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>> Okay. So we're going to be talking about equilibrium.
We're going to have a little bit more to say about mu
because mu is important and then I'm going to talk about the bulb
and then I'm going to talk about the equilibrium constant.
This is Chapter 17 material.
We've already tunneled through Chapter 16 leaving out lots
of stuff along the way.
Non-ideal solutions, we don't have time to talk about them.
They're very important.
And there's lots of other stuff there as well.
All right we're already in Chapter 17.
How do we know individual reactants
and products contribute to G?
There could be a whole slew of reactants there,
there could be a whole slew of products.
How do we attribute the G for reaction
to individual reactant and products?
As chemists, this is obviously an important thing for us.
We talked about this thought experiment right here.
We have 2 isotopes of hydrogen in 2 different bulbs separated
by a valve that's initially closed then we open it
and when we open it we can describe the change
in the free energy on the left side in terms
of what's happening with Component 1
and what's happening with Component 2.
In this case, Component 1 is H2 and Component 2 is D2.
Okay, so we can write 2 equations
for the left hand side,
2 equations for the right hand side, 2 equations, 2 terms.
Here's the left hand side, here's the right hand side,
here's Component 1, here's Component 2.
Same thing over here
on the right hand side Component 1 and Component 2.
Then if we consider that anything
that leaves the right hand right side has
to enter the left hand side and anything
that leaves the left hand side had
to enter the right hand side,
we can write a second pair of equations.
These two guys.
What this says is we're only talking
about Component 1 here let's say hydrogen, the partial derivative
of the free energy on the left side and the partial derivative
of the free energy on the right side those two things have
to be equal to one another at equilibrium.
The same thing has to be true in terms of Component 2.
So this could be for hydrogen, this could be for deuterium.
These partial derivatives have to be satisfied,
these equalities that involve these partial derivatives.
So these are so called partial molecular quantities.
Whenever you take the partial derivative
of a thermodynamic state function with respect
to the number of moles
of something it's called a partial molar quantity.
Molar because you're taking the partial derivative with respect
to the number of moles.
The chemical potential is this special case
of a partial molecular quantity that involves the Gibbs Energy.
The Gibbs Energy, the partial of the Gibbs Energy with respect
to Component 1 or Component 2
or Component 3 would be the chemical potential
of Component 1 or Component 2 or Component 3.
Mu describes how G is affected by changes
in the amount of 1 component.
It is in effect the G of a particular component.
It is the Gibbs Energy of a single component.
Why didn't they just call it the Gibbs Energy of Component 1?
I don't know.
It would have been a lot easier because that's what it is.
It's just the Gibbs Energy of Component 1;
hydrogen or deuterium.
Now, it's not obvious to me when I look
at this partial derivative it looks a lot more complicated
than just the Gibbs Energy of Component 1.
Why did they write it this way?
This really is the Gibbs Energy of component 1.
Do other people have a problem seeing that?
Because I know I always have.
So the way I like to think about it is shown here.
Here's a system that contains 3 different components
and the total G for this system you think about it
as just being a balance
and there's a scale here we call this the G meter we know the
free energy, excuse me, the Gibbs Energy,
is an extensive quantity don't we?
So this is a beautiful analogy for that.
We're weighing these components
like a G meter this is an extensive,
this going to be an extensive quantity isn't it?
It's just like mass.
Okay. So what we said is the partial derivative of G
with respect to N and here we're going
to pay attention to N blue.
All right we're going to try and think
about the chemical potential of blue here at constant green,
orange, P and T. All right the way to think
about that derivative is that if you remove a little bit of blue
or add to it a unit of blue here we've subtracted a unit of blue.
This partial derivative is just this difference
and what we're left with is the block that used to be here;
that's the chemical potential of the blue.
For me it's a lot easier to think about it this way.
Now it's clear, yes, this is just the chemical potential
of the blue.
Now it's not the chemical potential of the blue always.
It's the chemical potential blue exactly when the composition
of the system is as I'm showing it here.
There's green here, there's orange here,
there's a bunch of other blue here.
The chemical potential I'm talking about is specific
to the rest of the stuff being there, too,
the chemical potential could be different
if there was different amounts of stuff here.
It would be as though this balance could be different,
this arrow could point in different directions even
for a single blue block if the rest of these guys,
maybe if there's only 3 green guys or 6 orange guys,
the chemical potential could change a little bit
but were always talking about the chemical potential
of this pure component of whatever it is
that we're talking about and this case it's the blue blocks.
Does this make any more sense?
I don't know.
It does for me.
Okay. So what do we need to know about the chemical potential?
It's extensive just like G. It's got units
of energy per mole just like G. It's pressure dependence is the
same as G and so is the temperature dependence.
Okay so I can say the partial derivative
of the chemical potential of any component with respect
to its partial pressure is equal to the molar volume
of that component and if I integrate that I'm going
to get an equation that looks just like the equation
that we write for G except now it pertains
to a single component and same thing with the entropy.
But how does it depend on concentration?
How does the chemical potential depend on concentration?
We haven't talked about that even with respect
to G. Well, here's the equation.
The chemical potential of some species A is given
by the standard chemical potential that applies
under defined standard state conditions, which in the case
of a gas would be pressure of 1 bar and temperature
of 298.16 degrees Kelvin.
Right? In the solute, in a solution,
the standard state is an activity of 1 molar.
We haven't said what that activity is yet.
So it's equal to this mu 0, which is the chemical potential
for that pure compound in its standard state plus RT log
of the activity of A. What's that?
Well, to beg the question and not answer it, the activity
of A is equal to the activity coefficient
of A times its concentration.
Now, I haven't told you what the activity coefficient is;
that's that activity coefficient,
that's the concentration, so now we have a relationship
between the chemical potential and the concentration
of A. We all know what the concentration is
and the concentration could be a pressure.
There would still be an activity coefficient.
So if it's in the gas phase, it's a pressure,
it's the partial pressure, of course,
and if it's in the liquid phase, it would be a concentration
and this activity coefficient it turns out can vary between 0
and 1 and we're going to talk more about that later on
but it's sort of a distraction for us now so we're not going
to talk about it right now.
For our purposes we're just going to equate
for the time being the activity with the concentration.
We're just going to pretend for the time being
that this activity coefficient is always 1.
Okay? That's an assumption.
It's an incorrect assumption most of the time.
We'll talk about activity coefficients later.
Okay. So we can relate the chemical potential
to the concentration through an equation
that looks just like this.
Okay, now we already understand the G is minimized upon approach
to equilibrium.
We already talked about this.
Equilibrium is at the bottom of this curve but we didn't talk
about why this is bow shaped
because it's not intuitive that it would be.
I mean if this is the Gibbs Energy of the reactants,
this is the Gibbs Energy of the products,
why doesn't the Gibbs Energy of ever intermediate mixture
of reactants and products fall in this dashed line up here?
It's the simplest solution to this problem isn't it?
If that were the case,
there would never be any equilibrium would there?
Because there would never be any curvature to this thing.
No matter where you put this dot
and where you put this dot you're always going
to have a straight line and it's never going to have a minimum
where you would expect to find equilibrium is there?
So the fact that there's a bow to this curve is sort
of essential to how chemical systems work.
Chemical systems have equilibrium.
We need to understand that.
This is not what happens.
Okay, so let's do a simple calculation.
I've got 02 on the left side,
N2 on the right side; half a mole of each.
Now I'm going to open the valve.
This reaction coordinate is the mole fraction of N2.
When I open the valve, I'm going to have .25 moles of O2
and .25 moles of N2 after everything comes to equilibrium.
Okay. So the mole fraction is simple.
The moles of N2 is .25 over the total number of moles is .5
and so .25 over .5 is .5
and so we're right here on this diagram now.
We're right here after the mixing has happened.
Okay so let's calculate G. G is just equal to the number
of moles of N2 times the chemical potential of N2
and the number of moles of H2 except H2 is not in there;
that should be O2 times the chemical potential of O2;
it's all O2 and N2 here.
Okay. So what's the chemical potential of N2 that's given
by the, because we're talking
about a gas that's the partial pressure of N2, this is 1 bar
and we can sort of ignore it.
Your book usually does.
Just call it 1.
Okay, this is the analogous term for O2.
We multiply by the number of moles.
Yes, this is the chemical potential of N2 at 1 bar
and 298.16 degrees Kelvin and that's 1 bar.
Yes, your book just writes it gets rid
of the P0 and just calls it 1.
Don't be confused by that.
So since P is the same on both sides we'll just call it
P prime.
At equilibrium, the total pressure
on the two sides is the same.
We'll just call it P prime.
So after opening the valve, we're going to have P prime
over 2 for nitrogen and P prime over 2 for oxygen
because half the oxygen is going to go into the left bulb
and half the oxygen is going to go
into the, you know what I mean.
Half the nitrogen is going to go,
all right so the pressure is going to go
down by a factor of 2, right?
Yes. This is all for oxygen and this is all for nitrogen.
So the difference, the Delta G, [inaudible] is just going
to be P prime over 2 divided by 3P prime for nitrogen,
same thing for oxygen and if I put the numbers in,
half a mole total of nitrogen, 8.31451, 298.16 Kelvin.
All right there's a term for nitrogen, a term for oxygen
that is why there's 2 logs here.
All right when I plug it into my calculator,
I get minus 1,717 joules.
So I can look at my plot then and this Delta is 1,717 joules
when I'm at 0.5 that's not on the line.
It's nowhere close to the line.
The line is at 0.
Why is that?
The line is at 0 because the mu 0 for nitrogen and the mu 0
for oxygen are both 0.
Look in the back of your book; there's a big table
of standard free energy the formation.
Standard gives energy for the formation rather.
You can look up nitrogen and oxygen
and any other diatomic gas like that.
It gives energy of formation 0.
All right so we're measuring versus 0.
In other words, the curve has got to go
through this point right here and it's got to go
through that point and that point so it's got to be a curve.
There's no way it can be a straight line
between these 2 data points.
We've just proven that.
The question is why?
In other words, we know Delta G is Delta H minus T Delta S,
which of these 2 things is responsible for this curvature?
Well, there's no change in the entropy during this reaction
that these are [inaudible] gases is there?
There's no change in the entropy
that these are [inaudible] gases.
It's got to be in that term right there.
It's got to be the entropy and, of course, when you think
about it as you mix these 2 gases together, you allow each
of these 2 components to have more volume
to occupy the entropy of the system is going to go up.
That's the driving force for this curvature
that we're showing here.
All right?
In fact my plot, the entropy, that looks like that.
All right?
Now what I've plotted here is actually T times Delta S
because I wanted to show you
that it's the exact positive reflection of this guy.
If I put a mirror here, you get this.
That's multiplied by T so that is going to be numerically equal
to the depth of the Delta G well isn't it?
It's going to be numerically equal.
So entropy I think this is moderately profound, all right?
If you didn't have entropy acting in this process,
there wouldn't be an equilibrium at all because no curvature
of this G versus progress of reaction curve,
there is no equilibrium.
The universe would be fundamentally different
than it is.
We have this curvature in Delta G because of the entropy
of mixing; that's what this is the entropy of mixing.
Okay. Now, this is one of my favorite molecules.
It's called resveratrol [phonetic].
Why would it be one of my favorite molecules?
Where is resveratrol found in our environment?
One of the nice things about resveratrol is
that it makes things live longer and if you're my age,
you're very interested in that even though it's only been
tested in yeast, fruit flies and worms [laughter].
I'd like to think that human beings may also benefit
from the same biochemistry that's going on with these guys.
Mice do well on resveratrol.
They don't get cancer as often when they eat a lot of it.
Unfortunately, scientists feed them enormous quantities
and then they don't get cancer for gosh sakes.
Here's the mouse without eating huge amounts of resveratrol
in acetone mind you, happy day [Laughter].
So the reason we humans like it is because it's found
in red wine, California [inaudible], oh, my.
[Laughter] So, this could mean that red wine is good
for you but [inaudible].
[Laughter] Absolutely delicious.
Why do we even mention it?
Well, because there are isomers of resveratrol
and if you hit resveratrol
with 350 nanometer light you get isomerization [phonetic]
around that double bond right there.
So there's sis and trans versions and so this is
like the simplest chemical reaction.
Isomerization reaction is
like the world's simplest chemical reaction.
[Inaudible] Isomerization [inaudible] figure
out where the equilibrium generic version
of A [inaudible] to B.
Okay. Now, let's say I'm starting off
with [inaudible] that's why I'm calling it 0
because it's the starting point [inaudible] starting point.
If there's a change now and some reaction occurs,
some of this A is going to converted to B and some
of this B, so the reaction is going from left
to right here we're going to lose A and make B
and so this thing is called the progress of the reaction
or the extent of the reaction and it's going
to tell us how far the reaction is proceeded from left to right.
Okay, so that's minus .1, that's plus .1
and the units here are moles.
The extent of the reaction is units of moles.
It's not dimensionless.
Okay. So, we start off with 1 mole of A,
[inaudible] to B this we can convert this axis here
into an access in terms of the progress of the reaction.
If that's the case, this would be 1 and this would be 0.
The progress reaction is 0 we've got all A and it's 1
and we've got all B. Make sense?
One mole regardless
of the reaction it always has units of moles.
Okay. So we can find this quantity value of the progress
of the reaction as Delta R
of G. This is a notation your book uses.
This is the reaction Gibbs Free Energy.
So if I look at this point right here,
all right the reaction Gibbs Free Energy is the [inaudible]
derivative of the G function with respect
to the extent of the reaction.
It turns out you have to denominate everything in terms
of the extent of the reaction in order for this to work properly.
So here Delta R of G is less than 0, yes,
the slope is negative.
Here Delta G of R is equal to 0, yes, the slope is 0
and the Delta G, Delta R of G is greater than 0, yes,
the slope is positive.
Okay. So we've got this thing called the reactions Gibbs Free
Energy except we're not supposed to use the word free.
[inaudible].
Okay. So, [inaudible] change.
[ Pause ]
Okay. So here are conditions
of constant pressure and temperature.
The reaction advances by some small amount.
All right.
A will change by G moles of A, B will change by G moles
of B. Notice that the extend
of the reaction is positive by definition.
Okay. Here if we have a reaction from A to B,
D and A would be a negative quantity wouldn't it?
Because A would be getting smaller,
B would be getting bigger so that would be positive,
that would be negative and so in order for this to be equal
to this, I have to have a minus sign there because that's going
to be positive by definition.
Everyone see that?
Okay so these two things are equal to one another and so
if that's the case, I can factor out this, what is that ki?
Factor that out and I've got this expression right here.
So in other words, this Gibbs Reaction Energy is just going
to be proportional to the difference
in the chemical potentials.
So Delta G of R should be Delta R of G, of course,
it's just the difference between the chemical potentials.
It's just that.
When I take D, G, D ki, I get that guy.
Okay so it's obvious from this plot
but let's say it anyway the Gibbs Reaction Energy changes
continuously, monotonically with progress of the reaction
or the extent of the reaction, therefore, the composition
of the reaction mixture.
Yes, this is a smooth and continuous function.
Okay so that means that there's 3 types of reactions,
reactions that are exeronic, exergonic,
where Delta R of G is less than 0.
Endergonic, where Delta R of G is greater than 0
and equilibrium where it equals 0.
If you don't know what this word means, as always,
Wikipedia can help you.
It means the releasing energy in the form of work.
If it's an exergonic reaction,
it can release energy in the form of work.
You can do work if Delta G is, Delta R of G is less than 0.
If you want Delta R of G to be greater than 0,
you've got to put work into the system to make that happen.
Okay, but we still haven't learned anything more
about where equilibrium is located in terms
of the extent of the reactions.
To get there let's assume A and B are ideal gases.
Then you will recall from Lecture 17,
isn't this Lecture 17?
[Laughter] Let's assume A and B are ideal gases,
then you'll recall from Lecture 16, yes, that's right.
So the difference in the free Gibbs Energy
between some initial state and some final state is just given
by this integral where that's the molar volume
and we can convert that we know if we're talking
about an ideal gas, plug and chug, here's the expression
as a function of pressure.
Okay, and we can define a standard molar Gibbs Free Energy
as one that applies at a pressure of 1 bar
and 298.16 degrees Kelvin.
That's what that is right there.
Okay, and so if A and B are gases we can write the chemical
potentials in this form, for example,
because this is just a chemical potential of a single species
and it's going to follow exactly this form.
I can write this equation right here by analogy to that one
and this is the standard chemical potential,
which once again applies at 1 bar and 298.16 degrees Kelvin.
Okay, so the Gibbs Reaction Energy we said is the difference
between these two chemical potentials and so
if I write an equation like this for A and another one for B,
I can take the difference.
There's the B one, there's the A one,
and when I take the difference, yes, I already said that,
I've got mu B 0 minus mu A 0 that's that, and then I,
the P 0s are going to cancel and I'm going to end up with PB
over PA and this is the standard now Gibbs Reaction Energy.
It applies at 1 bar and 298.16 degrees Kelvin.
Okay. And we have a name for this, right, this quotient
between the partial pressure B and the partial pressure
of A is called the reaction quotient something
that you heard about in general chemistry.
Lo those many years ago.
We call it Q the reaction quotient.
Okay so for every value of the extent of the reaction
across the horizontal axis of G versus extent
of reaction diagram we calculate the Gibbs Reaction Energy
from the Gibbs Reaction Energy
of the product reactant species present at that extent
of the reaction, yes, and so finally
at equilibrium the reaction quotient is given by,
so when this is 0, at the bottom of that well
where you've got a horizontal line tangent to the curve,
we can call this K. The reaction quotient isn't Q anymore.
Here it's Q, all right, under the special case
where that's o0, we can call the reaction quotient K,
K or K equilibrium.
These are then equilibrium values
for these 2 partial pressures
and from this I can just move this guy to the left hand side
and I've got minus RT log K. So there's a relationship
between the free energy of the reaction,
the standard free energy of the reactions,
standard because this is a 0 and the equilibrium constant.
For example, let's say the Delta G of the reaction is 0.
If Delta G is 0, Y can solve for K and Delta G is 0,
exponential of 0 is 1, yes, all right.
In my extent in my equilibrium constant for the reaction is 1,
all right, which is going to be correspond
to equal partial pressures of B and A, right,
because the equilibrium constant is partial pressure of B divided
by the partial pressure of A. Okay, here's the reaction,
the standard reaction free energy difference between A
and B is 0 the equilibrium constant is right in the middle
and this would be a nice symmetrical curve isn't it?
Yes. And on the other hand
if this standard Gibbs Reaction Energy is less than 0,
in other words, the final Gibbs Energy minus the initial Gibbs
energy is as shown here this guy is less than 0
and now the curve is going to look like that
and now equilibrium has got to be over here, all right,
and that means that it's going to be greater than 1.
It has to be greater than 1.
What does that mean for the equilibrium constant?
Well, if we look at the equilibrium constant,
here is the equilibrium constant refresher
for a generic reaction, A moles of A plus B moles of B goes
to C moles to C and D moles of D. We can write
that equilibrium constant in terms of equilibrium activities
of each reactant in product species, yes,
it's just the activity of C taken
to the C power times the activity of D taken
to the D power and so on.
Products over reactants, right, we can write this in terms
of activity coefficients and concentrations.
Yes, these are activities and these are concentrations
and these are activity coefficients.
So what this means is
that you've got a product rich mixture.
You've got more product than you do reactant because you're
on the right hand side of 1.
If the opposite is true, if you're uphill from A to B,
you're going to end up with an equilibrium constant
that has the opposite, it's reactant rich.
All right we'll come back to this on Monday.
Have a good weekend. ------------------------------f995db2af038--