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-- sometimes called the van der Waal's interactions.
And we saw that we could make a molecule between two inert gas
atoms, like argon two or xenon two,
by virtue of these dispersion interactions,
where the instantaneous charge in one atom or molecule produces
a dipole. That dipole then induces a
dipole in the neighboring molecule, and the result is a
stabilization, an attraction.
And we saw, last time, we had a generic form for the
interaction potential for the dispersion interactions.
This Lennard-Jones interaction potential.
We talked about that in detail. And one of the parameters in
that Lennard-Jones potential was this quantity epsilon,
which actually was the well depth.
And I just wanted to talk a few moments, here,
about what determines what the well depth is,
what the strength of that interaction is.
And the bottom line is, what determines that is the
polarizability of the atoms or the molecules that are
interacting. And we give the symbol alpha to
the polarizability. And what that is,
is a measure of the ease with which a charge distribution can
be distorted. That is the polarizability of
the molecule. That is a term you will hear in
future courses quite a lot. And, in general,
the polarizability goes up with the number of electrons present.
As you go from -- And you can see on the side
slides, here. As you go from helium to argon
to xenon, and let me fix my pointer.
As we go from helium to argon to xenon, this polarizability
goes up because the number of electrons are going up.
If the number of electrons are going up, that means we have
electrons in outer shells, they are farther away from the
nucleus. They are more easily distorted.
As this polarizability goes up, helium, argon and xenon,
that well depth for the Lennard-Jones potential is going
up, 0.085, 0.996 and 1.8. And here, on this diagram,
I actually show you what the shape of the Lennard-Jones
potential is. Sometimes when I draw it on the
board, it is not so accurate. But you can see this
Lennard-Jones potentially actually has this repulsive wall
that goes up really very steeply.
And, likewise, on the next slide,
you see some interactions for at least one molecule here,
helium, nitrogen, and argon.
You can also have these dispersion interactions,
and you do have them, between molecules.
So, the polarizability is going up.
Helium, nitrogen, argon, the well depth is going
up. Again, these are the
Lennard-Jones potentials for those inert gases and for the
interactions between two nitrogen atoms,
dispersion interaction. All atoms and molecules have
these dispersion interactions. It is just that often times the
dispersion interactions can be so weak compared to some other
interactions, which we are going to look at
today, that we don't even think about them.
But they are actually there. But what I want to do now is to
talk about what happens as we lower the temperature even
further. In other words,
last time we were talking about the deviation from the inert gas
law as we lowered the temperature.
And what we said is that it is these dispersion interactions
that are the microscopic origin for the deviation from the inert
gas law, which is a macroscopic law, as you lower the
temperature. But we also know that if you
lower the temperature enough, that the gas condenses,
the gas liquefies. And it is these dispersion
interactions that are also responsible for the condensation
of these gases. But now, let's talk about,
a little bit more deeply, what exactly is going on as we
lower the temperature close to the liquification point.
Let me draw a Lennard-Jones potential again.
And now I am going to do it for two nitrogen molecules.
Here is nitrogen, here is nitrogen,
two separated nitrogens as a function of r,
the distance between the two nuclei.
This is our zero of interaction.
And, as I said from that slide, this bond association energy
measured from the bottom of the well is 0.79 kilojoules per
mole, the bottom of the well, where the molecules actually
can never be. But that is the well depth.
Now, let's think about this. At 300 degrees Kelvin,
what is the average energy of the molecules?
Well, the average energy is three-halves RT,
as we saw. And, if I substitute in this
temperature, I am going to get something on the order of 3.7
kilojoules per mole, if this has two significant
figures. 3.7 kilojoules per mole.
Well, let me do the following. Let me draw 3.7 kilojoules per
mole, here, on this Lennard-Jones potential.
When I do that, and I am going to start here at
the bottom of the well, and I draw 3.7 kilojoules per
mole, well, that is somewhere way up here, compared to the
interaction energy. At 300 degrees Kelvin,
the bottom line is that those two nitrogen molecules have
enough kinetic energy to totally ignore this interaction energy.
They are way up here. If they have 3.7 kilojoules per
mole of energy, they are not going to stick
together. They are not going to condense.
They are actually really going to ignore this small interaction
energy. They are just going to fly
apart. And they do.
You have a gas. However, let's lower the
temperature now. Say we lower the temperature to
100 degrees Kelvin. When we do that and calculate
the average energy at degrees Kelvin,
that is on the order of 1.2 or so kilojoules per mole average
energy. 1.2 kilojoules per mole,
we are somewhere here. And now, the relative energies
between these nitrogen molecules are much lower.
And it is beginning to be comparable to this well depth.
And those nitrogen molecules now are deviating from the inert
gas law. They are kind of hanging around
each other. They don't hit the walls as
often, the walls of the vessel that they are in,
because they are having an attractive interaction with
their neighboring nitrogen molecule.
And so, we are deviating, here, from the inert gas law.
The pressure is not as large if you are doing it under constant
volume conditions. And then, say we lower the
temperature to 77 degrees Kelvin, which is actually the
boiling point of nitrogen to make liquid nitrogen.
Well, at 77 degrees Kelvin, the kinetic energy is 0.96
kilojoules per mole. And now, in this case,
we are fairly comparable to the well depth here at 0.96
kilojoules per mole. What happens is the gas
condenses. The interaction energy between
those two nitrogen molecules is on the order of the kinetic
energy. And then these molecules stick
together, the gas condenses, and you have a liquid.
That is the origin, here, of this temperature
dependence and the relevance to the microscopic interactions
between the molecules. You can understand that.
And then vice versa, if you start to raise the
temperature, the molecules start to fly apart.
They ignore this interaction. Their energy is greater than
that attractive interaction. And so what you can see in the
macroscopic boiling points is the vestiges of this microscopic
interaction energy. I think I am going to put the
center screen down.
If you look at --
-- helium, neon, argon, krypton,
and xenon. What we said is that this
polarizability increases as we increase the number of
electrons. We see that the well depth
increases. Therefore, the boiling point
increases as we go down the inert gas column.
You can see how that macroscopic boiling point is a
reflection of what is happening on the microscopic scale between
two individual molecules. You can also see that in this
plot or in this chart. Here we have molecules,
hydrogen, nitrogen, and oxygen.
The polarizability alpha is also increasing as we go down.
Correspondingly, the well depth is increasing.
Correspondingly, that boiling point is also
getting larger and larger. That macroscopic quantity,
then, reflects the change in the individual interaction
energies between the molecules, the microscopic quantity.
But it also turns out that the shape of the molecules are
important. For example,
let's take this molecule, C five H twelve.
There are several ways I can draw a skeletal structure for C
five H twelve. One way is to make it pentane,
making a linear molecule. And another way is to make this
two, two dimethylpropane,
carbon in the center, some methyl groups around that
center carbon. It turns out that the boiling
point of pentane is 309 degrees Kelvin.
The boiling point of 2,2-dimethylpropane is
degrees Kelvin. These are two molecules that
have the same number of atoms, i.e., the same number of
electrons, should have the same polarizability.
However, one has a higher boiling point than the other.
And the reason for this is because of the different shapes
of these molecules. In the case of propane,
if we have an instantaneous fluctuation in our charge
distribution, it is going to be essentially
along a line. That instantaneous fluctuation
is essentially kind of rod-like because of the skeletal nature
of the pentane. So, one end is a little
positive, one end is a little negative, the other way around.
That is going to induce a dipole in a neighboring pentane
molecule. And then they are going to
align, positive-negative here, positive-negative there.
But then, in the case of dimethylpropane,
you are also going to have this charge fluctuation.
But in the case of propane, here, this is a more
spherical-looking molecule. And so, the charge fluctuation
is not going to be so rod-like. But, nevertheless,
this charge fluctuation, induced dipole here,
is going to induce another dipole in a neighboring
molecule, and you are going to have an attractive interaction.
But, in the case of the 2,2-dimethylpropane,
this is a much more spherical distribution than this.
In this case, these two dipoles are not as
close together as they are in the case of pentane,
meaning that the interaction energy, here,
between the dimethylpropane, is going to be less than it is
in the case of the pentane. And, again, that is reflected
in the macroscopic boiling points.
The boiling point of pentane is larger than that of
dimethylpropane because that microscopic interaction energy
is larger for propane because the induced dipoles can get
closer together. So, the shape is also important
in determining these boiling points, these energies of
interactions. That is going to take care of
our discussion of molecules or discussion of the induced dipole
- induced dipole interaction energy where we are talking
about molecules that do not have permanent dipole moments.
But now we are going to turn to molecules with permanent dipole
moments, such as HCl. And, of course,
in these molecules, the dispersion interaction is
also taking place. It is just that that is going
to be weak compared to the interaction between two
permanent dipoles. Here we have one HCl molecule,
permanent dipole. It is going to then,
in a collection of HCl molecules, attract another HCl
molecule. And that HCl molecule is going
to align in the opposite direction.
The alignment of those two dipoles is going to lower the
energy. That is the attractive
interaction. And then you might say,
you get this attractive interaction, but you also have
now this repulsive interaction between the two chlorines and
the two hydrogens. Doesn't this all cancel out?
And the answer is no. And that is because of this.
The positive end of one molecule and the negative end of
the other, this distance and this distance on the opposite
end are actually closer than the positive charges.
The yellow distance, here, is smaller than the
distance between the two hydrogens, is smaller than the
distance between the two chlorines.
And so, it is this attractive interaction that wins out,
actually. When you put those two dipoles
together, the repulsion actually is there, but it is the
attractive interaction that wins out, and the whole system is
stabilized because the distance between the unlike charges is
smaller than the distance between the like charges.
Now, it also turns out that we have a functional form for the
dipole-dipole attractive interaction.
And that attractive interaction turns out to be a minus one over
r cubed dependence. And this is exact.
You can actually derive this. You can show that this is the
interaction between two permanent dipoles is one over r
cubed. The quantity on the top,
mu, that is not a reduced mass, this time.
If we go back to when we started talking about dipole
moments, this is the dipole moment of the molecule.
We have to use our symbols for multiple quantities.
So, that is the dipole moment. We are talking about
dipole-dipole attractive interactions.
And that attractive interaction looks like this,
minus one over r cubed.
Now, you might say, where is the repulsive part of
this interaction? In the case of the
Lennard-Jones potential, remember we had two parts,
the attractive part and the repulsive part.
The repulsive part was one over r to the 12.
The attractive part was one over r to the 6.
But it turns out that for dipole-dipole interaction,
we do not have a general form for the repulsive part,
unlike induced dipole - induced dipole.
So, the best we can do, in general, is to tell you that
the dipole-dipole attractive interaction is one over r cubed.
But I want to compare this one over r cubed to the attractive
interaction due to the induced dipole - induced dipole
interaction. You see that the dipole-dipole
interaction is what we call longer range than the induced
dipole - induced dipole. What I mean by that is that the
value of r here, the distance between the two
dipoles, can be larger. In the case of a longer range
interaction, it can be larger and still have some non-zero
quantity for the interaction potential.
For example, if you are out here,
if this value of r, you can just see with the
shorter range interaction, one over r to the 6,
that the attractive interaction is only the
difference between the pink curve and the black curve,
which is zero. Whereas, with a longer range
interaction, we have more attractive interaction.
This is more negative. That energy difference is
larger. That is what we mean by a
longer range interaction. And I want to just compare,
in this diagram here, the strength of the dispersion
interaction to that of the permanent dipole-dipole
interaction by comparing the interaction between two argon
atoms to those between two HCl molecules.
Argon only has the dispersive interaction between it because
it does not have a dipole moment.
And you can see that this well depth, here, is one kilojoule
per mole, roughly speaking. But in the case of HCl,
that well depth, here, is three kilojoules per
mole because that has a permanent dipole.
And, in both cases, HCl and argon-argon,
we are talking about roughly the same number of electrons,
not exactly, but roughly,
meaning the polarizability is roughly the same.
The dispersion interaction energy is roughly the same for
HCl as it is for argon, but the difference is that HCl
has that permanent dipole moment.
Therefore, the deeper well depth, therefore,
in HCl, the higher boiling point, 239 degrees Kelvin as
opposed to 87 Kelvin for argon. What is on this slide is just
an example of how that dipole-dipole interaction energy
varies with the dipole moment. The larger the dipole,
of course, the larger the interaction energy.
Here, I show you several molecules that all have,
again, roughly the same number of atoms.
They have the same mass, they roughly have the same
number of electrons, so they roughly have the same
polarizability. The induced dipole - induced
dipole is the same, but what is changing here,
as I go down, is the dipole moment.
For propane, dipole moment really small,
0.1 debye. Dimethyl ethe,r 1.3.
Acetaldehyde,
Acetonitrile,
Dipole moment increases. The boiling point increases
because that attractive interaction is increasing.
It scales roughly as the dipole moment squared.
The dipole-dipole interaction is stronger.
Now, I just want to briefly then remind you about one other
attractive interaction that we have talked about before.
And that is between two ions. There we are talking about the
Coulomb interaction energy, where the dependence is minus
one over r. That is the longest range
interaction. Again, you can see this way out
here. If I choose this value of r,
the one over six interaction term would give me a
very small value for the attractive interaction,
-- -- because it is one over r to
the six. You take a large number for r
and raise it to the 6 power and put in the denominator.
You are going to have a small value for U of r.
The one over r to the three gives you some
interaction energy, but one over r gives you a lot
of attractive interaction. It is the longest range.
And you can also see here, in a moment,
that it is going to be the strongest interaction.
What I am just doing is comparing several molecules or
atoms. Here is argon 2 which has only
the dispersive interaction. Here is the plot,
well depth minus one kilojoule. Then, there is the HCl
interaction energy that has the dispersion interaction in it,
but also the dipole-dipole permanent interaction.
It is minus three kilojoules. And then we are talking about
chlorine two. This is a covalent bond.
This is no longer a Lennard-Jones potential energy,
but this well depth, here, is minus 200 kilojoules
per mole. And then, finally,
this ionic interaction between potassium and chlorine,
look at how strong that is, minus 450 kilojoules per mole.
So, these are the relative strengths here of these
interaction energies. That is all I that want to say
about these kinds of intermolecular interactions,
where we are dealing with some kind of dipole.
But before we move on, I want to talk about one other
kind of intermolecular interaction potential.
And that is something called hydrogen bonding.
All right. A final intermolecular
interaction, hydrogen bonding. Hydrogen bonding,
here, occurs between a hydrogen atom that is attached to an
electronegative atom and another molecule in the gas or in the
solution. The first requirement is that
you have hydrogen attached to a very electronegative atom.
Hydrogen bonding occurs for hydrogens attached to oxygen,
nitrogen, and chlorine. Those are all electronegative
atoms. Water is a good example.
In the case of water-- You have hydrogen bonded to this oxygen.
This oxygen is really very electronegative.
What that oxygen does is it pulls those electrons away from
the hydrogen. This hydrogen is kind of
partially unshielded. It is partially deshielded.
And then, there is a lot of electron density here on this
oxygen. Well, because that hydrogen is
partially deshielded and because it is really small,
this hydrogen actually will kind of see the oxygen atoms on
a neighboring water molecule. And since this is kind of
partially negative, this hydrogen will interact
with one of the lone pairs on this oxygen and will form a
bond. This is a little bit partially
positive, this is a little bit partially negative,
and the result, here, is a bond between the
hydrogen and the oxygen. And that bond is on the order
of 20 to 60 kilojoules per mole. This bond is not a covalent
bond. A covalent bond is
kilojoules per mole. This is 10% of it,
but it is still a very important quantity or a very
important phenomenon, this hydrogen bonding.
The hydrogen bonding is certainly responsible for the
peculiar properties of water, as you will learn more about in
5.60, but it is also responsible for the unique shape,
oftentimes, of biological molecules.
The helix in DNA owes its structure to hydrogen bonding.
The hydrogen bonding is what makes trees stand upright.
The long cellulose molecules in trees are actually bonded
together by hydrogen bonding. Nylon owes its strength to
hydrogen bonding. And hydrogen bonding is also
responsible for whether or not you have a bad hair day.
As an example of that, I want you to look at the slide
up there on the walls. What you see is a protein
molecule. This is the structure of hair.
The molecules that make up the strands of your hair look like
this. It is a polymer.
Well, it is a polypeptide. This unit right in here,
carbon-oxygen bound to carbon-hydrogen bound to
nitrogen-hydrogen is the peptide.
It is repeated. You see the next unit over?
CO-CH-NH. CO-CH-NH.
That keeps repeating. That is the repeat unit.
And, of course, this carbon-hydrogen right
there, you can see it has a line there, indicating that it is
bonded to something. And it is bonded to something.
If that carbon-hydrogen is bound to another hydrogen,
then you have a polypeptide which has been made from an
amino acid that you might know of as glycine.
If that carbon is bound to a CH three group,
then that peptide was made from an amino acid that you might
know of as alanine. And if it is bound to a CH two
S H group, well, that was an amino acid
known as cysteine. But what I want you to notice
here is that these hydrogens on the nitrogen --
That nitrogen is an electron negative atom.
And that hydrogen, if your hair is wet,
it is actually hydrogen bonded to a water molecule.
Here is that hydrogen bond. And then in the next strand
over, this oxygen is hydrogen bonded to a water molecule when
your hair is wet. And the bottom line is that
when your hair is wet, you have the strands of your
hair that kind of slip by each other.
There is no registry of one strand to another,
because each strand has this coating, here,
of water molecules due to hydrogen bonding.
Suppose you take your hair and put it in a very contorted
configuration like this. And then you drive off the
water molecules, you dry your hair.
What happens is that these water molecules then leave,
the hydrogen bonds are broken. This is not such a strong bond,
20 to 60 kilojoules per mole, those hydrogen bonds are
broken. And then this hydrogen on this
nitrogen looks around and sees the oxygen with its lone pairs
on this strand, and so you form a hydrogen bond
between this strand and this strand.
And so now the strands of your hair are in registry with each
other. They are actually stronger.
And they do tell you not to brush your hair when it is wet,
because it isn't so strong. Well, it is not so strong
because these water molecules are insulating each one of these
strands. And when it is dry these two
strands are bond to each other. They are in registry with each
other. And so, if you do this right,
and you then let go of the contorted configuration,
which I am having trouble doing, those strands are in
registry with each other now. And you have a good hair day.
So, that is the importance of hydrogen bonding.
I curled my hair just for this demo, a little asymmetric,
here. [LAUGHTER] But now,
if you are as fortunate, or unfortunate,
depending on your preference, to have naturally curly hair,
then what you have are a lot of sulfur-sulfur bonds.
You have a lot of cysteine peptide groups.
What happens there is this. On the CH groups,
you have CH two sulfur H.
And on the next strand over you have S, CH two,
and CH bonded to nitrogen, bonded to a CO.
And you actually have a covalent bond between these two
sulfurs, here. This is a strong bond.
The strands of your polymers in your hair are in registry all of
the time. And if you want to make your
natural curly hair straight you have to do drastic things like
use drastic chemicals to break this sulfur-sulfur bond.
You can do it, but it is not easy.
Likewise, if you have naturally straight hair and you want to
curl it and make it semi-permanently curly,
then you have to build in the sulfur-sulfur bonds.
And you have to do, again, some rather drastic
chemistry to make that happen. Hydrogen bonding is important,
especially if you go to do anything in biologically-
related sciences, you will see that.
Now, I am going to change topics here.
I am going to change topics, and we are going to talk a
little bit about thermodynamics in preparation to get up to
chemical equilibrium so that Professor Cummins can come in
next Wednesday and start talking about acid-base equilibrium.
He is great. You will love him.
We are going to review some thermodynamics today.
I am going to go kind of quickly because some of this I
think you know, but I want to make sure
everybody is on the same page. First of all,
bond energies. We talked about bond energies
as delta E sub D. And we measured it from the
bottom of the well. And I told you a few days ago,
I lied to you. The measured energies are
really from v equals zero, and they are.
But what I am going to do is change my language.
Instead of talking about energies, I am going to talk
about enthalpies. I am going to talk about delta
Hs rather than delta Es. The reason I am going to do
this is because it is easier for us to measure a bond enthalpy
than a bond energy. And that has to do with the
fact that we usually make measurements in bulk under
constant pressure conditions. And that is the quantity that
comes out. The relationship between delta
H and delta E is this. Delta H is equal to delta E
plus delta PV.
This is a relationship that you will learn
about in a lot of detail in 5.60, in Chemical
Thermodynamics. At this point,
we are going to take it as a given.
For gases, delta H differs on the order of 1% to 2% from delta
E. It is not much,
but if you are doing some precise calculation,
you need to be aware of that. For liquids and solids,
delta H and delta E are really the same for all intrinsic
purposes. The delta PV term is really
very small. And, in thermodynamics,
since we are most always looking at changes in energy,
we need what we call standard states.
And we are going to put a nought, here,
on all of our delta Hs to designate the standard state.
And our standard state that your book uses,
and will use, refers really to the pressure.
And the pressure is one bar. And one bar is equal to 10^5
Pascal. That is equal to 10^5 kilograms
per meter second squared. The delta Hs we are going to
talk about are also, just about all of them,
measured at 298.15 degrees Kelvin.
Delta H does depend on temperature, but we actually are
not going to look at that in the next few days.
You are going to do that in
Our delta Hs are going to be delta Hs at 298.15 degrees
Kelvin. On the first slide here,
I show you a bunch of bond enthalpies for CH bonds.
And, of course, those bond enthalpies are a
little bit different, depending on what molecule you
have. But they are not that
different. And so, what is often done,
and your book does this, is that somebody goes and
calculates the average of the bond energies for lots of CH
bonds and lots of molecules and they prepare a table that looks
like this. This is the mean bond enthalpy.
And they have CH, CC, carbon-carbon.
But these are average bond enthalpies.
Now, why are bond enthalpies important to us?
Well, they are important because they determine the
enthalpy of a chemical reaction. If the bonds are stronger in
the products than in the reactants, that is going to give
us an exothermic reaction. If the bonds are stronger in
the reactants than the products, that gives us an endothermic
reaction. And so let's look at this
reaction. This is an important reaction.
This is the oxidation of glucose.
This is a reaction very exothermic, minus
kilojoules per mole. It is a reaction that is being
carried out in every cell of your body as we speak.
It is the reaction that is providing the energy to maintain
your body temperature, the energy to move your
muscles, the energy to repair tissue, and the energy to think.
Important reaction. This is the reason why we eat.
This is the reason why we breathe, this is the reason why
we exhale, and this is the reason why we pee.
[LAUGHTER]
What do we need to do to calculate the enthalpy for this
reaction? We have to figure out how much
energy is required to break all of the bonds of the reactants
because that is how much energy we put in.
And then we have to figure out then how much energy we get back
when we form the product bonds. Bottom line is,
the enthalpy necessary to break all of the bonds,
you can calculate, is 12,452 kilojoules per mole.
That number comes from using these average bond energies I
told you about. We can then get back some
energy, minus 15,000, approximately,
when we form some new bonds. That number comes from those
average bond enthalpies. The difference between these
two energy levels is the exothermicity of the reaction.
Now, what did we do to get the enthalpy?
Well, what I did is took the bond enthalpies of each bond of
the reactants and summed them. Then, I took the bond
enthalpies for each one of the products, summed them,
and subtracted the two to get the enthalpy of the reaction.
I want you to notice something here, important.
This is reactants minus products.
In a moment, I am going to show you another
way to calculate the enthalpy for a reaction.
And it is going to be products minus reactants.
You have to know this. But you also see that the
calculated enthalpy is not the experimental enthalpy.
Why? Because we use the average bond
enthalpies. We did not use the exact bond
enthalpies because if we had to use the exact bond enthalpies,
can you imagine the size of the table of data that we would have
to have? We would have to have a bond
enthalpy for every bond for every known molecule.
That is a lot. What are we going to do,
then? Is there a more accurate way to
do that? Yes, with knowing the absolute
bond enthalpies. But that is too much data.
Is there some other way to do it?
Yes. We are going to use heats of
formation. A heat of formation,
delta H nought with an F as a subscript.
The heat of formation is the
enthalpy of a reaction that forms one mole of a compound
from the pure elements in their most stable form in their
standard state. For example,
here is water. We are forming one mole of
water from its elements, hydrogen and oxygen.
The enthalpy for this reaction is defined as the heat of
formation of water. Why is this the heat of
formation of water? Well, because we are forming
one mole, and that is important, from the elements that make up
water. What elements are they?
They are hydrogen. But notice that this hydrogen
is H two. It is not hydrogen atoms
because H two is the most stable form of hydrogen.
Oxygen is O two, not oxygen atoms because this
is the most stable form of oxygen at bar pressure.
Look at this here. What is the heat of formation
of oxygen? Well, the enthalpy change for
this reaction is zero. That is the heat of formation
of oxygen. Why?
Because we are forming one mole of oxygen from the elements in
their most stable form. For elements like oxygen,
hydrogen, nitrogen, chlorine, two
in the gas phase, those all have heats of
formation that are equal to zero.
And then, finally, here is the expression or the
reaction that gives us one mole of glucose.
The enthalpy for this reaction is the heat of formation of
glucose. We get it from its elements,
hydrogen, oxygen, and look at the elemental form,
the most stable form of the element carbon is graphite.
Is there an 18.0-something exam?
Yes? Okay.
See you on Monday.