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PROFESSOR: Last time,
you talked about
the Gibbs free energy.
And the fundamental equations.
And how powerful the fundamental
equations were in
being able to calculate anything
from pressure,
volume, temperature data.
And you saw that the Gibbs free
energy was especially
important for everyday
sort of processes.
Because of the constant
pressure constraint.
And the fact that the intrinsic
variables are
pressure and temperature.
Well, it turns out that the
Gibbs free energy is even more
important than that.
And this is something that it
took me a while to learn.
I had to take thermodynamics
many times to really
appreciate how important
that was.
Even when I was doing research,
at the beginning, I
was a theorist, and I was trying
to calculate different
quantities of liquids and
polymers and in these papers
the first thing they did was
to calculate the Gibbs free
energy and I didn't quite
appreciate why
they were doing that.
And the reason is, is if you've
got the Gibbs free
energy, you got really
everything you need to know.
Because you can get everything
from the Gibbs free energy.
And it really becomes the
fundamental quantity that you
want to have.
So let me give you an example
of how important that is, if
you have an equation that
describes the Gibbs free
energy as a function of pressure
and temperature,
number of moles of
different things.
Different things you
can calculate.
So let's just start from the
fundamental equation for the
Gibbs free energy, dG is
minus S dT plus V dp.
And let's say that you've gotten
some expression, G, as
a function of temperature and
pressure for your system.
It could be, as we're going to
see today that we're going to
increase the number of variables
here, by making the
system more complicated.
So what I'm saying, now what I'm
going to say now, is going
to be more general than just
these two variables here.
So you've gotten this.
So you have this.
You've got the fundamental
equation.
You've got all the other
fundamental equations, and
from there you can calculate
all these quantities.
For instance, you can calculate
an expression for S.
Because you know that S, from
the fundamental equation, is
just the derivative of G, with
respect to T, keeping the
pressure fixed.
So you've got your equation for
G. That translates into an
equation for S. You can get
volume, volume is not one of
the variables.
Temperature and pressure
are the two knobs.
But you can get volume out,
because volume is the
derivative of G, with
respect to pressure.
Keeping the temperature
constant.
And you've got S now,
you got volume.
Do you know where you,
how did you define G
in the first place?
We define G as H minus dS.
One of the many definitions
of g.
Reverse that, you've got now
H as a function of G and
temperature and entropy.
Well, you've got an expression
for G, we just calculated, we
just showed we could get an
expression for S, which is
sitting right here.
Temperature is a variable
here, so now we have an
expression for H. And you can
go on like that with every
variable that you've learned
in this class already.
For instance, u is H minus pV.
Well, there's the H here.
We have that equation for H.
We have an equation for V,
coming from here.
So there's nothing
unknown here.
If we have an equation for G
here in terms of temperature
and pressure.
Same thing for the Helmholtz
free energy.
You can even get the heat
capacities out.
Every single one of these
interesting, useful quantities
that one would want to calculate
falls out from the
Gibbs free energy here.
Any questions on that
important step?
And, really, I can't believe
how clueless I was when I
started doing research.
Because I would go through the
process of calculating G, and
getting G, and et cetera,
et cetera.
I wrote papers, you know,
G equals blah blah blah.
And I didn't realize that that's
why people wanted to
know G. Anyway, I
know better now.
So, there are a few things we
can say about G that are
fairly easy to calculate.
For instance, if I look
at liquids or solids.
And I want to know how G
changes with pressure.
So, I know that the volume here
dG/dp, that dG/dp is the
volume here.
So if I look under constant
temperature, I pick my
fundamental equation under
constant temperature, and I
want to know how G is changing,
I integrate.
So I have dG is equal to V dp.
So if I change my, and I look
at per mole, and if I change
my pressure from p1 to p2, I
integrate from p1 to p2, p1 to
p2 here, final state minus the
initial state is equal to the
integral from p1 to
p2, V dp per mole.
And what can I say, for a liquid
or a solid, the volume
per mole, over a liquid or a
solid, is small and it doesn't
change very much.
So V is small.
And usually these solids and
liquids, you can assume to be
incompressible.
Meaning, as you change
the pressure, the
volume doesn't change.
It's a good approximation.
So when you do your integral
here, you get that G at the
new pressure is G at the old
pressure, then if this isn't
changing very much with
pressure, or not at all, then
you can take it out.
It's just a constant.
Plus V times p2 minus p1.
And so, this is the
incompressible
part, you take it out.
The fact that it's small means
that you can assume that this
is zero, this whole thing is
zero, that it's small enough.
And then you see that G,
approximately doesn't change
with pressure.
Tells you that G, for a liquid
or solid, most of the time you
can assume that it's just a
function of temperature.
Just like we saw for an ideal
gas, that the energy and the
enthalpy were just functions
of temperature.
And that's a useful
approximation.
It's useful, but it's
not completely true.
And if it were true, then
there would not be any
pressure dependents to
phase transitions.
And we know that's
not the case.
We know that if you press on
water when it's close to the
water liquid-solid transition,
that you can lower the melting
point of ice.
You press on ice, and you press
hard enough, and ice
will melt, the temperature is
closer to melting point.
And we'll go through that.
So, that means that there has
to be some sort of pressure
dependence, eventually.
And we'll see that.
This is just an approximation.
What else can we do?
We can calculate, also,
for an ideal gas.
Liquid and solid, we can
do an ideal gas.
So for an ideal gas, again,
starting from the fundamental
equation, we have
dG equals V dp.
We can do it per mole.
So integrate both sides, G(T,
p2) is equal to G at the
initial pressure, plus
the integral from
p1 to p2, the volume.
So instead of putting the
volume, this is an ideal gas
now, we can put the
ideal gas law.
So V is really RT over p.
RT over p dp.
We can integrate this.
Get a log term out.
G(T, p1) plus RT
log p2 over p1.
And then it's very useful to
reference everything to the
center state. p1 is equal
to one bar, let's say.
So if you take p1 equals one
bar as our reference point,
and get rid of the little
subscript two here, we can
write G of T at some pressure
p, then is G and the little
naught on top here means
standard state one bar plus RT
log p divided by one bar.
And I put in a little dotted
line here because very often
you just write it without
the one bar and bar.
And you know that there has to
be a one bar, because inside a
log you can't have something
with units.
It has to be unitless.
So you know if you have
something with bar here,
you've got to divide with
something with bar, and there
happens to be one bar here.
So pressure p is G at its
standard state plus RT log p.
And this becomes a
very useful, very
useful, quantity to know.
OK, so G is so important.
And G per mole is so
fundamental, that we're going
to give it a special name.
Not to make your life more
complicated but just because
it's just so important.
We're going to call it the
chemical potential.
So G per mole, we're
going to call mu.
And that's going to be
a chemical potential.
We're going to do a
lot more with the
chemical potential today.
And the reason why we call
potential is because we
already saw that if you've got
something under constant
pressure, temperature, that
you want to use G as the
variable to tell you whether
something is going to be
spontaneous or not.
So you want G to go downhill.
And so, we're going
to be talking
about chemical species.
And instead of having a car up
and down mountains, trying to
go down to the valleys, we're
going to have chemical species
trying to find the valleys.
The potential valleys.
To get to equilibrium.
And so we're going to be looking
at the Gibbs free
energy, or the Gibbs free
energy per mole at that
particular species, and it's
going to want to be as small
as possible.
We're going to want to
minimize the chemical
potentials.
And that's why it's
called potential.
It's like an energy.
So, that's the end of the one
component, thermodynamic
background, before we get
to multi-components.
So it's a good time to
stop again and see
if there's any questions.
Any issues.
OK.
So, so far we've done everything
with one species.
One ideal gas, one liquid,
one solid.
We haven't done anything with
mixtures, except for maybe
looking at the entropy
of mixing.
We saw the entropy of mixing was
really important, because
it drove processes where
energy was constant.
But most of what we care about
in chemistry, at least in
chemical reactions,
species change.
They get destroyed.
New species get created.
There are mixtures.
It's pretty complicated.
For instance, if I take a
reaction of hydrogen gas plus
chlorine gas to form two moles
of HCl gas, I'm destroying
hydrogen, I'm destroying
chlorine, I'm making HCl in
the gas phase.
I get a big mixture
at the end.
I get three different kinds
of species at the end.
So the fundamental equations
that I've been talking about,
that we've been talking
about, they're too
simple for such a system.
Because they all care
about one species.
Even more complicated, for
instance, if I take hydrogen
gas and oxygen gas and I mix
them together to make water,
liquid, for instance, not only
do I have species that are
changing, that are getting
destroyed or created, in this
case here the total number
of moles is changing.
And the phase is changing.
Got all sorts of changes
going on here.
And so if I want to understand
equilibrium, if I want to
understand the direction
of time for these more
complicated processes, I have
to be able to take into
account, in an easy way, these
mixing processes, these phase
changes, these changes in
the number of moles.
And that's what we're going
to talk about today.
We're going to try to change our
fundamental equations to
make them a little bit more
complicated so that we can
deal with these sorts
of problems.
Because those are the real
problems we need
to keep track of.
And the ultimate goal, then,
of changing our fundamental
questions is to derive
equilibrium from first
principles.
To really understand chemical
equilibrium, which you've all
seen before.
You've all used the chemical
equilibrium constant K, you've
done problems.
But you've been given,
basically, the equilibrium
constant, and not really derived
it, understood where
it came from.
OK, another simple example here,
which is actually the
one that we're going to be
looking at in the first case.
Where there's a change
going on, is just to
look at a phase change.
H2O liquid going to H2O solid.
There's a phase change, you
can think of it as one
species, the H2O liquid
sort of changing
into an H2O a solid.
It's the same chemical in this
case here, there's no change
in the molecules.
But it's still a change that
we have to account for.
Another example that's also
simple like this, that you
all, I'm sure, have seen before,
suppose I take a cell.
My cell here, full of water.
And then I put my cell, let's
say I take a human cell.
My skin or something.
And I take it and I put
it in distilled water.
What's going to happen
to the cell?
Is it going to be happy?
What's going to happen to it?
It's going to burst, right?
Why is it going to burst?
Anybody have an idea why
it's going to burst?
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: That's right.
So the water wants to go from,
you're completely right.
But let me rephrase it in a
thermodynamic language here.
The water is going to go from
a place of high chemical
potential to low chemical
potential.
And the cell can't take all
that water in there.
The membrane's going
to try to swell.
And eventually burst, right?
Same thing if you if you take
a, go fishing, go to the
Atlantic Ocean and then get
a nice cod or something.
Bring it back and on your
sailboat, you dump it in a tub
of fresh water.
Is that cod going to be happy?
It's not going to be happy at
all, right, because its
biology is geared towards
living in salt water.
And turns out that the chemical
potential of water,
in salt water, is lower
than the chemical
potential of pure water.
And so when you put the cod in
there, the chemical potential
of the water and the cod, is
lower than the chemical
potential of the fresh water
you have on the outside.
And the fresh water
wants to be at a
lower chemical potential.
It rushes into the cod, and
well, the cod does what the
cell does, when you put
it in distilled water.
It sort of bloats.
It isn't very happy.
OK, so but all these things
are basically
the same idea here.
Where you have a complicated
change, where species are
mixing, and things like this.
And it turns out the chemical
potential is going to tell us
all about how to think
about that.
That's why the chemical
potential is so important.
So we're going to go
back to these two
examples here many times.
So let's take the simplest
example here.
Let's go back and derive
some equations.
Let's take our simplest example
that's not a one
species system, but
has two species.
Species 1 and 2.
And n1 and n2 are the number of
moles of species 1 and 2.
And then we're going
to see if I make a
perturbation in my system.
I change the number of
moles of 1, or the
number of moles of 2.
How does this affect the
Gibbs free energy?
That's the question we're
going to post.
And our goal is to find a new
fundamental equation for G
that includes the number of
moles of the different species
as they change.
Because in chemistry they're
going to be changing.
They're not going to be fixed.
So what we want is just purely
mathematically formally, take
the differential of the Gibbs
free energy, which we know is
dG/dT, keeping pressure, the
number of moles of 1, the
number of 2 constant, dT.
That is, dG/dp constant
temperature, n1 and n2 dp.
And then we have our two more
variables now, dG/dn1,
remember this is just a formal
statement keeping temperature
and pressure and n2 constant.
dn1 plus dG/dn2, dn2 keeping
temperature and pressure
and n1 constant here.
I'm not writing anything new
here, I'm just telling you
what the definition of the
differential is here, for G.
We already know what some
of these quantities are.
We know that this is the
entropy, minus the entropy.
This here is the volume.
And I know the answer already.
But I'm going to define
it anyways.
And we're going to prove it.
I'm going to define this
as the chemical
potential for species 1.
I'm going to define this, I'm
going to give it a symbol,
chemical potential mu,
for species 2.
So that I can write my new
fundamental equation as dG as
minus S dT.
Plus V dp plus, and if I have
more than two species present,
the sum of all species in my
mixture times the chemical
potential of that
species, dni.
The change in the number of
moles of that species.
So, this quantity mu, that I've
just defined, dG/dni,
keeping the temperature, the
pressure and all the n's,
except for the i'th one
constant, that is
an intensive quantity.
Because G scales with size,
scales, with size of system.
G is intensive. n, obviously,
scales with
the size of the system.
Also intensive, and you take
the ratio of two extensive
variables, you get an intensive
variable which
doesn't care about the
size of the system.
Which is a good thing.
For what we've been
talking about.
Intensive.
If I'm talking about putting a
freshwater fish and dumping it
in the, putting it in the
Atlantic Ocean, the chemical
potential of the water in the
Atlantic ocean better not care
whether the Atlantic Ocean
is huge or even huger.
It just cares about the
local environment.
Just cares that it that wants to
be in that freshwater fish.
So the chemical potential
is intensive.
Just as we've written
it down here for a
single species system.
It's the Gibbs energy,
free energy per mole.
Which we haven't proven yet.
We haven't proven yet here.
We've just defined it this way
and we're going to prove that
in fact the mu's are the Gibbs
free energies per mole for
each of the species
in our system.
OK, so this is now our first
new fundamental equation.
All we did was to add
the sum here.
Now, we started the lecture by
saying that if you have the
Gibbs free energy, you've
got everything.
And we wrote equations that are
covered, were we can get
S, we can get V, we can get H,
we can get u, we can get A. So
now that we have a fundamental
question for G, we've got our
new fundamental equations
for everything else.
Without really thinking
too much.
We go back to our definitions.
Enthalpy is G minus TS.
So dH is dG minus d of TS. dG,
we've got our new fundamental
equation for G. We
plug it in here.
Expand things out with a T dS
here, we can write immediately
a fundamental equation
for H. dH is T dS.
The beginning is going
to look just like
what you've seen before.
Plus V dp.
Plus an extra term, which is
exactly the same extra term
that we had in the fundamental
equation for
G. Exactly the same.
And every one of the other
fundamental questions can be
derived in the similar way from
G. And they're going to
be what you had before minus S
dT plus minus p dV plus the
sum of the mu i's, dni,
and du is T dS minus p
dV plus i mu i dni.
So immediately we can see that
this mu, this quantity mu that
we've defined as the derivative
of G with respect
to the n's, we can write many
other equations for mu.
There are many other
ways to derive it.
Because this is the differential
for H. This is
the first derivative of H
with respect to entropy.
This is the derivative of H
with respect to pressure.
And this is the derivative
of H with respect to n.
Just formally, that's
what it is.
When you write a differential.
So formally, this is also mu i,
is dH/dni, keeping, now, we
have to be very careful, keeping
the entropy and the
pressure constant.
Because those are
the variables.
Keeping the entropy, and the
pressure, and all the other
n's, constant.
We can also write it as
dA/dni, keeping the
temperature and the
volume constant.
And all the other n's, or we
can write it as du/dni,
keeping that this entropy and
the volume, and all the other
n's, constant.
So we have many ways to write
the chemical potential.
They give you all
the same result.
So this is the formal.
Sort of the formal part of
the chemical potential.
Now, what we really want to
show is that the chemical
potential really is connected
to the Gibbs
free energy per mole.
That's going to be
the useful part.
Let me get rid of this here.
So I said earlier, at the
beginning of the lecture, that
the Gibbs free energy per mole
was so important, we were
going to call it the
chemical potential.
And I said that here.
And then I said, well,
we're going to define
this here, the chemical.
But I haven't equated
the two yet.
I haven't proven to you that
in fact this quantity here,
which we've formally defined
as the derivative of G with
respect to n, is the Gibbs
free energy per
mole, for this species.
In fact, what we want to show
is that if I take the sum of
all the chemical potentials,
times the number of moles per
species, that that is the
total Gibbs free energy.
In other words, that the
chemical potential for one
species in the mixture is the
Gibbs free energy per mole for
that species.
Once we have that idea, then
we'll be able to talk about
the concept of chemical
potential as this thing that
we can use to look
at equilibrium.
To look at going downhill
for the species.
To see why the cell bursts
and all these things.
Because now we understand that
Gibbs free energy is so
important for equilibrium.
We don't understand that
quite yet, with
the chemical potential.
So we got to make that
relation here.
We need to go from the formal
definition to a relation that
we can understand better,
because it includes the Gibbs
free energy.
OK, so that's our goal now.
So let's see.
Let's formally do this now.
Let's define, let's
derive this.
OK.
So remember, our goal in this
derivation is to show that
this is true.
Or that this is true, here.
And again, we're going
to start with the
simplest system possible.
We're going to start with
a two component system.
And we can easily generalize
to multi-component.
And in our derivation, what
we're going to be after is,
we're going to start with the
Gibbs free energy, because
that's where we always
start with.
And we're going to remember that
by definition, mu i is
dG/dni, So if somehow in our
derivation dG/dni falls out,
that would be great.
Because we'll be able to replace
this derivative with
the chemical potential.
So the goal was to find
something where this falls
out, so we can replace it with
the chemical potential.
We're going to start with
the fact that G is
an extensive variable.
So if I take G at a temperature
and pressure times
some scaling factor for the
size of my system, lambda,
number of moles of n1, lambda
times the number of moles of
n2, if I double the size
of the system,
lambda is equal to 2.
If I half it, lambda
is equal to 1/2.
Because it's extensive, this
is the same thing as lambda
times G of temperature
and pressure, n1, n2.
Just rewriting the fact that
Gibbs free energy is an
extensive property.
And lambda is an arbitrary
number here.
Arbitrary variable.
Now I'm going to take the
derivative of both sides with
respect to lambda.
So I'm going to take d d lambda
of this side here.
And d d lambda of
that side here.
Now, lambda here is inside
the variable here.
So I'm going to have to
use the chain rule.
To do this properly.
So this is going to be dG/d
lambda n1, there's lambda
sitting in the variable
lambda n1 here, times
d lambda n1 d lambda.
Plus dG/d lambda n2 times
d lambda n2 d lambda.
Using the chain rule.
And on this side here,
lambda's sitting
straight out here.
So this is very easy.
This is G(T, p, n1, n2).
Now, this is good.
Because this is what
I'm looking for.
I'm looking for the derivative
of g with respect to the
number of moles.
Because that's the chemical
potential.
That was my goal up here, to
make sure in the derivation,
somehow, this came out.
And so it's coming
out right there.
Right here and right here.
Since the number of moles
is lambda n1, that first
derivative here is just
the chemical potential
of species 1 there.
So mu 1, then we have d
lambda n1 d lambda.
Well, d lambda n1 d lambda,
that's just n1.
It's lambda times the number n1
that doesn't have anything
to do with lambda.
So this is n1.
This is the chemical potential
of species 2.
Again, the derivative of lambda
n2 with respect to
lambda is just n2.
And there is G here.
It's a fairly simple derivation,
but it gets us
exactly what we want.
An association between this
formal definition of mu, up
here, directly from taking
the differential.
How much more formal can you
be, mathematically, here?
To associating this formal
definition to
the Gibbs free energy.
Number of moles times mu 1,
number of moles times mu 2,
this is the Gibbs free energy
per mole of species 1.
Gibbs free energy per
mole of species 2.
The sum of all the species of
the Gibbs free energy per mole
of species i times the number
of moles of species i
is G. Or, mu i.
Voila, we've done it.
This is what we wanted.
The chemical potential is the
Gibbs free energy per mole.
And in the mixture, it's the
Gibbs free energy per mole of
the individual species
in that mixture.
And if you want to know what
the total Gibbs free energy
is, because if you have an
equilibrium, what you care
about is the total Gibbs
free energy.
It's not the Gibbs free energy
for one particular species.
What's going to tell you whether
you have a minimum or
not in your system, whether
you're at equilibrium, where
you're at the lowest state
possible, is the total Gibbs
free energy.
Now we'll be able to
manipulate chemical
potentials, of the individual
species, to get
at this number here.
Any questions?
This is really, we're going
to see this over
and over again now.
This chemical potential.
This idea.
And it's not an easy concept.
OK, let me give you
an example, then,
of the water melting.
And how the chemical potential
comes in, now, instead of
using the chemical potential.
Instead of the Gibbs
free energy.
This is the phase transition.
But it's not very different than
the cell bursting when
you put it in distilled water.
So, we take a glass of water
with an ice cube in it.
H2O liquid.
This is H2O solid.
And I'm looking at the
melting process.
I'm looking at a process where
I take a small number of
molecules of water from
the solid phase.
And I bring them to
the liquid phase.
And I want to know, is this
process spontaneous or in
equilibrium, or not possible?
Is this process going
to go on?
Is the direction of time
that this is melting.
And I want to do this formally
thermodynamically.
In terms of the chemical
potentials.
That's going to be what we're
going to be talking about.
So, formally then, what's going
on is, I'm taking nl
moles of liquid water, H2O
liquid, which is in here.
Plus ns moles of solid water.
And I'm, this is my
initial state.
My final state is nl, plus a
small number of moles, dn, of
H2O liquids, H2O liquid, plus
ns minus dn, there's a
conservation of the number
of molecules here.
Whatever I add to the
liquid has to come
from the solid here.
Of H2O solid, of ice.
And to know if this is
spontaneous or not, if this is
done under constant temperature
and pressure, what
variable should we look at?
G, right.
We want to look at the
Gibbs free energy.
So what is G doing during
this process here?
What is delta G here?
Well, we have a way of doing
it now, in terms of the
chemical potentials.
Because we've just shown that
this is the case here.
So G is the sum of the chemical
potentials times the
number of moles in
the species.
Therefore, delta G is going to
be equal to mu for the number
of moles of l.
Liquid, dn, number of moles
of liquid, plus mu s dns.
So the change in G is going to
be equal to the chemical
potential times the change in
the species, which is in a
liquid form, plus the chemical
potential of the solid.
Times the change in the species
in the solid form.
Now, dns is equal
to minus dnl.
This is what we did
right here.
You take a certain number of
moles from the solid form.
You put it to the liquid form.
That the dn up here.
And you've got to have the
negative of it, up here.
So dns is minus dnl.
Which is minus dn.
Delta G is dn times mu
l minus mu solid.
So now we can rephrase this,
it's all rephrasing.
It's all basically
the same thing.
But, we can rephrase this
process by asking the
question, is the chemical
potential of the liquid
greater than, equal to, or
less than the chemical
potential of the solid?
Of the water in the solid.
So the chemical potential of the
water in the liquid phase
is greater than the chemical
potential of the water in the
solid phase, mu l is greater
than mu s, then delta G
becomes positive.
In that case, delta G is
greater than zero.
And that's not going
to happen.
On the other hand, if the
chemical potential of the
water molecules in the liquid
phase is smaller than the
chemical potential of the water
in the solid phase, mu s
is bigger than mu l, this
becomes a negative number.
Delta G is less than zero.
And this will happen
spontaneously.
So that illustrates this idea,
that the chemical potential of
a species will want to go, so
the species will want to go,
where it can minimize its
chemical potential.
So in this case here, when we
have the spontaneous process
of the water, of the ice cube,
melting, you can think of it
as these water molecules
that are in the ice
phase looking around.
They know what their chemical
potential here
is in the ice phase.
They're looking around, they're
looking to see the
water phase.
And they see that in the water
phase, those water molecules
have a smaller chemical
potential.
They're happier.
And so these solid water
molecules are jealous.
And they want to go in
the water phase.
And the ice cube's
going to melt.
And it all has to do with this
difference in chemical
potentials for the water.
And the same thing happens for
the water molecules that are
inside or outside of that
cell that you put in
the distilled water.
The water molecules in the
distilled water have a
chemical potential which is
higher than the water
molecules inside the cell.
And they don't want
to be like that.
They want to change, the system
wants to change, until
the water molecules couldn't
care less whether they're in
the water phase, or outside
or inside the cell.
The system is going to change
until the water molecules have
the same chemical potential
everywhere.
Where they don't have to choose
one place or the other.
And so that gives us,
immediately, what we're going
to need when we talk
about equilibrium.
Equilibrium, chemical
equilibrium, is going to be
where the chemical potential
of a species is the same
everywhere in the system.
So at 0 degrees Celsius, one
bar, which is the melting
point of water, the chemical
potential of a molecule of
water in the ice phase and in
the liquid phase is the same.
That's the definition of
the melting point.
It doesn't care.
It could go either way.
It's an equilibrium.
You take an ice cube, water,
liquid water, 0 degrees
Celsius, one bar.
You come back three
days later.
It's the same.
Come back a week later,
it's the same.
It's an equilibrium.
Chemical potential of the water
species is the same
everywhere.
It's an equilibrium.
And I'm just repeating that
because this is so important.
OK, any questions?
The last thing we're going to
do is to illustrate also the
importance of mixing to the
chemical potential.
So I'm going to set up sort of
an arbitrary system here.
This is kind of like the cell,
or the fish, also, idea.
I'm going to put a system where
on one side I have a
pure gas, A. On the other side,
I have a mixture of A
and B. And here is going to be
a membrane that only allows A
to go through.
Only A can go through
that membrane.
They're going to be partial
pressures in here, p prime B,
and p prime A. For the gas
pressures on that side, and
pressure on that side
is p sub A.
And my goal in this example
here is to show that if I
compare the chemical potential
of a species in a mixture,
where the temperature and the
pressure total are T and p,
and I compare that to the
chemical potential of the same
species when it's pure, when
it's not mixed with anything
else, under the same temperature
and pressure
conditions, that, in fact that
that equals sign is not there.
That's not what I'm
trying to show.
I'm trying to show
that there's a
less-than sign right here.
To show that if you take, again,
this is the cell idea.
If you take the water and the
distilled water, under
constant temperature
and pressure
conditions, it's pure.
And it's looking at the cell.
And inside the cell is
there, boy, is there
a mixture of things.
There's salts, there are
proteins, there are all sorts
of things, right?
It's a mixed system.
The water in that mixed
system, under the same
temperature and pressure
conditions, the chemical
potential of that water
molecule is less.
And that's just an
entropy thing.
Entropy wants to increase.
It just wants to be in a
place with high energy.
It's the Gibbs free energy.
Gibbs free energy has
enthalpy and entropy
incorporated into it.
The enthalpy's not
doing anything.
It's all driven by entropy.
So this is what we're going
to try to show.
And I'm not going to
get to it today.
We'll start with it
on Wednesday.
And I'll let your ruminate on
this for the next few days.
