connected reactants and products, there would be no possibility for equilibrium to exist. Or if you would, this system is just going to go downhill
40:42
until you hit products. That's where equilibrium exists, all right? Well, we know that's not true, all right? Equilibrium exists in some intermediate mixture of reactants and products that depends on what the Gibbs energy is. And so we did this little thought experiment where we mixed oxygen and nitrogen.
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And when we did that, we concluded that we're going to, when we're done mixing these two things, we're going to be halfway between them in terms of the composition of the system. And we can then calculate what the change in the Gibbs energy is for that one-to-one mixture of nitrogen and oxygen. And we got minus 1,717 joules.
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That's nowhere near the line between these two guys. It's way the heck down here. That means that there's a lot of bowing in this system. And where it comes from is not delta H because for an ideal gas, delta H of mixing is zero, right?
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There are no forces acting between gas molecules, all right? So there's no heat released or formed when you expand or contract an ideal gas. There are no forces acting between those molecules. So delta H is zero, but the entropy of mixing is not, right? This is T times delta S. That's the entropy
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of mixing acting on the system, all right? And delta G is delta H minus T delta S, isn't it? Delta G is delta H minus T delta S. And so consequently, that's where this comes from.
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It's the entropy that's driving this bowing of the Gibbs energy, all right? It's the entropy that allows there to be an equilibrium state that is not all products or all reactants, all right? It's somewhere in between. So we then recast the Gibbs energy in terms
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of this extent of the reaction. And we concluded that we can derive an expression for the reaction Gibbs energy, all right, which is just the slope of this curve at any point, all right? And we've redenominated the horizontal axis here in terms
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of the extent of the reaction, this symbol that I can't even figure out what it is. All right, we've got equilibrium down here. We've got positive reaction delta G's here, negative reaction delta G's here. So there's three possibilities. If the reaction Gibbs energy is less than zero,
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if the reaction Gibbs energy is less than zero, that means the slope is negative, the forward reaction is spontaneous. Yes, if we're here, we're spontaneously going to go in this direction. If we're here, we're spontaneously going to go in this direction. And if we're here, we're not going anywhere. And that's all this says.
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Okay, so we have not learned any more about where equilibrium is located in terms of the progress of the reaction. And so we derived an expression that allows us to say more about that.
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We derived this very important expression, all right. The standard reaction Gibbs energy is equal to minus RT over log K. That's a big K. That's an equilibrium constant. All right, the equilibrium constant of the system is related to the standard reaction Gibbs energy.
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So if the reaction, the standard reaction Gibbs energy is zero, we expect equilibrium constant to be right in between reactants and products. In other words, halfway between reactants and products.
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In other words, the numerator of the equilibrium constant equals the denominator in the equilibrium constant. The Q for the reaction is equal to one. There's many different ways that I can say the same thing. Okay, equilibrium constant would be one in this case. All right, and if we skew, if B is more stable
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than A and we get the skewing, we still have this bowing that's brought about by the entropy of mixing. All right, and we can conclude if the standard reaction Gibbs energy is less than zero. Right, here the standard Gibbs energy is less than zero.
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Equilibrium constant is going to be greater than one. And if it's less than zero, it's going to be less than one. All right, so qualitatively we want this to make sense. All right, we want it to make sense that this curve tells us something
45:42
about what the equilibrium constant is going to be. All right, qualitatively we want to be able to connect this picture with the equilibrium constant. Okay, so we can also do that quantitatively. So, ten more methanol will be formed and we can make predictions
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about which way a reaction is going to go. If we know what the composition of the system is and we know what the equilibrium constant is, we can say the reaction is going to shift this way, it's going to shift this way. We can even say what the final concentrations of each reactant and product will be at equilibrium. We have to go all the way back to Chem 1
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to remember how to do that. All right, but this might be the first time that you've been asked to do that since Chem 1. I don't know. Cl plus 2H2 gives methanol, write an expression for Q, calculate it, Q is just the equilibrium constant expression where these partial pressures are not necessarily equilibrium
46:42
partial pressures. That's all Q is. Q is the equilibrium constant, not at equilibrium. Right? So, you have to take the partial pressures that you're given, calculate Q from them, all right? You're not calculating the equilibrium constant unless these
47:01
are equilibrium pressures, in which case you are. If they're equilibrium pressures, then this would actually be the equilibrium constant. Right? In this case, when we plug all the numbers in, we get .01, equilibrium constant 6 times 10 to the minus 3, so if we compare Q to K, Q is bigger. What does that mean?
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It means that where we are now is 2 product rich. If Q is higher than K, because products are in the numerator of Q, right, if Q is higher than K, it means we're product rich relative to where we want to be at equilibrium. In other words, we're here, equilibrium constants here,
47:42
we're product rich. We're to the right of equilibrium. Okay? And so, obviously, what's going to happen is, we're going to want to go downhill in this direction. We're not going to make more methanol. We're going to make less methanol. We're going to make more of these guys. That's the answer to the question.
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And if we wanted to, we could calculate the reaction Gibbs energy at that point, all right? At every point along this curve, we can, if we want, if we know the composition of the system, we can calculate the reaction Gibbs energy. It's just equal to this, all right? We just plug in for Q. If we know the standard reaction
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Gibbs energy, why then we can calculate the Gibbs energy at any point. That's what we did here. And we concluded that it's positive. Well, it would have to be, all right? The slope of this tangent here is obviously positive at that point. If we didn't get a positive number, we knew we would know we made a mistake.
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Okay, and so this just says we can change the conditions. We can move the position of equilibrium. Instead of here, we can be over here. Now, the reaction is going to proceed towards making a little bit more methanol. We're going to go from here to here. Not a huge amount, but a little bit.
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Okay, this is what problem two is going to look like, all right? Some version of these types of problems testing these concepts right here, all right? And the extra credit problem is right off of mid-term one. Okay? Good luck.