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PROFESSOR: So let's talk about reacting gas equilibrium.
We started setting this up last time.
And now we're going to derive the key equation.
So we're going to equilibrate.
It's really hard to say the word equilibrate while writing
the word equilibrium.
I somehow got that right.
Gas phase reaction at equilibrium at fixed T and P.
So we mixed gases, and we let them react.
We're going to do this in general terms.
Reaction A plus B goes to 2C.
And of course, everything we're going to do is general.
Doesn't have to be for a reaction of this form.
OK.
So we know that the condition of equilibrium
at fixed temperature and pressure,
exactly Gibbs energy at an extrema.
And so we care about its differential.
So let's use what we know.
OK.
So here, we have the total differential
of the Gibbs free energy for changes of temperature,
changes of the pressure, and changes of mole number.
And again, the independent variables for Gibbs
are temperature, pressure, and mole number.
You can see that right here in the differential form.
And why do we choose that free energy?
Why do we choose Gibbs?
Why is this convenient?
Because we can ignore 2/3 of the right-hand side
of the equation.
For fixed temperature and pressure,
right, those are fixed.
Fixed and stubborn and independent
are, like, synonymous in thermo, right?
If they're independent, that means
they can be stubborn and not move.
So those go to 0.
And we care only about this term.
So we're going to be spending the lecture
analyzing that term.
And so equilibrium is determined by the chem potentials mu
i and the mole numbers.
So that's going to be it.
Our entropy and our volume are not
going to be explicitly part of this.
Get rid of that.
All right.
So at equilibrium, dG equals--
now, that's the equilibrium expression.
So, all right.
How many independent variables are there in this case?
I'll remind you, we have a general reaction
A plus B goes to 2C.
That's a reaction.
Let me break that down a little bit.
Somebody tell me, what are the components?
Let's list the components.
And those are the i labels, components i.
What are the components in the system?
AUDIENCE: A, B, and C.
PROFESSOR: Components are A, B, and C. Right.
So this label runs over A, B, and C. OK.
So we have three mole numbers, n of A, n of B, and n of C.
How many independent variables do we have?
This is carried from the previous lecture.
AUDIENCE: That's five?
PROFESSOR: Five.
And how do you get five?
AUDIENCE: Number of components, and then
two independent variables of thermodynamics.
PROFESSOR: So I see you're taking the numbers three
and two, but you're adding them, and then
you should be subtracting them.
AUDIENCE: Oh.
PROFESSOR: And the reason is the following.
We have three components.
We have three ni's, right?
But reaction balance means there's
only one degree of freedom.
How do we know that?
Because we have two constraints.
What are they?
The n of A over nu of A equals the n of B,
remember this from last time?
The nu of B equals the n of C over nu of C.
There's two independent equations in here.
You can count the equal signs.
There are two equal signs.
So we have three components, but two constraints,
which means there's only one degree of freedom.
And we introduced this reaction, extent equals d xi.
You don't have to use xi to answer any problem you're
going to find in this class, but it is a nice, general form.
So we have-- for one reaction, univariance, it can be called.
We have only one degree of freedom.
All right.
So now we're going to solve the reaction equilibrium condition
by applying those constraints.
So apply constraints.
And here, I'm going to arbitrarily choose
the product mole number as the independent variable.
So let's see.
We have d n of A equals nu A over nu of C d n of c
equals minus 1/2 d n of C. d n of B equals nu B over nu C.
d n of C equals minus 1/2 d n of C. You got this?
This sort of makes sense?
A plus B going to C. For every mole of C that's created--
sorry.
Two C's.
My apologies.
For every 2 moles of C that are created, there's 1 mole of A
and 1 mole of B that are destroyed.
So it works out, right?
For every 2 moles of C are created,
you destroy one each mole of A and B.
So what I'm going to do is I'm going to plug these
into my expression for dG.
dG equals, then repeating from the previous board, mu of i,
the n of i, and now I'm going to eliminate the n of e--
A and the n of B and write everything
in terms of the the n of C. And you get the following.
Mu of A, nu of A over nu C plus mu of B,
nu of B over nu of C plus mu of C.
And here is my equilibrium condition.
OK.
Now, this should have a certain similarity
to what we did with the case of unary systems,
when we derived the equilibrium condition for two phase
coexistence.
We had a chemical.
We had an entropy, I think, in that case,
on the left-hand side.
And we needed it to be at an extrema or optimized, that
is, its differential being 0.
So we have one independent variable here, the n of C.
Write down some points here.
dnC is unconstrained.
It's unconstrained.
Can't-- we can't assume that this is 0.
Why is that?
Why can't we assume that's 0?
We call this an unconstrained internal variable.
Why is it unconstrained?
AUDIENCE: Because even in equilibrium,
the reaction doesn't stop.
PROFESSOR: Right.
Thanks, So we can constrain pressure.
We have pressure gauges and pressure regulators.
Those exist.
You can buy them.
We can constrain temperature.
We have thermostats.
Those exist.
You can buy them.
And you can think of different experimental setups
to constrain the temperature and constrain the pressure.
However, I want you to think of a box of gas.
And you have molecules of A, B, and C.
And they're zooming around in here.
And this reaction can take place.
You can't stop it.
So even at equilibrium, this is going to fluctuate.
The reaction is going to run a little to the right, a little
to the left.
And if you're away from equilibrium,
the reaction will definitely either run to the right
or to the left, right, depending.
That's the meaning of unconstrained.
You can't fix it.
You can't go in there with your atomic tweezers
and stop the atoms from reacting.
That's an important concept.
So if we have--
and this is rather like the case in unary, two phase system when
we had a phase boundary and we couldn't stop the phase
boundary from fluctuating.
We couldn't stop a little bit of solid from evaporating,
for instance, or a little bit of gas from condensing.
That would be an equivalent example.
The math worked out in a very similar way.
So if we need this expression to be zero,
and this, in general, is nonzero,
how do we enforce equilibrium?
AUDIENCE: So we have to assume that the coefficient is zero,
then, since--
PROFESSOR: Coefficient
AUDIENCE: Yeah, zero?
PROFESSOR: Thank you.
I didn't see who said that, but that's exactly right.
If you have a function and you're optimizing it-- here,
I draw for maximum.
Same thing for a minimum.
How do you enforce the condition of being optimized?
You make the slope 0.
That's what we're doing here.
Coefficient.
The coefficient dG dnC must be 0 at equilibrium.
That becomes the equilibrium condition.
Right.
So there's some calculation here.
There's some, in this case, one-dimensional function
optimization in here.
That's calculus.
And then there's also some science, right?
You have to recognize that this reaction is going to happen.
You're not going to stop it.
And so if dnC can't be enforced to be 0,
then you've got to call this thing equals to 0
the equilibrium condition.
So there's a lot in here, a lot to Unpack
All right.
Now, fortunately, we have expressions
of those chemical potentials.
So we're going to substitute ideal gas mixture expressions.
This is from last time.
For the mu i's.
We have expressions for the mu i's from last time.
So I'm going to substitute those in.
And I get the following.
dG equals mu A0.
That is component A in tis reference state, pure
at 1 atmosphere, plus RT natural log partial pressure of A over
standard pressure, this quantity times mu A over mu
C plus mu B0 plus RT natural log partial pressure
B over standard pressure, nu B over nu C plus mu C0,
plus RT natural log pressure of C over standard pressure.
This whole thing minus dn of C equals 0.
And as a reminder, mu i of 0 equals
the reference or standard chem potential of component A--
I should say, from point i, at pressure P0 and temp T.
And for P0, we almost always choose one atmosphere.
So we substitute.
Now we're going to collect terms.
Collect terms and multiply through by nu of C.
I'm going to get the following.
Nu of A, mu A0, plus nu of B, mu B0, plus nu of C, mu C not.
I group these, plus RT natural log P of A over P0
to the power of mu A, P of B over P0
to tau of nu B, P of C over P0 to the power nu of C equals 0.
Just manipulating the thing which we wrote down previously.
We're going to continue to simplify this.
We're going to define something.
Defined delta G0 equals this thing in parentheses
equals the sum over i of the nu of i times the reference
potentials.
Why does that make sense to define it that way?
In this case, it's the following.
2 mu of C0 minus mu of A0 minus mu of B0,
recalling, again, what our reaction is,
it's A plus B going to 2C.
So what is this thing?
It's the free energy change of the reaction
when all components are in their standard state.
So for instance, if we assume that our standard pressure is
1 atmosphere, if you take one atmosphere of A
and you react with one atmosphere of B
and you produce C, where C is also one atmosphere,
the change in Gibbs free energy is this delta G0.
And that doesn't have to ever correspond
to a physically practical reaction.
You might not ever realize that particular reaction in the lab.
But thermo is all about state functions and reference data.
And we're picking convenient reference data.
And you can imagine that reaction,
even if you don't carry it out.
That is, 1 atmosphere of A, 1 atmosphere of B,
and 1 atmosphere C at equilibrium.
So--
AUDIENCE: And--
PROFESSOR: --you will find this delta G-- one second, sorry,
You will find these delta G0's in databases.
That's the sort of thing you're going to find.
If you're trying to engineer something
and you need to design a reactor,
you're going to be looking at these delta G0's and going
forward from there.
OK.
AUDIENCE: Yeah, sorry.
I just had a quick question.
Are all those partial pressures in the natural log?
PROFESSOR: Yes, they're all in the natural.
That's a natural log.
AUDIENCE: Yep, OK.
PROFESSOR: Let me make that really explicit.
Does that help?
AUDIENCE: Yeah, thanks.
PROFESSOR: And that just comes from the property
of the natural logs, you know.
Natural log x plus natural log y equals natural log xy.
So we just apply that sum product rule three--
multiple times.
OK.
Let me continue simplifying.
We're simplifying.
I was just at Walden Pond with my kids over the weekend.
I think that was Thoreau who said above all, simplify.
It's like he was describing what someone
should do in Mathematica.
So we're going to define this thing called
the equilibrium constant, which some of you
may have encountered in other classes.
But here, we've derived it.
K of P equals product operator over components P
of i, partial pressure, normalized to standard pressure
to the power of nu of i.
So that is the equilibrium constant in terms
of the pressure, so that's K sub P.
So that's kind of a big deal because it shows up in--
all over the place, including in some introductory chemistry
classes.
All right.
We're going to further simplify.
Simplify by assuming that total pressure,
which we're simplifying here, but the total pressure
is the reference pressure so that P of i over P0
equals x sub i, making that even simpler, again,
using Dalton's rule.
And if we make that simplification,
we get a cleaner expression for the equilibrium constant, which
is the one that we're going to use most often in this class.
And for this reaction, again, for the reaction of A plus B
going to 2C, this equilibrium constant
is the mole fraction of C squared over the mole fraction
of A, mole fraction of B. Yeah.
So one more simplification, and then I'm
going to stop and ask for questions.
We're going to collect terms to write
concise expression for reaction equilibrium constant.
And this is a sort of expression that you
may have been given in a chemistry class
or you might find in Wikipedia for Kp equals e minus delta
G0 over RT.
OK.
So this is our final answer.
This is what we're going to--
this is what's going to be useful for you
after this class, when you're analyzing reacting gases.
And so now you've seen it derived,
and you know where it comes from.
I want to pause now.
I went through that basically without interruption, or hardly
without interruption, I want to pause and take questions.
At Walden Pond I saw a great demonstration of thermo.
There was one person out on the ice.
This is on Saturday.
I mean, it was warm.
And one person was out on the ice.
And everyone who wasn't on the ice was watching this person.
And the ice was--
it was sunny.
And they fell through.
So it was thermo in action.
I saw someone fall through the ice.
They were fine.
I think people were laughing at them.
It wasn't very deep where they were.
And they struggled back onto the shore with their friends
and I'm sure were laughed at for the rest of the day.
So there's thermodynamics in action at Walden Pond.
I don't have a picture to share with you, though.
All right.
Questions on how we got this?
Because we're going to be using this a lot.
AUDIENCE: Yeah, I have a question
about the mole fractions.
Are those-- like, I guess, where did the mole fractions
come from?
Are they the mole fractions that the components find themselves
in at equilibrium?
PROFESSOR: Right, so first of all,
let me answer your question in two parts.
First of all, this is Dalton's rule, right?
It's the definition of partial pressures, almost.
So that's Dalton's rule.
What mole fractions are we talking about here, right?
That's a great point.
So we're not going to do this a whole bunch in this class,
but you're set up to do this.
Let's imagine nonequilibrium and equilibrium.
Nonequilibrium-- there's something called--
sometimes as written as reaction quotient.
Sometimes it's written as reaction constant.
It has different forms.
But it's written-- it's the exact same thing
as the reaction constant.
But in general, in general, it's not
equal to is equilibrium value.
In general, it's not equal to its equilibrium value.
So let's see what that means.
Say that Q greater than K sub p.
So the numerator is too large or the denominator is too small
relative to equilibrium, right?
So if the numerator is too large or the denominator is too small
relative to equilibrium, what direction
will the reaction run?
And we can answer the question thinking specifically
about this case.
AUDIENCE: Would it go to the left, towards the reactants?
PROFESSOR: Right.
Let's look at this.
If this thing is too big relative to equilibrium,
that means there's too much C or there's not
enough A times B. Sorry, I was really messy with that x sub a.
So that's telling you is that you're out of equilibrium,
and you're out of equilibrium to the right.
And the reaction will run spontaneously to the left.
And the math work out.
You will find in this situation that running to the left
decreases the Gibbs free energy.
In other words, dG, dn of C. What's our situation here?
Let me try to draw that.
This is a situation where Gibbs free energy versus n of C--
you want to run to the left to decrease Gibbs.
So you have a curve like that.
You run to the left to decrease Gibbs.
The math will work out.
I know I'm being qualitative here.
The G d nfC is greater than 0, positive slope.
And so it's telling you the reaction will run to the left.
And the difference between the reaction quotient
and equilibrium content.
That can be called reaction infinity or driving force.
And there's all sorts of analyses of this.
And it couples to rate-- and there's a lot
of engineering in here, right?
You want to figure out-- you want to design a reactor
or how far away from equilibrium should you design it to be,
stuff like that.
On the contrary, right, let's say Q is less than K of p.
I mean, it's going to be the opposite.
This is a case where there's not enough numerator.
There's too much denominator.
The quotient is small, smaller than equilibrium.
So this is a case where you're going to decrease Gibbs
energy by going to the right.
dG d n of C less than 0.
And the reaction will drive to the right.
At equilibrium, n of C, G--
they're right there, where they Gibbs free energy is minimized.
So you can write down this expression out of equilibrium,
right?
And you might-- and compare this to this expression, which
you get from the thermodynamics of the reaction
and determine whether your reaction will run spontaneously
to the right or to the left, or is it actually at equilibrium?
did that help answer your question?
I'm not sure.
AUDIENCE: Yeah.
That makes sense.
Thank you.
PROFESSOR: So basically, you can evaluate this instantaneously.
Whether or not it's equilibrium depends,
but you can always write that thing down.
And again, we're not doing really--
we don't really have the time in this class
to talk about a bunch of out-of-equilibrium reactions
and talk about rates and affinities and stuff.
But that will come later in course 3 and other courses.
And so you're well set up.
Because if we understand equilibrium,
we can start analyzing out-of-equilibrium situations.
And nonequilibrium thermos is-- exist.
It's a fascinating topic, I think.
And I teach it as part of the advanced thermo class,
but we have no time for it here.
Other questions?
I'd like to make sure that people-- this
is a point from last lecture, but I'd
like to make people-- sure that people understand
how we get this numerator and denominator
thing out of this expression.
Remember, the reaction coefficients
are positive for products and negative for reactants.
So products end up with a positive power,
and reactants end up in the basement
here with a negative power.
So often, if you've seen this before, let's say
an introductory chemistry, you were probably
told just to put products on the top and reactants
on the bottom.
Well, this is where that comes from.
So it's good to know it comes from somewhere.
Maybe everyone understood that.
I don't know.
But worth mentioning.
So on the PSAT, you're asked to evaluate the equilibrium
condition for a couple of reactions.
And that often boils down to finding the delta G's.
So I want to make this really explicit here before moving on.
Delta G's-- delta G0 equals G0 products minus G0 reactants.
So that's G0 products at some temperature, let's say T2,
equals G0 products at T1 plus some integrals.
We've done those integrals already.
We know how to do those integrals.
For instance, you integrate entropy or temperature
or something like that.
G0 reactants at some temperature 2
equals G0 reactants at some temperature 1 plus integrals.
So we're seeing that delta G0 at some T2
equals delta G0 at some T1 plus integrals.
And these depend on the heat capacity differences.
So I'm doing this at a very high level
and quickly here because I want to highlight
the similarity to the case of unary transformations.
We had-- in unary systems, we had
transformations between phases.
And we can analyze things like the change of Gibbs
free energy or any other state function, change of h or change
of s, across that transformation.
And we could calculate the temperature dependence of that
by integrating things that depended on heat capacity
differences.
It's all formally the same right now.
We're having a transformation from reactants and products.
The delta G0 is important.
That helps us determine equilibrium condition.
And in general, it's temperature dependent.
And you're going to find things like delta G0's
or their equivalents tabulated at standard temperatures
in databases.
And you might be asked to calculate reaction balance
at some other temperature.
And to do that you're going to need the cp's and the delta
cp's.
So formally, it's very similar to the unary case.
I want to highlight that.
AUDIENCE: And if-- sorry.
If a problem is just happening at standard temperature,
can you then just use the standard Gibbs values?
And--
PROFESSOR: Yep.
AUDIENCE: --I guess you wouldn't need
it to integrate in that case?
PROFESSOR: Yes, indeed.
Yep.
h0 for elements at standard temperature and pressure
is defined at 0.
s0 for elements at standard temperature and pressure
is a number, often in the back of the Hoff.
Delta h0, delta s0, delta g0 for compounds--
those are numbers often found in the back of the Hoff
for common reactions.
That's right.
And so if you're a standard temperature and pressure,
you normally have everything you need, even in Wikipedia,
for standard reactions.
And I mean, we're analyzing, like, ammonia in this,
so it's very common substances, easy to find the data.
Thank you for that.
All right.
Let me move on.
At risk of repeating myself, notes reaction equilibrium.
OK.
A negative delta G0 drives reaction to the right.
That's a straightforward point.
Adding more of a given component shifts reaction balance.
It shifts the reaction balance to maintain equilibrium.
That's Chatelier again.
Any change in the status quo prompts an opposing reaction
in the responding system.
And one more thing.
For three or more components, there is no unique equilibrium
condition.
So let's analyze again this case,
x sub C squared, x sub A, x sub B. And this equals--
OK.
So this thing here is just a number,
let's say at a given temperature.
It's a number.
And we have three variables.
We know that x sub A plus x sub B plus x sub C equals 1
by definition.
They're mole fractions.
We still have one degree of freedom at equilibrium,
so there's an infinite number of compositions that
would satisfy equilibrium here.
You can play with the method.
If you want to parameterize--
fix-- you can fix x sub C and then
calculate x sub A as a function of x sub B here, if you like.
There's lots of ways to slice this.
But basically, you could shift your equilibrium balance.
So how does that work?
Let's see how this works with Chatelier.
Let's say you're at equilibrium.
And you add a little bit more x sub
B. You inject a little bit more component B into your system.
So the denominator ticked up a little bit.
How's the reaction going to respond?
How is the reaction going to respond?
We just add a little bit more B. We kicked it out
of equilibrium.
How will the reaction respond?
AUDIENCE: It'll flow to the right to make more C.
PROFESSOR: Yeah, it's going to swerve to the right
to try to counteract that change and bring this expression back
to equilibrium.
So you add more B, the reaction shifts to the right.
Likewise, if you add more C, the reaction
will shift to the left.
If you take away some A, reaction
will shift to the left, and so forth and so on.
You cannot get away from the Chatelier.
It's always going to be there trying to restore equilibrium.
It's in the math.
It bears pointing out from time to time.
All right.
Let's talk about temperature dependence.
This gets to one of the problems in the homework.
Temperature dependence of reaction equilibrium.
Now, I'm just going to do some math here,
and we'll talk about the meaning.
Let's calculate the partial with temperature at fixed
pressure of reaction balance.
And I'll do log of K sub p because the math is easier
this way.
This is d/dt at fixed pressure of minus delta G0 over RT.
And by chain rule, this equals delta h0 over RT
squared minus 1 over RT d delta h0 dt
at a fixed pressure plus 1 over R
d delta s0 dt at a fixed pressure.
So that's just the chain rule.
d delta h0 over dt at fixed pressure
equals delta Cp, again, coming back
to transformation quantities, and dx delta s0 dt
at fixed pressure equals delta cp over T.
We rely on this a lot.
And you might want to remind yourself of this.
You might want to show this just in the minutes after class.
Go back to when we introduce these potentials.
So using these expressions, we find d log of k of p,
dt at fixed pressure equals delta h0 over RT squared.
And this has a name.
This is called the van 't Hoff equation.
OK.
You're going to use a form of this in the homework
because you are implicitly asked to calculate how the reaction
balance changes with temperature.
Let's see how this works out.
Let's see how the math works.
So for example, let's consider an endothermic reaction.
m, endothermic reaction.
Q and delta h0 are greater than 0.
That's an endothermic reaction.
For example, that's the cold pack, right?
So an instant cold pack.
Delta h equals reversible heat for the reaction.
Delta h greater than 0 means dx log kp dt
pressure is greater than 0.
What does that mean?
Reaction moves to the right or left
with increasing temperature?
With increasing temperature, do I move to the right or the left
if this condition is true?
AUDIENCE: To the right.
PROFESSOR: Move to the right.
k of p is products over reactants.
So if k of p is increasing, if this slope is positive,
that means I'm moving to the right.
Reaction moves to the right with increasing temperature.
What does that mean?
This system tries, doesn't actually know what it's doing,
to oppose the temp rise by taking up heat
in an endothermic reaction.
If you try to raise the temperature,
it's going cold pack on you.
That's what van 't Hoff says.
You cannot get away from the Chatelier.
And the converse, of course, is going
to happen for an exothermic reaction.
Let me ask you.
If I just rip that-- wrote that backwards, products--
I just flip it, make product reactants
and reactants products, everything I just said
would be--
have a sign flipped.
And it works out.
Because nature doesn't know how I wrote this.
Nature doesn't care how I wrote this.
It doesn't know what the difference between products
and reactants are.
So an exothermic reaction moves to the left
with increasing temp.
Right?
OK.