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PROFESSOR: So again, this
is one more question

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on hybridization.
 

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This is the last question
you'll get on hybridization

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before you see the exam.
 

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So now we're talking about
hybridization of two different

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atoms in a molecule.
 

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So we're looking at this carbon
atom here and this nitrogen.

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So using the rules that we've
learned in terms of doing this

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quickly, let's see if you can
get these answers in quick.

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So let's go ahead and just
take 10 more seconds

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on this question here.
 

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Your parents are free to help
if they're here with you.

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OK.
 

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Excellent. excellent
job, 98% of you.

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You're a lot louder with the
parents here, so let's make

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sure we're still listening.
 

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OK, so let's go over what we
just we just established here.

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Carbon a is going to
be s p 3 hybridized.

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What is the geometry, what's
the vsper geometry there?

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Good, it's tetrahedral.
 

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And this nitrogen b here,
that's s p 3 hybrid as well.

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What's the geometry
around that nitrogen?

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Trigonal pyramidal, right.
 

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We have to take into account
the fact that there's

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that lone pair there.
 

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All right, so let's go ahead
and switch over to the

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lecture notes and see
what we just did here.

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And essentially actually what
we just did was I had you

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identify two of four components
that make up what's

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called the morphine rule.
 

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And the morphine rule is a set
of structural elements that are

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responsible for the biological
activity of morphine and other

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morphine-like drugs that have a
similar pharmacological

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activity in terms of
how they're active.

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So specifically, the morphine
rule is a set of four

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components, which start with
the phenyl ring here,

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so, an aromatic ring.
 

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The second thing you
identified, which is an

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s p 3 hybrid carbon.
 

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And then following that we have
a c h 2 c h 2 group here.

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And the last part of
the rule is this s p 3

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hybridized nitrogen.
 

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So, in terms of thinking about
what this means, I said this is

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responsible for the biological
activity of morphine.

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If we take a look at the
morphine molecule here, what

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you can see is that it's a
little bit more complicated

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than this structure that
we see right here.

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But if I go ahead and highlight
the residues in purple, you can

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see that it, in fact, does
follow the morphine rule.

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And what this means is this
is what allows morphine to

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actually bind into its
receptor, which is a

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pain receptor and block
the feeling of pain.

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So that's a very important
thing in terms of thinking

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about hybridization in
geometry, because we've

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actually established a very
important structure and shape,

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which is this morphine rule
here, that if it's found in a

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molecule, as long as there
aren't other parts of the

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molecule that are messing up
that interaction, we can

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actually form a very tight
binder into these

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pain receptors.
 

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So morphine, as you know, is
a very potent pain killer.

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But you also, I'm sure, know
that it's also very addictive,

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so it's only used sparingly in
terms of treating pain in a

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general sense, and it's always
used in very closely-monitored

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situations in terms
of hospital care.

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But what's interesting to take
note of is that the bioaction

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of morphine is very similar
to endorphins that we

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naturally biosynthesize.
 

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Endorphins have a structure
that's very similar to this

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structure up here -- endorphins
are small peptide molecules

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that we biosynthesize and have
in very low concentrations

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in our brain.
 

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They bind to pain receptors and
block those pain receptors.

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So, when someone talks about
an endorphin high, which you

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sometimes get -- one example
would be the runners' high, if

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you run for a long way running
fast, eventually you'll hit

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that runner's high where you
get this burst of endorphins.

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All of a sudden your feet don't
hurt, your shins don't hurt,

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the pain goes away and you got
this feeling of eurphoria.

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It's the exact same interaction
and it's because of

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that structure there.
 

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It's not exactly the morphine
rule in endorphins but

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it's very similar.
 

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So, if we look at other
derivatives of morphine, such

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as codeine and
diacetylmorphine, these also

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you can see have this morphin
rule within them, this

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structure that you've
identified using your vsper

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rules and thinking about
hybridization.

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Codeine, as you may know, is
less of a potent pain killer

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compared to morphine, but
it's also less addictive.

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So for that reason it's
prescribed with slightly

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less oversight.
 

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So if you get your wisdom teeth
removed or something like that,

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some of you might have had
taken codeine before, and now

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you can think about by looking
at the structure, a little bit

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more about how that worked in
terms of blocking your pain.

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One that I just want to mention
because it's so interesting is

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this derivative, which
is diacetylmorphine.

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The only difference between
morphine and diacetylmorphine

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is the changing of an alcohol
group, or two actually,

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two acetyl groups here.
 

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This was synthesized by the
company Bayer, and you

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probably know Bayer
from Bayer's aspirin.

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Bayer had a huge success with
aspirin where they took

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salicylic acid and replaced an
o h group with an acetyl

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group here, and made
acetylsalicylic acid.

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Does anyone know the
common name for that?

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Yup.
 

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So that's aspirin, and we often
think of Bayer's aspirin.

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We associate it really closely.
 

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They worked really hard in
marketing to make that really

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close association and with
copywriting, so that when we

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think about aspirin we
think about Bayers.

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Bayers also made this change
here, and they did find that

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diacetylmorphine is much, much
more potent than morphine is.

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They could use just a tiny,
tiny amount of this compound

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and get the same pain-killing
properties that they

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saw with morphine.
 

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The problem is that this masked
some of the side effects that

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they didn't realize initially,
which was the extreme, extreme

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ability for you to become
addicted to diacetylmorphine.

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And this was first thought of
as a hero drug because it was

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such a strong pain killer, and
this was called a

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Bayer's heroin.
 

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We don't usually associate the
name Bayers with heroin, like

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we did with aspirin -- that's
because Bayers did not work

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so hard to keep that
association there.

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But it's an interesting story
from the history of drugs.

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One more example I want to
show you is demerol, which is

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used clinically quite often.
 

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Demerol, if we look at the
structure here, actually looks

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nothing like morphine at all.
 

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So to an untrained eye,
you might not see the

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relationship there.
 

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But because you all know some
of these basic principles of

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chemistry, you should be able
to pick out that morphine rule

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right in the middle of demerol.
 

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And, in fact, demerol is a pain
killer, it's an alternative

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to using morphine.
 

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Again, not quite as strong as
morphine in terms of pain

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killing abilities, but also not
as addictive either, and there

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are some other side effects,
such as nausea that are limited

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with demerol compared
to morphine.

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Demerol has its own set of
problems, so it's not the

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perfect solution to a different
alternative to morphine, but

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you do see it used
in certain cases.

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All right, so that's thinking
about vsper theory, and that's

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thinking about hybridization.
 

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We're now going to shift
gears to talking about

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some new topics today.
 

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So, we ended in Wednesday's
lecture with giving that set

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of rules to very quickly
determine hybridization.

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And then after we'd established
that, we moved on to starting

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to talk about a little
bit of thermochemistry.

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Specifically what we were
talking about is the enthalpies

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of chemical reactions, so
either the amount of heat

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that's released, or the amount
of heat that's required

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in order to have a
chemical reaction go.

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And we'd established so far two
ways to think about how we can

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measure what the overall
enthalpy of a reaction is.

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And I just wanted to introduced
one more technique that

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we can use to do that.
 

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And this relies on the fact
that when we talk about

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enthalpy or we talk about the
heat change of a reaction, this

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is a state function, and any
time we're talking about

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state function, it's
independent of path.

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So essentially, all that
we're worried about is the

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state that we started in.
 

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So, for example, when we were
talking about the oxidation of

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glucose, the state that we
start in are the reactants,

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so it's glucose plus oxygen.
 

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And what we ended up with were
six moles each of carbon

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dioxide and water.
 

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So, we don't care how we get
there in terms of making this

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calculation, all we're
interested in is the difference

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between those two states, to
think about the overall

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change in enthalpy.
 

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So, for example, when we talked
about how we could calculate

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this, one way that we can do
it, and we'll look at these

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numbers even more specifically
in a second, is thinking about

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instead of going straight down
here, which we might not have

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the numbers available to us to
calculate directly, we can

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think about well, what happens
if instead we break apart this

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glucose molecule and decompose
it into its elements, because

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using tables in the back of our
book, for example, we can

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figure out what the change in
enthalpy is of this

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reaction here.
 

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That doesn't quite get us to
where we need to go though, so

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then we can think about the
enthalpy of formation of six

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carbon dioxide molecules.
 

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And then we can go one step
further and think about the

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enthalpy of formation for
six water molecules.

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So this is just a round about
way of going from here to here.

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Instead we went up and then
down and then down, but because

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it's a state function, we're
absolutely allowed to do this.

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And to formalize this, we
can talk about what's

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called Hess's law.
 

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And Hess's law just tell us
that if we have two or three or

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00:10:01 --> 00:10:04
any number of chemical
reactions that we can add

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00:10:04 --> 00:10:07
together in order to get the
chemical equation that we're

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actually interested in, we can
figure out the change in

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enthalpy of that reaction that
we're interested in, just by

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adding together all of the
other enthalpies of reaction

212
00:10:18 --> 00:10:20
for every other reaction that
we had to add together

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in order to get there.
 

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So, let's take a look at what
we're talking about here.

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00:10:26 --> 00:10:29
So when I just showed you that
example graphically, I'm just

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writing it out more like an
equation here, because the nice

217
00:10:31 --> 00:10:34
thing about Hess's law, is it
allows you to think about

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00:10:34 --> 00:10:37
chemical reactions kind of as
if they're algebraic

219
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expressions.
 

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So we can just add all of these
different expressions together,

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and if we're thinking about
algebra, we can just think

222
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about crossing things out
that are in the reactants

223
00:10:46 --> 00:10:48
versus in the products.
 

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So if we do this, what we end
up getting is the reaction that

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we're interested in, the
oxidation of glucose to form

226
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carbon dioxide in water.
 

227
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So let's look pretty carefully
at exactly the fact that these

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all cancel out to give us
the reaction I say we get.

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So, for example, we can think
about the oxygen molecules --

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we can cancel out 6 here and 6
here, as long as we multiply

231
00:11:10 --> 00:11:12
this whole reaction by 6.
 

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Again, we can get rid of those
last 3 oxygen molecules there.

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Similarly, we can cross out our
6 hydrogens from the products

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00:11:20 --> 00:11:23
with 6 molecular hydrogens
from the reactants.

235
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And then the last thing we can
cancel out are our carbon atoms

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-- we have 6 in the products,
and we have 6 in the

237
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reactants here.
 

238
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So you'll see that the only
things that are not canceled

239
00:11:34 --> 00:11:37
out is what's left in
our bottom reaction.

240
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So we can go ahead and just
add up all of these different

241
00:11:41 --> 00:11:42
reaction enthalpies.
 

242
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In the first case what we're
going to have is an enthalpy of

243
00:11:45 --> 00:11:53
1260, and that's in kilojoules
per mole of glucose.

244
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And the reason that this is
positive is because it's

245
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essentially the reverse
reaction of the change in

246
00:12:01 --> 00:12:02
enthalpy of formation.
 

247
00:12:02 --> 00:12:04
So it's going to be a
positive number there.

248
00:12:04 --> 00:12:08
We add to that the enthalpy
for this reaction here,

249
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which is negative 393 .
 

250
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5, but we need to remember to
multiply that by 6, because

251
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we're actually adding
together six of those

252
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individual reactions.
 

253
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And lastly, we also add
together six of this final

254
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reaction here, which has
an enthalpy change

255
00:12:23 --> 00:12:25
of negative 285 .
 

256
00:12:25 --> 00:12:26
8.
 

257
00:12:26 --> 00:12:30
So if we go ahead and just add
up all of our individual

258
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enthalpy changes, what we're
going to end up with for our

259
00:12:33 --> 00:12:37
entire reaction here, is a
change in enthalpy of negative

260
00:12:37 --> 00:12:41
2816, and that's kilojoules
per mole of glucose.

261
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Does anyone remember from
Wednesday's class, does

262
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this match what we had
seen experimentally?

263
00:12:45 --> 00:12:47
STUDENT: [INAUDIBLE]
 

264
00:12:47 --> 00:12:50
PROFESSOR: Yeah, so this is the
exact number that we calculated

265
00:12:50 --> 00:12:53
when we used heats of
formation, and it's also the

266
00:12:53 --> 00:12:57
exact number that's seen in
terms of experimental.

267
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All right.
 

268
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So there's actually three ways
we now have to calculate heats

269
00:13:02 --> 00:13:06
of the change in enthalpy
of an overall reaction.

270
00:13:06 --> 00:13:10
So, we have these in our notes
from last time and from

271
00:13:10 --> 00:13:14
finishing the end of this unit
right now, but I'm just going

272
00:13:14 --> 00:13:16
to summarize them for you on
the board, because there are a

273
00:13:16 --> 00:13:19
couple things that can get
confusing that I want to make

274
00:13:19 --> 00:13:21
sure no one is confusing,
particularly in the

275
00:13:21 --> 00:13:23
upcoming exam.
 

276
00:13:23 --> 00:13:25
So what was the first way that
we learned to think about

277
00:13:25 --> 00:13:28
calculating enthalpies
of reaction?

278
00:13:28 --> 00:13:30
STUDENT: [INAUDIBLE]
 

279
00:13:30 --> 00:13:30
PROFESSOR: Um-hmm.
 

280
00:13:30 --> 00:13:37
So, it's talking about
bond enthalpies.

281
00:13:37 --> 00:13:40
And when we talk about bond
enthalpy, the symbol that we

282
00:13:40 --> 00:13:43
see for a bond enthalpy
is usually delta h.

283
00:13:43 --> 00:13:45
And this is where all the
confusion starts, because

284
00:13:45 --> 00:13:48
everything's delta h, it just
depends on the subscript here,

285
00:13:48 --> 00:13:50
right, what kind of delta
h we're talking about.

286
00:13:50 --> 00:13:52
In the case of bond enthalpy
often you'll see no

287
00:13:52 --> 00:13:54
subscript at all.
 

288
00:13:54 --> 00:13:57
But sometimes you do see a
subscript, which would then

289
00:13:57 --> 00:14:01
just be delta h sub b here.
 

290
00:14:01 --> 00:14:03
So if we're trying to do this
calculation based on bond

291
00:14:03 --> 00:14:09
enthalpies, for the delta h for
an overall reaction, what we

292
00:14:09 --> 00:14:15
want to do is take the sum of
all of those bond enthalpies of

293
00:14:15 --> 00:14:22
our bonds that are broken, and
what we want to do is subtract

294
00:14:22 --> 00:14:28
from that the sum of all our
bond enthalpies of our

295
00:14:28 --> 00:14:29
bonds that are formed.
 

296
00:14:29 --> 00:14:36
All right, so let's
think a little bit more

297
00:14:36 --> 00:14:38
about what this means.
 

298
00:14:38 --> 00:14:41
So, if we're talking about
bonds broken, are we talking

299
00:14:41 --> 00:14:42
about reactants or products?
 

300
00:14:42 --> 00:14:44
STUDENT: [INAUDIBLE]
 

301
00:14:44 --> 00:14:44
PROFESSOR: .
 

302
00:14:44 --> 00:14:44
Reactants.
 

303
00:14:44 --> 00:14:47
So, essentially we're summing
up the bond enthalpy of

304
00:14:47 --> 00:14:51
reactants and subtracting
it from the products.

305
00:14:51 --> 00:14:54
All right, great.
 

306
00:14:54 --> 00:14:57
So that's our first
strategy there.

307
00:14:57 --> 00:15:01
The second strategy that we
learned is thinking about bond

308
00:15:01 --> 00:15:06
enthalpies of reaction by using
the enthalpy change in terms

309
00:15:06 --> 00:15:08
of the enthalpy of formation.
 

310
00:15:08 --> 00:15:12
So this one is sometimes a
little bit more intuitive,

311
00:15:12 --> 00:15:15
because we're talking about the
enthalpy change it takes in

312
00:15:15 --> 00:15:17
order to form a given molecule.
 

313
00:15:17 --> 00:15:20
So, if we're trying to do a
calculation using enthalpies of

314
00:15:20 --> 00:15:25
formation, what we find is that
delta h of the reaction is

315
00:15:25 --> 00:15:30
equal to the sum of the
delta h of formation of

316
00:15:30 --> 00:15:31
products or reactants?
 

317
00:15:31 --> 00:15:32
STUDENT: Products.
 

318
00:15:32 --> 00:15:36
PROFESSOR: Products, good.
 

319
00:15:36 --> 00:15:42
Of products minus the delta
h of formation, and this is

320
00:15:42 --> 00:15:45
the sum again of reactants.
 

321
00:15:45 --> 00:15:52
All right, and our third
strategy is what we just talked

322
00:15:52 --> 00:15:55
about, which is Hess's law, and
in fact, both one and two are

323
00:15:55 --> 00:15:59
applications of Hess's law --
specific applications, but any

324
00:15:59 --> 00:16:02
application of Hess's law you
can use any way of adding up

325
00:16:02 --> 00:16:06
different equations to get the
final enthalpy of the reaction.

326
00:16:06 --> 00:16:09
The reason I wanted to put
these two strategies, however,

327
00:16:09 --> 00:16:11
on the board right next to each
other, is so we can just

328
00:16:11 --> 00:16:14
confront a point of confusion
that happens a lot with

329
00:16:14 --> 00:16:17
students, which is why in the
first case do we have reactants

330
00:16:17 --> 00:16:20
minus product, and in
the second case we have

331
00:16:20 --> 00:16:22
products minus reactants.
 

332
00:16:22 --> 00:16:25
So, you need to keep these two
straight when you're doing your

333
00:16:25 --> 00:16:28
calculations in terms of
enthalpies, and the reason is

334
00:16:28 --> 00:16:31
because the definition of a
bond enthalpy when we're

335
00:16:31 --> 00:16:35
talking about bonds, is the
enthalpy that it takes in order

336
00:16:35 --> 00:16:38
to break the bond, the amount
of heat that goes in

337
00:16:38 --> 00:16:39
to breaking a bond.
 

338
00:16:39 --> 00:16:42
So it's typically a positive
number, because if you have a

339
00:16:42 --> 00:16:44
stable bond you have to
put heat into it in

340
00:16:44 --> 00:16:45
order to break it.
 

341
00:16:45 --> 00:16:48
So you end up having a
positive bond enthalpy here.

342
00:16:48 --> 00:16:52
In contrast, when we're talking
about enthalpies of formation,

343
00:16:52 --> 00:16:55
now we're talking about is
the change of enthalpy in

344
00:16:55 --> 00:16:57
order to form a molecule.
 

345
00:16:57 --> 00:17:00
So, for example, if we have a
nice stable molecule with

346
00:17:00 --> 00:17:04
strong bonds in it, what we're
going to find is that the delta

347
00:17:04 --> 00:17:08
h of formation, would that
be positive or negative?

348
00:17:08 --> 00:17:10
So it's negative.
 

349
00:17:10 --> 00:17:13
So you'd end up with a negative
delta h of formation.

350
00:17:13 --> 00:17:16
So because you're talking about
a term that's negative here

351
00:17:16 --> 00:17:20
where it would be positive in
the first case, you end up

352
00:17:20 --> 00:17:22
flipping signs to
have it work out.

353
00:17:22 --> 00:17:24
So, if this doesn't make sense
as I'm saying it right now,

354
00:17:24 --> 00:17:27
just go back in your notes and
look through and make sure that

355
00:17:27 --> 00:17:29
you aren't going to get mixed
up when you're using bond

356
00:17:29 --> 00:17:32
enthalpies versus enthalpies
of formation here.

357
00:17:32 --> 00:17:36
All right, so that's pretty
much all we're going

358
00:17:36 --> 00:17:39
to say on enthalpy.
 

359
00:17:39 --> 00:17:41
So I really just want to
stress, this is going to be the

360
00:17:41 --> 00:17:44
end of exam 2 material here, I
said I'd be clear about

361
00:17:44 --> 00:17:45
it in class today.
 

362
00:17:45 --> 00:17:46
This is it.
 

363
00:17:46 --> 00:17:49
It's also written in your notes
when you go ahead to study.

364
00:17:49 --> 00:17:52
So remember, in terms of exam
2, which I'll remind you is on

365
00:17:52 --> 00:17:55
Wednesday, it's going to be all
the way from the material on

366
00:17:55 --> 00:17:58
lecture 10 up to
this point here.

367
00:17:58 --> 00:18:01
So only through enthalpy, and
it's also going to be on

368
00:18:01 --> 00:18:05
problem-sets four and five, and
I'll mention that this

369
00:18:05 --> 00:18:08
afternoon, as I said, I'll post
a bunch of extra practice

370
00:18:08 --> 00:18:12
problems, which are more review
of these same concepts, so you

371
00:18:12 --> 00:18:16
can get even more practice this
weekend and early next

372
00:18:16 --> 00:18:17
week to prepare.
 

373
00:18:17 --> 00:18:20
All right, so that's
enthalpy, that's the

374
00:18:20 --> 00:18:22
end of exam 2 material.
 

375
00:18:22 --> 00:18:25
But there's actually another
really important concept that

376
00:18:25 --> 00:18:27
we need to talk about -- we
really don't want to stop with

377
00:18:27 --> 00:18:30
enthalpy, luckily it's exam 2
and not the end of the class,

378
00:18:30 --> 00:18:32
because a really important
concept to think about

379
00:18:32 --> 00:18:34
is spontaneous change.
 

380
00:18:34 --> 00:18:37
So thinking about whether
a reaction is spontaneous

381
00:18:37 --> 00:18:39
or non-spontaneous.
 

382
00:18:39 --> 00:18:42
And when we think about a
spontaneous reaction, I think

383
00:18:42 --> 00:18:45
this is a term many of you are
familiar with, spontaneous just

384
00:18:45 --> 00:18:47
means that the reactions going
to proceed in the forward

385
00:18:47 --> 00:18:50
direction without any kind
of outside intervention.

386
00:18:50 --> 00:18:53
So thinking of a spontaneous
process, can talk about a

387
00:18:53 --> 00:18:55
chemical reaction, but you
could just picture, for

388
00:18:55 --> 00:19:00
example, putting a round rock
on the top of a hill -- it

389
00:19:00 --> 00:19:03
will spontaneously just
roll right down that hill.

390
00:19:03 --> 00:19:05
So a spontaneous process also
has direction, right, because

391
00:19:05 --> 00:19:08
the rock won't spontaneously
roll back up the hill

392
00:19:08 --> 00:19:10
without putting in work.
 

393
00:19:10 --> 00:19:12
So essentially we're talking
about a spontaneous process

394
00:19:12 --> 00:19:15
when something's going to
happen without actually having

395
00:19:15 --> 00:19:20
to do anything else to
force it to happen.

396
00:19:20 --> 00:19:23
So let's think about in terms
of chemistry what we're talking

397
00:19:23 --> 00:19:26
about is a spontaneous reaction
-- that's the specific type of

398
00:19:26 --> 00:19:29
spontaneous process that
we're interested in.

399
00:19:29 --> 00:19:32
So let's think about a few
different types of spontaneous

400
00:19:32 --> 00:19:35
reactions, and see if we can
come up with some idea of

401
00:19:35 --> 00:19:38
what's going to cause
them to be spontaneous.

402
00:19:38 --> 00:19:42
So one spontaneous reaction is
written here, so this is just

403
00:19:42 --> 00:19:45
the oxidation of iron, or
the formation of rust.

404
00:19:45 --> 00:19:49
This turns out to also be an
exothermic reaction, it has a

405
00:19:49 --> 00:19:55
negative delta h of 824
kilojoules per mole.

406
00:19:55 --> 00:19:58
Another spontaneous reaction is
written here, the combination

407
00:19:58 --> 00:20:00
of an acid and a base, which
neutralizes each other.

408
00:20:00 --> 00:20:03
So we have a hydronium
ion and a hydroxide ion

409
00:20:03 --> 00:20:06
interacting to form water.
 

410
00:20:06 --> 00:20:08
This is spontaneous, and
again, it's exothermic.

411
00:20:08 --> 00:20:12
So, its delta h
is negative 55 .

412
00:20:12 --> 00:20:13
9 kilojoules per mole.
 

413
00:20:13 --> 00:20:17
Let's think about
some really relevant

414
00:20:17 --> 00:20:19
reactions in our bodies.
 

415
00:20:19 --> 00:20:21
So one incredibly important
reaction, of course, is ATP

416
00:20:21 --> 00:20:27
hydrolysis where we have
adenosine here, a triphosphate,

417
00:20:27 --> 00:20:29
so we have three
phosphate groups here.

418
00:20:29 --> 00:20:30
That's called ATP.
 

419
00:20:30 --> 00:20:33
It has a total charge
of minus four.

420
00:20:33 --> 00:20:36
So if we hydrolize this, one of
the phosphate bonds here and

421
00:20:36 --> 00:20:39
lose a phosphate, we end up
with adenosine diphosphate

422
00:20:39 --> 00:20:42
or ADP and that has a
charge of minus three.

423
00:20:42 --> 00:20:43
Yup?
 

424
00:20:43 --> 00:20:50
STUDENT: On our notes, this
equation [INAUDIBLE].

425
00:20:50 --> 00:20:50
PROFESSOR: Oh, no.
 

426
00:20:50 --> 00:20:51
OK, thank you.
 

427
00:20:51 --> 00:20:52
Which equation are
we talking about?

428
00:20:52 --> 00:21:03
STUDENT: [INAUDIBLE]
 

429
00:21:03 --> 00:21:05
PROFESSOR: I'm sorry,
what page are you on?

430
00:21:05 --> 00:21:09
Page two.
 

431
00:21:09 --> 00:21:14
So the one with h 3 -- OK.
 

432
00:21:14 --> 00:21:19
All right, let's
go back to that.

433
00:21:19 --> 00:21:24
OK, so if you can fix this
reaction in your notes here.

434
00:21:24 --> 00:21:26
I'll also fix it on the
website, so if you don't want

435
00:21:26 --> 00:21:29
to fix it now I'll just
re-post the notes.

436
00:21:29 --> 00:21:35
So it should be four irons
plus three oxygens is

437
00:21:35 --> 00:21:38
two f e 2 o 3 solid.
 

438
00:21:38 --> 00:21:39
Thanks for pointing that out.
 

439
00:21:39 --> 00:21:41
All right, does everyone
have that down?

440
00:21:41 --> 00:21:46
I know you had to
flip a page there.

441
00:21:46 --> 00:21:48
All right, so I'll post
it in the notes if you

442
00:21:48 --> 00:21:49
didn't get it there.
 

443
00:21:49 --> 00:21:52
Let's go back to ATP
hydrolysis here.

444
00:21:52 --> 00:21:58
So we're going from ATP to ADP,
and that is a spontaneous

445
00:21:58 --> 00:22:02
process, and it's an
exothermic reaction as well.

446
00:22:02 --> 00:22:05
So we find that it's negative
24 kilojoules per mole in

447
00:22:05 --> 00:22:08
terms of the change
in enthalpy there.

448
00:22:08 --> 00:22:11
All right, so at this point I
just showed you three reactions

449
00:22:11 --> 00:22:13
where we have something that's
spontaneous and it's

450
00:22:13 --> 00:22:14
also exothermic.
 

451
00:22:14 --> 00:22:16
And I could show you a
countless number of other

452
00:22:16 --> 00:22:18
reactions where it's
the same thing.

453
00:22:18 --> 00:22:22
It's actually quite common if
you have an exothermic reaction

454
00:22:22 --> 00:22:24
that it's also spontaneous
at room temperature.

455
00:22:24 --> 00:22:27
So we might start to draw the
conclusion that, in fact, it's

456
00:22:27 --> 00:22:30
enthalpy that's responsible for
whether or not a reaction is

457
00:22:30 --> 00:22:33
spontaneous or non-spontaneous.
 

458
00:22:33 --> 00:22:35
So it's very easy to get that
impression, but let me show you

459
00:22:35 --> 00:22:38
a few more reactions before
we draw any conclusions.

460
00:22:38 --> 00:22:41
Let me show you some more
spontaneous reactions.

461
00:22:41 --> 00:22:45
So, for example, the conversion
of solid h 2 o to liquid h 2 o

462
00:22:45 --> 00:22:50
at room temperature, I think we
all know this is spontaneous.

463
00:22:50 --> 00:22:53
Ice melts at room temperature,
but it turns out that the

464
00:22:53 --> 00:22:58
enthalpy change is positive,
it's 7 kilojoules per mole.

465
00:22:58 --> 00:23:01
Similarly, if we look at
this reaction here, which

466
00:23:01 --> 00:23:03
is ammonium nitrate.
 

467
00:23:03 --> 00:23:05
This is a very commonly used
fertilizer, a very commonly

468
00:23:05 --> 00:23:08
used nitrogen source
in agriculture.

469
00:23:08 --> 00:23:12
If we think about solvating
this and forming the two ions,

470
00:23:12 --> 00:23:14
ammonium ion and nitrate ion.
 

471
00:23:14 --> 00:23:17
Again, this is a spontaneous
reaction, but what we find

472
00:23:17 --> 00:23:20
here is that delta h again
is positive, it's

473
00:23:20 --> 00:23:21
also endothermic.
 

474
00:23:21 --> 00:23:25
So thinking about all of this,
would you say that enthalpy

475
00:23:25 --> 00:23:26
is the key to spontaneity?
 

476
00:23:26 --> 00:23:27
STUDENT: No.
 

477
00:23:27 --> 00:23:30
PROFESSOR: No, it's definitely
not the key to spontaneity.

478
00:23:30 --> 00:23:33
It tends to correlate in many
cases at room temperature, but

479
00:23:33 --> 00:23:36
it is not the key
to spontaneity.

480
00:23:36 --> 00:23:39
And the key to spontaneity is
instead something called Gibbs

481
00:23:39 --> 00:23:43
free energy or delta g.
 

482
00:23:43 --> 00:23:46
And if we think about what
delta g is, we can relate it to

483
00:23:46 --> 00:23:49
enthalpy, and this will sort of
show us why there's often this

484
00:23:49 --> 00:23:54
correlation, because delta g is
equal delta h minus this term

485
00:23:54 --> 00:23:58
here, which is temperature, and
that's temperature times

486
00:23:58 --> 00:23:59
a change in entropy.
 

487
00:23:59 --> 00:24:04
And we'll talk very in-depth in
just a few minutes about what

488
00:24:04 --> 00:24:07
entropy actually is, but for
now I just want you to think

489
00:24:07 --> 00:24:10
about the fact that in addition
to thinking about enthalpy

490
00:24:10 --> 00:24:13
there's this other term
here that comes into play.

491
00:24:13 --> 00:24:16
So when we're talking about
what the sign about delta g is

492
00:24:16 --> 00:24:19
or free energy is, some of you
might already be familiar with

493
00:24:19 --> 00:24:22
this, when we're talking about
a negative delta g, is this

494
00:24:22 --> 00:24:27
spontaneous or non-spontaneous.
 

495
00:24:27 --> 00:24:30
So, maybe not so familiar,
which is totally fine.

496
00:24:30 --> 00:24:33
So in terms of thinking about a
negative delta g, that's going

497
00:24:33 --> 00:24:35
to be a spontaneous process.
 

498
00:24:35 --> 00:24:38
Any time you see a negative
free energy, which just like

499
00:24:38 --> 00:24:40
when we talk about entropy,
it's a similar idea, it means

500
00:24:40 --> 00:24:44
that free energy is released
as the reaction progresses.

501
00:24:44 --> 00:24:47
That's going to be a
spontaneous process.

502
00:24:47 --> 00:24:50
So that means that if delta g
is greater than zero, then

503
00:24:50 --> 00:24:53
what we're going to see a
non-spontaneous process.

504
00:24:53 --> 00:24:56
And finally in the case where
delta g is equal to zero, at

505
00:24:56 --> 00:24:59
this point we're at
equilibrium, which basically

506
00:24:59 --> 00:25:02
means that there's no net
change in either the forward or

507
00:25:02 --> 00:25:05
the reverse reaction in terms
of thinking about whether

508
00:25:05 --> 00:25:08
something's spontaneous or not
spontaneous, it's just at

509
00:25:08 --> 00:25:11
equilibrium we see
no net change.

510
00:25:11 --> 00:25:14
And something I want to point
out is that this reaction is

511
00:25:14 --> 00:25:16
valid only when where at
constant temperature and

512
00:25:16 --> 00:25:20
pressure, which we will be
throughout the discussions in

513
00:25:20 --> 00:25:22
this class in terms of
using this equation.

514
00:25:22 --> 00:25:25
And also, it turns out that
when you do most chemical

515
00:25:25 --> 00:25:29
equations, you are, in fact, at
a pretty constant temperature,

516
00:25:29 --> 00:25:32
and a constant pressure in
terms of most things are done

517
00:25:32 --> 00:25:34
in the open atmosphere, so
that's really not

518
00:25:34 --> 00:25:35
going to change.
 

519
00:25:35 --> 00:25:39
All right, so let's just talk
very briefly about why it is

520
00:25:39 --> 00:25:42
that free energy is what
determines spontaneity, and not

521
00:25:42 --> 00:25:45
this enthalpy here, not what
the change in enthalpy is.

522
00:25:45 --> 00:25:48
And it really comes in terms
remembering it in terms

523
00:25:48 --> 00:25:49
of the definition.
 

524
00:25:49 --> 00:25:52
When we think about delta g,
what that is is that it's

525
00:25:52 --> 00:25:55
energy that's released or used
in the reaction, but if it

526
00:25:55 --> 00:25:58
is released, that it can
directly be used to do work.

527
00:25:58 --> 00:26:00
It's what we call free
energy -- it's free

528
00:26:00 --> 00:26:01
to do other things.
 

529
00:26:01 --> 00:26:04
Whereas when we were talking
about enthalpy here, well, this

530
00:26:04 --> 00:26:07
is the amount of heat that's
released when we break and form

531
00:26:07 --> 00:26:09
the bonds in the reaction.
 

532
00:26:09 --> 00:26:12
But some of it actually gets
stuck in the molecules, and

533
00:26:12 --> 00:26:15
this is just a way to think
about it, this term here is one

534
00:26:15 --> 00:26:17
way we can think about it,
it's just some energy

535
00:26:17 --> 00:26:18
that's getting stuck.
 

536
00:26:18 --> 00:26:21
So, for example, molecules,
which we don't really go into

537
00:26:21 --> 00:26:24
in this class, have vibrational
movements, they have rotational

538
00:26:24 --> 00:26:27
movements, there's energy
associated with that that can

539
00:26:27 --> 00:26:31
sort of get stuck, as we could
say, some of the enthalpy

540
00:26:31 --> 00:26:34
that's released in terms of
forming and breaking the

541
00:26:34 --> 00:26:35
bonds in a reaction.
 

542
00:26:35 --> 00:26:38
So that's our case where we
could see that we have a

543
00:26:38 --> 00:26:42
negative delta h, but we still
end up with a positive delta g.

544
00:26:42 --> 00:26:44
And again, I haven't really
gone into what entropy

545
00:26:44 --> 00:26:49
is here yet, and we will
just in a few minutes.

546
00:26:49 --> 00:26:52
But first, I want to take a
look at one of the examples

547
00:26:52 --> 00:26:55
that we talked about, which is
the conversion of ammonium

548
00:26:55 --> 00:26:58
nitrate when its solvated,
breaking up into its ions.

549
00:26:58 --> 00:27:03
So what I had told you is that,
in fact, this is a positive

550
00:27:03 --> 00:27:07
delta h, but still that the
reaction is spontaneous.

551
00:27:07 --> 00:27:10
So let's just do this
calculation and see if

552
00:27:10 --> 00:27:13
we can confirm that
in terms of delta g.

553
00:27:13 --> 00:27:16
So the reaction again that
we're going to use is delta g

554
00:27:16 --> 00:27:19
equals delta h minus t delta s.
 

555
00:27:19 --> 00:27:25
So if we're talking about room
temperature here, so we're

556
00:27:25 --> 00:27:29
going to say that delta g in
this case is going to be equal

557
00:27:29 --> 00:27:37
to 28 kilojoules per mole,
minus -- we're at room

558
00:27:37 --> 00:27:41
temperature or 298 kelvin -- in
this reaction, you always put

559
00:27:41 --> 00:27:42
temperature into kelvin.
 

560
00:27:42 --> 00:27:46
And then we need to multiply
it by the entropy, which is

561
00:27:46 --> 00:27:48
109 joules per kelvin mole.
 

562
00:27:48 --> 00:27:51
I want to point out that
entropy tends to be a much

563
00:27:51 --> 00:27:54
smaller value than enthalpy,
so it's reported in joules

564
00:27:54 --> 00:27:55
instead of kilojoules.
 

565
00:27:55 --> 00:27:58
When you solve these problems,
it's really important to

566
00:27:58 --> 00:28:01
convert it into kilojoules so
that your units work out

567
00:28:01 --> 00:28:04
and you don't get a
completely crazy answer.

568
00:28:04 --> 00:28:06
So we want to
convert this to 0 .

569
00:28:06 --> 00:28:12
109 kilojoules per kelvin mole.
 

570
00:28:12 --> 00:28:15
So we can just re-write this,
so we have 28 kilojoules per

571
00:28:15 --> 00:28:19
mole, then this term
here will be 32 .

572
00:28:19 --> 00:28:21
5 kilojoules per mole.
 

573
00:28:21 --> 00:28:26
So what we end up with our
delta g for this reaction

574
00:28:26 --> 00:28:30
is going to be negative
4 kilojoules per mole.

575
00:28:30 --> 00:28:38
All right, so talking about
this as a negative delta

576
00:28:38 --> 00:28:40
g, would you say this
is a spontaneous or

577
00:28:40 --> 00:28:43
non-spontaneous reaction.
 

578
00:28:43 --> 00:28:44
Yeah, it's spontaneous.
 

579
00:28:44 --> 00:28:48
So what we're going to see is
delta g is negative 4, so this

580
00:28:48 --> 00:28:51
reaction is spontaneous
even though our enthalpy

581
00:28:51 --> 00:28:54
change was positive.
 

582
00:28:54 --> 00:28:56
Let's take a look at one more
reaction since we spent so

583
00:28:56 --> 00:29:00
much time discussing the
oxidation of glucose.

584
00:29:00 --> 00:29:03
So again, what we know about
this reaction, we've already

585
00:29:03 --> 00:29:05
calculated the delta
h is negative 2816

586
00:29:05 --> 00:29:07
kilojoules per mole.
 

587
00:29:07 --> 00:29:10
And you could look up and we'll
figure out how to calculate it

588
00:29:10 --> 00:29:13
soon, the change in
entropy, which is plus 233

589
00:29:13 --> 00:29:16
joules per k mole.
 

590
00:29:16 --> 00:29:18
So before we actually do this
calculation, let's go to a

591
00:29:18 --> 00:29:21
clicker question and I
want you to think about

592
00:29:21 --> 00:29:23
what's happening here.
 

593
00:29:23 --> 00:29:27
So if you're thinking about the
oxidation of glucose and what

594
00:29:27 --> 00:29:29
the sign is of the change in
enthalpy and the change in

595
00:29:29 --> 00:29:32
entropy, which of these
statements is true.

596
00:29:32 --> 00:29:35
Is it going to be spontaneous
at all temperatures,

597
00:29:35 --> 00:29:38
non-spontaneous at all
temperatures, or will it depend

598
00:29:38 --> 00:29:41
on the temperature whether this
reaction is spontaneous

599
00:29:41 --> 00:29:42
or not spontaneous?
 

600
00:29:42 --> 00:29:45
So, remember this is
our relationship here.

601
00:29:45 --> 00:30:02
So let's go ahead and just
take 10 seconds on this one.

602
00:30:02 --> 00:30:06
OK, so we have a pretty
mixed response here.

603
00:30:06 --> 00:30:09
So this could have made or
broke the chances of your team

604
00:30:09 --> 00:30:11
on the competition today, but
they'll be a few more left

605
00:30:11 --> 00:30:14
to redeem yourselves if
you got it incorrect.

606
00:30:14 --> 00:30:17
So what we're going to find is
that it's spontaneous at all

607
00:30:17 --> 00:30:20
temperatures, and we can
actually just look at this

608
00:30:20 --> 00:30:23
equation here to figure
out why that is.

609
00:30:23 --> 00:30:26
If our reaction is
spontaneous, that means

610
00:30:26 --> 00:30:28
that delta g is negative.
 

611
00:30:28 --> 00:30:31
So in terms of delta h being
negative, well, that's going to

612
00:30:31 --> 00:30:34
always contribute to delta g
being negative no matter

613
00:30:34 --> 00:30:35
what the temperature is.
 

614
00:30:35 --> 00:30:40
And if delta s is positive,
since it's minus t delta s, and

615
00:30:40 --> 00:30:42
our temperature is always
positive because we're on the

616
00:30:42 --> 00:30:47
kelvin scale, so it starts at
zero there, then this component

617
00:30:47 --> 00:30:49
here is always going to
be negative as well.

618
00:30:49 --> 00:30:52
So we're always going to have
two negative numbers so that

619
00:30:52 --> 00:30:55
our combination, our delta g's
always going to be

620
00:30:55 --> 00:30:58
negative, it's always going
to be spontaneous.

621
00:30:58 --> 00:31:00
All right, so let's make sure
that's what we do see when

622
00:31:00 --> 00:31:07
we calculate this for the
oxidation of glucose here.

623
00:31:07 --> 00:31:10
So if we do this and we plug in
our numbers, we see that delta

624
00:31:10 --> 00:31:15
g is going to be equal to
negative 2816 times the

625
00:31:15 --> 00:31:19
temperature, room temperature,
times again, remember to put 0.

626
00:31:19 --> 00:31:23
233, because we need to convert
from joules to kilojoules

627
00:31:23 --> 00:31:24
for our entropy term.
 

628
00:31:24 --> 00:31:27
So what we find out is
that this reaction has a

629
00:31:27 --> 00:31:31
delta g of negative 2885
kilojoules per mole.

630
00:31:31 --> 00:31:33
This is a spontaneous reaction.
 

631
00:31:33 --> 00:31:37
All right, so let's get
back to that entropy term.

632
00:31:37 --> 00:31:40
I introduced it without
explaining at all

633
00:31:40 --> 00:31:41
what it really is.
 

634
00:31:41 --> 00:31:44
So let's take a few moments
and think about entropy.

635
00:31:44 --> 00:31:48
Entropy is actually a very
easy concept to think about,

636
00:31:48 --> 00:31:50
it's a measure of disorder.
 

637
00:31:50 --> 00:31:53
I think most of us have a very
good concept of how things tend

638
00:31:53 --> 00:31:56
to go to disorder, so it's
something we can conceptualize

639
00:31:56 --> 00:31:59
just in an infinite
number of ways.

640
00:31:59 --> 00:32:02
And entropy is simply a measure
of the disorder of a system.

641
00:32:02 --> 00:32:05
And when we're talking about
chemical reactions, what

642
00:32:05 --> 00:32:07
we're talking about is
a change in entropy.

643
00:32:07 --> 00:32:11
So whether, as a reaction goes
forward, are we becoming more

644
00:32:11 --> 00:32:16
ordered or are we becoming
more disordered?

645
00:32:16 --> 00:32:20
So just like we saw for
enthalpy, entropy also is a

646
00:32:20 --> 00:32:23
state function, so it doesn't
matter in terms of path how we

647
00:32:23 --> 00:32:25
got from point a to point b.
 

648
00:32:25 --> 00:32:27
All we need to worry about is
the current state of our

649
00:32:27 --> 00:32:30
system, what the current
entropy is and take

650
00:32:30 --> 00:32:32
the difference.
 

651
00:32:32 --> 00:32:35
So, an example that I like to
use in terms of conceptually

652
00:32:35 --> 00:32:39
thinking about what's going on
in terms of disorder, and which

653
00:32:39 --> 00:32:41
is especially relevant in New
England is thinking

654
00:32:41 --> 00:32:43
about stone wall.
 

655
00:32:43 --> 00:32:46
So you can think about a
stone wall in terms of being

656
00:32:46 --> 00:32:47
very, very ordered, right.
 

657
00:32:47 --> 00:32:51
If you just built your stone
wall, every stone is in place,

658
00:32:51 --> 00:32:53
it has a high degree of order.
 

659
00:32:53 --> 00:32:56
So in other words, the
disorder is very low.

660
00:32:56 --> 00:33:01
So it has a low entropy,
a low level of disorder.

661
00:33:01 --> 00:33:03
But if we're looking at a New
England stone wall, which are

662
00:33:03 --> 00:33:06
typically found if you're
hiking in the woods you see

663
00:33:06 --> 00:33:10
ones that are just ancient and
are completely crumbling down,

664
00:33:10 --> 00:33:12
you'll find that the stone wall
is no longer quite so

665
00:33:12 --> 00:33:14
ordered as it was.
 

666
00:33:14 --> 00:33:17
In fact, let's say there's a
few stones that have fallen

667
00:33:17 --> 00:33:21
off, our disorder has
increased, so we end up

668
00:33:21 --> 00:33:23
increasing the entropy.
 

669
00:33:23 --> 00:33:26
And I do just want to point
out with this analogy that

670
00:33:26 --> 00:33:27
it is a state function.
 

671
00:33:27 --> 00:33:31
So for thinking about the
change in entropy, which is

672
00:33:31 --> 00:33:34
this distance right here, we
can either calculate it

673
00:33:34 --> 00:33:38
directly from here to here. but
it also doesn't matter, let's

674
00:33:38 --> 00:33:40
say the wall crumbled down
completely and we ended up

675
00:33:40 --> 00:33:44
having a very high degree of
disorder, if then someone put

676
00:33:44 --> 00:33:47
most of the stones back in
place and we went back down to

677
00:33:47 --> 00:33:50
this degree of disorder here,
it doesn't matter, it's a state

678
00:33:50 --> 00:33:54
function, all we care about is
the actual difference between

679
00:33:54 --> 00:33:58
our starting point and
our ending point.

680
00:33:58 --> 00:34:01
And I just can't resist putting
in a little quote, "something

681
00:34:01 --> 00:34:04
there is that doesn't love a
wall." Does anybody know where

682
00:34:04 --> 00:34:07
this quote is from -- very
famous poem, very famous poet.

683
00:34:07 --> 00:34:13
Good guesses, parents
can help with this.

684
00:34:13 --> 00:34:15
Yes, excellent.
 

685
00:34:15 --> 00:34:18
I've never heard the correct
answer this question before.

686
00:34:18 --> 00:34:18
Excellent.
 

687
00:34:18 --> 00:34:21
So this is Robert Frost
in The Mending Wall.

688
00:34:21 --> 00:34:25
He talks about, it sounds like
a lot of you did know this,

689
00:34:25 --> 00:34:29
this makes me full of joy.
"Something there is that

690
00:34:29 --> 00:34:33
doesn't love a wall," there's
many interpretations for this

691
00:34:33 --> 00:34:35
poem, but maybe it's
about entropy.

692
00:34:35 --> 00:34:38
Entropy doesn't love a wall,
we're going from order to

693
00:34:38 --> 00:34:41
disorder, that wall
wants to come down.

694
00:34:41 --> 00:34:44
All right.
 

695
00:34:44 --> 00:34:46
So let's formalize these
thoughts on entropy

696
00:34:46 --> 00:34:49
and in terms of what
we're talking about.

697
00:34:49 --> 00:34:52
Again, what I said on that's
chart there is as we

698
00:34:52 --> 00:34:55
increase entropy, we're
increasing disorder.

699
00:34:55 --> 00:35:00
So if you saw a positive change
in entropy in a reaction,

700
00:35:00 --> 00:35:03
would you think about that as
being more or less ordered?

701
00:35:03 --> 00:35:06
STUDENT: [INAUDIBLE]
 

702
00:35:06 --> 00:35:08
PROFESSOR: Yeah, so it's
actually going to be less

703
00:35:08 --> 00:35:11
ordered, and, in fact, I should
stop saying order, I should be

704
00:35:11 --> 00:35:13
just saying disorder because
disorder's actually what we're

705
00:35:13 --> 00:35:16
measuring, and there's an
increase in disorder here

706
00:35:16 --> 00:35:19
when we have a positive
change in enthalpy.

707
00:35:19 --> 00:35:22
So if we have a negative change
in enthalpy, we're going to

708
00:35:22 --> 00:35:26
see a decrease in disorder.
 

709
00:35:26 --> 00:35:30
So in terms of considering
different types of states that

710
00:35:30 --> 00:35:33
we can be in, whether we're in
a gas state or a liquid state

711
00:35:33 --> 00:35:38
or a solid state for any given
molecule or any given compound,

712
00:35:38 --> 00:35:41
we can think about how
much order or disorder

713
00:35:41 --> 00:35:42
these states have.
 

714
00:35:42 --> 00:35:45
So which of these three
states would you call

715
00:35:45 --> 00:35:46
the most disordered?
 

716
00:35:46 --> 00:35:48
STUDENT: Gas.
 

717
00:35:48 --> 00:35:48
PROFESSOR: Yeah,
it's the gas state.

718
00:35:48 --> 00:35:51
So it turns out that the
disorder gas is greater

719
00:35:51 --> 00:35:54
than liquid., which is
greater than solid.

720
00:35:54 --> 00:35:56
This makes sense and the gas
molecules are free to bounce

721
00:35:56 --> 00:35:58
around, go wherever they want.
 

722
00:35:58 --> 00:36:00
Once you're in a liquid,
they have a little bit

723
00:36:00 --> 00:36:02
more limitation, they
can't go anywhere.

724
00:36:02 --> 00:36:05
But the molecules can still
slide all around and

725
00:36:05 --> 00:36:06
pass each other.
 

726
00:36:06 --> 00:36:09
Once you're in a solid state,
this molecule is not going

727
00:36:09 --> 00:36:12
to flip places with the
other molecules there.

728
00:36:12 --> 00:36:16
It's going to be kind of stuck
in its place within the solid.

729
00:36:16 --> 00:36:20
So, just understanding this
explanation of entropy that

730
00:36:20 --> 00:36:24
we're going to have an increase
in entropy if we have an

731
00:36:24 --> 00:36:27
increase in disorder, allows us
to make predictions about

732
00:36:27 --> 00:36:31
reactions even without doing
any calculations, and we always

733
00:36:31 --> 00:36:33
like when that happens, because
then once we do calculations,

734
00:36:33 --> 00:36:36
we can check our calculations
and make sure they make sense

735
00:36:36 --> 00:36:38
with what we would
be predicting.

736
00:36:38 --> 00:36:41
So before we talk about how to
actually do a calculation,

737
00:36:41 --> 00:36:44
let's go to a clicker question
here and make sure everyone is

738
00:36:44 --> 00:36:48
on the same page in terms of
thinking about changes in

739
00:36:48 --> 00:36:52
entropy or changes or
delta s for the reaction.

740
00:36:52 --> 00:36:56
So let's say we're talking
about the decomposition of

741
00:36:56 --> 00:36:59
hydrogen peroxide into
water and oxygen.

742
00:36:59 --> 00:37:01
I want you to tell me if you
think it's going to have

743
00:37:01 --> 00:37:06
a positive delta s, a
negative, or zero, or maybe

744
00:37:06 --> 00:37:08
this will change with
temperature as well.

745
00:37:08 --> 00:37:12
All right, so let's
go ahead and take 10

746
00:37:12 --> 00:37:25
seconds on this one.
 

747
00:37:25 --> 00:37:26
OK, excellent.
 

748
00:37:26 --> 00:37:28
Great job with the
questions today.

749
00:37:28 --> 00:37:33
So we are actually going to see
a positive delta s. this should

750
00:37:33 --> 00:37:36
be very clear because we're
going from two moles of liquid,

751
00:37:36 --> 00:37:41
to now having two moles of
liquid, plus a mole of gas, so

752
00:37:41 --> 00:37:43
we're going to be
increasing the disorder of

753
00:37:43 --> 00:37:44
our system here.
 

754
00:37:44 --> 00:37:50
All right, so let's think
about how we actually can

755
00:37:50 --> 00:37:51
calculate, though, delta s.
 

756
00:37:51 --> 00:37:54
So far I've just been giving
you the delta s's for the

757
00:37:54 --> 00:37:57
reactions, but, as I'm sure you
can guess, you won't be given

758
00:37:57 --> 00:37:59
that information, for example,
on upcoming problem-sets,

759
00:37:59 --> 00:38:02
you'll actually need to
calculate the entropy.

760
00:38:02 --> 00:38:06
So what we can do is we can
actually calculate the entropy

761
00:38:06 --> 00:38:10
of a reaction from absolute
entropies of individual

762
00:38:10 --> 00:38:13
molecules that
we're discussing.

763
00:38:13 --> 00:38:17
So this is very similar to how
we calculated the enthalpy of a

764
00:38:17 --> 00:38:21
reaction by taking the change
of enthalpy of formation,

765
00:38:21 --> 00:38:24
except we don't even have to
worry about a change in entropy

766
00:38:24 --> 00:38:27
because we can talk about
entropy as an absolute value.

767
00:38:27 --> 00:38:31
There's an absolute zero in
terms of entropy where there

768
00:38:31 --> 00:38:33
is no disorder at all.
 

769
00:38:33 --> 00:38:38
And this is described for any
molecule as being its perfect

770
00:38:38 --> 00:38:41
crystal at absolute zero.
 

771
00:38:41 --> 00:38:43
So this is what we're going to
say there's absolute complete

772
00:38:43 --> 00:38:46
order, there's no disorder
at all in this system.

773
00:38:46 --> 00:38:49
So what we can take to
calculate the entropy in a

774
00:38:49 --> 00:38:52
reaction, is just to take the
sum of the entropy of all the

775
00:38:52 --> 00:38:55
products, and subtract from
that the entropy from

776
00:38:55 --> 00:38:59
all of the reactants.
 

777
00:38:59 --> 00:39:02
So let's try this for the
example that you just did on

778
00:39:02 --> 00:39:05
the clicker question, and make
sure that our calculation

779
00:39:05 --> 00:39:07
matches up with our
prediction here.

780
00:39:07 --> 00:39:11
So again, we're just going to
take the sum of the entropy

781
00:39:11 --> 00:39:15
of the products minus
out of the reactants.

782
00:39:15 --> 00:39:18
So if we talk about this, we're
starting with our product, so

783
00:39:18 --> 00:39:20
our first product is water.
 

784
00:39:20 --> 00:39:22
And what I want to point out is
you need to make sure you

785
00:39:22 --> 00:39:26
multiply this by 2, because, in
fact, we have two moles of

786
00:39:26 --> 00:39:27
water that are formed here.
 

787
00:39:27 --> 00:39:31
STUDENT: [INAUDIBLE]
 

788
00:39:31 --> 00:39:33
PROFESSOR: OK, I'm so sorry.
 

789
00:39:33 --> 00:39:37
Yeah, so yikes, this
should be water.

790
00:39:37 --> 00:39:43
You know what, let's just
write this on the board.

791
00:39:43 --> 00:39:49
So we're going to talk about
delta s of the reaction.

792
00:39:49 --> 00:39:53
So that's to be sum of the
entropy of the products.

793
00:39:53 --> 00:40:02
So we're going to be talking
about 2 times entropy of water,

794
00:40:02 --> 00:40:07
and we'll add to that our other
product, which is molecular

795
00:40:07 --> 00:40:12
oxygen here, and we don't need
to put a number out here

796
00:40:12 --> 00:40:14
because it's just one mole.
 

797
00:40:14 --> 00:40:20
So that's of o 2, and we'll
subtract from that, the entropy

798
00:40:20 --> 00:40:24
of hydrogen peroxide here.
 

799
00:40:24 --> 00:40:27
And do we need to put
anything out here for this?

800
00:40:27 --> 00:40:28
STUDENT: 2.
 

801
00:40:28 --> 00:40:30
PROFESSOR: So, yeah, because
there's two moles there.

802
00:40:30 --> 00:40:35
So water plus oxygen
minus hydrogen peroxide.

803
00:40:35 --> 00:40:46
Let's hope I at least
got the values right.

804
00:40:46 --> 00:40:48
Yeah, and actually I think
this is right, I believe this

805
00:40:48 --> 00:40:49
is the entropy of water.
 

806
00:40:49 --> 00:40:52
So OK, so the actual numbers
are right here for what is

807
00:40:52 --> 00:40:56
written on the board and
hopefully in your notes.

808
00:40:56 --> 00:41:01
So what we find out is that the
entropy of this reaction here,

809
00:41:01 --> 00:41:03
the change in entropy, the
delta s, is 125 joules

810
00:41:03 --> 00:41:05
per k per mole.
 

811
00:41:05 --> 00:41:11
All right, so we can also --
well, first, we already thought

812
00:41:11 --> 00:41:13
about why delta s is positive.
 

813
00:41:13 --> 00:41:17
Delta s is positive because
we're going from a liquid to

814
00:41:17 --> 00:41:20
a liquid and a gas -- we're
increasing our disorder.

815
00:41:20 --> 00:41:24
So let's think about whether
this reaction is spontaneous

816
00:41:24 --> 00:41:27
or not, and to do that, we
actually need to go ahead and

817
00:41:27 --> 00:41:29
completely calculate delta g.
 

818
00:41:29 --> 00:41:33
So to do this, we can take our
values here, and delta g is

819
00:41:33 --> 00:41:37
equal to delta h
minus t delta s.

820
00:41:37 --> 00:41:42
So if we plug all of this in,
we have our delta h is negative

821
00:41:42 --> 00:41:46
196 kilojoules per mole, minus
our temperature, and I wanted

822
00:41:46 --> 00:41:48
you to write this again,
because I want to make sure you

823
00:41:48 --> 00:41:51
get in the habit of converting
our joules to kilojoules.

824
00:41:51 --> 00:41:52
It's 0 .
 

825
00:41:52 --> 00:41:57
125 kilojoules per kelvin mole.
 

826
00:41:57 --> 00:42:00
So we end up with a
delta g of negative 233

827
00:42:00 --> 00:42:02
kilojoules per mole.
 

828
00:42:02 --> 00:42:03
Is that spontaneous
or non-spontaneous?

829
00:42:03 --> 00:42:05
STUDENT: [INAUDIBLE]
 

830
00:42:05 --> 00:42:06
PROFESSOR: Great.
 

831
00:42:06 --> 00:42:09
So the reaction here
is spontaneous.

832
00:42:09 --> 00:42:11
So actually, this is probably a
reaction that you're familiar

833
00:42:11 --> 00:42:14
with because a lot of us do
have hydrogen peroxide just

834
00:42:14 --> 00:42:16
in our medicine cabinet.
 

835
00:42:16 --> 00:42:18
And if you go and look at the
expiration date on your

836
00:42:18 --> 00:42:22
hydrogen peroxide, you will see
that it does, in fact, expire.

837
00:42:22 --> 00:42:25
And you might think oh
no, what happens when it

838
00:42:25 --> 00:42:27
expires, when it goes bad?
 

839
00:42:27 --> 00:42:30
All that's happening is
it's turning into water.

840
00:42:30 --> 00:42:33
So I wouldn't drink it after
its expiration date, but

841
00:42:33 --> 00:42:36
probably there's more water
than hydrogen peroxide

842
00:42:36 --> 00:42:37
at that point.
 

843
00:42:37 --> 00:42:40
But it also actually brings out
a really important point about

844
00:42:40 --> 00:42:41
thermodynamics, which is that
thermodynamics, in fact, tells

845
00:42:41 --> 00:42:47
us if a reaction happening in
the forward direction is

846
00:42:47 --> 00:42:49
spontaneous or non-spontaneous.
 

847
00:42:49 --> 00:42:51
So, for example, we
see that yes, it is a

848
00:42:51 --> 00:42:53
spontaneous reaction.
 

849
00:42:53 --> 00:42:55
But you'll note that
thermodynamics tells us

850
00:42:55 --> 00:42:58
absolutely nothing about the
timeframe of that reaction.

851
00:42:58 --> 00:43:02
So some reactions that are
spontaneous take place in a

852
00:43:02 --> 00:43:04
matter of seconds or
microseconds or less.

853
00:43:04 --> 00:43:08
Other reactions, such as this,
to have an appreciable amount

854
00:43:08 --> 00:43:11
of that reaction go forward,
it actually takes years.

855
00:43:11 --> 00:43:13
You'll notice that it's not
just one day before that

856
00:43:13 --> 00:43:15
hydrogen peroxide expires.
 

857
00:43:15 --> 00:43:18
It's takes a really long time
to get a significant amount

858
00:43:18 --> 00:43:19
of this reaction to go.
 

859
00:43:19 --> 00:43:23
So just make sure, we will get
to talking about kinetics,

860
00:43:23 --> 00:43:25
which does tell us
about the timeframe of

861
00:43:25 --> 00:43:26
chemical reactions.
 

862
00:43:26 --> 00:43:28
But don't confuse
thermochemistry or

863
00:43:28 --> 00:43:30
thermodynamics with kinetics.
 

864
00:43:30 --> 00:43:34
It doesn't matter how huge our
delta g is in terms of being if

865
00:43:34 --> 00:43:37
it's negative two or if it's
negative two million, that

866
00:43:37 --> 00:43:40
still doesn't tell us how fast
the reaction's going to go.

867
00:43:40 --> 00:43:42
All right.
 

868
00:43:42 --> 00:43:45
So let's think about one more
example here in terms of

869
00:43:45 --> 00:43:47
thinking about entropy.
 

870
00:43:47 --> 00:43:49
So another example we talked
about was the melting of

871
00:43:49 --> 00:43:51
ice at room temperature.
 

872
00:43:51 --> 00:43:55
I think all of us agree that
this is spontaneous, but

873
00:43:55 --> 00:43:58
that the enthalpy change
is positive, it's an

874
00:43:58 --> 00:44:01
endothermic reaction.
 

875
00:44:01 --> 00:44:07
So we can think about
calculating the entropy here,

876
00:44:07 --> 00:44:11
and I'm happy to see that I
did, in fact, put the products

877
00:44:11 --> 00:44:15
h 2 o here as a liquid,
subtracting the reactant,

878
00:44:15 --> 00:44:19
which is h 2 o as a solid.
 

879
00:44:19 --> 00:44:22
If we go ahead and plug in
those values here, what we

880
00:44:22 --> 00:44:26
end up with is a delta s
of this reaction of 28 .

881
00:44:26 --> 00:44:28
59 joules.
 

882
00:44:28 --> 00:44:31
This makes sense because we're
increasing the disorder of the

883
00:44:31 --> 00:44:37
system, we're seeing a
positive delta s here.

884
00:44:37 --> 00:44:38
And I just answered
that question for you.

885
00:44:38 --> 00:44:42
We're going why is
delta s positive?

886
00:44:42 --> 00:44:44
We're going from a
solid to liquid, we're

887
00:44:44 --> 00:44:46
increasing disorder.
 

888
00:44:46 --> 00:44:50
So, we can calculate delta g
as well, and if we do this we

889
00:44:50 --> 00:44:51
plug in these numbers here.
 

890
00:44:51 --> 00:44:55
What we end up with is that
delta g equals negative 1 .

891
00:44:55 --> 00:44:57
57 kilojoules per mole.
 

892
00:44:57 --> 00:45:01
So it is spontaneous, delta
g is just barely negative.

893
00:45:01 --> 00:45:04
And what actually is the case,
which we'll talk about more on

894
00:45:04 --> 00:45:07
Monday and thinking more and
more about how temperature

895
00:45:07 --> 00:45:09
actually affects these
reactions, but what we know

896
00:45:09 --> 00:45:12
about melting of ice is that it
is, in fact, temperature

897
00:45:12 --> 00:45:13
dependent, right.
 

898
00:45:13 --> 00:45:16
Ice doesn't melt at just
any old temperature.

899
00:45:16 --> 00:45:18
So that's where temperature
actually comes into play here.

900
00:45:18 --> 00:45:21
All right.
 

901
00:45:21 --> 00:45:23
So we have a case --
spontaneous, even though

902
00:45:23 --> 00:45:24
delta h is positive.
 

903
00:45:24 --> 00:45:27
All right, so let's think also
about how we can go about

904
00:45:27 --> 00:45:30
calculating the free
energy of formation.

905
00:45:30 --> 00:45:34
So we know that we can use the
equation that we saw in terms

906
00:45:34 --> 00:45:36
of the free energy the
reaction, but what if we're

907
00:45:36 --> 00:45:39
actually thinking about the
free energy of the formation of

908
00:45:39 --> 00:45:41
a particular molecule?
 

909
00:45:41 --> 00:45:43
So that's actually completely
analogous to thinking about the

910
00:45:43 --> 00:45:48
enthalpy of formation, or it's
basically the standard Gibbs

911
00:45:48 --> 00:45:49
free energy of formation.
 

912
00:45:49 --> 00:45:52
So this is talking about
forming a mole of a compound

913
00:45:52 --> 00:45:55
from its most stable elements,
from its elements in their most

914
00:45:55 --> 00:45:58
stable states at pressure
of one bar and at

915
00:45:58 --> 00:46:01
room temperature.
 

916
00:46:01 --> 00:46:04
So we know that we can tabulate
it must like we tabulated delta

917
00:46:04 --> 00:46:08
h formation, but we can also
tabulate it from this reaction

918
00:46:08 --> 00:46:11
here, which it looks like I
already told you this reaction,

919
00:46:11 --> 00:46:14
which I did but I told you a
more general form, which was

920
00:46:14 --> 00:46:16
for the free energy
of a reaction.

921
00:46:16 --> 00:46:19
This is the free energy of a
reaction, which is the

922
00:46:19 --> 00:46:22
formation of a certain
compound, and this is the free

923
00:46:22 --> 00:46:25
energy of formation equals the
enthalpy of formation

924
00:46:25 --> 00:46:29
minus t delta s.
 

925
00:46:29 --> 00:46:33
So let's think about why it is
important to be thinking about

926
00:46:33 --> 00:46:38
these changes of free energy in
terms of forming a molecule,

927
00:46:38 --> 00:46:41
and it's important because the
delta g information is actually

928
00:46:41 --> 00:46:45
a measure of how stable or
unstable a molecule is

929
00:46:45 --> 00:46:48
relative to its element.
 

930
00:46:48 --> 00:46:51
So, for example, if something
is really stable relative to

931
00:46:51 --> 00:46:54
its elements, we can think
about whether delta g

932
00:46:54 --> 00:46:56
of formation should be
negative or positive.

933
00:46:56 --> 00:47:00
So let's have you do one last
clicker question here, and tell

934
00:47:00 --> 00:47:03
me what do you think, if delta
g of formation, of actually

935
00:47:03 --> 00:47:07
forming a molecule is going to
be negative, would you say that

936
00:47:07 --> 00:47:11
the compound is going to be
stable or unstable relative

937
00:47:11 --> 00:47:14
to the elements that that
compound is made up of?

938
00:47:14 --> 00:47:37
All right, let's take
10 seconds on this.

939
00:47:37 --> 00:47:38
OK, interesting.
 

940
00:47:38 --> 00:47:40
So this is going to be
another tie breaker for us.

941
00:47:40 --> 00:47:46
So let's go back to the
notes and think about this.

942
00:47:46 --> 00:47:50
So it turns out that it's going
to be stable, relative to its

943
00:47:50 --> 00:47:55
-- the compound is stable
relative to its elements.

944
00:47:55 --> 00:47:58
And the reason for this is if
you think of the reaction of

945
00:47:58 --> 00:48:03
the elements forming the
compound, if the delta g of

946
00:48:03 --> 00:48:06
that reaction is negative,
that means it's going to

947
00:48:06 --> 00:48:08
spontaneously form
the compound.

948
00:48:08 --> 00:48:11
It's going to go -- release
energy when it actually

949
00:48:11 --> 00:48:12
forms that compound.
 

950
00:48:12 --> 00:48:15
So that's another way of
telling us that the compound is

951
00:48:15 --> 00:48:17
more stable than its elements.
 

952
00:48:17 --> 00:48:22
So conversely, if we have delta
g of formation being greater

953
00:48:22 --> 00:48:25
than zero, so delta g of
formation is just delta g of

954
00:48:25 --> 00:48:29
the reaction where the compound
is formed, then what we find is

955
00:48:29 --> 00:48:32
that it's thermodynamically
unstable relative to its

956
00:48:32 --> 00:48:35
elements, because instead of
releasing energy in that

957
00:48:35 --> 00:48:38
reaction, you actually have to
put energy into that reaction

958
00:48:38 --> 00:48:40
in order to make it happen.
 

959
00:48:40 --> 00:48:43
So, delta g of formation can
give us an indication of

960
00:48:43 --> 00:48:46
whether we have a stable
or an unstable compound.

961
00:48:46 --> 00:48:46