Lecture 18. Equilibrium in action.
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Transcript: English(auto-generated)
00:05
Okay, so quiz five went better than quiz four.
00:30
It went pretty well in fact so that's sort of an A minus right there. And we only have two of these things left.
00:41
There's one Friday and then a week from Friday we have the midterm two and then there's one more after midterm two and then we're done with these quizzes. So in principle if you already have an A on these quizzes you don't need to take quizzes six and seven
01:05
because you can drop two quizzes. All right, so you only need to take them if you want to improve your quiz average. I know some of you will want to but okay so what we're going to do today is we're going
01:21
to talk just briefly about some of the things that we talked about on Friday and then we're going to do some examples. We're in the middle of chapter 17. We're trying to close out this thermodynamic stuff. We should be done with it by the end of the lecture on Wednesday.
01:42
And then we're going to start talking about kinetics. Okay, so we've been talking about the chemical potential. All it is is the partial derivative of the Gibbs energy with respect to one of the components, the moles of the, number of moles of one of the components
02:03
that serves as either a reactant or a product in the reaction that we're talking about at constant temperature and pressure where we hold the number of moles of all the other components constant. All right, so you described how the Gibbs energy is affected by changes in the amount of one component.
02:23
That's what we've been saying. It is in effect the Gibbs energy of a single component. And the way I like to think about that, because I find this equation to be confusing, is if you have a bunch of different components
02:42
and you've got some total Gibbs energy that we can think of as just being the mass of all of these elements of these components, blue, green, orange. All right, if we were to remove one unit of the blue and weigh again, that difference is tantamount
03:02
to taking that partial derivative with respect to the blue while holding green and orange constant. And I think you can see intuitively when you look at it this way that what you're going to end up with is the weight of this unit of blue. Consequently, this is telling us, right, this is in effect,
03:23
the Gibbs energy of a single component if you think about it that way, right? So, what we know about the chemical potential is that it's extensive. It's got units of energy per mole just like G.
03:42
It's pressure dependence and it's temperature dependence mimics that of the Gibbs energy. The concentration dependence is something that we haven't talked about for the Gibbs energy, all right? For the chemical potential, here's what it looks like for some component A, all right? There's a standard state for A, all right?
04:02
And RT log and that is the activity of A and the activity is just the concentration multiplied by a factor called the activity coefficient which we haven't talked about yet, but for the time being, we're just going to assume that this activity coefficient is 1. So, this activity closely approximates the concentration.
04:25
This is true, this is rigorously true in the limit of dilute A, all right? If the concentration of A is very low, then its activity will equal its concentration, okay? And we'll have more to say about the activity later.
04:44
Okay, so we have this thing called the chemical potential. Now, with this chemical potential, we can begin to understand why G is bowed. In other words, why we can't just connect the Gibbs energy of reactants and the Gibbs energy
05:00
of products with a straight line, right? Seems logical that you could do that. Isn't the Gibbs energy of these intermediate states just a linear combination of the reactant and product concentrations? And if that's true, wouldn't we get a straight line that just connects these two things? And if that's true, there really wouldn't be an equilibrium, would there? Because there'd be no way for there to be an equilibrium.
05:22
There'd be no dip, no bowing. The bowing is essential in order for there to be equilibrium. And so what we showed is that if we calculate the G
05:45
when we've half converted O2 to N2. In other words, we have vessels of O2 and N2. We've allowed them to intermix until equilibrium is obtained. All right, what we calculate is minus 1,717 joules,
06:04
right, even though the free energy of formation, the Gibbs energy of formation of O2 and of N2 are both zero. All right, and so if mixtures of these two things were just a linear combination of zero and zero, we would just have a dash line here, but that's not what we see, right?
06:23
There's this bowing of the Gibbs energy that occurs as we mix these two components, and the reason that happens is because the entropy of mixing is positive. All right, and so if we take T delta S, that equals delta G,
06:43
except that it's positive. All right, so that peak would be at plus, well, I think I brought it here, yes, plus 1717 joules. Here it's minus 1717 joules because the delta H
07:00
of mixing is zero for ideal gases. All right, we know delta G is delta H minus T delta S, don't we? Okay, so this positive entropy of mixing is the source of bowing in all types of chemical reactions, right? It's a general feature of chemical reactions
07:21
that this bowing occurs, and consequently because there's a minimum introduced by this bowing, that's where the equilibrium state is, and so this, when I say the positive entry of mixing is the reason for equilibrium, you get what I'm talking about.
07:41
Okay, the last thing that we talked about on Friday was this thing called the extent of reaction or the progress of the reaction. I call it both things. It's got this funky symbol here which is called psi, I think, is that right?
08:03
I'm not very good with Greek letters. I'm just going to refer to it as the E because it looks like an E. All right? If the delta E is 0.1 moles, then as this reaction here progresses from A to B,
08:25
A is going to get smaller by minus 0.1, B is going to get bigger by that amount. And so we can recast our Gibbs energy versus reaction progress plot instead of having a reaction coordinate here, we can call it the extent of reaction.
08:47
And what we pointed out is that the slope of this plot at any value of this E is the reaction Gibbs energy.
09:02
Now you see this delta here? We don't say that. We just call it the reaction Gibbs energy, but when you say the reaction Gibbs energy, you're right, the delta sub R. Okay? So the reaction Gibbs energy is just the chemical potential sort
09:24
of evaluated with respect to E instead of the number of moles. All right? It looks like the chemical potential except we've substituted E for N. All right? Now we call this thing the reaction Gibbs energy. And there's three possibilities.
09:40
It can be less than 0, 0, or greater than 0, okay? And so under conditions of constant E, so we understand that a change in the Gibbs energy is given by this expression, the chemical potential of A times the change
10:00
in its number of moles plus the chemical potential of B times the change in its number of moles, right? And if we want to exchange E for N, we can recast this expression in terms of the extent of reaction. Okay? And to do that we have to recognize that DNA is going to be negative because A is getting smaller, all right?
10:23
And the extent of the reaction is always positive, all right? And so we have to introduce a minus sign here in order to convert from moles to the extent of reaction. Okay? And once we've done that, we can write this expression. And so the reaction Gibbs energy is just a difference in the chemical potential, right?
10:43
DG DE, DG DE, that divided by that is just that right there. It's the difference in the chemical potentials. So there are three possibilities as we proceed from A to B, all right?
11:01
Initially the reaction is going to be the Gibbs, the reaction Gibbs energy is going to be less than 0. In other words, the slope is going to be less than 0 as we proceed from here to here. And then when we get down here it's going to be exactly 0 at equilibrium, and then when we go further it's going to be greater than 0, and we have these funny words that describe that.
11:25
Okay. Now we haven't said where equilibrium is with respect to the extent of reaction. To get there let's assume that A and B are ideal gases, then you recall from Lecture 16 that we can take the difference between the molar Gibbs energy of the final state and the initial state, that's just equal
11:40
to this integral right here. And for an ideal gas because that's just RT over P, we get RT log P final over P initial. And so we can then define a molar, a standard molar Gibbs energy, no free, as one that applies
12:02
at 1 bar and 298.16 degrees Kelvin. Okay, and so instead of thinking about an initial state and a final state we're going to convert one of these two guys into a standard state that we reference our Gibbs energy against all the time. Right, and that standard state is characterized by pressure
12:22
of 1 bar and a temperature of 298.16 degrees Kelvin. Right, it's that guy, we're going to move him over to the right side, now we're going to call him G superscript 0. Right, the standard molar Gibbs energy, okay, and so if A
12:47
and B are gases we can write their chemical potentials in this form, alright, parallel of this form, now we're just talking about the Gibbs energy of a single component, A, and so if we look
13:04
at this difference here I can substitute this expression in for A and then write an analogous expression to it for B and I get this guy. Alright, and now I'm going to collect these two mu's together and call that the reaction, standard reaction Gibbs energy,
13:27
standard reaction Gibbs energy, okay, and the P zeros cancel, we just end up with PB over PA. This quotient here is in general going to be called the reaction quotient.
13:43
Okay, so now for every value of the extent of reaction across the horizontal axis of our G versus extent of reaction diagram we can calculate the reaction Gibbs energy from the Gibbs energy of formation, right,
14:01
there's an F there for the product and reaction species, so we can look up these Gibbs energies of formation for the reactants and the products, that allows us to calculate what the reaction Gibbs energy is. At equilibrium the reaction quotient is given by a special name, so in other words at equilibrium
14:22
where this guy is zero, okay, we have this, we're going to call the reaction quotient K instead of Q, where did we call it Q? Here, all right, that guy is equal to Q but when this is equal to zero and we're at equilibrium Y now we're going
14:45
to call it K, K only applies when you're at equilibrium so you have to be talking about equilibrium concentrations or pressures of every reactant and product, okay, and so that being the case we can just move this guy over to the left hand side or move that
15:01
over to the left hand side, put a minus sign in front of it, now we have a relationship between the standard reaction Gibbs energy and the equilibrium constant. This is an important equation, okay, what does it mean? Well, let's say that the standard reaction Gibbs energy
15:22
is zero, right, in other words the reaction in the product Gibbs energies are the same, all right, if that's the case, if I plug zero in for this guy, all right, I get zero over RT, I get the EXP of zero,
15:42
that's just one, all right, and what that means is the equilibrium constant is just one, in other words if we look at a diagram like this one where we're plotting the Gibbs energy as a function of Q, all right, we're going to end up with an equilibrium constant that is here right
16:01
at the bottom of this well, symmetrically located between A and B and this curve is also perfectly symmetrical. On the other hand, what if this standard Gibbs reaction
16:22
energy is negative, all right, we've got a minus sign here and if that thing is also minus we're going to have a minus times a minus is a plus, right, and the EXP of any positive number is going to be greater than one, right, and if it's greater than one, all right, here's that case, all right,
16:44
standard Gibbs, standard reaction Gibbs energy is positive, sorry, negative, negative, negative, negative, all right, final minus initial,
17:01
and so now our curve is going to be skewed, all right, here's one, all right, our, we're going to be greater than one, we're on the right hand side of one, we're over here, all right, so qualitatively the equilibrium constant is going
17:22
to be greater than one, what does that mean? What does that tell us about the equilibrium state? More of this than that, right, so we can be a little bit more specific about that
17:41
if we have some generic reaction, this is just a review from Chem 1, A moles of A, B moles of B, C moles of C and D moles of D, the equilibrium constants given by this expression where we've written everything in terms of activities, right, these are the product species, these are the reactant species, and I've taken in every case the activity to the stoichiometric,
18:02
to the power of the stoichiometric coefficient, everybody remember how to write these things, all right, and these, because these are activities, each activity, for example, the activity of C is its activity coefficient times its concentration, and for D, and for A,
18:22
and for B we have the analogous expressions, now what I told you earlier is for the time being we're going to assume these gammas are all one, so this whole thing is going to go away, isn't it, it's just going to be one, okay, and that's our equilibrium constant expression written
18:40
out in terms of concentrations, that's the activity of A, that's the activity coefficient for A, that's the concentration of A, you get the picture, okay, so what this tells us is that we're going to have a product rich state, these are product concentrations, these are reactant concentrations, if K is greater
19:02
than one we've got more of this than this, that's what this is telling us, we have a product rich state for our system, conversely if the standard reaction Gibbs energy is
19:22
greater than zero, in other words, B is up here and A is down here, that's the positive standard Gibbs energy, now the curve is going to be skewed the other way, K is going to be less than one, okay, and we're going to be reactant rich by virtue of the same thinking,
19:47
okay, so this picture is self-consistent, we can think of it in terms of K, G, standard Gibbs reaction energy, any of those things, let's do an example, CO, hydrogen,
20:08
and methanol, all right, are mixed together at 500 degrees K with these concentrations, partial pressure of CO is 10 bar, H21 bar, methanol 0.1 bar, we pass this
20:22
over a catalyst, this mixture, all right, the question we're going to ask is can more methanol be formed, can we adjust the reaction conditions to form more methanol, or are we maxed out, okay,
20:41
we're given the standard reaction Gibbs energy, 21.21 kilojoules per mole, all right, now the first thing to say is catalyst, what does that word mean, how does that affect our calculation,
21:00
what's the catalyst doing, what's the role of the catalyst here, anybody know, speed up the reaction, that's exactly right, it's going
21:20
to accelerate the reaction, it's going to accelerate the approach to equilibrium, but it does nothing to alter the position of the equilibrium, in other words, inserting the catalyst doesn't change where we're going to end up in terms of the mix of products and reactants, right, it's not going to do that by definition, okay,
21:42
so here's what our picture looks like, we're told the reaction, standard reaction Gibbs energy is 21.21 kilojoules per mole, here's what that looks like, all right, we're uphill compared to where we're starting, it's a positive number, all right, so what do we expect, well we expect K is going to be less than 1, right,
22:05
now I've denominated this axis in terms of the reaction quotient, right, the reaction quotient is 0 when we're starting the reaction because the partial pressure of these guys would be 0 at this point, right, that's not where we are
22:22
in this problem, in this problem we're over here somewhere, right, but hypothetically if we were here the reaction quotient would be 0 and if we're here the reaction quotient would be infinite wouldn't it because we've got no, none of, we've got only products, no reactants, okay, so we're here, we've got none of this
22:40
so the reaction quotient would be infinite, okay, so how do you approach this problem, the first thing you have to do is write a balanced chemical reaction, boom, CO plus 2 hydrogens gives methanol, then you write an expression for the reaction quotient and you calculate it, the reaction quotient
23:02
for where you are right now, the reaction quotient for the conditions that you're given, that's the one I'm talking about, there are an infinite number of reaction quotients, right, all different number, these are all different reaction quotients that vary between 0 and infinity, all right, so what we want
23:25
to calculate is the reaction quotient that corresponds to the initial conditions that we're given, I should have called it something special but I didn't, I just called it the reaction quotient, it's the partial pressure of methanol divided by the partial pressure of CO multiplied by the partial pressure
23:40
of hydrogen squared because there's a 2 in front of the hydrogen, okay, I plug my numbers in, I get 0.01, reaction quotients and equilibrium constants are always dimensionless, okay, reaction quotients
24:02
and equilibrium constants are always dimensionless, I'll show you why in a moment. Okay, so now we can calculate K, K is just the XP of minus standard reaction Gibbs energy divided by RT
24:20
and we were given this in the problem, it's 21.21 kilojoules per mole, so that's 21,210 joules per mole and I can calculate K, sure enough it's less than 1, okay, then I compare Q to K, this is where I am now, this is where K wants me to be,
24:41
Q is bigger than K, what does that mean? Well, we're not at equilibrium, that's the first thing to say because to be at equilibrium, Q would have to equal K, right? It's bigger, what does it mean that it's bigger?
25:07
It's on the product side of equilibrium because remember what Q is, Q is products over reactants, right, okay, so if Q is bigger, that means we've got more products
25:20
with respect to reactants than we will have at equilibrium, so we are on the right hand side compared to where equilibrium is, right, we're here and here's where equilibrium is, right, qualitatively, we're at .01, equilibrium is at here, okay, and so we can immediately start to make predictions
25:42
about what's going to happen, can't we? Just looking at this diagram, okay, but let's just do a couple other things, let's calculate the reaction Gibbs energy for this progress of reaction, all right,
26:00
the reaction Gibbs energy is just equal to the standard reaction Gibbs energy plus RT over Q, we calculated Q already, we said it was .01, that's the temperature, that's R, that's the standard reaction Gibbs energy I can calculate now and the reaction Gibbs energy at the Q that we care about, .01 and that's the number I get, 2.06 kilojoules per mole,
26:25
right, the fact that it's positive means that the slope is positive, right, means that the slope is positive, it means that we're here, right, that's why it's positive,
26:42
right, we already concluded that we were here, we know we're at Q equals 0.01, all right, and so what this means of course is that at this mix of reactant and product pressures the reaction towards methanol is not spontaneous, in fact we're going to go back
27:00
in the other direction, right, methanol is going to be consumed and we're going to make CO and hydrogen, that's not what we wanted to do, all right, so that we were asked, all right, can we make more methanol under these conditions, no, you can't, you're going to go backwards, you're going to consume methanol
27:23
and make these two reactants. Now, what if I increase the concentration of the reactants, I'm going to increase these concentrations and I can just recalculate, now the Q is .001 instead
27:42
of .01, it's gotten smaller, I can recalculate the reaction Gibbs energy and when I do that now I'm getting a negative number, the reaction Gibbs energy is the slope remember,
28:03
the slope of that green curve, right, so if that's negative that tells us that we went from being on the right hand side to being on the left hand side, what did I write here? Minus value means that the reaction is now spontaneous and we can make more methanol, we were here
28:22
under these conditions and now we've moved over to here under these conditions right here, right, and now the slope is negative, it was positive before and so now we will make a little more methanol, we're going to go from here down to here, it's not going to be a huge amount
28:42
but we should make a little more methanol under these conditions by adding more, what did I add, more CO or more H2, 1, 10, oh 10, 1, I just changed, you get a different Q because the hydrogen partial pressure squared, okay,
29:05
so we can make predictions and everything is self consistent, that is self consistent with that is self consistent with this thing right here, all right, it all makes sense, it's a little intricate but it all fits together like a puzzle,
29:26
consider the following reaction, N2O4 reacts unimolecularly, in other words by itself to give 2 NO2's, it's a unimolecular reaction, calculate delta GR, that should be delta RG and R should be
29:45
over there at 298.15, calculate equilibrium constant, calculate the equilibrium total pressure of the system, consisting of 1 mole of N2O4 and 1 liter at 298.15 degrees Kelvin, got to do all three things,
30:05
very similar to a problem that you're going to have on mid-term too because it rolls together a lot of things doesn't it? Okay, that R should be right there, it's called the reaction, standard reaction Gibbs energy, all right,
30:24
now we're not given that here for goodness sakes but fortunately you've got your book and there's a table of Gibbs energies of formations in the back of your book and you can just open up that table and calculate this, okay?
30:43
So the standard reaction Gibbs energy where that R is actually underneath right there is 2 times that because that's the product, 2 times the NO2, the standard Gibbs energy of formation for the NO2 minus the standard Gibbs energy
31:03
of formation of the N2O4, that's going to give us the standard reaction Gibbs energy. I'm really working hard not to wrap these words around my throat today, all right, I looked that up in the back of your book, I looked that up in the back of your book and I just plug them in here
31:20
and what I get is that number right there, that's the standard reaction Gibbs energy where that R is actually right there, all right, so it's thermodynamically uphill, that's what we mean when we say thermodynamically uphill, that's a positive number.
31:43
Okay, this reaction is thermodynamically uphill, we expect KP to be less than 1, right, because if I think about where the products and the reactants are the green curve is going to be skewed towards the reactant side and K is going to be less than 1, that's my chemical intuition kicking in, I look at that number
32:01
and I know K is going to be less than 1, all right, we want to develop this chemical intuition that maybe we haven't had, calculate KP at this temperature, we just calculated that, the standard reaction Gibbs energy,
32:20
now I can just rearrange this equation and calculate K, that's pretty easy for me to do, I just plug in that number that I just calculated, there it is right there and I can calculate everything else, oops that's my equilibrium constant, sure enough it's less than 1, that reassures me that I might not have made a mistake, all right, I knew it was going to be less than 1 before I calculated it,
32:44
okay, what else, calculate the total equilibrium total pressure of a system consisting of 1 mole of N2O4 and 1 liters of 298.15 degrees Kelvin, remember G Chem,
33:02
remember doing these problems, you're going to have to be able to do problems like this again, it's the second time you've had to do this probably, maybe you didn't like it the first time, I hope you like it better this time, I like these problems, myself, 1 mole of this, no moles of this, here's the change,
33:27
minus X is the number of moles that are going to be consumed as the reaction proceeds from left to right, plus 2X because there's a 2 here, all right, so the total is going to be 1 minus X and plus 2X, remember this,
33:44
okay here's our equilibrium constant expression, where did that P0 come from? I know I've got products over reactants, so I've got the partial pressure of N2O squared
34:01
over the partial pressure of N2O4 but then I got an extra P here, where did that come from? We need to just talk about this one time and then we never need to talk about it again hopefully. Every one of these pressures is really the partial pressure
34:24
divided by the pressure that characterizes the standard state, every one of these pressures, every time you write a pressure, an equilibrium constant expression, did anybody ever tell you this in G-chem? No, all right, it's left out and you're just told that equilibrium constants are always going
34:42
to be dimensionless, all right, but the reason they're dimensionless is because this is not the partial pressure of NO2, it's the partial pressure of NO2 over P0, all right, and so if you work out the algebra, if you write PNO2 over P0 squared and PN2O4
35:01
over P0, all right, you end up with an extra factor of P0 in the denominator that makes this expression dimensionless, you see you got P squared in the numerator and you got P squared in the denominator and all the units are going to cancel, all right, so it's not magical that equilibrium constant expressions have no units,
35:22
it's because every pressure or every concentration that you input into an equilibrium constant expression is really a ratio between the concentration of the thing that you're talking about and the concentration or pressure of the standard state, okay, now as you'll see in a second it becomes important that you not forget
35:43
about this P0, all right, and you can always insert the right number of P0s just by looking at the expression, if you leave out the P0s and you write the expression the same way that you would in GCAM you can go back and look at it and go, oh, I can see there's going to have to be extra factors
36:01
of P0 either in the numerator or the denominator to balance the units, you can stick them in after without thinking about them too hard, that's the pressure corresponding to the standard state 1 bar, okay, so now I'm just going to make substitutions from the ideal gas equation,
36:21
I haven't forgotten my P0, I'm going to square this just like that squared right there and then I'm going to plug for this N, I'm just going to plug in from my expression that I generated right here and so in the numerator I got 2X squared and in the denominator I got 1 minus X right there, yes, that's from there and that's from there, right, okay,
36:46
and then now look at this cluster of variables here, I've got R times T over V multiplied by P0, P0 is 1 bar, V is 1 liter, T is 298.15 degrees Kelvin and R in units
37:02
of liter bars per Kelvin per mole, which is not a factor that I can remember, alright, that's R in units of liter bars per Kelvin per mole, all of these units cancel but this thing is not equal to 1, it's equal to 24.789, so you've got
37:24
to divide your equilibrium constant by that factor to get the right answer, alright, the P0 is important, if without that P0 you're not going to get a dimensionless quantity in this pink rectangle here, it's not going
37:44
to be dimensionless, you see, bar cancels with bar, liters cancel with liters, moles cancel with moles, K cancels with K, okay, so now I can solve for X
38:01
in the normal way, I cross multiply, generate a quadratic equation, use the quadratic formula, this C parenthetically will always be negative in these problems, C will always be negative, that's important because that means that this operation underneath this square root sign here is
38:23
always going to be an addition, okay, if you remember that it's somewhat helpful, your hands fly across the keys of your calculator because you've done this a billion times, alright, and it's always a plus, okay,
38:42
and here's what you get, of course there's also a negative root that you neglect because it's plus or minus, okay, so that is the X and now we can calculate finally what these two, what the number of moles are going to be, the final number
39:02
of moles for both of these two chemical species, 1 minus X, 2X, here's the final number of moles, but we weren't asked for the final number of moles, we were asked for the final pressure and the final pressure is just the sum of the moles times RT over V, right,
39:21
because there's two species, we've got to add them up to get the total number of moles of gas and then RT over V gives us total pressure which is that number right there. This is just taking you back to GChem, alright, but that stuff is important, we do need to remember how
39:42
to do these problems. Okay, so French guy, you know how this can be important in the future, you remember that name, Le Chatelier,
40:01
that's a name that you know from GChem, if a chemical system at equilibrium experiences a change in concentration, temperature, volume or partial pressure than equilibrium should counteract the imposed change in new equilibrium is established, that's what he's famous for. He didn't win a Nobel Prize though, he just missed out on that.
40:22
At the end of his career he wrote, I let the discovery of ammonia synthesis slip through my hands, it was the greatest blunder of my scientific career, nobody could figure this reaction out in the early 1900s, everybody was working on it and Le Chatelier even had the right catalyst, right,
40:41
anybody know what the right catalyst is? Silver works, but that's not the catalyst that was actually used, it was iron. The reason Le Chatelier didn't understand this reaction is this is one of the rare exceptions to Le Chatelier's principle,
41:02
it doesn't follow his principle, he was thinking about his principle and this reaction doesn't follow it, why, that's the extra credit question. That's the extra credit question for midterm two
41:22
so if you know the answer and between now and then you can try and figure it out. Let me say it again, why does this reaction which is catalyzed by iron not follow Le Chatelier's
41:46
principle, one sentence, right, it's not a long drawn out thing, there are no equations, I want you to just tell me and who solved this problem?
42:12
Well it turns out there's two guys, Fritz Haber and Karl Bosch, right, Fritz Haber figured out the chemistry first
42:24
and won the Nobel Prize in 1918 for doing that and then Karl Bosch commercialized the process and those two guys are the most influential, not scientists,
42:41
not the most influential scientists, they are the most influential people of the 20th century because they figured this out according to Nature Magazine, a magazine about scientists, right? Alright, here's, what am I plotting here? This is the population of the world in units of billions,
43:06
alright, and this is the year, alright, and here is what the population of the world was doing and here's when the Haber Bosch process was invented, early, about 19, early 1900s, alright, here's what happened
43:23
to the population of the world, this could never have happened without that, alright that's why they're the most influential people of the 20th century, what's so important about this reaction? Fertilizer, alright, you need to fix nitrogen,
43:44
you have to separate nitrogen from its super strong triple bond in order to make proteins because proteins have amino acids in them that don't have triple bonds, right? Half of the protein in your body was prepared synthetically
44:01
by this reaction, half of it, half of every protein in your body came from this reaction right here, isn't that phenomenal? That's called impact, alright, that's a huge impact on your life, where did that fixed nitrogen come from before the 1920s?
44:25
You can't see it here but these are cows, right, that's a person and this is a mountain of exactly what you think it is, alright,
44:41
on the most of the world's seagull guano was on the coast of Chile, Peru and Bolivia, right, there's a long coastline on the west coast of South America where there are mountains of seabird guano and that is
45:03
where all of fixed nitrogen came from in the world, alright, at this time in the early 1920s. There were wars fought over it, alright, war of the Pacific, read about it, alright, it was fought over bird guano,
45:21
alright it was a super important commodity, it's called saltpeter, saltpeter, right, potassium nitrate, alright, critical component in the manufacture of explosives and gunpowder too, alright, so very important strategically in the early, late 1800s
45:41
and early 1900s, alright, very interesting. Okay, so Le Chatelier did get the principle right, there are reactions that are, that run counter to what his principle predicts, alright, almost all cases
46:01
of reactions that don't obey Le Chatelier's principle have something to do with catalysts, catalysts mess up the principle, alright, so that's part of a clue, alright, if you want to get the extra credit problem right, you have to think about the catalyst, why doesn't the catalyst allow Le Chatelier's principle to be obeyed for the Haber-Bosch process,
46:23
what does the catalyst do to prevent that from happening, it's about the catalyst, okay, Le Chatelier's principle says there's a, position of this equilibrium should shift to the left because this shift will result in reduction in total, so in other words, if you increase the total pressure
46:42
of a reaction vessel that contains a mixture of these gases, you increase the total pressure, you expect equilibrium to shift to the left because the pressure of the system will be reduced if you do that, right, if I increase the pressure of this reaction vessel, we expect equilibrium to shift
47:02
to the left because here we've got two moles of NO2 and only one mole of N2O4 and so the pressure of the system will go down if equilibrium shifts from right to left, that's what the Le Chatelier's principle predicts and that's what happens most of the time, if there's no catalyst involved.
47:23
Can we determine the relationship between X and P total for this reaction and verify the prediction of Le Chatelier, can you work it out, right, in detail and in the one minute that we have left we don't really have time to do this
47:41
but let me just click through these slides very quickly and show you, this is what we did a minute ago and we have to calculate the mole fraction of each of these two components, we're not going to get through this today so. Alright, so we'll see you on Wednesday.