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PROFESSOR: So let's start with
some of the things we
learned last time.
So there are two things
that were important.
We learned that the chemical
potential for a species is the
Gibbs free energy for that
species divided by the number
of moles, this is Gibbs
free energy per mole.
We learned that the pressure
dependence of the Gibbs free
energy gives you the pressure
dependence for
the chemical potential.
That it's equal to the chemical
potential at one bar
for an ideal gas
plus RT log p.
We also learned that a species
will want to go to minimize
its chemical potential, and we
saw that for the cell bursting
in salt, in distilled water.
Or an ice cube melting
at a temperature
greater than 0 degrees.
And what we want to prove right
now is that if I take a
species A, in a mixture, some
temperature T, some pressure
p, and I compare its chemical
potential to the same species,
A, when it's pure, the same
temperature, the same
pressure, what I want to argue
is that the chemical potential
in the mixture is always less
than the chemical potential
when it's pure.
The same conditions of pressure
and temperature.
And, so we can do a little
thought experiment.
Let's do a little thought
experiment.
Let's make a box, and
in our box we're
going to have a partition.
And a flexible membrane here.
And on one side of the partition
we're going to have
a gas, B. A gas, A,
on this side here.
And gas A on this side here.
So let me just red chalk for
A, And I don't think I have
any other colors listed.
Yellow chalk for B here.
And everything's one bar.
Everything's one bar.
So one bar B here.
One bar A here, one
bar A here .
And this membrane here
only lets A through.
And this membrane is deformable,
but it's elastic.
You can't deform it forever.
It has some strength
to it, right?
So if you push on,
it'll push back.
There'll be pressure associated
with that.
So the next thing I do, then,
in my experiment, shouldn't
have done it here.
Next thing I do in my experiment
is to break this
partition here.
I'm going to break
the partition.
And this will cause
A and B to mix.
So now in my box I have my
partition, my membrane here.
I've got A at one bar here,
total pressure of one bar.
And on the other side I have A
plus B, with a total pressure
of one bar.
That's my initial point now.
What's going to happen?
What's going to happen is that
molecules of A here are going
to want to go through
the membrane to go
in this area here.
And there's two ways
to look at it.
You can look at it the
thermodynamic way, which is
the way that we're going to
want to be looking at it.
Which is that, from what we're
going to prove, is that the
chemical potential in the
mixture is always less than
for the pure substance.
Here we have a mixture.
One bar.
Here we have the pure substance
at one bar.
So these molecules are going
to look around and say hey,
you know, I'm much
happier here.
And they're all going to want
to go in this area here.
As a result, the volume of this
area, if you want to keep
the same pressure on both sides,
you're going to deform
the membrane.
The membrane's going
to get deformed.
It's going to bloat
on that side here.
It's going to cause
an increase in
pressure on that side.
And the increase in pressure
from the membrane sort of
deforming and pushing back, is
going to increase until the
partial pressure of A, here,
equals the pressure of A here.
At which point the flow of A
from either side is going to
be the same and you're going
to be in equilibrium.
And at that point the chemical
potentials of both sides are
going to be the same.
So I sort of gave you the other
way of looking at it.
Which is just purely in
terms of pressures.
At equilibrium, the partial
pressure of A here has to be
the same as the pressure
on this side here.
So that the flow of A on either
side of that membrane,
going from right to left or
left to right is the same.
And that's going to cause
this mixing to happen.
If you look at it from partial
pressure perspective.
But really, it's a chemical
potential idea.
Now, the chemical potential,
as we saw, was
the Gibbs free energy.
And the Gibbs free energy, you
can write it as H minus TS.
So basically, in this process
that I described, the
enthalpy's not doing anything.
These are ideal gases.
They're not interacting
with each other.
The only thing that's changing,
that's driving the
chemical potential, which is
basically Gibbs free energy
per mole, the only thing that's
driving the chemical
potential to be lower on
this side here, is
that entropy term.
It's the entropy of mixing.
So entropy of mixing is really
super important.
When we're talking about
systems where you have
multiple components.
That's going to drive
a lot of things.
And in fact that's going to
drive equilibrium, as we're
going to see a little
bit later today.
Alright, let's quickly
go through the math
and prove this here.
So our goal, then, is to have a
mixture, chemical potential
of the mixture on one side, and
the chemical potential of
the pure material on
the other side.
And so we're going to start by
sort of a similar thing here.
We're going to have a box, let
me redo my box here, we're
going to have a box with,
let me get rid of
these one bars here.
We're going to put our
membrane in the box.
And we're going to have
a pressure, pA,
on this side here.
And we're going to have a
pressure pA prime for the
partial pressure of A. And pB
prime for the partial pressure
of B. And the p total is going
to be pA prime plus pB prime.
And I'm going to see at
equilibrium, I'm going to
write everything I know
about equilibrium.
At equilibrium, I know that the
partial pressure of A on
that side here has to be equal
to the pressure of A here.
The partial pressure, the
pressure is basically the
force of these molecules hitting
that membrane per unit
time, times the number
of molecules hitting.
So we have the same flux of
molecules going this way, is
equal to the flux of molecules
going the other way.
So at equilibrium, pA
prime equals pA.
So.
What else can I say?
At equilibrium, in terms of
the chemical potentials, I
know that the chemical potential
of the mixture, mu A
in the mixture, temperature
under pressure p total.
Is equal to the chemical
potential of the pure system,
same temperature under
pressure is p
sub a on that side.
These are the two things that
I know at equilibrium.
So let's start to
turn the crank.
And see if we can come up
with, well, we basically
already have this.
If we can massage the right side
of our equation so that
the pressure term,
pA, is p total.
And then we'll have an equation
that will compare the
chemical potential of the
mixture under the same
conditions as the chemical
potential of
A in the pure state.
So, what can we use?
We can use Dalton's law here,
which tells us that pA prime
is equal to xA p total.
That's from Dalton.
And now, pA prime
is equal to pA.
So this is also just pA.
And we can plug that in here,
and suddenly we've got p total
included in here.
Let's just pass these out.
If you want to, thank you.
Thank you.
So, this is equal to mu A pure
temperature xA p total.
So what else do I know?
I know that I've written
something here, that mu is
equal to mu naught, to temp,
so this is at one bar.
RT log p.
So I'm going to rewrite
this as mu A pure
temperature T at one bar.
Plus RT log xA p total.
Now, the log of xA times p total
is the log of xA plus
the log of p total.
It's equal to plus RT log
p total plus RT log xA.
So I can lump these two
things together.
I have mu A pure T
plus RT log p.
Well, that's just the chemical
potential of A, in the pure
state, at temperature
T and pressure pT.
That's the equation here to
relate pressure, at some
variable pressure p, to what
it would be at one bar.
Which is that.
Then we have the plus RT
log xA sitting here.
Plus RT log xA.
So we're done.
We're done because on that side
here I have mu A, the
chemical potential of A in the
mixture, temperature T,
pressure p total.
I've got mu A pure temperature
T pressure p
total, plus a term.
Plus RT log xA.
Can xA be bigger than one? xA's
the mole fraction. xA's
always less than one.
Log xA is always
less than zero.
This term here is always less
than or equal to zero.
Therefore, this term is always
less than that term.
Therefore the chemical potential
in the mixture is
always less than the
chemical potential
inside the pure material.
This is going to be important
for the next part of the
presentation.
Any questions?
This is what drives the death
of saltwater fish in fresh
water, right?
Because osmotic pressure
is basically given by
this basic idea here.
And it's all driven by
entropy of mixing.
Sure you don't have
any questions?
Speak up.
So now we have all the tools
we need to look at chemical
equilibrium.
When we have a mixture, and
we're going to start with
ideal gases.
But everything I'm going to
say about equilibrium and
ideal gases is valid
for solutions.
An ideal gas, and we're
going to be
talking about ideal solutions.
And molecules in an ideal
solution, an ideal solvent,
are not very different than
molecules in an ideal gas.
They don't interact
with each other.
You use concentration instead
of partial pressures, but
pretty much all the ideas
are the same.
All the concepts are the same.
The equations are basically
the same.
You just do a little bit of
replacement of variables.
But it pretty much is
the same thing.
It's easier to think about it,
to learn first in terms of the
ideal gas, but it applies
equally well to what you're
more likely to use.
Which is solutions.
So let's look at the
prototypical gas phase
reaction that everybody writes
down when they first do this
sort of problems.
The Haber process.
Gas, temperature T,
you take nitrogen.
You react it with hydrogen,
gas, temperature T, p.
And you make ammonia.
NH3 gas T, p.
And we're going to ask the
question, so I take nitrogen,
I take some, hydrogen, I mix
them together in a container.
And I make ammonia.
And I'm going to ask, after I
reach equilibrium, what is the
partial pressure of nitrogen
hydrogen and ammonia?
Standard equilibrium problem.
You all, I'm sure you've all
seen equations about
equilibrium.
Equilibrium constant K, log
delta G of the reaction, et
cetera, et cetera.
But you probably don't have a
really good intuition as to
why it is what it is.
So the point of the class here
is not to relearn log K is
equal to minus delta G, reaction
divided by RT.
The point is to learn
how we get there.
How we get to that equation.
And what parts are important.
Specifically, how entropy of
mixing really becomes key to
equilibrium.
But before we get there, let
me give you a little bit of
history as to why this is such
an important reaction.
This reaction, which you see in
every chemistry class, was
developed by Mr. Haber
and Mr. Bosch.
Mr. Bosch made it large-scale
around 1910.
And it's the reaction that you
use to make fertilizer.
Ammonia is the feed stock
to make nitrogen-based
fertilizers.
So today, there are a hundred
million tons of fertilizer, of
nitrogen fertilizer
made, using,
essentially, the Haber process.
It's a huge, huge, commodity.
1% of the world's energy
is taken up
to make this reaction.
Almost 1% is about 3/4
of the world's
energy is used on this.
In World War 1, Germany was
making explosives out of
nitrogen feed stock.
And it was getting its feed
stock from Chile.
From saltpeter mines in Chile.
Chile was under British
hands at the time.
Well, the British didn't
let Chile sell
saltpeter to Germany.
Germany had to find another
way to make ammunitions.
And the Haber process, which
had just been invented by
Bosch-Haber, became the way that
Germany made explosives.
Without this, Germany would have
stopped the war in 1916.
Or even before then.
Way before then.
And this process was basically,
and Haber and Bosch
got the Nobel Prize,
essentially, for this process.
For showing how to take
chemistry, and doing chemistry
and large scale processes under
high pressure and high
temperature conditions.
Haber got the Nobel Prize in
1918 and Bosch got it in 1931.
This process, arguably, you
could say, was the birth of
the dominance of Germany
as a chemical industry.
Based on that, and the fact that
they had to make liquid
fuels out of coal.
The syn-gas, or the
syn-fuel process.
Also invented during World War
1, where they had to use, they
couldn't find any oil and they
had to use their coal mines.
And obviously this process,
the syn-fuel process, is
coming back in vogue with the
energy crisis around now.
So they developed a lot of
knowledge about how to do
large-scale chemical
reactions.
And that was the birth of, or
the explosion of, companies
like Merck and Bayer and Bosch,
and BASF, and all these
German companies that dominate,
basically, the
chemical industry.
And when Germany lost the war,
the US government confiscated
the American divisions of
these German companies.
And I'm sure you've heard
of Merck as the company.
And you think of Merck
as an American
company that makes drugs.
Well, that's true.
It's a very large American
company that makes drugs.
But there's another Merck, which
is the German Merck.
Which also makes drugs.
But it makes liquid crystals.
And liquid crystals is
where it makes all of
its money right now.
And it's very confusing because
they're both Merck.
And they both came from the
Merck family from the 1600s
that were pharmacists.
But after World War 1, Merck
Germany was allowed to use the
name Merck everywhere in the
world except for the US.
In the US it's called EMD.
Merck USA is allowed to use the
name Merck in the US and
outside of the US it's called,
let me remind myself, it's
called MSD.
It's very confusing.
Bayer is the same way.
So Bayer was split up.
And there was Buyer in Germany
and there was Bayer in the US.
You've got Bayer Aspirin
from the US term Bayer.
And you've got all the other
pharmaceuticals from Bayer Germany.
All the chemicals.
And a few years ago, Buyer
bought Bayer, and now we have
one Buyer-Bayer, depending
on where you live.
And so it's very interesting
to see the history of all
these big chemical companies.
And that's why Switzerland and
Germany are still the home of,
it's all due to that reaction.
That's why you always
see that reaction.
Fertilizers, at the birth of the
chemical company, and it's
a great example for
equilibrium.
So that's the example.
Let's generalize it.
We're actually not going
to work on this
until the very end.
And we'll see it again.
But let's generalize.
Let's just take a mixture of
gases with stoichiometries nu
A, a gas, pressure, temperature
plus nu B, B, gas,
pressure, temperature, goes to
nu sub C moles of C, it's a
gas, with a temperature and
pressure, plus nu sub D moles
of D, which is a gas,
temperature, pressure.
All ideal gases.
For now.
So this the setup here.
Now, when we write that reaction
up here, what we're
really writing is, in terms
of the process, is take a
container that contains A, plus
a container that contains
B, and go to a container that
contains C, plus a container
that contains D. And when we
write delta G for that
reaction here, the process that
we're talking about is
taking the reactions separately
from each other.
That's the initial state.
And the final state
is the product.
Separated in their own
containers from each other.
And when we say delta G is
less than zero, for this
process, which means it's
spontaneous, we mean the
process to go from the separated
reactions to the
separated products.
But in reality, that's not what
you do when you do an
experiment.
In reality, you take these
two containers and
you mix them together.
A plus B. And you
let that react.
And at the end of the day, you
have a big container with A
plus B plus C plus D inside,
all mixed up together in
equilibrium.
And this is not the
same as that.
It's not the same as that.
Because you've got
entropy of mixing
happening in all of this.
Forget about A and B interacting
with each other.
Entropy of mixing is going
to dominate equilibrium.
You've got this problem
to deal with.
So that means that we're going
to have to worry about, if
we're going to want to know at
which state the process is in
equilibrium, you're going to
have to worry about this issue
right here.
It's not enough to know what
delta G of the reaction is.
So for instance, if I plot, as
function of the reaction, I've
got the reactants on
that side here.
And the products on
this side here.
And I want to plot delta G as
a function of the reaction.
Well, my initial delta G that
I would write, to calculate
delta G of the reaction, which
is the delta G of the
products, minus the delta G of
the reactants, so initially I
have delta G of the reactants.
That's, in those two boxes,
that aren't mixed.
So I'm up here somewhere.
Delta G of the reactants.
And at the end of the process,
in these two boxes, when I
calculate this delta G naught
reaction, I'm sitting
somewhere here.
Let's say it's a downhill
process.
Delta G naught of
the products.
Well, the first thing that
happens, when I take these two
boxes and mix them together,
is delta G naught of the
reaction's going to go
up or going to go
down, or stay the same.
Anybody want to guess?
When I go from here to here,
before there's any reaction
that happens.
Shall we vote?
How many say that it's
going to go down?
How many say that it's
going to go up?
How many say that it's going
to stay the same?
We have about three people that
said it's going to go up.
And a lot of people that say
it's going to go down.
And a few people that
are abstaining.
OK.
We're mixing things.
Anybody want to change
their votes?
Something is happening here.
And this entropy term here, when
A and B come together and
start mixing up.
You're changing your vote.
Alright.
It's going to go down.
Now we have near-unanimity.
So the first thing that
happens is, you're
going to go down here.
You're going to have delta
G naught of the
reactants in the mixture.
And if it were to go all the
way, if A and B were to
disappear, it still
would be mixed.
And so the end product here
would actually be down here.
It wouldn't be up
in the products.
It would be the mixture.
So that's the first
thing we know.
That this delta G naught
reaction is
not the full story.
And along the way, here I have
two species to begin with.
I've got two species
to end with.
But I've got three different
species here in the middle.
So as soon as I form a little
bit of products, in this case
here, or if I start from
products and go the other way
in the reaction and form the
reactants, the first thing
that's going to happen is I'm
going to decrease the delta G.
Just from the entropy
of mixing.
And so if I plot my delta G
as a function of reaction
conditions, I'm going to get
a bowing curve like that.
If the entropy wasn't there,
then it would just be a
straight line from
one to the other.
The entropy of mixing
of reactants and
products wasn't there.
Actually, let me
put it up here.
If entropy of mixing wasn't
there, I would
start from up here.
And as my stoichiometry changed,
I would have a linear
curve from here to there
as a function of
the process of reaction.
But the entropy of mixing is
causing my initial state and
my final state to go down.
And it's causing a
bow in this here.
Because if I have equal amounts
of A, B, C, and D,
that's a lot of entropy
of mixing there.
So equilibrium is actually
somewhere down here.
It's where delta G of the
mixture is at its lowest.
Delta G of the mixture
is at its lowest.
Any questions?
Now we going to do the math.
We're going to see how
that comes about.
Let me do it here.
So the question we're going to
ask is, suppose that I've got
my mixture here.
And I'm sitting somewhere
on this curve here.
So I've got pA for partial
pressure of A, partial
pressure of B, partial pressure
of C, and partial
pressure of D in my mixture.
And I want to know, I've
got this mixture of
reactants and products.
Which way is the reaction
going to go?
Is it going to go towards
the products?
Is it going to go towards
the reactants?
Or is it at equilibrium?
And to answer that question, I'm
going to let the reaction
react a little bit
more to create a
little bit more products.
Remove a little bit
of reactants.
And see what the sign of delta
G is for that process.
So I'm going to go from a moles
of A, b moles of B, c
moles of C and d moles of D in
my mixture, which is that
point on my graph here.
Partial pressure A, partial
pressure of B, partial
pressure of C, and partial
pressure D, and I'm going to
react it a little bit more. a
plus, a minus epsilon times nu
A, where epsilon is a
very small number.
Of A, b minus epsilon times nu
B times B. And I'm going to
have, now, products being
formed. c moles plus epsilon
times nu C, I've going to have
the stoichiometry in there.
For every nu A moles of A that I
lose, I create mu C moles of
C. Times epsilon as the
scaling factor.
And d plus epsilon times nu D
moles of D. That's going to be
my new mixture.
And I'm going to ask, as I go
from this initial state to
that final state, I'm sitting
on that curve, which sign is
delta G. How do I know what
the sign of delta G is?
What is delta G for
this process?
Not the reaction with the
isolated reactants and
products, but the
reaction where
everything is mixed together.
Where I'm going to have to
worry about the chemical
potentials of mixtures.
And I'm going to go from one
mixture to another mixture,
and that's going to be the key
to telling me if I'm in
equilibrium or not.
But we know how to do this.
We know how to do this.
So I want to calculate delta G
for the process is delta G
after minus delta G before.
That's delta G after minus
delta G before.
And I know that delta G, rather,
let's call it G after
minus G before, the Gibbs free
energy after minus the Gibbs
free energy before.
And I know the Gibbs free energy
is just the sum of the
chemical potentials, right?
You take all the species, and
you take all the chemical
potentials.
You add it up together times
the stoichiometry.
And that gives you the
Gibbs free energy.
That's what we learned
last time.
So if I want to find what the
Gibbs free energy at the end
here is, I look at the chemical
potentials of A, B,
and C and D times their
number of moles.
So I look at a minus epsilon
times nu A times the chemical
potential of A, plus b minus
epsilon times nu B times the
chemical potential of B plus
c, plus epsilon times nu C,
the chemical potential at C,
plus d, plus epsilon times nu
D, the chemical potential
of D.
And then I subtract
what G was before.
This infinitesimally
small process.
Was a times mu A plus b times mu
B plus c times mu C, plus d
times mu D. And we're assuming
that my small change, the
small amount of A and B that get
destroyed to form C and D
is small enough that the
chemical potential basically
stays the same during this
infinitesimally small change.
That's why I can use
the same chemical
potentials before and after.
OK, a lot of things drop out.
This term drops out
from that term.
This term drops out
from this term.
This term drops out
from this term.
This is a minus, no,
this is all plus.
That's fine.
There's the minus
sign right here.
So then this term drops
out from that term.
And I am left with epsilon times
nu C mu C, the chemical
potentials of the products minus
the sum of the chemical
potentials of the reactants,
nu A, mu A plus nu B mu B.
That's the delta G for
this small change.
Now, you remember way back,
maybe from even just last
semester if you've taken 5.112
last semester, or from last
year, or from high school, that
the partial pressure is
going to the equilibrium
constant.
Somehow we're going to
have to get partial
pressures in there.
But we know now how to
go from chemical
potentials to partial pressures.
It's written right here.
The chemical potentials of the,
and we also know how to
go from the chemical potential
in the mixed species, in the,
mixture to the chemical
potential in a pure.
We saw that mu A in the mixture,
temperature, pressure
was equal to mu A pure
temperature,
pressure plus RT log xA.
So those are things that we're
going to be using, to go from
something that has chemical
potential to something where
we'll be able to get back
pressures, partial pressures,
and delta G of the reaction,
because we're going to need
the chemical potential
of the pure stuff.
So let me go forward
just a little bit.
Remind you, if I look at the
delta G of the reaction, delta
G of the reaction, in terms
of chemical potentials.
Delta G naught of
the reaction.
This is the delta G of the
products minus the delta G of
the reactants when they're pure,
not when they're mixed.
So delta G naught of the
reaction is nu C mu C at one
bar pure plus nu D mu D pure.
Minus nu A mu A pure minus nu
B mu B in the pure state.
That's what delta G naught of
the reaction is in terms of
the chemical potentials
of all these species.
Everything's at one bar,
and everything is pure.
Somehow this is going to have
to come out of this.
So let's keep going.
And see how it falls out.
So for each one of these
chemical potentials, I'm going
to write it in terms
of the pressure.
What I just covered up here.
Mu(T, p) is mu naught
of T times RT log p.
So now, delta G is going to be
equal to epsilon, and I'm
going to do a little
massaging quickly.
And I'll let you take this, go
home and see how I went from
one state to the other.
The secret is to put in here
the pressure dependence.
Nu C mu C naught, plus nu D mu
D naught, minus nu A mu A
naught, plus nu B mu B naught,
plus RT log pC to the nu C, pD
to the nu D, divided by pA to
the nu A, pB to the nu B.
These log partial pressures all
come from expanding out
the chemical potential as
mu naught plus RT log p.
And then we recognize that this
part right here, this
part right here is delta
G naught reaction.
So I have delta G now, for
this process of taking
reactants to products just a
little bit, let me take it as
a function of epsilon here.
Delta G naught of the reaction
plus RT log of this ratio of
partial pressures.
I'm going to call that Q. I'm
going to call this thing here
Q, which you've seen before.
The reaction quotient.
And this tells me that for
this very small process,
epsilon, very small, if delta
G naught, delta G, of this
process, is less than zero, then
the reaction will keep
going forward.
It means that I'll be on the
side of the curve here.
I'm going to go down a little
bit. delta G naught is going
to be great, it's going
to be spontaneous.
I'm not in equilibrium.
I'm going to go towards
the products.
If delta G is equal to zero,
then I'm at the bottom of this
curve here.
I'm here.
If I go forward a little
bit, the slope is zero.
Delta G is zero, I'm
in equilibrium.
And if delta G is greater than
zero, then I'm going to back
to reactants, and I'm sitting,
then, on that part of the
curve here.
And if I try to make more
products I'm just going uphill
a little bit.
So now, I've done this
all for epsilon is
equal to very small.
What you usually find written in
books is where the epsilon
equals to one.
Basically taking one mole
of this process.
So you'll see it per-mole
of this reaction.
But it's the same idea.
It's the same thing.
And in fact it doesn't really
matter, because the quantity
that matters is this delta G
naught reactions plus RT log
Q. It's the sign of this
quantity here.
This epsilon is just an
arbitrary number.
I can pick it, really
whatever I want.
So it's the sign of this
thing that's important.
Let me go back.
So what you'll see, then, is
delta G is equal to delta G
naught reaction plus RT log Q.
Basically, taking epsilon is
going to zero.
As determining where the
equilibrium is going to be.
OK, any questions?
All driven by entropy
of mixing.
Without entropy of mixing,
we would be sitting
on this curve here.
Delta G naught of the reaction
would tell us that everything
should go to completion.
Things don't go to completion,
and that's a good thing.
Otherwise there would not
be life on Earth.
We're basically a set of
equilibria in a big membrane.
Which is our skin, right?
All these biochemical
cycles are in
equilibrium with each other.
And it's a complicated
process.
And it's all driven
by entropy.
Ultimately.
And other things, but entropy
is very important.
Alright, so equilibrium now.
Equilibrium is when we have this
delta G equal to zero.
That's when delta G naught of
the reaction equals RT log Q.
And at that point, we replace Q
with to equilibrium, and we
call that the equilibrium
constant.
And we're going to put a little
p here, because it's in
terms of the partial
pressures.
And it's equal to the partial
pressures of the products
raised to their stoichiometry.
And divided by the partial
pressures of the reactants,
raised to the power of
their stoichiometry.
And this, you've seen
before, I'm sure.
At equilibrium.
And there's a minus sign
somewhere here.
Because it's this plus that
that's equal to zero, and
there's the minus sign that
I forgot to write down.
So we define, then, equilibrium
constant this way.
And one of the things to note is
that Kp as written here is
actually unitless.
It doesn't look like it.
The way that I've written it.
Because I did a shorthand way of
writing the pressures, when
I wrote RT log p here.
There's the assumption, when
you have the log of the
pressure, that there's always
one bar sitting behind there.
Underneath.
It's always referenced to
a reference pressure.
There's always a reference
pressure. p naught dividing it
by, because you don't want to
have any units inside the log.
And this reference pressure,
we took as one bar.
And it's pretty common to just
ignore the fact that you've
got one bar sitting in
the denominator.
And so, actually, all these
pressures here are divided by
p reference divided by p
reference, divided by p
reference, divided by
p reference, which
happens to be 1 bar.
But it's there.
And that means that the bars
on top and the bars on the
bottom cancel out.
Which means that K sub p
doesn't have any units.
It is unitless.
It's a number.
Straight number.
Very common mistake to make,
to write it and forget that
there's one bar sitting
on the bottom here.
And you take all the bars to the
new powers on top, and the
bars to the new powers
on the bottom.
And make units there.
And get Kp is equal to some
number, to the, times bar to
some power.
That would be wrong.
I know I've made that
mistake before.
But you are not going to make
that mistake, right?
Because you've been warned.
That this is a common mistake.
This is unitless.
Now, you can invert this to get
Kp as a function of delta
G. e to the minus delta
G naught of the
reaction divided by RT.
And those are the things that
you use to go back and forth
between the thermodynamic
quantities, like G that you
calculate, to equilibrium
quantities like the K, to
finding equilibria
between something
like the Haber process.
Now, there's other equilibrium
constant that is used a lot.
Which is the form of the
equilibrium constant not in
terms of the partial pressures,
but in terms of the
mole fraction.
That's also an important one
when you're looking at
solution cases.
Because we can rewrite the
partial pressures using
Dalton's law. pC is equal to the
mole fraction of species C
times the total pressure,
I'll call it p.
So if I replace every one of
these partial pressures using
Dalton's law, I get K sub p is
equal to xC pC times p to the
nu C, times xD pD to the nu D,
divided by xA pA - no, not pA,
total pressure.
Dalton's law.
To the nu A, xB p to the nu B.
OK, so I've got p, p, p, p.
They all come out.
And that's p to the
minus delta nu.
Where delta nu is the difference
in the number of
moles of reactants minus the
number of moles of products.
So delta nu is nu C plus nu D
minus nu A minus nu B. It's
the moles of products minus
moles of reactants.
And then I have xA to the nu A,
xC to the nu C, xD to the
nu D, xA to the nu A, xB to
the nu B. And that ratio,
which is in terms of mole
fractions, we call K sub x.
The mole fractions.
Because p, to the minus
delta nu, times Kx.
Now, this was unitless.
This pressure to the minus delta
nu sitting here, this
has bars to the minus
delta nu.
Kx has units.
It may not look like it because
it's a bunch of mole
fractions, and it certainly
doesn't look
like it has any units.
Mole fractions or ratio,
but it's got units.
So let's rewrite Kx
in terms of Kp.
Rewrite Kx as equal to p to the
minus delta nu, K sub p.
And the units are in terms of
bars, let's say, bar to the
minus delta nu.
Always want to check your units
at the end of the day.
If your units don't work
out after you've done a
calculation, you're
in trouble.
Any questions?
Because we're going to
end today, at here.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: In the
notes it says Kp
and Kx are both unitless.
That is a mistake.
I'm going to fix that, and
put it on the Web.
So it may be that there's a
special case where the number
of moles before and after
are the same.
But generally that
is not true.
Where did I say that?
Oh, yeah.
Both unitless.
Well, that is actually true.
Let me think through this.
STUDENT: [INAUDIBLE]
PROFESSOR: About one bar,
one bar, one bar.
That is actually true.
They're both, that is true.
They are both unitless.
Well, this is a big boob
on my part here.
Because you are absolutely
right.
Because there's one bar sitting
here. one bar, one
bar, one bar, one bar, and
the bars cancel out.
Good catch.
OK, the notes are right.
My notes are right.
Alright.
Next time we'll talk about the
temperature dependence and the
pressure dependence of
equilibrium constants.
