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PROFESSOR: So again, this is one more question on
hybridization.
This is the last question you'll get on hybridization
before you see the exam.
So now we're talking about hybridization of two different
atoms in a molecule.
So we're looking at this carbon atom
here and this nitrogen.
So using the rules that we've learned in terms of doing this
quickly, let's see if you can get these answers in quick.
So let's go ahead and just take 10 more seconds on this
question here.
Your parents are free to help if they're here with you.
OK.
Excellent. excellent job, 98% of you.
You're a lot louder with the parents here, so let's make
sure we're still listening.
OK, so let's go over what we just we just established here.
Carbon a is going to be s p 3 hybridized.
What is the geometry, what's the vsper geometry there?
Good, it's tetrahedral.
And this nitrogen b here, that's s p 3 hybrid as well.
What's the geometry around that nitrogen?
Trigonal pyramidal, right.
We have to take into account the fact that there's that
lone pair there.
All right, so let's go ahead and switch over to the lecture
notes and see what we just did here.
And essentially actually what we just did was I had you
identify two of four components that make up what's
called the morphine rule.
And the morphine rule is a set of structural elements that
are responsible for the biological activity of
morphine and other morphine-like drugs that have
a similar pharmacological activity in terms of how
they're active.
So specifically, the morphine rule is a set of four
components, which start with the phenyl ring here, so, an
aromatic ring.
The second thing you identified, which is an s p 3
hybrid carbon.
And then following that we have a c h 2 c h 2 group here.
And the last part of the rule is this s p
3 hybridized nitrogen.
So, in terms of thinking about what this means, I said this
is responsible for the
biological activity of morphine.
If we take a look at the morphine molecule here, what
you can see is that it's a little bit more complicated
than this structure that we see right here.
But if I go ahead and highlight the residues in
purple, you can see that it, in fact, does follow the
morphine rule.
And what this means is this is what allows morphine to
actually bind into its receptor, which is a pain
receptor and block the feeling of pain.
So that's a very important thing in terms of thinking
about hybridization in geometry, because we've
actually established a very important structure and shape,
which is this morphine rule here, that if it's found in a
molecule, as long as there aren't other parts of the
molecule that are messing up that interaction, we can
actually form a very tight binder
into these pain receptors.
So morphine, as you know, is a very potent pain killer.
But you also, I'm sure, know that it's also very addictive,
so it's only used sparingly in terms of treating pain in a
general sense, and it's always used in very closely-monitored
situations in terms of hospital care.
But what's interesting to take note of is that the bioaction
of morphine is very similar to endorphins that we naturally
biosynthesize.
Endorphins have a structure that's very similar to this
structure up here -- endorphins are small peptide
molecules that we biosynthesize and have in very
low concentrations in our brain.
They bind to pain receptors and
block those pain receptors.
So, when someone talks about an endorphin high, which you
sometimes get -- one example would be the runners' high, if
you run for a long way running fast, eventually you'll hit
that runner's high where you get this burst of endorphins.
All of a sudden your feet don't hurt, your shins don't
hurt, the pain goes away and you got
this feeling of eurphoria.
It's the exact same interaction and it's because
of that structure there.
It's not exactly the morphine rule in endorphins but it's
very similar.
So, if we look at other derivatives of morphine, such
as codeine and diacetylmorphine, these also
you can see have this morphin rule within them, this
structure that you've identified using your vsper
rules and thinking about hybridization.
Codeine, as you may know, is less of a potent pain killer
compared to morphine, but it's also less addictive.
So for that reason it's prescribed with
slightly less oversight.
So if you get your wisdom teeth removed or something
like that, some of you might have had taken codeine before,
and now you can think about by looking at the structure, a
little bit more about how that worked in terms of
blocking your pain.
One that I just want to mention because it's so
interesting is this derivative, which is
diacetylmorphine.
The only difference between morphine and diacetylmorphine
is the changing of an alcohol group, or two actually, two
acetyl groups here.
This was synthesized by the company Bayer, and you
probably know Bayer from Bayer's aspirin.
Bayer had a huge success with aspirin where they took
salicylic acid and replaced an o h group with an acetyl group
here, and made acetylsalicylic acid.
Does anyone know the common name for that?
Yup.
So that's aspirin, and we often
think of Bayer's aspirin.
We associate it really closely.
They worked really hard in marketing to make that really
close association and with copywriting, so that when we
think about aspirin we think about Bayers.
Bayers also made this change here, and they did find that
diacetylmorphine is much, much more potent than morphine is.
They could use just a tiny, tiny amount of this compound
and get the same pain-killing properties that
they saw with morphine.
The problem is that this masked some of the side
effects that they didn't realize initially, which was
the extreme, extreme ability for you to become addicted to
diacetylmorphine.
And this was first thought of as a hero drug because it was
such a strong pain killer, and this was
called a Bayer's heroin.
We don't usually associate the name Bayers with heroin, like
we did with aspirin -- that's because Bayers did not work so
hard to keep that association there.
But it's an interesting story from the history of drugs.
One more example I want to show you is demerol, which is
used clinically quite often.
Demerol, if we look at the structure here, actually looks
nothing like morphine at all.
So to an untrained eye, you might not see the
relationship there.
But because you all know some of these basic principles of
chemistry, you should be able to pick out that morphine rule
right in the middle of demerol.
And, in fact, demerol is a pain killer, it's an
alternative to using morphine.
Again, not quite as strong as morphine in terms of pain
killing abilities, but also not as addictive either, and
there are some other side effects, such as nausea that
are limited with demerol compared to morphine.
Demerol has its own set of problems, so it's not the
perfect solution to a different alternative to
morphine, but you do see it used in certain cases.
All right, so that's thinking about vsper theory, and that's
thinking about hybridization.
We're now going to shift gears to talking about some new
topics today.
So, we ended in Wednesday's lecture with giving that set
of rules to very quickly determine hybridization.
And then after we'd established that, we moved on
to starting to talk about a little bit of thermochemistry.
Specifically what we were talking about is the
enthalpies of chemical reactions, so either the
amount of heat that's released, or the amount of
heat that's required in order to have a
chemical reaction go.
And we'd established so far two ways to think about how we
can measure what the overall enthalpy of a reaction is.
And I just wanted to introduced one more technique
that we can use to do that.
And this relies on the fact that when we talk about
enthalpy or we talk about the heat change of a reaction,
this is a state function, and any time we're talking about
state function, it's independent of path.
So essentially, all that we're worried about is the state
that we started in.
So, for example, when we were talking about the oxidation of
glucose, the state that we start in are the reactants, so
it's glucose plus oxygen.
And what we ended up with were six moles each of carbon
dioxide and water.
So, we don't care how we get there in terms of making this
calculation, all we're interested in is the
difference between those two states, to think about the
overall change in enthalpy.
So, for example, when we talked about how we could
calculate this, one way that we can do it, and we'll look
at these numbers even more specifically in a second, is
thinking about instead of going straight down here,
which we might not have the numbers available to us to
calculate directly, we can think about well, what happens
if instead we break apart this glucose molecule and decompose
it into its elements, because using tables in the back of
our book, for example, we can figure out what the change in
enthalpy is of this reaction here.
That doesn't quite get us to where we need to go though, so
then we can think about the enthalpy of formation of six
carbon dioxide molecules.
And then we can go one step further and think about the
enthalpy of formation for six water molecules.
So this is just a round about way of going
from here to here.
Instead we went up and then down and then down, but
because it's a state function, we're absolutely
allowed to do this.
And to formalize this, we can talk about what's called
Hess's law.
And Hess's law just tell us that if we have two or three
or any number of chemical reactions that we can add
together in order to get the chemical equation that we're
actually interested in, we can figure out the change in
enthalpy of that reaction that we're interested in, just by
adding together all of the other enthalpies of reaction
for every other reaction that we had to add together in
order to get there.
So, let's take a look at what we're talking about here.
So when I just showed you that example graphically, I'm just
writing it out more like an equation here, because the
nice thing about Hess's law, is it allows you to think
about chemical reactions kind of as if they're algebraic
expressions.
So we can just add all of these different expressions
together, and if we're thinking about algebra, we can
just think about crossing things out that are in the
reactants versus in the products.
So if we do this, what we end up getting is the reaction
that we're interested in, the oxidation of glucose to form
carbon dioxide in water.
So let's look pretty carefully at exactly the fact that these
all cancel out to give us the reaction I say we get.
So, for example, we can think about the oxygen molecules --
we can cancel out 6 here and 6 here, as long as we multiply
this whole reaction by 6.
Again, we can get rid of those last 3 oxygen molecules there.
Similarly, we can cross out our 6 hydrogens from the
products with 6 molecular hydrogens from the reactants.
And then the last thing we can cancel out are our carbon
atoms -- we have 6 in the products, and we have 6 in the
reactants here.
So you'll see that the only things that are not canceled
out is what's left in our bottom reaction.
So we can go ahead and just add up all of these different
reaction enthalpies.
In the first case what we're going to have is an enthalpy
of 1260, and that's in
kilojoules per mole of glucose.
And the reason that this is positive is because it's
essentially the reverse reaction of the change in
enthalpy of formation.
So it's going to be a positive number there.
We add to that the enthalpy for this reaction here, which
is negative 393 .
5, but we need to remember to multiply that by 6, because
we're actually adding together six of
those individual reactions.
And lastly, we also add together six of this final
reaction here, which has an enthalpy change of
negative 285 .
8.
So if we go ahead and just add up all of our individual
enthalpy changes, what we're going to end up with for our
entire reaction here, is a change in enthalpy of negative
2816, and that's kilojoules per mole of glucose.
Does anyone remember from Wednesday's class, does this
match what we had seen experimentally?
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, so this is the exact number that we
calculated when we used heats of formation, and it's also
the exact number that's seen in terms of experimental.
All right.
So there's actually three ways we now have to calculate heats
of the change in enthalpy of an overall reaction.
So, we have these in our notes from last time and from
finishing the end of this unit right now, but I'm just going
to summarize them for you on the board, because there are a
couple things that can get confusing that I want to make
sure no one is confusing, particularly in
the upcoming exam.
So what was the first way that we learned to think about
calculating enthalpies of reaction?
STUDENT: [INAUDIBLE]
PROFESSOR: Um-hmm.
So, it's talking about bond enthalpies.
And when we talk about bond enthalpy, the symbol that we
see for a bond enthalpy is usually delta h.
And this is where all the confusion starts, because
everything's delta h, it just depends on the subscript here,
right, what kind of delta h we're talking about.
In the case of bond enthalpy often you'll see no
subscript at all.
But sometimes you do see a subscript, which would then
just be delta h sub b here.
So if we're trying to do this calculation based on bond
enthalpies, for the delta h for an overall reaction, what
we want to do is take the sum of all of those bond
enthalpies of our bonds that are broken, and what we want
to do is subtract from that the sum of all our bond
enthalpies of our bonds that are formed.
All right, so let's think a little bit more about what
this means.
So, if we're talking about bonds broken, are we talking
about reactants or products?
STUDENT: [INAUDIBLE]
PROFESSOR: .
Reactants.
So, essentially we're summing up the bond enthalpy of
reactants and subtracting it from the products.
All right, great.
So that's our first strategy there.
The second strategy that we learned is thinking about bond
enthalpies of reaction by using the enthalpy change in
terms of the enthalpy of formation.
So this one is sometimes a little bit more intuitive,
because we're talking about the enthalpy change it takes
in order to form a given molecule.
So, if we're trying to do a calculation using enthalpies
of formation, what we find is that delta h of the reaction
is equal to the sum of the delta h of formation of
products or reactants?
STUDENT: Products.
PROFESSOR: Products, good.
Of products minus the delta h of formation, and this is the
sum again of reactants.
All right, and our third strategy is what we just
talked about, which is Hess's law, and in fact, both one and
two are applications of Hess's law -- specific applications,
but any application of Hess's law you can use any way of
adding up different equations to get the final enthalpy of
the reaction.
The reason I wanted to put these two strategies, however,
on the board right next to each other, is so we can just
confront a point of confusion that happens a lot with
students, which is why in the first case do we have
reactants minus product, and in the second case we have
products minus reactants.
So, you need to keep these two straight when you're doing
your calculations in terms of enthalpies, and the reason is
because the definition of a bond enthalpy when we're
talking about bonds, is the enthalpy that it takes in
order to break the bond, the amount of heat that goes in to
breaking a bond.
So it's typically a positive number, because if you have a
stable bond you have to put heat into it in
order to break it.
So you end up having a positive bond enthalpy here.
In contrast, when we're talking about enthalpies of
formation, now we're talking about is the change of
enthalpy in order to form a molecule.
So, for example, if we have a nice stable molecule with
strong bonds in it, what we're going to find is that the
delta h of formation, would that be positive or negative?
So it's negative.
So you'd end up with a negative delta h of formation.
So because you're talking about a term that's negative
here where it would be positive in the first case,
you end up flipping signs to have it work out.
So, if this doesn't make sense as I'm saying it right now,
just go back in your notes and look through and make sure
that you aren't going to get mixed up when you're using
bond enthalpies versus enthalpies of formation here.
All right, so that's pretty much all we're
going to say on enthalpy.
So I really just want to stress, this is going to be
the end of exam 2 material here, I said I'd be clear
about it in class today.
This is it.
It's also written in your notes when
you go ahead to study.
So remember, in terms of exam 2, which I'll remind you is on
Wednesday, it's going to be all the way from the material
on lecture 10 up to this point here.
So only through enthalpy, and it's also going to be on
problem-sets four and five, and I'll mention that this
afternoon, as I said, I'll post a bunch of extra practice
problems, which are more review of these same concepts,
so you can get even more practice this weekend and
early next week to prepare.
All right, so that's enthalpy, that's the
end of exam 2 material.
But there's actually another really important concept that
we need to talk about -- we really don't want to stop with
enthalpy, luckily it's exam 2 and not the end of the class,
because a really important concept to think about is
spontaneous change.
So thinking about whether a reaction is spontaneous or
non-spontaneous.
And when we think about a spontaneous reaction, I think
this is a term many of you are familiar with, spontaneous
just means that the reactions going to proceed in the
forward direction without any kind of outside intervention.
So thinking of a spontaneous process, can talk about a
chemical reaction, but you could just picture, for
example, putting a round rock on the top of a hill -- it
will spontaneously just roll right down that hill.
So a spontaneous process also has direction, right, because
the rock won't spontaneously roll back up the hill without
putting in work.
So essentially we're talking about a spontaneous process
when something's going to happen without actually having
to do anything else to force it to happen.
So let's think about in terms of chemistry what we're
talking about is a spontaneous reaction -- that's the
specific type of spontaneous process that
we're interested in.
So let's think about a few different types of spontaneous
reactions, and see if we can come up with some idea of
what's going to cause them to be spontaneous.
So one spontaneous reaction is written here, so this is just
the oxidation of iron, or the formation of rust. This turns
out to also be an exothermic reaction, it has a negative
delta h of 824 kilojoules per mole.
Another spontaneous reaction is written here, the
combination of an acid and a base, which
neutralizes each other.
So we have a hydronium ion and a hydroxide ion interacting to
form water.
This is spontaneous, and again, it's exothermic.
So, its delta h is negative 55 .
9 kilojoules per mole.
Let's think about some really relevant
reactions in our bodies.
So one incredibly important reaction, of course, is ATP
hydrolysis where we have adenosine here, a
triphosphate, so we have three phosphate groups here.
That's called ATP.
It has a total charge of minus four.
So if we hydrolize this, one of the phosphate bonds here
and lose a phosphate, we end up with adenosine diphosphate
or ADP and that has a charge of minus three.
Yup?
STUDENT: On our notes, this equation [INAUDIBLE].
PROFESSOR: Oh, no.
OK, thank you.
Which equation are we talking about?
STUDENT: [INAUDIBLE]
PROFESSOR: I'm sorry, what page are you on?
Page two.
So the one with h 3 --
OK.
All right, let's go back to that.
OK, so if you can fix this reaction in your notes here.
I'll also fix it on the website, so if you don't want
to fix it now I'll just re-post the notes.
So it should be four irons plus three oxygens is two f e
2 o 3 solid.
Thanks for pointing that out.
All right, does everyone have that down?
I know you had to flip a page there.
All right, so I'll post it in the notes if you
didn't get it there.
Let's go back to ATP hydrolysis here.
So we're going from ATP to ADP, and that is a spontaneous
process, and it's an exothermic reaction as well.
So we find that it's negative 24 kilojoules per mole in
terms of the change in enthalpy there.
All right, so at this point I just showed you three
reactions where we have something that's spontaneous
and it's also exothermic.
And I could show you a countless number of other
reactions where it's the same thing.
It's actually quite common if you have an exothermic
reaction that it's also spontaneous at room
temperature.
So we might start to draw the conclusion that, in fact, it's
enthalpy that's responsible for whether or not a reaction
is spontaneous or non-spontaneous.
So it's very easy to get that impression, but let me show
you a few more reactions before we draw any
conclusions.
Let me show you some more spontaneous reactions.
So, for example, the conversion of solid h 2 o to
liquid h 2 o at room temperature, I think we all
know this is spontaneous.
Ice melts at room temperature, but it turns out that the
enthalpy change is positive, it's 7 kilojoules per mole.
Similarly, if we look at this reaction here, which is
ammonium nitrate.
This is a very commonly used fertilizer, a very commonly
used nitrogen source in agriculture.
If we think about solvating this and forming the two ions,
ammonium ion and nitrate ion.
Again, this is a spontaneous reaction, but what we find
here is that delta h again is positive, it's also
endothermic.
So thinking about all of this, would you say that enthalpy is
the key to spontaneity?
STUDENT: No.
PROFESSOR: No, it's definitely not the key to spontaneity.
It tends to correlate in many cases at room temperature, but
it is not the key to spontaneity.
And the key to spontaneity is instead something called Gibbs
free energy or delta g.
And if we think about what delta g is, we can relate it
to enthalpy, and this will sort of show us why there's
often this correlation, because delta g is equal delta
h minus this term here, which is temperature, and that's
temperature times a change in entropy.
And we'll talk very in-depth in just a few minutes about
what entropy actually is, but for now I just want you to
think about the fact that in addition to thinking about
enthalpy there's this other term here
that comes into play.
So when we're talking about what the sign about delta g is
or free energy is, some of you might already be familiar with
this, when we're talking about a negative delta g, is this
spontaneous or non-spontaneous.
So, maybe not so familiar, which is totally fine.
So in terms of thinking about a negative delta g, that's
going to be a spontaneous process.
Any time you see a negative free energy, which just like
when we talk about entropy, it's a similar idea, it means
that free energy is released as the reaction progresses.
That's going to be a spontaneous process.
So that means that if delta g is greater than zero, then
what we're going to see a non-spontaneous process.
And finally in the case where delta g is equal to zero, at
this point we're at equilibrium, which basically
means that there's no net change in either the forward
or the reverse reaction in terms of thinking about
whether something's spontaneous or not
spontaneous, it's just at equilibrium
we see no net change.
And something I want to point out is that this reaction is
valid only when where at constant temperature and
pressure, which we will be throughout the discussions in
this class in terms of using this equation.
And also, it turns out that when you do most chemical
equations, you are, in fact, at a pretty constant
temperature, and a constant pressure in terms of most
things are done in the open atmosphere, so that's really
not going to change.
All right, so let's just talk very briefly about why it is
that free energy is what determines spontaneity, and
not this enthalpy here, not what the
change in enthalpy is.
And it really comes in terms remembering it in terms of the
definition.
When we think about delta g, what that is is that it's
energy that's released or used in the reaction, but if it is
released, that it can directly be used to do work.
It's what we call free energy -- it's
free to do other things.
Whereas when we were talking about enthalpy here, well,
this is the amount of heat that's released when we break
and form the bonds in the reaction.
But some of it actually gets stuck in the molecules, and
this is just a way to think about it, this term here is
one way we can think about it, it's just some energy that's
getting stuck.
So, for example, molecules, which we don't really go into
in this class, have vibrational movements, they
have rotational movements, there's energy associated with
that that can sort of get stuck, as we could say, some
of the enthalpy that's released in terms of forming
and breaking the bonds in a reaction.
So that's our case where we could see that we have a
negative delta h, but we still end up with a
positive delta g.
And again, I haven't really gone into what entropy is here
yet, and we will just in a few minutes.
But first, I want to take a look at one of the examples
that we talked about, which is the conversion of ammonium
nitrate when its solvated, breaking up into its ions.
So what I had told you is that, in fact, this is a
positive delta h, but still that the reaction is
spontaneous.
So let's just do this calculation and see if we can
confirm that in terms of delta g.
So the reaction again that we're going to use is delta g
equals delta h minus t delta s.
So if we're talking about room temperature here, so we're
going to say that delta g in this case is going to be equal
to 28 kilojoules per mole, minus -- we're at room
temperature or 298 kelvin -- in this reaction, you always
put temperature into kelvin.
And then we need to multiply it by the entropy, which is
109 joules per kelvin mole.
I want to point out that entropy tends to be a much
smaller value than enthalpy, so it's reported in joules
instead of kilojoules.
When you solve these problems, it's really important to
convert it into kilojoules so that your units work out and
you don't get a completely crazy answer.
So we want to convert this to 0 .
109 kilojoules per kelvin mole.
So we can just re-write this, so we have 28 kilojoules per
mole, then this term here will be 32 .
5 kilojoules per mole.
So what we end up with our delta g for this reaction is
going to be negative 4 kilojoules per mole.
All right, so talking about this as a negative delta g,
would you say this is a spontaneous or
non-spontaneous reaction.
Yeah, it's spontaneous.
So what we're going to see is delta g is negative 4, so this
reaction is spontaneous even though our
enthalpy change was positive.
Let's take a look at one more reaction since we spent so
much time discussing the oxidation of glucose.
So again, what we know about this reaction, we've already
calculated the delta h is negative 2816
kilojoules per mole.
And you could look up and we'll figure out how to
calculate it soon, the change in entropy, which is plus 233
joules per k mole.
So before we actually do this calculation, let's go to a
clicker question and I want you to think about what's
happening here.
So if you're thinking about the oxidation of glucose and
what the sign is of the change in enthalpy and the change in
entropy, which of these statements is true.
Is it going to be spontaneous at all temperatures,
non-spontaneous at all temperatures, or will it
depend on the temperature whether this reaction is
spontaneous or not spontaneous?
So, remember this is our relationship here.
So let's go ahead and just take 10 seconds on this one.
OK, so we have a pretty mixed response here.
So this could have made or broke the chances of your team
on the competition today, but they'll be a few more left to
redeem yourselves if you got it incorrect.
So what we're going to find is that it's spontaneous at all
temperatures, and we can actually just look at this
equation here to figure out why that is.
If our reaction is spontaneous, that means that
delta g is negative.
So in terms of delta h being negative, well, that's going
to always contribute to delta g being negative no matter
what the temperature is.
And if delta s is positive, since it's minus t delta s,
and our temperature is always positive because we're on the
kelvin scale, so it starts at zero there, then this
component here is always going to be negative as well.
So we're always going to have two negative numbers so that
our combination, our delta g's always going to be negative,
it's always going to be spontaneous.
All right, so let's make sure that's what we do see when we
calculate this for the oxidation of glucose here.
So if we do this and we plug in our numbers, we see that
delta g is going to be equal to negative 2816 times the
temperature, room temperature, times again,
remember to put 0.
233, because we need to convert from joules to
kilojoules for our entropy term.
So what we find out is that this reaction has a delta g of
negative 2885 kilojoules per mole.
This is a spontaneous reaction.
All right, so let's get back to that entropy term.
I introduced it without explaining at all
what it really is.
So let's take a few moments and think about entropy.
Entropy is actually a very easy concept to think about,
it's a measure of disorder.
I think most of us have a very good concept of how things
tend to go to disorder, so it's something we can
conceptualize just in an infinite number of ways.
And entropy is simply a measure of the
disorder of a system.
And when we're talking about chemical reactions, what we're
talking about is a change in entropy.
So whether, as a reaction goes forward, are we becoming more
ordered or are we becoming more disordered?
So just like we saw for enthalpy, entropy also is a
state function, so it doesn't matter in terms of path how we
got from point a to point b.
All we need to worry about is the current state of our
system, what the current entropy is and take the
difference.
So, an example that I like to use in terms of conceptually
thinking about what's going on in terms of disorder, and
which is especially relevant in New England is thinking
about stone wall.
So you can think about a stone wall in terms of being very,
very ordered, right.
If you just built your stone wall, every stone is in place,
it has a high degree of order.
So in other words, the disorder is very low.
So it has a low entropy, a low level of disorder.
But if we're looking at a New England stone wall, which are
typically found if you're hiking in the woods you see
ones that are just ancient and are completely crumbling down,
you'll find that the stone wall is no longer quite so
ordered as it was.
In fact, let's say there's a few stones that have fallen
off, our disorder has increased, so we end up
increasing the entropy.
And I do just want to point out with this analogy that it
is a state function.
So for thinking about the change in entropy, which is
this distance right here, we can either calculate it
directly from here to here. but it also doesn't matter,
let's say the wall crumbled down completely and we ended
up having a very high degree of disorder, if then someone
put most of the stones back in place and we went back down to
this degree of disorder here, it doesn't matter, it's a
state function, all we care about is the actual difference
between our starting point and our ending point.
And I just can't resist putting in a little quote,
"something there is that doesn't love a wall." Does
anybody know where this quote is from -- very famous poem,
very famous poet.
Good guesses, parents can help with this.
Yes, excellent.
I've never heard the correct answer this question before.
Excellent.
So this is Robert Frost in The Mending Wall.
He talks about, it sounds like a lot of you did know this,
this makes me full of joy. "Something there is that
doesn't love a wall," there's many interpretations for this
poem, but maybe it's about entropy.
Entropy doesn't love a wall, we're going from order to
disorder, that wall wants to come down.
All right.
So let's formalize these thoughts on entropy and in
terms of what we're talking about.
Again, what I said on that's chart there is as we increase
entropy, we're increasing disorder.
So if you saw a positive change in entropy in a
reaction, would you think about that as being more or
less ordered?
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, so it's actually going to be less
ordered, and, in fact, I should stop saying order, I
should be just saying disorder because disorder's actually
what we're measuring, and there's an increase in
disorder here when we have a positive change in enthalpy.
So if we have a negative change in enthalpy, we're
going to see a decrease in disorder.
So in terms of considering different types of states that
we can be in, whether we're in a gas state or a liquid state
or a solid state for any given molecule or any given
compound, we can think about how much order or disorder
these states have. So which of these three states would you
call the most disordered?
STUDENT: Gas.
PROFESSOR: Yeah, it's the gas state.
So it turns out that the disorder gas is greater than
liquid., which is greater than solid.
This makes sense and the gas molecules are free to bounce
around, go wherever they want.
Once you're in a liquid, they have a little bit more
limitation, they can't go anywhere.
But the molecules can still slide all around
and pass each other.
Once you're in a solid state, this molecule is not going to
flip places with the other molecules there.
It's going to be kind of stuck in its place within the solid.
So, just understanding this explanation of entropy that
we're going to have an increase in entropy if we have
an increase in disorder, allows us to make predictions
about reactions even without doing any calculations, and we
always like when that happens, because then once we do
calculations, we can check our calculations and make sure
they make sense with what we would be predicting.
So before we talk about how to actually do a calculation,
let's go to a clicker question here and make sure everyone is
on the same page in terms of thinking about changes in
entropy or changes or delta s for the reaction.
So let's say we're talking about the decomposition of
hydrogen peroxide into water and oxygen.
I want you to tell me if you think it's going to have a
positive delta s, a negative, or zero, or maybe this will
change with temperature as well.
All right, so let's go ahead and take 10
seconds on this one.
OK, excellent.
Great job with the questions today.
So we are actually going to see a positive delta s. this
should be very clear because we're going from two moles of
liquid, to now having two moles of liquid, plus a mole
of gas, so we're going to be increasing the disorder of our
system here.
All right, so let's think about how we actually can
calculate, though, delta s.
So far I've just been giving you the delta s's for the
reactions, but, as I'm sure you can guess, you won't be
given that information, for example, on upcoming
problem-sets, you'll actually need to calculate the entropy.
So what we can do is we can actually calculate the entropy
of a reaction from absolute entropies of individual
molecules that we're discussing.
So this is very similar to how we calculated the enthalpy of
a reaction by taking the change of enthalpy of
formation, except we don't even have to worry about a
change in entropy because we can talk about entropy as an
absolute value.
There's an absolute zero in terms of entropy where there
is no disorder at all.
And this is described for any molecule as being its perfect
crystal at absolute zero.
So this is what we're going to say there's absolute complete
order, there's no disorder at all in this system.
So what we can take to calculate the entropy in a
reaction, is just to take the sum of the entropy of all the
products, and subtract from that the entropy from all of
the reactants.
So let's try this for the example that you just did on
the clicker question, and make sure that our calculation
matches up with our prediction here.
So again, we're just going to take the sum of the entropy of
the products minus out of the reactants.
So if we talk about this, we're starting with our
product, so our first product is water.
And what I want to point out is you need to make sure you
multiply this by 2, because, in fact, we have two moles of
water that are formed here.
STUDENT: [INAUDIBLE]
PROFESSOR: OK, I'm so sorry.
Yeah, so yikes, this should be water.
You know what, let's just write this on the board.
So we're going to talk about delta s of the reaction.
So that's to be sum of the entropy of the products.
So we're going to be talking about 2 times entropy of
water, and we'll add to that our other product, which is
molecular oxygen here, and we don't need to put a number out
here because it's just one mole.
So that's of o 2, and we'll subtract from that, the
entropy of hydrogen peroxide here.
And do we need to put anything out here for this?
STUDENT: 2.
PROFESSOR: So, yeah, because there's two moles there.
So water plus oxygen minus hydrogen peroxide.
Let's hope I at least got the values right.
Yeah, and actually I think this is right, I believe this
is the entropy of water.
So OK, so the actual numbers are right here for what is
written on the board and hopefully in your notes.
So what we find out is that the entropy of this reaction
here, the change in entropy, the delta s, is 125 joules per
k per mole.
All right, so we can also -- well, first, we already
thought about why delta s is positive.
Delta s is positive because we're going from a liquid to a
liquid and a gas -- we're increasing our disorder.
So let's think about whether this reaction is spontaneous
or not, and to do that, we actually need to go ahead and
completely calculate delta g.
So to do this, we can take our values here, and delta g is
equal to delta h minus t delta s.
So if we plug all of this in, we have our delta h is
negative 196 kilojoules per mole, minus our temperature,
and I wanted you to write this again, because I want to make
sure you get in the habit of converting our joules to
kilojoules.
It's 0 .
125 kilojoules per kelvin mole.
So we end up with a delta g of negative 233
kilojoules per mole.
Is that spontaneous or non-spontaneous?
STUDENT: [INAUDIBLE]
PROFESSOR: Great.
So the reaction here is spontaneous.
So actually, this is probably a reaction that you're
familiar with because a lot of us do have hydrogen peroxide
just in our medicine cabinet.
And if you go and look at the expiration date on your
hydrogen peroxide, you will see that it
does, in fact, expire.
And you might think oh no, what happens when it expires,
when it goes bad?
All that's happening is it's turning into water.
So I wouldn't drink it after its expiration date, but
probably there's more water than hydrogen
peroxide at that point.
But it also actually brings out a really important point
about thermodynamics, which is that thermodynamics, in fact,
tells us if a reaction happening in the forward
direction is spontaneous or non-spontaneous.
So, for example, we see that yes, it is
a spontaneous reaction.
But you'll note that thermodynamics tells us
absolutely nothing about the timeframe of that reaction.
So some reactions that are spontaneous take place in a
matter of seconds or microseconds or less.
Other reactions, such as this, to have an appreciable amount
of that reaction go forward, it actually takes years.
You'll notice that it's not just one day before that
hydrogen peroxide expires.
It's takes a really long time to get a significant amount of
this reaction to go.
So just make sure, we will get to talking about kinetics,
which does tell us about the
timeframe of chemical reactions.
But don't confuse thermochemistry or
thermodynamics with kinetics.
It doesn't matter how huge our delta g is in terms of being
if it's negative two or if it's negative two million,
that still doesn't tell us how fast the
reaction's going to go.
All right.
So let's think about one more example here in terms of
thinking about entropy.
So another example we talked about was the melting of ice
at room temperature.
I think all of us agree that this is spontaneous, but that
the enthalpy change is positive, it's
an endothermic reaction.
So we can think about calculating the entropy here,
and I'm happy to see that I did, in fact, put the products
h 2 o here as a liquid, subtracting the reactant,
which is h 2 o as a solid.
If we go ahead and plug in those values here, what we end
up with is a delta s of this reaction of 28 .
59 joules.
This makes sense because we're increasing the disorder of the
system, we're seeing a positive delta s here.
And I just answered that question for you.
We're going why is delta s positive?
We're going from a solid to liquid,
we're increasing disorder.
So, we can calculate delta g as well, and if we do this we
plug in these numbers here.
What we end up with is that delta g equals negative 1 .
57 kilojoules per mole.
So it is spontaneous, delta g is just barely negative.
And what actually is the case, which we'll talk about more on
Monday and thinking more and more about how temperature
actually affects these reactions, but what we know
about melting of ice is that it is, in fact, temperature
dependent, right.
Ice doesn't melt at just any old temperature.
So that's where temperature actually comes into play here.
All right.
So we have a case -- spontaneous, even though delta
h is positive.
All right, so let's think also about how we can go about
calculating the free energy of formation.
So we know that we can use the equation that we saw in terms
of the free energy the reaction, but what if we're
actually thinking about the free energy of the formation
of a particular molecule?
So that's actually completely analogous to thinking about
the enthalpy of formation, or it's basically the standard
Gibbs free energy of formation.
So this is talking about forming a mole of a compound
from its most stable elements, from its elements in their
most stable states at pressure of one bar and at room
temperature.
So we know that we can tabulate it must like we
tabulated delta h formation, but we can also tabulate it
from this reaction here, which it looks like I already told
you this reaction, which I did but I told you a more general
form, which was for the free energy of a reaction.
This is the free energy of a reaction, which is the
formation of a certain compound, and this is the free
energy of formation equals the enthalpy of formation
minus t delta s.
So let's think about why it is important to be thinking about
these changes of free energy in terms of forming a
molecule, and it's important because the delta g
information is actually a measure of how stable or
unstable a molecule is relative to its element.
So, for example, if something is really stable relative to
its elements, we can think about whether delta g of
formation should be negative or positive.
So let's have you do one last clicker question here, and
tell me what do you think, if delta g of formation, of
actually forming a molecule is going to be negative, would
you say that the compound is going to be stable or unstable
relative to the elements that that compound is made up of?
All right, let's take 10 seconds on this.
OK, interesting.
So this is going to be another tie breaker for us.
So let's go back to the notes and think about this.
So it turns out that it's going to be stable, relative
to its -- the compound is stable
relative to its elements.
And the reason for this is if you think of the reaction of
the elements forming the compound, if the delta g of
that reaction is negative, that means it's going to
spontaneously form the compound.
It's going to go -- release energy when it actually forms
that compound.
So that's another way of telling us that the compound
is more stable than its elements.
So conversely, if we have delta g of formation being
greater than zero, so delta g of formation is just delta g
of the reaction where the compound is formed, then what
we find is that it's thermodynamically unstable
relative to its elements, because instead of releasing
energy in that reaction, you actually have to put energy
into that reaction in order to make it happen.
So, delta g of formation can give us an indication of
whether we have a stable or an unstable compound.