Gibbs free energy and spontaneity

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Kind: captions
Language: en

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We've learned over the last
several videos that if we have

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a system undergoing constant
pressure, or it's in an

00:00:06.120 --> 00:00:08.490
environment with constant
pressure, that its change in

00:00:08.490 --> 00:00:11.870
enthalpy is equal to the heat
added to the system.

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And I'll write this little
p here, because that's at

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constant pressure.

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So if you have a reaction, let's
say A plus B yields C

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and our change in enthalpy-- so
our enthalpy in this state

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minus the change in the enthalpy
in that state-- so

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let's say our change in enthalpy
is less than 0, we

00:00:28.860 --> 00:00:30.520
know that this is exothermic.

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Why is that?

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And once again, I'm assuming
constant pressure.

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How do we know this is
exothermic, that we're

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releasing energy?

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Because change in enthalpy,
when we're dealing with a

00:00:38.950 --> 00:00:42.360
constant pressure system, is
heat added to the system.

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If the heat added to the system
is negative, we must be

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releasing heat.

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So we're releasing
heat or energy.

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So plus energy.

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And we learned in the last
video, I think it was either

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the last video or a couple of
videos ago, we call this an

00:00:57.860 --> 00:00:59.110
exothermic reaction.

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And then if you have a reaction
that needs energy--

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so let's say you have A plus B
plus some energy yields C,

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then what does that mean?

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Well, that means that the
system absorbed energy.

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The amount of energy
you absorb is

00:01:24.560 --> 00:01:26.040
your change in enthalpy.

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So your delta H is going
to be positive.

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Your change in enthalpy
is positive.

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You've absorbed energy
into the system.

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And we call these endothermic
reactions.

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You're absorbing heat.

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Now, if we wanted to figure
out whether reaction just

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happens by itself-- whether it's
spontaneous-- it seems

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like this change in enthalpy
is a good candidate.

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Obviously, if I'm releasing
energy, I didn't need any

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energy for this reaction to
happen, so maybe this reaction

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is spontaneous.

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And likewise, since I'd have to
somehow add energy into the

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system, my gut tells me that
this maybe isn't spontaneous.

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But there's a little part of me
that says, well, you know,

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what if the particles are
running around really fast,

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and they have a lot of kinetic
energy that can be used to

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kind of ram these particles
together, maybe all of a

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sudden this would
be spontaneous.

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So maybe enthalpy by itself
wouldn't completely describe

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what's going to happen.

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So in order to get a little
intuition, and to maybe build

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up our sense of whether
a reaction happens

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spontaneously, let's think about
the ingredients that

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probably matter.

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We already know that delta
H probably matters.

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If we release energy-- you know,
delta H less than 0,

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that tends to make me think
it might be spontaneous.

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But what if our delta S, what
if our entropy goes down?

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What if things become
more ordered?

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We've already learned from the
second law of thermodynamics,

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that that doesn't tend
to be the case.

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And just from personal
experience, we know that

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things on their own just don't
kind of go to the macro state

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that has fewer micro
states, you know.

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An egg doesn't just put itself
together, and bounce, and kind

00:03:10.710 --> 00:03:13.080
of jump off the floor on its
own, although there's some

00:03:13.080 --> 00:03:14.330
probability it would happen.

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So it seems like entropy
matters somewhat.

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And then there's the idea
of temperature.

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Because I already talked about,
when I talked about

00:03:22.710 --> 00:03:23.360
energy here.

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I was like, well, you know, even
if this requires energy,

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maybe if the temperature is
high enough, maybe I could

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actually ram these particles
together in some way, and kind

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of create energy to go here.

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So let's think about--
so let's see.

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let's think about the
ingredients, and let's think

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about what the reactions would
look like depending on

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different combinations
of the ingredients.

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So the ingredients I'm going to
deal with-- delta H seems

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to definitely matter,
whether or not we

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absorb energy or not.

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We have delta S, our
change in entropy.

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Does the system take on more
states or fewer states?

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Does it become more
or less ordered?

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And then there's temperature,
which is

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average kinetic energy.

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So let's just think about a
whole bunch of situations.

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So let's think of
the first case.

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Let's think of the situation
where our delta H is less than

00:04:13.810 --> 00:04:21.200
0, and our entropy is
greater than 0.

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I mean, my gut already
tells me that

00:04:22.630 --> 00:04:23.490
this is going to happen.

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This is a situation where
we're going to be more

00:04:27.400 --> 00:04:29.380
entropic after the reaction.

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So one way of looking
at entropy,

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you could more states.

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Maybe we have more particles.

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We've seen that entropy is
related to the number of

00:04:35.590 --> 00:04:40.770
particles we have. So this
could be a reaction where

00:04:40.770 --> 00:04:46.440
let's say we have this--
see, we want

00:04:46.440 --> 00:04:47.250
to have more particles.

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So let's say I have that guy.

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And say he's got one guy like
that there, and then I have

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another guy like this, and
let's say he's got a

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molecule like this.

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Let's say that a more-- well,
I won't say stable or not.

00:05:04.390 --> 00:05:11.240
But let's say that when these
guys bump into each other, you

00:05:11.240 --> 00:05:14.730
end up with this.

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And I'm making things
up on the fly.

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Maybe one of these molecules
bonds with this molecule, so

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you have one of the
dark blues.

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I'll draw all the dark blues.

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Bonds with this light blue
molecule, one of the dark

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blues bonds with the
magenta molecule.

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And maybe that brown
molecule just gets

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knocked off all by himself.

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So we went from having
two molecules to

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having three molecules.

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We have more disorder,
more entropy.

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This can obviously take
on more states.

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And I'm telling you that
delta H is less than 0.

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So by doing this, these guys,
their electrons are in a lower

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potential, or they're in a more
stable configuration.

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So when the electrons go from
their higher potential

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configurations over here, and
they become more stable, they

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release energy.

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So you have plus-- and then I
just know that, because I said

00:06:04.350 --> 00:06:06.590
from the beginning that
my change in enthalpy

00:06:06.590 --> 00:06:07.470
is less than 0.

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So plus some energy.

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So it seems pretty obvious to me
that this reaction is going

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to be spontaneous in this
rightward direction.

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Because there's no reason why--
first of all, it's much

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easier for two particles to
bump into each other just

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right to go in that direction
than it is for three

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particles-- if you just think of
it from a probability point

00:06:29.520 --> 00:06:32.940
of view-- for three particles to
get together just right and

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go in that direction.

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And even more, these guys
are more stable.

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Their electrons are in a
lower potential state.

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So there's no even kind of
enthalpic reason for them to

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move in this direction, or you
know, kind of a energy reason

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for them to move in
this direction.

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So this, to me, I kind of have
the intuition that regardless

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of what the temperature is,
we're going to favor this

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forward reaction.

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So I would say that this is
probably spontaneous.

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Now, what happens-- let's do
something that's maybe a

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little less intuitive.

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What happens if my delta
H is less than 0?

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But let's say I lose entropy.

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And this seems, you know, with
second law of thermodynamics,

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if the entropy of the
universe goes up.

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I'm just talking about
my system.

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But let's say I lose entropy.

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So that would be a situation
where I go from, let's say,

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two particles.

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Let's say I got that particle,
and then I have this particle.

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And then, if they bump into each
other just right, their

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electrons are going to be more
stable, and maybe they form

00:07:39.230 --> 00:07:40.480
this character.

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And when they do that, the
electrons can enter into lower

00:07:45.420 --> 00:07:48.350
potential states, and when they
do, the electrons release

00:07:48.350 --> 00:07:52.020
energy, so you have some
plus energy here.

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And we know that, because this
was the change in enthalpy was

00:07:57.390 --> 00:07:58.230
less than zero.

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We have lower energy in this
state than that one, and the

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difference is released
right here.

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Now will this reaction happen?

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Well, it seems like-- let's
introduce our temperature.

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What's going to happen
at low temperatures?

00:08:12.160 --> 00:08:16.020
At low temperatures, these guys
have a very low average

00:08:16.020 --> 00:08:16.770
kinetic energy.

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They're just drifting
around very slowly.

00:08:20.510 --> 00:08:22.740
And as they drift around
very slowly--

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And remember, when I talk about

00:08:24.670 --> 00:08:29.200
spontaneity-- I wrote sponteous.

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This is spontaneous.

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Sponteous should be another
thermodynamic.

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It's a fun word.

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When I talk about spontaneity,
I'm just talking about whether

00:08:49.440 --> 00:08:51.320
the reaction is just going
to happen on its own.

00:08:51.320 --> 00:08:53.180
I'm not talking about
how fast, or

00:08:53.180 --> 00:08:54.370
the rate of the reaction.

00:08:54.370 --> 00:08:55.550
That's a key thing to know.

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You know, is this
going to happen.

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I don't care if it takes, you
know, a million years for the

00:08:59.720 --> 00:09:00.203
thing to happen.

00:09:00.203 --> 00:09:02.890
I just want to know, is it going
to happen on its own?

00:09:02.890 --> 00:09:06.920
So if the temperature is slow,
these guys might be really

00:09:06.920 --> 00:09:10.670
creeping along, barely bumping
into each other.

00:09:10.670 --> 00:09:12.790
But they will eventually
bump into each other.

00:09:12.790 --> 00:09:16.640
And when they do, they're just
drifting past each other.

00:09:16.640 --> 00:09:20.310
And when they drift past each
other, they will configure

00:09:20.310 --> 00:09:22.510
themselves in a way-- things
want to go to a lower

00:09:22.510 --> 00:09:23.270
potential state.

00:09:23.270 --> 00:09:25.880
I'm just trying to give you kind
of a hand-wavey, rough

00:09:25.880 --> 00:09:27.090
intuition of things.

00:09:27.090 --> 00:09:29.590
But because this will release
energy, and it will go to a

00:09:29.590 --> 00:09:31.690
lower potential state, the
electrons kind of configure

00:09:31.690 --> 00:09:33.820
themselves when they get near
each other, and enter into

00:09:33.820 --> 00:09:34.510
this state.

00:09:34.510 --> 00:09:35.990
And they release energy.

00:09:35.990 --> 00:09:38.510
And once the energy is gone, and
maybe it's in the form of

00:09:38.510 --> 00:09:41.290
heat or whatever it is, it's
hard to kind of get it back

00:09:41.290 --> 00:09:42.350
and go on in other direction.

00:09:42.350 --> 00:09:45.620
So it seems like this would
be spontaneous if the

00:09:45.620 --> 00:09:47.450
temperature is low.

00:09:47.450 --> 00:09:48.210
So let me write that.

00:09:48.210 --> 00:09:57.250
Spontaneous if the
temperate is low.

00:09:57.250 --> 00:10:00.750
Now what happens if the
temperature is high?

00:10:00.750 --> 00:10:02.340
Remember, these aren't the
only particles here.

00:10:02.340 --> 00:10:03.000
We have more.

00:10:03.000 --> 00:10:05.090
You know, I'll have another
guy like that, and

00:10:05.090 --> 00:10:07.430
another guy like that.

00:10:07.430 --> 00:10:09.020
And then this, on this
side, I'll have,

00:10:09.020 --> 00:10:10.330
you know, more particles.

00:10:10.330 --> 00:10:12.040
There's obviously not
just one particle.

00:10:12.040 --> 00:10:14.500
Then all of these macro
variables really make no

00:10:14.500 --> 00:10:16.360
sense, if we're just talking
about particular molecules.

00:10:16.360 --> 00:10:17.450
We're talking about
entire systems.

00:10:17.450 --> 00:10:20.440
But what happens here if
the temperature of

00:10:20.440 --> 00:10:22.150
our system is high?

00:10:22.150 --> 00:10:26.280
So let's think of a situation
where the temperature is high.

00:10:26.280 --> 00:10:29.520
Now all of a sudden-- so on the
side, people are going to

00:10:29.520 --> 00:10:32.550
be knocking into each other
super fast. You know, if this

00:10:32.550 --> 00:10:36.610
guy bumps into this guy super
fast, you can almost view it

00:10:36.610 --> 00:10:41.540
as a car collision.

00:10:41.540 --> 00:10:43.580
Well, even better, this could
be car collision.

00:10:43.580 --> 00:10:46.310
If these were each individual
cars, and the atoms were the

00:10:46.310 --> 00:10:48.870
components of the cars, if
they're like smashing into

00:10:48.870 --> 00:10:51.750
each other, even though they
want to be attached to each

00:10:51.750 --> 00:10:54.220
other, they have screws and
whatever else that are holding

00:10:54.220 --> 00:10:57.440
it together-- if two cars run
into each other fast enough,

00:10:57.440 --> 00:11:00.130
all that screws and the glue and
the welding won't matter.

00:11:00.130 --> 00:11:01.720
They're just going
to blow apart.

00:11:01.720 --> 00:11:04.350
So high kinetic energy--
let me draw that.

00:11:04.350 --> 00:11:07.910
So if they have high kinetic
energy, my gut tells me that

00:11:07.910 --> 00:11:09.920
on the side of the reaction,
these guys are just going to

00:11:09.920 --> 00:11:11.830
blow each other apart
to this side.

00:11:11.830 --> 00:11:14.800
And these guys, since these guys
also have high kinetic

00:11:14.800 --> 00:11:17.340
energy, they're going to be
moving so fast past each

00:11:17.340 --> 00:11:19.640
other, and they're going to
ricochet off of each other so

00:11:19.640 --> 00:11:25.870
fast, that the counteracting
force, or the contracting

00:11:25.870 --> 00:11:28.380
inclination for their electrons
to get more stably

00:11:28.380 --> 00:11:30.170
configured, won't matter.

00:11:30.170 --> 00:11:34.720
It's like, imagine trying to
attach a tire to something

00:11:34.720 --> 00:11:37.860
while you're running
past the car.

00:11:37.860 --> 00:11:40.380
You kind of have to do it-- even
though that's a more--

00:11:40.380 --> 00:11:42.320
well, maybe the analogy
is getting weak, here.

00:11:42.320 --> 00:11:45.290
But I think you get the idea
that if the temperature's

00:11:45.290 --> 00:11:48.590
really high, it seems less
likely that these guys are

00:11:48.590 --> 00:11:51.730
going to kind of drift near each
other just right to be

00:11:51.730 --> 00:11:53.740
able to attach to each other,
and their electrons to get

00:11:53.740 --> 00:11:57.020
more stable, and to do this
whole exothermic thing.

00:11:57.020 --> 00:12:00.230
So my sense is that if the
temperature is high enough-- I

00:12:00.230 --> 00:12:01.640
mean, you know, maybe say, oh,
that's not high enough.

00:12:01.640 --> 00:12:03.880
But what if it's super
high temperatures?

00:12:03.880 --> 00:12:05.750
If it's super high temperatures,
then maybe even

00:12:05.750 --> 00:12:06.860
this guy will bump into that.

00:12:06.860 --> 00:12:10.590
Instead of forming that, he'll
knock this other blue guy off,

00:12:10.590 --> 00:12:12.840
and then he'll be over here.

00:12:12.840 --> 00:12:14.820
I should do the blue
guy in blue.

00:12:14.820 --> 00:12:17.010
And maybe he'll knock this
guy into his constituent

00:12:17.010 --> 00:12:20.200
particles, if there's enough
kinetic energy.

00:12:20.200 --> 00:12:24.330
So here I get the idea that
it's not spontaneous.

00:12:24.330 --> 00:12:27.625
And even more, the reverse
reaction, if the temperature

00:12:27.625 --> 00:12:29.870
is high enough, is probably
going to be spontaneous.

00:12:29.870 --> 00:12:31.950
If the temperature is high
enough, these guys are going

00:12:31.950 --> 00:12:34.130
to react, are going to bump
into each other, and the

00:12:34.130 --> 00:12:36.380
reaction is going
to go that way.

00:12:36.380 --> 00:12:39.890
So temperature is high, you go
that way, temperature is low,

00:12:39.890 --> 00:12:41.300
you go that way.

00:12:41.300 --> 00:12:43.580
So let's see if we can put
everything together that we've

00:12:43.580 --> 00:12:47.560
seen so far and kind of come up
with a gut feeling of what

00:12:47.560 --> 00:12:50.890
a formula for spontaneity
would look like.

00:12:50.890 --> 00:12:54.170
So we could start
with enthalpy.

00:12:54.170 --> 00:12:58.000
So we already know that look, if
this is less than 0, we're

00:12:58.000 --> 00:13:00.256
probably dealing with something
that's spontaneous.

00:13:00.256 --> 00:13:03.840
Now let's say I want a whole
expression, where if the whole

00:13:03.840 --> 00:13:07.350
expression is less than 0, it
tells me that it's going to be

00:13:07.350 --> 00:13:08.600
spontaneous.

00:13:11.300 --> 00:13:18.760
So we know that positive entropy
is something good for

00:13:18.760 --> 00:13:19.510
spontaneity.

00:13:19.510 --> 00:13:21.770
We saw that in every
situation here.

00:13:21.770 --> 00:13:24.790
That if we have more states,
it's always a good thing.

00:13:24.790 --> 00:13:29.100
It's more likely to make
something spontaneous.

00:13:29.100 --> 00:13:32.670
Now, we want our whole
expression to be negative if

00:13:32.670 --> 00:13:34.670
it's spontaneous, right?

00:13:34.670 --> 00:13:38.420
So positive entropy should make
my whole expression more

00:13:38.420 --> 00:13:41.350
negative, so maybe we should
subtract entropy.

00:13:45.590 --> 00:13:45.800
Right?

00:13:45.800 --> 00:13:48.170
If this is positive, then
my whole expression

00:13:48.170 --> 00:13:49.550
will be more negative.

00:13:49.550 --> 00:13:51.590
Which tells me, hey, this
is spontaneous.

00:13:51.590 --> 00:13:54.240
So if this is negative, we're
releasing energy.

00:13:54.240 --> 00:13:56.860
And then if this is positive,
we're getting more disordered,

00:13:56.860 --> 00:13:58.650
so this whole thing
will be negative.

00:13:58.650 --> 00:14:00.360
So that seems good.

00:14:00.360 --> 00:14:04.530
Now what if entropy
is negative?

00:14:04.530 --> 00:14:07.340
If entropy is negative, this
also kind of speaks to the

00:14:07.340 --> 00:14:10.720
idea that if entropy is
negative, it kind of makes the

00:14:10.720 --> 00:14:13.360
reaction a little less
spontaneous.

00:14:13.360 --> 00:14:13.610
Right?

00:14:13.610 --> 00:14:16.100
In this situation, entropy
was negative.

00:14:16.100 --> 00:14:18.130
We went from more
disorder to less

00:14:18.130 --> 00:14:20.560
disorder, or fewer particles.

00:14:20.560 --> 00:14:22.530
And what did we say?

00:14:22.530 --> 00:14:29.070
When temperature is high,
entropy matters a lot.

00:14:29.070 --> 00:14:32.680
When temperature is high, this
less entropic state, they ram

00:14:32.680 --> 00:14:34.940
into each other, and they'll
become more entropic.

00:14:34.940 --> 00:14:38.560
When temperature is low, maybe
they'll drift close to each

00:14:38.560 --> 00:14:43.030
other, and then the enthalpy
part of the equation will

00:14:43.030 --> 00:14:44.300
matter more.

00:14:44.300 --> 00:14:45.550
So let's see if we
can weight that.

00:14:45.550 --> 00:14:49.110
So when temperature is high,
entropy matters.

00:14:49.110 --> 00:14:51.630
When temperature is low,
entropy doesn't matter.

00:14:51.630 --> 00:14:54.880
So what if we just scaled
entropy by temperature?

00:14:54.880 --> 00:14:59.000
What if I just took a
temperature variable here?

00:14:59.000 --> 00:15:02.850
Now, my claim, or my intuition,
based on everything

00:15:02.850 --> 00:15:05.800
we've experimented so far, is
that if this expression is

00:15:05.800 --> 00:15:08.260
less than 0, we should
be dealing with

00:15:08.260 --> 00:15:09.510
a spontaneous reaction.

00:15:12.650 --> 00:15:15.680
And let's see if it gels with
everything we say here.

00:15:15.680 --> 00:15:19.050
If the temperature is high--
so this reaction right here

00:15:19.050 --> 00:15:21.630
was exothermic, in the
rightwards direction.

00:15:21.630 --> 00:15:23.750
When we go to the right, from
more of these molecules to

00:15:23.750 --> 00:15:26.490
these fewer ones, I told
you it's exothermic.

00:15:26.490 --> 00:15:29.900
Now, at low temperatures, my gut
told me, hey, this should

00:15:29.900 --> 00:15:30.510
be spontaneous.

00:15:30.510 --> 00:15:32.230
These guys are going to drift
close to each other, and get

00:15:32.230 --> 00:15:34.520
into this more stable
configuration.

00:15:34.520 --> 00:15:35.370
And that makes sense.

00:15:35.370 --> 00:15:39.380
At low temperatures, this term
isn't going to matter much.

00:15:39.380 --> 00:15:40.300
You can imagine the extreme.

00:15:40.300 --> 00:15:42.370
At absolute 0, this term
is going to disappear.

00:15:42.370 --> 00:15:44.190
You can't quite reach
there, but it would

00:15:44.190 --> 00:15:45.390
become less and less.

00:15:45.390 --> 00:15:47.330
And this term dominates.

00:15:47.330 --> 00:15:51.020
Now, at high temperatures, all
of a sudden, this term is

00:15:51.020 --> 00:15:52.040
going to dominate.

00:15:52.040 --> 00:15:56.890
And if our delta S is less than
0, then this whole term

00:15:56.890 --> 00:15:59.190
is going to dominate and
become positive.

00:15:59.190 --> 00:15:59.500
Right?

00:15:59.500 --> 00:16:02.320
And even if this is negative,
we're subtracting.

00:16:02.320 --> 00:16:03.510
So our delta S is negative.

00:16:03.510 --> 00:16:04.590
We put a negative here.

00:16:04.590 --> 00:16:06.540
So this is going to
be a positive.

00:16:06.540 --> 00:16:08.530
So this positive, if the
temperature is high enough--

00:16:08.530 --> 00:16:10.260
and remember, we're dealing with
Kelvin, so temperature

00:16:10.260 --> 00:16:11.610
can only be positive.

00:16:11.610 --> 00:16:14.500
If this is positive enough,
it will overwhelm

00:16:14.500 --> 00:16:15.810
any negative enthalpy.

00:16:15.810 --> 00:16:17.790
And so it won't be spontaneous
anymore.

00:16:17.790 --> 00:16:21.720
And so if the temperature is
high enough, this direction

00:16:21.720 --> 00:16:22.940
won't be spontaneous.

00:16:22.940 --> 00:16:24.980
And this equation
tells us this.

00:16:24.980 --> 00:16:32.960
And then if we go to the
negative enthalpy, positive

00:16:32.960 --> 00:16:36.380
entropy, so we're releasing
energy, so this is negative,

00:16:36.380 --> 00:16:38.550
and our entropy is increasing--
our entropy,

00:16:38.550 --> 00:16:40.350
we're getting more disordered--
then this becomes

00:16:40.350 --> 00:16:41.500
a negative as well.

00:16:41.500 --> 00:16:43.330
So our thing is definitely
going to be negative.

00:16:43.330 --> 00:16:46.300
And we already had the sense
that look, if this is negative

00:16:46.300 --> 00:16:48.950
and this is positive, we're
getting more entropic and

00:16:48.950 --> 00:16:51.060
we're releasing energy, that
should definitely be

00:16:51.060 --> 00:16:51.610
spontaneous.

00:16:51.610 --> 00:16:53.840
And this equation also
speaks to that.

00:16:53.840 --> 00:16:56.890
So so far, I feel pretty good
about this equation.

00:16:56.890 --> 00:16:59.520
And as you can imagine,
I didn't think of

00:16:59.520 --> 00:17:00.730
this out of the blue.

00:17:00.730 --> 00:17:03.850
This actually is the equation
that predicts spontaneity.

00:17:03.850 --> 00:17:06.030
And I'm going to show it to
you in a slightly more

00:17:06.030 --> 00:17:08.709
rigorous way in the future,
maybe going back to some of

00:17:08.709 --> 00:17:12.710
our fundamental formulas for
entropy and things like that.

00:17:12.710 --> 00:17:15.040
But this is the formula for
whether something is

00:17:15.040 --> 00:17:15.589
spontaneous.

00:17:15.589 --> 00:17:16.940
And what I wanted to do in this
video is just give you an

00:17:16.940 --> 00:17:19.440
intuition why this formula
kind of makes sense.

00:17:19.440 --> 00:17:23.630
And this quantity right here
is called the delta G, or

00:17:23.630 --> 00:17:26.180
change in Gibbs free energy.

00:17:26.180 --> 00:17:28.850
And this is what does predict
whether a reaction is

00:17:28.850 --> 00:17:29.390
spontaneous.

00:17:29.390 --> 00:17:31.250
So in the next video, I'll
actually apply this formula a

00:17:31.250 --> 00:17:32.090
couple of times.

00:17:32.090 --> 00:17:34.800
And then a few videos after
that, we'll do a little bit

00:17:34.800 --> 00:17:37.640
more of how you can actually
get this from some of our

00:17:37.640 --> 00:17:40.120
basic thermodynamic
principles.


Hess's law and reaction enthalpy change

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:03.520
Now that we know a little bit
about the formation and

00:00:03.520 --> 00:00:06.190
enthalpy change, and what
enthalpy is, we can talk a

00:00:06.190 --> 00:00:07.970
little bit about Hess's Law.

00:00:11.520 --> 00:00:16.020
And what this tells us is that
the energy change of a process

00:00:16.020 --> 00:00:20.690
is independent of how we get
from one state to another.

00:00:20.690 --> 00:00:24.340
And really, that's a by-product
of the fact that

00:00:24.340 --> 00:00:26.430
energy is a state variable.

00:00:26.430 --> 00:00:28.740
Whether we're talking about
enthalpy or internal energy,

00:00:28.740 --> 00:00:29.470
they're state variables.

00:00:29.470 --> 00:00:32.180
And we've talked multiple times
that it's independent of

00:00:32.180 --> 00:00:35.250
how many steps it takes to
get there, or what path

00:00:35.250 --> 00:00:36.400
you happen to take.

00:00:36.400 --> 00:00:38.730
But how is that useful to us
when we're dealing with

00:00:38.730 --> 00:00:39.495
everyday reactions?

00:00:39.495 --> 00:00:46.990
So let me just make up some
reaction where I have A plus B

00:00:46.990 --> 00:00:49.640
yields, oh, I don't know,
let's just say this

00:00:49.640 --> 00:00:52.290
yields C plus D.

00:00:52.290 --> 00:00:55.350
And I wanted to figure out
what was the change in

00:00:55.350 --> 00:00:57.230
enthalpy of this reaction?

00:00:57.230 --> 00:01:00.400
Or essentially, how much heat
is absorbed or released by

00:01:00.400 --> 00:01:00.950
this reaction.

00:01:00.950 --> 00:01:01.960
I don't know what it is.

00:01:01.960 --> 00:01:03.000
I haven't measured it.

00:01:03.000 --> 00:01:06.290
And all I have are the
heats of formation.

00:01:06.290 --> 00:01:09.630
So all I know is, how do you
go-- so I know the heat of

00:01:09.630 --> 00:01:11.490
formation of A-- so let me call

00:01:11.490 --> 00:01:13.620
that the heat of formation.

00:01:13.620 --> 00:01:14.985
Remember, H isn't for heat.

00:01:14.985 --> 00:01:17.180
Even though we kept calling
it heat of formation, it's

00:01:17.180 --> 00:01:19.300
actually the change
in enthalpy.

00:01:19.300 --> 00:01:21.670
And it's the standard
change in enthalpy.

00:01:21.670 --> 00:01:23.630
But the change in enthalpy
we know as heat.

00:01:23.630 --> 00:01:26.155
So it's heat, change in enthalpy
of formation was the

00:01:26.155 --> 00:01:27.280
same thing as heat
of formation.

00:01:27.280 --> 00:01:28.620
This little naught sign
tells us it's a

00:01:28.620 --> 00:01:30.130
standard heat the formation.

00:01:30.130 --> 00:01:32.500
We can look up that in a table,
and let's say that

00:01:32.500 --> 00:01:33.670
that's some number.

00:01:33.670 --> 00:01:38.190
And then we have our heat of
formation of B-- delta heat of

00:01:38.190 --> 00:01:41.460
formation, let me
call it, of B.

00:01:41.460 --> 00:01:43.130
This Is heat of formation
of A, and it's a

00:01:43.130 --> 00:01:44.750
standard heat of formation.

00:01:44.750 --> 00:01:47.300
And we could look up in a table
that heat of formation

00:01:47.300 --> 00:01:52.700
of C, which is change
in enthalpy.

00:01:52.700 --> 00:01:56.870
And then the heat of
formation for D.

00:01:56.870 --> 00:01:59.680
So all of these things we can
look up in a table, right?

00:01:59.680 --> 00:02:01.330
And we'll do that in a second.

00:02:01.330 --> 00:02:03.730
Now, what has Hess's law tells
us is that the change in

00:02:03.730 --> 00:02:07.340
energy, the change in-- and
enthalpy is what we're

00:02:07.340 --> 00:02:09.820
measuring here-- the change in
enthalpy here is independent

00:02:09.820 --> 00:02:10.509
of what we're doing.

00:02:10.509 --> 00:02:12.380
So instead of saying this
reaction, we could say hey,

00:02:12.380 --> 00:02:15.365
let's go from this reaction, and
go back-- let me do it in

00:02:15.365 --> 00:02:17.530
a different color.

00:02:17.530 --> 00:02:20.430
Let's go back to our constituent
products, so kind

00:02:20.430 --> 00:02:22.120
of the elemental
form of these.

00:02:22.120 --> 00:02:25.080
So you know, if this was like
carbon dioxide, you'd be going

00:02:25.080 --> 00:02:28.100
back to the carbon and
the oxygen molecules.

00:02:28.100 --> 00:02:30.870
So you'd go back to the
elemental form.

00:02:30.870 --> 00:02:34.000
And how much energy, or what's
the change in enthalpy, as you

00:02:34.000 --> 00:02:36.040
go back to the elemental form?

00:02:36.040 --> 00:02:38.480
The heat of formation is what
you get from the element of

00:02:38.480 --> 00:02:41.600
form to A, or the elemental
form to B.

00:02:41.600 --> 00:02:44.580
So to get A and B back to the
elemental form is going to be

00:02:44.580 --> 00:02:45.920
the minus of those.

00:02:45.920 --> 00:02:46.920
You're going to take
the reaction

00:02:46.920 --> 00:02:49.470
in the other direction.

00:02:49.470 --> 00:02:54.740
So this change is going to take
minus delta-- the heat

00:02:54.740 --> 00:03:00.040
of, I guess, of forming A, or it
could be the minus the heat

00:03:00.040 --> 00:03:02.670
of deconstructing A, you
can almost view it.

00:03:02.670 --> 00:03:08.790
And it would also be minus
the same thing for B.

00:03:08.790 --> 00:03:10.730
And then, this is just
the elemental form.

00:03:10.730 --> 00:03:12.020
And now we can go from
the elemental

00:03:12.020 --> 00:03:14.530
form back to the products.

00:03:14.530 --> 00:03:16.010
Because we have the
same atoms here.

00:03:16.010 --> 00:03:18.700
They're just rearranging
themselves into two different

00:03:18.700 --> 00:03:20.300
sets of molecules.

00:03:20.300 --> 00:03:23.880
So now we can go back from the
elemental form and go up here.

00:03:23.880 --> 00:03:25.090
And we know what those are.

00:03:25.090 --> 00:03:27.550
We know how much energy it takes
to go from the elemental

00:03:27.550 --> 00:03:29.030
form to C and D.

00:03:29.030 --> 00:03:30.800
That's their heats
of formation.

00:03:30.800 --> 00:03:39.090
So Hess's Law tells us that
delta H of this reaction, the

00:03:39.090 --> 00:03:41.840
change in enthalpy of this
reaction, is essentially going

00:03:41.840 --> 00:03:45.780
to be the sum of what it takes
to decompose these guys, which

00:03:45.780 --> 00:03:48.930
is the minus heat of formations
of these guys, plus

00:03:48.930 --> 00:03:51.750
what it takes to reform
these guys over here.

00:03:51.750 --> 00:03:58.270
So we can just write it as delta
H of formation for C

00:03:58.270 --> 00:04:00.990
plus delta H of formation
for D.

00:04:00.990 --> 00:04:04.400
So the heat of formations for
these guys minus these guys.

00:04:04.400 --> 00:04:07.290
This is what it took you to
get to the elemental form.

00:04:07.290 --> 00:04:12.630
So minus delta heat of formation
of A, minus delta

00:04:12.630 --> 00:04:14.980
heat of formation of B.

00:04:14.980 --> 00:04:16.720
And then you'll have the
heat of the reaction.

00:04:16.720 --> 00:04:19.790
And if it's negative, we would
have released energy.

00:04:19.790 --> 00:04:22.710
And if this number is positive,
then that means that

00:04:22.710 --> 00:04:25.310
there's more energy here than on
this side, so we would have

00:04:25.310 --> 00:04:27.430
to absorb energy for this
reaction to happen, and it

00:04:27.430 --> 00:04:28.760
would be endothermic.

00:04:28.760 --> 00:04:31.540
So this is all abstract and
everything, and I've told you

00:04:31.540 --> 00:04:32.640
about Hess's Law.

00:04:32.640 --> 00:04:36.000
Let's actually apply it
to some problems.

00:04:36.000 --> 00:04:39.740
So let's say I have this
reaction right here, where I

00:04:39.740 --> 00:04:41.830
start with ammonia.

00:04:41.830 --> 00:04:44.010
And it's ammonia gas.

00:04:44.010 --> 00:04:50.050
And I'm going to react that with
molecular oxygen to yield

00:04:50.050 --> 00:04:53.500
some nitrogen monoxide, 4 moles
of it, and some water.

00:04:53.500 --> 00:04:56.790
So what's the heat of this
reaction right here?

00:04:56.790 --> 00:04:59.360
So what we do, is we just look
up the heats of formation of

00:04:59.360 --> 00:04:59.990
each of these.

00:04:59.990 --> 00:05:01.440
So let's just look them up.

00:05:01.440 --> 00:05:03.160
Let's start with the ammonia.

00:05:03.160 --> 00:05:05.420
What's the heat of formation
of ammonia?

00:05:05.420 --> 00:05:09.010
And it's always given in
kilojoules per mole, so

00:05:09.010 --> 00:05:11.800
they'll say to form one
mole of ammonia.

00:05:11.800 --> 00:05:14.640
So to form 1 mole of ammonia--
let's look up here.

00:05:14.640 --> 00:05:17.540
This is all cut and paste
from Wikipedia.

00:05:17.540 --> 00:05:20.430
And am I starting in the gaseous
or the aqueous state?

00:05:20.430 --> 00:05:22.540
Well, I think I just-- see, I'm

00:05:22.540 --> 00:05:23.650
starting the gaseous state.

00:05:23.650 --> 00:05:25.590
I've added that G there.

00:05:25.590 --> 00:05:29.510
So ammonia in the gaseous state
has a heat of formation

00:05:29.510 --> 00:05:34.660
of minus 45.9 per joule.

00:05:34.660 --> 00:05:37.190
So what is that going--
so minus 45.9

00:05:37.190 --> 00:05:39.420
kilojoules per mole.

00:05:43.480 --> 00:05:45.170
That's just for one mole of

00:05:45.170 --> 00:05:46.770
ammonia, the heat of formation.

00:05:46.770 --> 00:05:48.240
It's in kilojoules.

00:05:48.240 --> 00:05:51.210
I'll just look them
all up right now.

00:05:51.210 --> 00:05:52.725
Now what's the heat of
formation of oxygen?

00:05:52.725 --> 00:05:55.760
And I'm not going to look it up
right now, because oxygen

00:05:55.760 --> 00:05:57.680
is in its elemental form.

00:05:57.680 --> 00:05:59.970
So if you see something in the
form that it just always

00:05:59.970 --> 00:06:02.700
takes, before you do anything
to it, its heat of

00:06:02.700 --> 00:06:04.570
formation is 0.

00:06:04.570 --> 00:06:06.730
So if you see O2, its heat
of formation is 0.

00:06:06.730 --> 00:06:09.970
If you see hydrogen, if you
see H2, its heat of

00:06:09.970 --> 00:06:10.450
formation is 0.

00:06:10.450 --> 00:06:12.940
If you see carbon by itself,
heat of formation is 0.

00:06:12.940 --> 00:06:15.340
Carbon in the solid state, heat
of formation is 0, at

00:06:15.340 --> 00:06:17.210
standard temperature
and pressure.

00:06:17.210 --> 00:06:18.840
Now what about nitrogen
monoxide?

00:06:18.840 --> 00:06:20.090
Let's look that up.

00:06:24.490 --> 00:06:25.250
I have it right here.

00:06:25.250 --> 00:06:26.680
Nitrogen monoxide.

00:06:26.680 --> 00:06:27.770
Heat of formation.

00:06:27.770 --> 00:06:36.810
It's positive, 90.29.

00:06:36.810 --> 00:06:40.370
And finally, what's the heat
of formation of water?

00:06:40.370 --> 00:06:41.620
Well, let me see.

00:06:46.890 --> 00:06:47.910
Liquid water.

00:06:47.910 --> 00:07:00.000
Minus 285.83.

00:07:00.000 --> 00:07:01.430
Now you might tempted
to say, OK.

00:07:01.430 --> 00:07:04.900
Hess's Law says that if if we
want the delta H for this

00:07:04.900 --> 00:07:08.850
reaction, we just take this plus
this, and subtract that.

00:07:08.850 --> 00:07:11.900
And you'd be almost right, but
you'd get the problem wrong.

00:07:11.900 --> 00:07:16.000
Because these are the heat
of formation per mole.

00:07:16.000 --> 00:07:18.052
But we notice in this reaction,
we have 4 moles of

00:07:18.052 --> 00:07:21.190
this, plus 5 moles of this,
yields 4 moles of this plus 6

00:07:21.190 --> 00:07:22.090
moles of that.

00:07:22.090 --> 00:07:25.310
So we have to multiply this
times the number of moles.

00:07:25.310 --> 00:07:30.124
So here I have to multiply this
times 4, 4 here, and I

00:07:30.124 --> 00:07:32.080
have to multiply it times 4
here, and I have to multiply

00:07:32.080 --> 00:07:33.120
it times 6 here.

00:07:33.120 --> 00:07:35.620
I don't even worry about
multiplying 0 times 5, because

00:07:35.620 --> 00:07:36.970
it's just going to be 0.

00:07:36.970 --> 00:07:41.090
So now we can apply Hess's Law
to figure out the delta H of

00:07:41.090 --> 00:07:41.830
this reaction.

00:07:41.830 --> 00:07:47.660
So the delta H of this reaction
is going to be equal

00:07:47.660 --> 00:07:51.560
to, 4 times the heat of
formation of nitrogen

00:07:51.560 --> 00:07:58.400
monoxide-- so 4 times 90.29,
plus 6 times the heat of

00:07:58.400 --> 00:07:59.860
formation of water.

00:07:59.860 --> 00:08:06.330
So plus, I'll switch colors,
6 times minus 285.83.

00:08:06.330 --> 00:08:08.800
And just as a side note, given
that the heat of formation of

00:08:08.800 --> 00:08:12.570
nitrogen monoxide is positive,
that means that you have to

00:08:12.570 --> 00:08:16.320
add heat to a system to get this
to its elemental form.

00:08:16.320 --> 00:08:19.650
So it has more energy than
its elemental form.

00:08:19.650 --> 00:08:21.940
So it won't just happen
by itself.

00:08:21.940 --> 00:08:24.460
And water, on the other hand,
it releases energy when you

00:08:24.460 --> 00:08:26.230
form it from its
elemental form.

00:08:26.230 --> 00:08:27.900
So in some ways, it's
more stable.

00:08:27.900 --> 00:08:28.580
But anyway let me--

00:08:28.580 --> 00:08:32.280
So these are the heats of
formations of the products.

00:08:32.280 --> 00:08:36.590
And then we want to subtract out
the heats of formation of

00:08:36.590 --> 00:08:39.470
the reactants in our reaction.

00:08:39.470 --> 00:08:44.280
So here it's 4 times 45.9--
Let me make sure.

00:08:44.280 --> 00:08:48.770
It's a minus 45.9.

00:08:48.770 --> 00:08:49.120
Right?

00:08:49.120 --> 00:08:54.220
That ammonia had a minus
45.9 heat of formation.

00:08:54.220 --> 00:08:56.780
So what did we end up with?

00:08:56.780 --> 00:08:59.680
Let me get the calculator out.

00:08:59.680 --> 00:09:03.522
So I have-- let me make sure
I put it over here.

00:09:03.522 --> 00:09:04.640
I have to be able to read it.

00:09:04.640 --> 00:09:06.840
Well, I'll just do it off the
screen, because my screen is

00:09:06.840 --> 00:09:08.020
getting filled up.

00:09:08.020 --> 00:09:10.610
So I have-- let me
just do it here.

00:09:10.610 --> 00:09:28.600
4 times 90.29 plus 6 times
285.83 negative is equal to--

00:09:28.600 --> 00:09:33.090
so so far, we're
at minus 1,353.

00:09:33.090 --> 00:09:34.570
Does that sound about right?

00:09:34.570 --> 00:09:35.370
That looks about right.

00:09:35.370 --> 00:09:41.060
And now we want to subtract from
that 4 times minus 45.9.

00:09:41.060 --> 00:09:51.710
So we want to subtract-- so
minus 4 times 45.9 negative is

00:09:51.710 --> 00:09:54.870
equal to minus 1,170.

00:09:54.870 --> 00:09:57.770
So our delta h of this reaction
is equal to minus

00:09:57.770 --> 00:10:04.490
1,170 kilojoules for
this reaction.

00:10:04.490 --> 00:10:06.270
And all we did is, we took the
heat of formation of the

00:10:06.270 --> 00:10:08.400
products, multiply it times
the number of moles, and

00:10:08.400 --> 00:10:10.320
subtracted out the
heat of formation

00:10:10.320 --> 00:10:12.570
of the actual reactants.

00:10:12.570 --> 00:10:13.120
There you go.

00:10:13.120 --> 00:10:15.770
Let's do one more of these.

00:10:15.770 --> 00:10:18.010
Let's say I had some propane.

00:10:18.010 --> 00:10:20.440
I had some propane, and I'm
going to combust it.

00:10:20.440 --> 00:10:24.290
I'm going to oxidize the propane
to yield some carbon

00:10:24.290 --> 00:10:25.710
dioxide in water.

00:10:25.710 --> 00:10:26.830
Well, it's the same drill.

00:10:26.830 --> 00:10:29.480
What's the heat of formation
of propane?

00:10:29.480 --> 00:10:30.230
Look it up here.

00:10:30.230 --> 00:10:33.330
It is amazing how exhaustive
these lists really are.

00:10:33.330 --> 00:10:35.850
Propane is down here in
its liquid state.

00:10:35.850 --> 00:10:39.976
Heat of formation minus 104.7.

00:10:39.976 --> 00:10:41.600
So let me write that down.

00:10:41.600 --> 00:10:45.470
Minus 104.7.

00:10:45.470 --> 00:10:48.010
Heat of formation of oxygen
in its elemental state.

00:10:48.010 --> 00:10:49.840
That's how you always
find oxygen.

00:10:49.840 --> 00:10:51.000
So it's just 0.

00:10:51.000 --> 00:10:54.700
Heat of formation of carbon
dioxide-- Let's see.

00:10:54.700 --> 00:11:04.170
Carbon dioxide, and as
a gas, minus 393.5.

00:11:04.170 --> 00:11:04.800
And water.

00:11:04.800 --> 00:11:05.900
We already figured that out.

00:11:05.900 --> 00:11:12.960
It's minus 285.83.

00:11:12.960 --> 00:11:17.120
So how much heat is formed when
we combust one mole of

00:11:17.120 --> 00:11:18.750
propane right here?

00:11:18.750 --> 00:11:19.850
So let's see.

00:11:19.850 --> 00:11:23.120
We have to figure out the heat
of the products, the heat of

00:11:23.120 --> 00:11:27.290
formation of the products-- so
it's going to be 3 times this.

00:11:27.290 --> 00:11:28.760
Because we formed
3 moles of this.

00:11:28.760 --> 00:11:31.520
For every mole, we release
this much energy.

00:11:31.520 --> 00:11:34.610
And then plus 4 times
this, and then

00:11:34.610 --> 00:11:37.280
subtract out 1 times this.

00:11:37.280 --> 00:11:39.440
So what do we get?

00:11:39.440 --> 00:11:45.900
We get 3 times 393.5
and that's a

00:11:45.900 --> 00:11:48.320
negative, is equal to that.

00:11:48.320 --> 00:12:01.180
Plus 4 times 285.83 negative
is equal to minus 2,300

00:12:01.180 --> 00:12:02.280
kilojoules, roughly.

00:12:02.280 --> 00:12:05.680
And then we have to subtract
out 1 times this.

00:12:05.680 --> 00:12:08.400
Or we could just add 104.7.

00:12:08.400 --> 00:12:09.820
So let me just do that.

00:12:09.820 --> 00:12:18.150
So plus 104.7 is equal
to minus 2,200.

00:12:18.150 --> 00:12:25.450
So here my heat of this
reaction, is equal to minus

00:12:25.450 --> 00:12:31.810
2,219 kilojoules as we
go in this direction.

00:12:31.810 --> 00:12:36.050
For every mole of propane that
I combust, I will actually

00:12:36.050 --> 00:12:38.630
produce this much energy
on the other side.

00:12:38.630 --> 00:12:45.650
Because this right here has
roughly 2,200 less kilojoules

00:12:45.650 --> 00:12:47.390
than this side right there.

00:12:47.390 --> 00:12:49.860
So I could actually rewrite this
reaction where I write

00:12:49.860 --> 00:12:52.140
all that, and I could
have added--

00:12:52.140 --> 00:12:53.440
actually, let me do it.

00:12:53.440 --> 00:12:58.870
I could rewrite this reaction
is C3H8 propane, plus 5

00:12:58.870 --> 00:13:07.060
oxygens, yields 3 carbon
dioxides plus 4 waters plus

00:13:07.060 --> 00:13:09.880
2,219 kilojoules.

00:13:09.880 --> 00:13:12.160
That's actually what's released
by this reaction.

00:13:12.160 --> 00:13:13.680
It's exothermic.

00:13:13.680 --> 00:13:17.610
This side of the reaction has
less heat than this side, and

00:13:17.610 --> 00:13:18.890
that-- it didn't
just disappear.

00:13:18.890 --> 00:13:20.090
It got released.

00:13:20.090 --> 00:13:22.390
And this is where
it got released.

00:13:22.390 --> 00:13:25.430
Now sometimes you'll see a
question where they say, hey.

00:13:25.430 --> 00:13:25.970
Fair enough.

00:13:25.970 --> 00:13:27.850
You figured out the heat
of this reaction.

00:13:27.850 --> 00:13:30.540
How much heat is going to be
released if I were to hand

00:13:30.540 --> 00:13:32.210
you, I don't know, let's
say I were to hand

00:13:32.210 --> 00:13:38.980
you 33 grams of propane?

00:13:38.980 --> 00:13:40.690
Well, then you just start
thinking, oh, well, how many

00:13:40.690 --> 00:13:41.960
moles of propane this is.

00:13:41.960 --> 00:13:44.890
Because if I combust one
mole of propane, I

00:13:44.890 --> 00:13:46.180
get this much heat.

00:13:46.180 --> 00:13:49.240
So how many moles of propane
is 33 grams?

00:13:49.240 --> 00:13:51.530
Well, how much does
1 mole weigh?

00:13:51.530 --> 00:14:02.520
The 1 mole of carbon weighs 12
grams. 1 mole of hydrogen

00:14:02.520 --> 00:14:04.320
weighs 1 gram.

00:14:04.320 --> 00:14:08.990
So 1 mole of propane is going to
be 3 times 12-- so times 3,

00:14:08.990 --> 00:14:12.417
because we have 3 carbons there
and 8 hydrogens, so

00:14:12.417 --> 00:14:18.870
times 8-- so it's going to
be equal to 36 plus 88.

00:14:18.870 --> 00:14:20.900
So it's going to be 44.

00:14:20.900 --> 00:14:24.750
So this is going to be 44
grams per mole, right?

00:14:24.750 --> 00:14:26.930
This is, let me write
that down.

00:14:26.930 --> 00:14:30.420
44 grams per mole.

00:14:30.420 --> 00:14:32.680
Now, if I give you 33
grams, how many

00:14:32.680 --> 00:14:34.050
moles am I giving you?

00:14:34.050 --> 00:14:44.030
Well, 33 grams times, I guess we
could say, 1 over 44 moles

00:14:44.030 --> 00:14:47.840
per gram-- I don't have to write
the whole gram there.

00:14:47.840 --> 00:14:49.290
And then the grams cancel.

00:14:49.290 --> 00:14:52.180
I'm giving you 33 over
44 of a mole, or I'm

00:14:52.180 --> 00:14:56.140
giving you 0.75 moles.

00:14:56.140 --> 00:15:00.510
So if one mole produces this
much energy, 3/4 of a mole is

00:15:00.510 --> 00:15:02.340
going to produce 3/4 of this.

00:15:02.340 --> 00:15:08.960
So we just multiply
that times 0.75.

00:15:08.960 --> 00:15:12.330
And you get 1,664.

00:15:12.330 --> 00:15:17.990
So times 0.75 is
equal to 1,664.

00:15:17.990 --> 00:15:20.710
So if I were to give you 1 mole
of propane, and I were to

00:15:20.710 --> 00:15:25.020
combust it with enough oxygen,
I'll produce 2,200 kilojoules

00:15:25.020 --> 00:15:26.450
that's released from
the system.

00:15:26.450 --> 00:15:29.310
So this side of system has
less energy left over.

00:15:29.310 --> 00:15:33.050
But if I were to only give you
33 grams, which is 3/4 of a

00:15:33.050 --> 00:15:36.530
mole, then you're going to
release roughly 1,600

00:15:36.530 --> 00:15:37.550
kilojoules.

00:15:37.550 --> 00:15:37.970
Anyway.

00:15:37.970 --> 00:15:39.980
Hopefully you found
that helpful.


Heat of formation

WEBVTT
Kind: captions
Language: en

00:00:00.600 --> 00:00:05.260
We saw in the last video that if
we defined enthalpy, H, as

00:00:05.260 --> 00:00:08.160
being equal to the internal
energy of a system plus the

00:00:08.160 --> 00:00:10.860
pressure of the system times the
volume of the system-- and

00:00:10.860 --> 00:00:13.060
this is an almost arbitrary
definition.

00:00:13.060 --> 00:00:16.100
But we know that this is
a valid state variable.

00:00:16.100 --> 00:00:18.910
That no matter what you do in
terms of how you get there,

00:00:18.910 --> 00:00:20.330
you're always going to
have the same value.

00:00:20.330 --> 00:00:21.770
Because it's the sum
and product of

00:00:21.770 --> 00:00:23.510
other valid state variables.

00:00:23.510 --> 00:00:26.100
But this by itself isn't that
useful or that intuitive.

00:00:26.100 --> 00:00:29.330
But we saw in the last video
that if you assume constant

00:00:29.330 --> 00:00:32.720
pressure-- and that's a big
assumption, but it's not an

00:00:32.720 --> 00:00:35.290
unreasonable assumption for
most chemical reactions.

00:00:35.290 --> 00:00:36.780
Because most chemical reactions,
you know, we're

00:00:36.780 --> 00:00:39.480
sitting at the beach with our
beakers, and they're exposed

00:00:39.480 --> 00:00:43.640
to just standard temperature and
pressure, or at least some

00:00:43.640 --> 00:00:47.420
pressure, that's not changing
as the reaction occurs.

00:00:47.420 --> 00:00:50.200
If we assume constant pressure,
we saw that the

00:00:50.200 --> 00:00:57.570
change in enthalpy becomes the
heat added to the system at

00:00:57.570 --> 00:00:58.670
that constant pressure.

00:00:58.670 --> 00:01:04.090
That P there is just to show you
that hey, this is assuming

00:01:04.090 --> 00:01:05.870
that we're dealing with
heat being added

00:01:05.870 --> 00:01:07.480
at a constant pressure.

00:01:07.480 --> 00:01:08.090
Fair enough.

00:01:08.090 --> 00:01:12.820
So how can we use these concepts
in any useful way?

00:01:12.820 --> 00:01:16.380
Let's say that I had some carbon
in its elemental form

00:01:16.380 --> 00:01:19.150
as graphite, and I add
to that-- say I

00:01:19.150 --> 00:01:20.280
have a mole of carbon.

00:01:20.280 --> 00:01:23.010
And I add to that 2 moles
of hydrogen in

00:01:23.010 --> 00:01:24.820
its elemental form.

00:01:24.820 --> 00:01:26.810
It's going to be a gas, and
it's going to be as a

00:01:26.810 --> 00:01:27.930
molecule, right?

00:01:27.930 --> 00:01:32.360
If I just have a bunch of
hydrogen in its gaseous state,

00:01:32.360 --> 00:01:34.670
let's say, in a balloon, I'm not
going to have individual

00:01:34.670 --> 00:01:35.760
atoms of hydrogen.

00:01:35.760 --> 00:01:38.890
They're going to bond and form
these diatomic molecules.

00:01:38.890 --> 00:01:45.800
And if I react them, I'm going
to produce a mole of methane.

00:01:45.800 --> 00:01:47.530
CH4.

00:01:47.530 --> 00:01:49.020
But that's not all I'm
going to produce.

00:01:49.020 --> 00:01:51.110
I'm also going to produce
some heat.

00:01:51.110 --> 00:02:01.250
I'm going to produce 74
kilojoules of heat-- plus 74

00:02:01.250 --> 00:02:06.460
kilojoules of heat-- when
I produce that one mole.

00:02:06.460 --> 00:02:08.699
I'll do a lower case
k for the kilo.

00:02:08.699 --> 00:02:12.330
When I produce that one
mole of methane.

00:02:12.330 --> 00:02:14.240
So what's happening here?

00:02:14.240 --> 00:02:17.960
So first of all, how much heat
is being added to the system?

00:02:17.960 --> 00:02:20.740
And let's assume that this heat
just gets released from

00:02:20.740 --> 00:02:21.230
the system.

00:02:21.230 --> 00:02:23.480
That this isn't an adiabatic
process.

00:02:23.480 --> 00:02:25.520
I haven't insulated the
system from anything.

00:02:25.520 --> 00:02:27.050
That this just gets released.

00:02:27.050 --> 00:02:28.650
It just goes away.

00:02:28.650 --> 00:02:30.200
It gets released.

00:02:30.200 --> 00:02:33.830
So my question is, how much--
you know, I started off with

00:02:33.830 --> 00:02:37.550
this container, I guess we could
call it, that's at a

00:02:37.550 --> 00:02:39.700
standard, fixed pressure.

00:02:39.700 --> 00:02:42.270
And maybe I had a bunch
of-- well, I

00:02:42.270 --> 00:02:43.090
wanted to do the carbon.

00:02:43.090 --> 00:02:44.430
I'll do it in gray.

00:02:44.430 --> 00:02:47.210
I have a bunch of solid
carbon lying around.

00:02:47.210 --> 00:02:51.230
Maybe some type of dust. And
then I have some molecular

00:02:51.230 --> 00:02:52.470
hydrogen gas.

00:02:52.470 --> 00:02:54.910
Each of those dots is actually
two hydrogen atoms. And then,

00:02:54.910 --> 00:02:56.700
I don't know, maybe I shake it
up or something to make them

00:02:56.700 --> 00:03:05.430
react, and then I get a
bunch of methane gas.

00:03:05.430 --> 00:03:07.120
I'll do that in green.

00:03:07.120 --> 00:03:11.080
So now I just have a bunch of
methane gas, and I released 74

00:03:11.080 --> 00:03:12.330
kilojoules.

00:03:15.160 --> 00:03:18.980
So how much heat was added
to the system?

00:03:18.980 --> 00:03:21.150
Well, we released heat
from the system.

00:03:21.150 --> 00:03:22.980
We released 74 kilojoules.

00:03:22.980 --> 00:03:29.700
So the heat added to the
system was minus 74

00:03:29.700 --> 00:03:32.170
kilojoules.

00:03:32.170 --> 00:03:32.460
Right?

00:03:32.460 --> 00:03:34.870
If I asked you the heat
released, then I

00:03:34.870 --> 00:03:35.840
would have said 74.

00:03:35.840 --> 00:03:38.930
But remember, we care about the
heat added to the system,

00:03:38.930 --> 00:03:41.420
is 74 kilojoules.

00:03:41.420 --> 00:03:43.820
And I just showed you that
that's the exact same thing as

00:03:43.820 --> 00:03:45.070
the change in enthalpy.

00:03:50.280 --> 00:03:51.990
So how can we think
about this?

00:03:51.990 --> 00:03:54.510
What is the enthalpy
of this system

00:03:54.510 --> 00:03:56.280
relative to this system?

00:03:56.280 --> 00:03:58.650
Well, it's going to
be lower, right?

00:03:58.650 --> 00:04:01.550
Because if you take enthalpy--
so the change in enthalpy is

00:04:01.550 --> 00:04:06.480
the enthalpy of your final
system, minus the enthalpy of

00:04:06.480 --> 00:04:08.720
your initial system.

00:04:08.720 --> 00:04:09.970
And we got a negative number.

00:04:09.970 --> 00:04:12.600
We got minus 74 kilojoules.

00:04:12.600 --> 00:04:17.589
So this has to be lower than
this by 74 kilojoules.

00:04:17.589 --> 00:04:21.680
So this enthalpy right here
is less than this

00:04:21.680 --> 00:04:22.640
enthalpy right here.

00:04:22.640 --> 00:04:25.850
So if we were to actually draw
it on a diagram, if I were to

00:04:25.850 --> 00:04:30.490
draw the reaction-- let's say
that this is just time or

00:04:30.490 --> 00:04:34.290
something, as the reaction
proceeds, that axis.

00:04:34.290 --> 00:04:38.340
And on the y-axis, I'll
draw enthalpy.

00:04:38.340 --> 00:04:43.150
So the reaction starts off at
your initial enthalpy, Hi and

00:04:43.150 --> 00:04:44.970
that's this state right here.

00:04:44.970 --> 00:04:45.760
So you start there.

00:04:45.760 --> 00:04:47.615
I'll do it in the yellow
of that container.

00:04:50.680 --> 00:04:52.060
You start there.

00:04:52.060 --> 00:04:54.270
And then, I don't know, you
shake it up, or I'm not going

00:04:54.270 --> 00:04:56.180
to go into the activation
energies, so it might have

00:04:56.180 --> 00:04:58.160
little hump and all of
that, but who knows.

00:04:58.160 --> 00:05:01.320
But then we end up at
our final enthalpy.

00:05:01.320 --> 00:05:05.090
We have this final enthalpy,
right here, after the reaction

00:05:05.090 --> 00:05:05.510
has occured.

00:05:05.510 --> 00:05:06.910
That's this state right here.

00:05:06.910 --> 00:05:08.790
This is H final.

00:05:08.790 --> 00:05:11.680
So you can see, you've had this
drop off in enthalpy.

00:05:11.680 --> 00:05:14.530
And what's interesting here is,
is not so much what the

00:05:14.530 --> 00:05:17.240
absolute value of this enthalpy
is here, or what the

00:05:17.240 --> 00:05:20.150
absolute value of this
enthalpy here is.

00:05:20.150 --> 00:05:23.580
But now that we have enthalpy,
we can kind of have a

00:05:23.580 --> 00:05:26.260
framework for thinking about
how much heat energy is in

00:05:26.260 --> 00:05:27.970
this system relative
to this system.

00:05:27.970 --> 00:05:30.310
And given that there's less
heat energy in this system

00:05:30.310 --> 00:05:33.100
than that energy system, we
must have released energy.

00:05:33.100 --> 00:05:35.960
And you know, to some degree,
I told you that from the

00:05:35.960 --> 00:05:37.120
beginning, right?

00:05:37.120 --> 00:05:38.990
I told you that energy
is released.

00:05:38.990 --> 00:05:40.910
And the word for this we
use is exothermic.

00:05:45.860 --> 00:05:47.920
Now, if you want to go the
other way-- let's say we

00:05:47.920 --> 00:05:51.000
wanted to go from methane and
go back, you'd have to add

00:05:51.000 --> 00:05:52.510
heat into the reaction.

00:05:52.510 --> 00:05:55.500
If you wanted to go backwards
through this reaction, go

00:05:55.500 --> 00:06:00.290
upwards, you would have to add
that heat content to get that

00:06:00.290 --> 00:06:01.850
positive delta H, and
then you would have

00:06:01.850 --> 00:06:03.430
an endothermic reaction.

00:06:03.430 --> 00:06:06.350
So if a reaction releases
energy, exothermic.

00:06:06.350 --> 00:06:09.940
If a reaction needs energy to
occur, it's endothermic.

00:06:12.560 --> 00:06:15.390
Now you might be asking, Sal,
where'd that energy come from?

00:06:15.390 --> 00:06:17.400
So I started at this
enthalpy here.

00:06:17.400 --> 00:06:19.836
And enthalpy has this weird
definition right here, and

00:06:19.836 --> 00:06:21.510
then'll ended up as that
other enthalpy here.

00:06:21.510 --> 00:06:23.520
And as you see, enthalpy,
the pressure

00:06:23.520 --> 00:06:25.090
we're assuming is constant.

00:06:25.090 --> 00:06:27.250
Let's say the volume isn't
changing much in this

00:06:27.250 --> 00:06:29.360
situation, or maybe doesn't
change at all.

00:06:29.360 --> 00:06:32.030
So most of the change is going
to come from the change in

00:06:32.030 --> 00:06:33.680
internal energy, right?

00:06:33.680 --> 00:06:37.030
There's some higher internal
energy here, and some lower

00:06:37.030 --> 00:06:39.080
internal energy here
that's causing the

00:06:39.080 --> 00:06:40.990
main drop in enthalpy.

00:06:40.990 --> 00:06:44.620
And that change in internal
energy is really a conversion

00:06:44.620 --> 00:06:47.630
from some potential energy,
up here, into

00:06:47.630 --> 00:06:48.870
the heat that's released.

00:06:48.870 --> 00:06:52.830
So there was some heat that was
released, 74 kilojoules,

00:06:52.830 --> 00:06:55.360
and so our internal
energy dropped.

00:06:55.360 --> 00:06:58.750
And what all of this does is,
it gives us a framework so

00:06:58.750 --> 00:07:02.380
that if we know how much heat
it takes to form or not form

00:07:02.380 --> 00:07:06.370
certain products, then we can
kind of predict how much heat

00:07:06.370 --> 00:07:09.450
will either be released, or how
much heat will be absorbed

00:07:09.450 --> 00:07:10.880
by different reactions.

00:07:10.880 --> 00:07:13.190
And so here I'm going to touch
on another notion.

00:07:13.190 --> 00:07:16.920
The notion of heat of formation,
or sometimes it's

00:07:16.920 --> 00:07:18.770
change in enthalpy
of formation.

00:07:18.770 --> 00:07:21.640
So the way they talk about
it is, the change

00:07:21.640 --> 00:07:23.390
in enthalpy of formation.

00:07:23.390 --> 00:07:25.840
And it's normally given
at some standard

00:07:25.840 --> 00:07:26.990
temperature and pressure.

00:07:26.990 --> 00:07:29.190
So you put a little, usually
it's a naught, sometimes it's

00:07:29.190 --> 00:07:31.270
just a circle in there.

00:07:31.270 --> 00:07:37.310
And what that is is, what is the
change in enthalpy to get

00:07:37.310 --> 00:07:40.480
to some molecule from
its elemental form?

00:07:40.480 --> 00:07:45.420
So for example, if we want it
for methane-- if we have

00:07:45.420 --> 00:07:51.150
methane there, and we want
to figure out its heat of

00:07:51.150 --> 00:07:56.630
formation, we say, look, if
we form methane from its

00:07:56.630 --> 00:08:00.080
elemental forms, what is the
delta H of that reaction?

00:08:00.080 --> 00:08:02.480
Well, we just learned what the
delta H of that reaction was.

00:08:02.480 --> 00:08:05.750
It was minus 74 kilojoules.

00:08:05.750 --> 00:08:10.080
Which means that if you form
methane from its elemental, I

00:08:10.080 --> 00:08:13.500
guess, building blocks, you're
going to release 74

00:08:13.500 --> 00:08:14.900
kilojoules of energy.

00:08:14.900 --> 00:08:16.625
That this is an exothermic
reaction.

00:08:19.590 --> 00:08:23.940
Because you've released heat,
you can say that the methane

00:08:23.940 --> 00:08:27.180
is in a lower energy state, or
it has a lower potential

00:08:27.180 --> 00:08:29.170
energy, than these guys did.

00:08:29.170 --> 00:08:30.620
And because it has
lower potential

00:08:30.620 --> 00:08:32.990
energy, it's more stable.

00:08:32.990 --> 00:08:37.460
I mean, one way to think of it
is, if you have a guy, you

00:08:37.460 --> 00:08:41.840
have a mountain here, and down
here, you have a ball.

00:08:41.840 --> 00:08:44.650
And this isn't, you know, a
complete, direct analogy.

00:08:44.650 --> 00:08:47.090
But the analogy to potential
energy is, look.

00:08:47.090 --> 00:08:50.000
When you're down at a lower
potential energy state, you

00:08:50.000 --> 00:08:51.670
tend to be more stable.

00:08:51.670 --> 00:08:55.990
And so, in the everyday world,
if you have a bunch of methane

00:08:55.990 --> 00:09:01.830
sitting around, the fact that
it has a negative heat of

00:09:01.830 --> 00:09:06.020
formation, or a standard heat
of formation, because I have

00:09:06.020 --> 00:09:09.490
that naught here, or a negative
standard change in

00:09:09.490 --> 00:09:12.270
enthalpy of formation-- those
are all the same things--

00:09:12.270 --> 00:09:15.480
tells me that methane is
stable relative to its

00:09:15.480 --> 00:09:17.030
constituent compounds.

00:09:17.030 --> 00:09:18.690
And actually, you can look
these things up.

00:09:18.690 --> 00:09:20.500
You don't have to memorize them,
but it's good to know

00:09:20.500 --> 00:09:21.010
what they are.

00:09:21.010 --> 00:09:24.850
And I copied all of this stuff--
actually, let me get

00:09:24.850 --> 00:09:30.390
the actual tables from
Wikipedia down here.

00:09:30.390 --> 00:09:33.270
I copied all of these directly
from Wikipedia.

00:09:33.270 --> 00:09:36.750
These give you the standard
heat of formation

00:09:36.750 --> 00:09:37.760
of a bunch of things.

00:09:37.760 --> 00:09:43.050
And if you look down here, for--
let's see if they have

00:09:43.050 --> 00:09:47.110
methane-- right there, this is
what we were dealing with.

00:09:47.110 --> 00:09:51.030
They're telling us essentially
the delta H of the reaction

00:09:51.030 --> 00:09:52.280
that forms methane.

00:09:55.240 --> 00:09:58.000
This point table right there
tells us that if we start off

00:09:58.000 --> 00:10:02.030
with some carbon in a solid
state, plus two moles of

00:10:02.030 --> 00:10:05.860
hydrogen in a gaseous state,
and we form one mole of

00:10:05.860 --> 00:10:09.860
methane, that if you take the
enthalpy here minus the

00:10:09.860 --> 00:10:12.930
enthalpy here-- so the change
in enthalpy for this

00:10:12.930 --> 00:10:17.320
reaction-- at standard
temperature and pressure, is

00:10:17.320 --> 00:10:21.100
going to be equal to minus
74 kilojoules.

00:10:21.100 --> 00:10:21.990
Per mole.

00:10:21.990 --> 00:10:23.170
And this is all given
per mole.

00:10:23.170 --> 00:10:26.580
So if you have a mole of this,
two moles of this, to form one

00:10:26.580 --> 00:10:29.340
mole of methane, you're
going to release 74

00:10:29.340 --> 00:10:30.880
kilojoules of heat.

00:10:30.880 --> 00:10:32.370
So this is a stable reaction.

00:10:32.370 --> 00:10:33.850
Now there's a couple of
interesting things here, and

00:10:33.850 --> 00:10:37.050
we'll keep using this table
over the next few videos.

00:10:37.050 --> 00:10:46.200
You see here, monoatomic oxygen
has a positive standard

00:10:46.200 --> 00:10:47.590
heat of formation.

00:10:47.590 --> 00:10:50.300
Which means it takes
energy to form it.

00:10:50.300 --> 00:10:50.620
Right?

00:10:50.620 --> 00:10:54.360
That if you have a reaction, let
me just say, the reaction,

00:10:54.360 --> 00:10:55.180
I'll write it this way.

00:10:55.180 --> 00:11:05.980
One half of molecular oxygen as
a gas to go to one mole of

00:11:05.980 --> 00:11:10.290
oxygen in its gaseous state.

00:11:10.290 --> 00:11:15.060
This tells us that this
state has more

00:11:15.060 --> 00:11:16.330
potential than this state.

00:11:16.330 --> 00:11:18.980
And in order for this reaction
to occur, you have to add

00:11:18.980 --> 00:11:20.190
energy to it.

00:11:20.190 --> 00:11:21.860
You have to put the energy
on the other side.

00:11:21.860 --> 00:11:29.340
So you'd have to say,
plus 249 joules.

00:11:29.340 --> 00:11:31.500
So you might say, hey, Sal,
that doesn't make sense.

00:11:31.500 --> 00:11:33.000
Oxygen is just oxygen.

00:11:33.000 --> 00:11:35.800
Why is there a heat of
formation of oxygen?

00:11:35.800 --> 00:11:39.260
And that's because you always
use the elemental form as your

00:11:39.260 --> 00:11:40.500
reference point.

00:11:40.500 --> 00:11:43.870
So oxygen, if you just had
a bunch of oxygen sitting

00:11:43.870 --> 00:11:46.790
around, it's going to
be in the O2 form.

00:11:46.790 --> 00:11:49.020
If you have a bunch of hydrogen,
it's going to be H2.

00:11:49.020 --> 00:11:51.590
If you have a bunch of nitrogen,
it's going to be N2.

00:11:51.590 --> 00:11:54.830
Carbon, on the other hand, is
just C, and it tends to be in

00:11:54.830 --> 00:11:57.710
its solid form as graphite.

00:11:57.710 --> 00:12:01.620
So all heats of formation are
relative to the form that you

00:12:01.620 --> 00:12:04.050
find that element when you have
a pure version of it.

00:12:04.050 --> 00:12:08.020
Not necessarily its atomic form,
although sometimes it is

00:12:08.020 --> 00:12:09.400
its atomic form.

00:12:09.400 --> 00:12:12.450
Now, in the next video, we're
going to use this table, which

00:12:12.450 --> 00:12:15.690
is a very handy table-- I cut
and pasted parts of it-- to

00:12:15.690 --> 00:12:18.470
actually solve problems. In this
last video, I gave you

00:12:18.470 --> 00:12:20.550
the heat of formation,
and we just thought

00:12:20.550 --> 00:12:21.490
about it a little bit.

00:12:21.490 --> 00:12:23.720
In the next few videos, we're
going to use this table that

00:12:23.720 --> 00:12:26.580
gives us standard heats of
formation to actually figure

00:12:26.580 --> 00:12:29.800
out whether reactions are
endothermic-- meaning they

00:12:29.800 --> 00:12:32.040
absorb energy-- or exothermic--
meaning they

00:12:32.040 --> 00:12:34.750
release energy-- and we'll
figure out how much.


Thermodynamic entropy definition clarification

WEBVTT
Kind: captions
Language: en

00:00:00.350 --> 00:00:03.280
In the video where I first
introduced the concept of

00:00:03.280 --> 00:00:05.430
entropy, I just tried
something out.

00:00:05.430 --> 00:00:12.700
I defined my change in entropy
as being equal to the heat

00:00:12.700 --> 00:00:16.250
added to a system, divided by
the temperature at which it

00:00:16.250 --> 00:00:17.700
was added to the system.

00:00:17.700 --> 00:00:19.270
And then I tested to
see if this was

00:00:19.270 --> 00:00:20.660
a valid state variable.

00:00:20.660 --> 00:00:22.890
And when I did that, I looked
at the Carnot cycle.

00:00:25.850 --> 00:00:28.370
And this is a bit of a review.

00:00:28.370 --> 00:00:29.340
Never hurts to review.

00:00:29.340 --> 00:00:31.240
Let me draw the PV
diagram here.

00:00:31.240 --> 00:00:35.670
We saw that we start at this
state here, and then we

00:00:35.670 --> 00:00:37.320
proceed isothermically.

00:00:37.320 --> 00:00:40.020
We removed little pebbles
off the piston.

00:00:40.020 --> 00:00:43.190
So we increased the volume
and lowered the pressure.

00:00:43.190 --> 00:00:48.170
Then we proceed adiabatically,
where we isolated things and

00:00:48.170 --> 00:00:49.750
we moved like that.

00:00:49.750 --> 00:00:51.650
That was adiabatically.

00:00:51.650 --> 00:00:54.880
Then at this other isotherm,
we added the pebbles back.

00:00:58.280 --> 00:01:01.590
And then we isolated
the system again.

00:01:01.590 --> 00:01:06.160
So adiabatically, we continued
to add more pebbles, and we

00:01:06.160 --> 00:01:08.230
got back to our original
state.

00:01:08.230 --> 00:01:11.430
And I did a couple of videos
where I show that if you take

00:01:11.430 --> 00:01:14.750
the heat added here-- so this is
all being done at some high

00:01:14.750 --> 00:01:16.720
temperature, T1.

00:01:16.720 --> 00:01:20.320
This is being done at some
low temperature, T2.

00:01:20.320 --> 00:01:23.910
There's some heat being added
here, Q1, and that there's

00:01:23.910 --> 00:01:27.760
some heat being released
here, Q2.

00:01:27.760 --> 00:01:31.110
And since these are adiabatic,
there's no transfer of heat to

00:01:31.110 --> 00:01:32.650
and from the system.

00:01:32.650 --> 00:01:35.050
And when I looked at this, and
when I looked at the Carnot

00:01:35.050 --> 00:01:37.970
cycle, and I used this
definition of entropy, I saw

00:01:37.970 --> 00:01:42.400
that the total change in S, when
I go from this point all

00:01:42.400 --> 00:01:47.780
the way around and got back, the
change in S, was equal to

00:01:47.780 --> 00:01:55.770
Q1 over T1 plus Q2 over T2.

00:01:55.770 --> 00:01:58.220
And then I actually showed you
that this was equal to 0,

00:01:58.220 --> 00:02:00.880
which is exactly the result
that I wanted to see.

00:02:00.880 --> 00:02:03.690
Because in order for this to be
a state variable, in order

00:02:03.690 --> 00:02:08.800
for S to be a state variable, it
should not be dependent on

00:02:08.800 --> 00:02:09.615
how I got there.

00:02:09.615 --> 00:02:12.490
It should only be dependent
on my state variables.

00:02:12.490 --> 00:02:15.980
So even if I go on some crazy
path, at the end of the day,

00:02:15.980 --> 00:02:18.560
it should get back to 0.

00:02:18.560 --> 00:02:22.650
But I did something, I guess,
a little bit-- what I did

00:02:22.650 --> 00:02:26.870
wasn't a proof that this is
always a valid state variable.

00:02:26.870 --> 00:02:29.320
It was only a proof that it's
a valid state variable if we

00:02:29.320 --> 00:02:30.850
look at the Carnot cycle.

00:02:30.850 --> 00:02:34.070
But it turns out that it was
only valid because the Carnot

00:02:34.070 --> 00:02:36.600
cycle was reversible.

00:02:36.600 --> 00:02:40.340
And this is a subtle but super
important point, and I really

00:02:40.340 --> 00:02:41.920
should've clarified this
on the first video.

00:02:41.920 --> 00:02:44.450
I guess I was too caught up
showing the proof of the

00:02:44.450 --> 00:02:47.080
Carnot cycle to put the
reversibility there.

00:02:47.080 --> 00:02:50.720
And before I even show you why
it has to be reversable, let

00:02:50.720 --> 00:02:53.210
me just review what
reversibility means.

00:02:53.210 --> 00:02:57.600
Now, we know that in order to
even define a path here, the

00:02:57.600 --> 00:03:00.640
system has to be pretty close to
equilibrium the whole time.

00:03:00.640 --> 00:03:04.230
That's the whole reason why
throughout these videos, I've

00:03:04.230 --> 00:03:07.330
been drawing this piston, you
know, have the gases down

00:03:07.330 --> 00:03:09.780
here, and then always-- instead
of having one big

00:03:09.780 --> 00:03:12.660
weight on top that I took off
or took on, because it would

00:03:12.660 --> 00:03:14.550
throw the system out of
equilibrium-- I did it in

00:03:14.550 --> 00:03:16.310
really small increments.

00:03:16.310 --> 00:03:19.860
I just moved grains of sand, so
that the system was always

00:03:19.860 --> 00:03:21.640
really close to equilibrium.

00:03:21.640 --> 00:03:23.430
And that's called quasistatic.

00:03:23.430 --> 00:03:24.680
and I've defined that before.

00:03:29.320 --> 00:03:31.440
And that means that you're
always in kind of a

00:03:31.440 --> 00:03:32.720
quasi-equilibrium.

00:03:32.720 --> 00:03:34.910
So your state variables
are always defined.

00:03:34.910 --> 00:03:38.280
But that, by itself, does not
give you reversability.

00:03:38.280 --> 00:03:47.480
You have to be quasistatic and
frictionless in order to be

00:03:47.480 --> 00:03:48.650
reversible.

00:03:48.650 --> 00:03:50.110
Now, what do we mean
by frictionless?

00:03:50.110 --> 00:03:52.720
Well, I think you know what
frictionless means.

00:03:52.720 --> 00:03:56.120
Is that like you see in this
system right here, if I make

00:03:56.120 --> 00:04:00.590
this piston a little bit bigger,
that when this piston

00:04:00.590 --> 00:04:04.330
rubs against the side of this
wall, in kind of our real

00:04:04.330 --> 00:04:05.950
world, there's always a little
bit of friction.

00:04:05.950 --> 00:04:08.730
Those molecules start bumping
against each other, and then

00:04:08.730 --> 00:04:11.050
they start making them vibrate,
so they transfer some

00:04:11.050 --> 00:04:12.160
kinetic energy.

00:04:12.160 --> 00:04:15.000
From just by rubbing into each
other, they start generating

00:04:15.000 --> 00:04:17.399
some kinetic energy,
or some heat.

00:04:17.399 --> 00:04:21.350
So you normally have some heat
generated from friction.

00:04:21.350 --> 00:04:23.900
Now, if you have some heat
generated from friction, when

00:04:23.900 --> 00:04:28.550
I remove a pebble-- first all,
when I remove that first

00:04:28.550 --> 00:04:30.010
pebble, it might not
even do anything.

00:04:30.010 --> 00:04:33.430
Because it might not even
overcome-- you can kind of

00:04:33.430 --> 00:04:35.140
view it as the force
of friction.

00:04:35.140 --> 00:04:37.400
But let's say I remove some
pebbles, and this thing moves

00:04:37.400 --> 00:04:38.650
up a little bit.

00:04:41.310 --> 00:04:44.210
But because some of the, I guess
you could say, the force

00:04:44.210 --> 00:04:46.280
differential, the pressure
differential between the

00:04:46.280 --> 00:04:50.800
pebbles and the gas inside, and
the pressure of the gas,

00:04:50.800 --> 00:04:54.480
was used to generate heat as
opposed to work, when I add

00:04:54.480 --> 00:04:57.360
the pebbles back, if I have
friction, I'm not going to get

00:04:57.360 --> 00:05:00.930
back to the same point
that I was before.

00:05:00.930 --> 00:05:03.940
Because friction is always
resisting the movement.

00:05:03.940 --> 00:05:07.440
So in order for something to be
reversible, when I remove a

00:05:07.440 --> 00:05:10.110
couple of pebbles, if I removed
ten pebbles and add

00:05:10.110 --> 00:05:13.810
the ten pebbles back, I should
be at the exact same state.

00:05:13.810 --> 00:05:15.550
But as, you know, you can just
do the thought experiment.

00:05:15.550 --> 00:05:18.520
If there's friction, I won't
be at the exact same state.

00:05:18.520 --> 00:05:24.470
My piston won't move as much as
you would expect if it was

00:05:24.470 --> 00:05:24.960
frictionless.

00:05:24.960 --> 00:05:30.140
So this is a key assumption
for reversability.

00:05:30.140 --> 00:05:35.330
Now, the Carnot cycle, by
definition, is reversible.

00:05:35.330 --> 00:05:38.500
And that's why no one could
actually really implement an

00:05:38.500 --> 00:05:43.490
engine that fully does
the Carnot cycle.

00:05:43.490 --> 00:05:46.750
And we even showed that, that
the Carnot cycle is the most

00:05:46.750 --> 00:05:48.130
efficient potential engine.

00:05:48.130 --> 00:05:50.970
That if anyone made a more
efficient engine, you could

00:05:50.970 --> 00:05:54.220
have a perpetual motion
machine, or a

00:05:54.220 --> 00:05:56.190
perpetual energy machine.

00:05:56.190 --> 00:05:58.820
And the reason why the Carnot
engine is the most efficient

00:05:58.820 --> 00:06:01.600
engine-- and there's no secret
here-- it's because it's

00:06:01.600 --> 00:06:02.770
frictionless.

00:06:02.770 --> 00:06:04.860
Any engineer who makes engines
could tell you, wow, if I

00:06:04.860 --> 00:06:08.460
could just remove all of the
friction from my system, I

00:06:08.460 --> 00:06:11.410
would get a lot more
efficient.

00:06:11.410 --> 00:06:14.610
Now, with that said-- so I've
told you that look.

00:06:14.610 --> 00:06:18.570
The definition doesn't have to
be-- it happened to work, Q

00:06:18.570 --> 00:06:20.680
divided by T, because
I was dealing with

00:06:20.680 --> 00:06:22.380
a reversible system.

00:06:22.380 --> 00:06:24.750
And just to hit the point home,
let me show you that it

00:06:24.750 --> 00:06:28.520
would not have worked if I had
just defined it Q divided by T

00:06:28.520 --> 00:06:31.320
on an irreversible system.

00:06:31.320 --> 00:06:34.495
So let's say that I have-- let
me draw another PV diagram.

00:06:38.370 --> 00:06:43.550
And I'm going to do almost
a very simple thought

00:06:43.550 --> 00:06:45.830
experiment.

00:06:45.830 --> 00:06:47.520
Now I'm going to have an
irreversible system.

00:06:47.520 --> 00:06:51.410
And I start here at some
point on my PV diagram.

00:06:51.410 --> 00:06:55.900
And you know, this could be some
type of cylinder, and it

00:06:55.900 --> 00:07:00.320
has a piston on top, and I have
my rocks, like always.

00:07:00.320 --> 00:07:02.210
But this time, there's a
little bit of friction.

00:07:02.210 --> 00:07:05.030
When this thing moves, a
little bit of heat is

00:07:05.030 --> 00:07:06.040
generated in this.

00:07:06.040 --> 00:07:07.440
And when it moves in
either direction,

00:07:07.440 --> 00:07:08.910
some heat is generated.

00:07:08.910 --> 00:07:11.500
So we can call that the
heat from friction.

00:07:11.500 --> 00:07:15.330
When it moves either up
or when it moves down.

00:07:15.330 --> 00:07:16.320
So let's do something.

00:07:16.320 --> 00:07:19.590
Let's stick this on a big
reservoir, like we tend to do.

00:07:19.590 --> 00:07:21.370
So it's an isothermal system.

00:07:21.370 --> 00:07:24.780
So let's call this T1.

00:07:24.780 --> 00:07:28.220
And let's just start
removing pebbles.

00:07:28.220 --> 00:07:29.610
And we'll move along
an isotherm.

00:07:34.070 --> 00:07:37.010
Maybe to that point there.

00:07:37.010 --> 00:07:38.900
And then we're going to-- and
I want to make a very

00:07:38.900 --> 00:07:40.400
important point here.

00:07:40.400 --> 00:07:44.440
Because this is has friction,
I'm not going to get quite as

00:07:44.440 --> 00:07:47.830
far along the isotherm than
if I didn't have friction.

00:07:47.830 --> 00:07:49.500
If this was a frictionless
system, I would've gotten a

00:07:49.500 --> 00:07:52.430
little bit further along
the isotherm.

00:07:52.430 --> 00:07:55.460
So the number of rocks isn't
going to be the same as it was

00:07:55.460 --> 00:07:55.960
frictionless.

00:07:55.960 --> 00:07:57.390
But let's just say I
move from here to

00:07:57.390 --> 00:07:59.480
here on the PV diagram.

00:07:59.480 --> 00:08:03.050
And then we go and add a bunch
of rocks back, and we want to

00:08:03.050 --> 00:08:04.450
go all the way back.

00:08:04.450 --> 00:08:05.910
And I'm not even saying whether
we have the same

00:08:05.910 --> 00:08:07.010
number of rocks or
different rocks.

00:08:07.010 --> 00:08:09.860
You're probably going to have to
add a few more rocks to get

00:08:09.860 --> 00:08:11.260
back to this point.

00:08:11.260 --> 00:08:14.140
But the idea here is, is that
we've gotten back to the same

00:08:14.140 --> 00:08:16.090
point on the state diagram.

00:08:16.090 --> 00:08:22.270
So our delta U total should be
equal to 0, which is equal to

00:08:22.270 --> 00:08:25.150
the delta U of expansion.

00:08:25.150 --> 00:08:28.350
So the delta U of expansion is
the delta u to go that way,

00:08:28.350 --> 00:08:33.059
delta U of expansion, plus the
delta U of contraction, which

00:08:33.059 --> 00:08:36.450
is the delta U of going
back like this.

00:08:36.450 --> 00:08:40.909
Those have to be equal to
0 by definition, right?

00:08:40.909 --> 00:08:43.400
Because internal energy is a
state variable, and if we get

00:08:43.400 --> 00:08:46.920
to that same point, our delta
U has to be equal to 0.

00:08:46.920 --> 00:08:49.290
So what's our delta
U of expansion?

00:08:49.290 --> 00:08:53.230
What's our change in internal
energy as we expand?

00:08:53.230 --> 00:08:59.680
Our delta U of expansion is
equal to the heat added to the

00:08:59.680 --> 00:09:06.965
system minus the work
done by the system.

00:09:12.280 --> 00:09:13.870
And we know how much
work was done, this

00:09:13.870 --> 00:09:15.640
whole area right there.

00:09:15.640 --> 00:09:19.670
And then plus the heat added
by the friction.

00:09:19.670 --> 00:09:21.550
There's some heat added
by the friction.

00:09:21.550 --> 00:09:22.800
Let me do that in brown.

00:09:27.210 --> 00:09:29.620
What's that?

00:09:29.620 --> 00:09:34.005
I was on some random website off
the screen, and all of a

00:09:34.005 --> 00:09:35.870
sudden that cartoon
sound started up.

00:09:35.870 --> 00:09:37.660
I have no idea what that was.

00:09:37.660 --> 00:09:39.100
But anyway, where was I?

00:09:39.100 --> 00:09:43.120
So I said our change in internal
energy from expansion

00:09:43.120 --> 00:09:45.310
is going to be the heat added
to the system from our

00:09:45.310 --> 00:09:51.360
reservoir minus the work done
by the system, as we expand,

00:09:51.360 --> 00:09:54.520
plus the heat added to the
system or generated by the

00:09:54.520 --> 00:09:57.220
system-- I guess you could say,
it's not being added.

00:09:57.220 --> 00:10:00.650
The system is creating this heat
itself, as it expands.

00:10:00.650 --> 00:10:02.300
There's this friction
right there.

00:10:02.300 --> 00:10:02.920
Fair enough.

00:10:02.920 --> 00:10:04.750
So this is the one variation.

00:10:04.750 --> 00:10:06.860
Now that we're not dealing with
the reversible process,

00:10:06.860 --> 00:10:08.080
we have this friction.

00:10:08.080 --> 00:10:11.510
Now what's our change in energy
from contraction?

00:10:11.510 --> 00:10:14.740
So our change in internal energy
from contraction is

00:10:14.740 --> 00:10:19.150
going to be the heat that leaves
the system, that has to

00:10:19.150 --> 00:10:21.940
go back into the reservoir
as we contract.

00:10:21.940 --> 00:10:23.660
Because otherwise, if we didn't
have the reservoir, the

00:10:23.660 --> 00:10:24.950
temperature would go up.

00:10:24.950 --> 00:10:26.450
But we want to release heat.

00:10:26.450 --> 00:10:28.640
So we want to say, heat
released-- and

00:10:28.640 --> 00:10:29.430
I'm going to do something.

00:10:29.430 --> 00:10:31.940
Let's just assume that all
of the Q's are positive.

00:10:31.940 --> 00:10:39.000
So if I'm releasing heat, it's
going to be a minus Q release.

00:10:39.000 --> 00:10:41.450
Let's just say that this is a
positive number, and if I'm

00:10:41.450 --> 00:10:43.850
releasing it, it's going to
be a minus right there.

00:10:43.850 --> 00:10:46.470
And I want to just do that, just
to hopefully make things

00:10:46.470 --> 00:10:48.030
little bit clearer.

00:10:48.030 --> 00:10:50.350
Plus the work performed
on the system.

00:10:50.350 --> 00:10:53.040
I'm assuming that work is always
positive, so if we're

00:10:53.040 --> 00:10:54.790
doing work, it'll
be minus work.

00:10:54.790 --> 00:10:57.930
If work is being done to
us, it'll be plus work.

00:10:57.930 --> 00:11:00.290
But in this situation
as well, we're still

00:11:00.290 --> 00:11:01.880
adding heat from friction.

00:11:01.880 --> 00:11:04.840
Or heat from friction is still
being generated in the system.

00:11:04.840 --> 00:11:06.540
This is still positive.

00:11:06.540 --> 00:11:09.230
In either direction, when we
move upwards or downwards, the

00:11:09.230 --> 00:11:11.430
system is generating friction.

00:11:11.430 --> 00:11:14.370
Now, we always said, we went all
the way here, we went all

00:11:14.370 --> 00:11:15.300
the way back.

00:11:15.300 --> 00:11:18.990
So the sum of these has to be
equal to 0, because this is a

00:11:18.990 --> 00:11:20.120
state variable.

00:11:20.120 --> 00:11:23.080
So if the sum of all of this
has to be equal to 0, let's

00:11:23.080 --> 00:11:23.750
sum this up.

00:11:23.750 --> 00:11:28.630
So this gets us to
Qa minus Qr.

00:11:28.630 --> 00:11:31.900
So the heat accepted minus
the heat released.

00:11:31.900 --> 00:11:34.440
The W's cancel out.

00:11:34.440 --> 00:11:41.870
Plus-- let me see right here--
plus 2 times the heat of

00:11:41.870 --> 00:11:43.250
friction in either direction.

00:11:43.250 --> 00:11:46.670
All of that has to
be equal to 0.

00:11:46.670 --> 00:11:47.290
Let's see.

00:11:47.290 --> 00:11:51.120
What we can do is, we can
rewrite this as the heat

00:11:51.120 --> 00:11:56.210
accepted minus the heat released
is equal to minus 2

00:11:56.210 --> 00:11:59.760
times the amount of heat
generated from friction.

00:11:59.760 --> 00:12:03.240
And then if we just switch these
around, we'll get the

00:12:03.240 --> 00:12:08.070
heat released minus the heat
accepted is equal to-- well, I

00:12:08.070 --> 00:12:10.580
just wanted to get all positive
numbers-- 2 times the

00:12:10.580 --> 00:12:11.560
heat of friction.

00:12:11.560 --> 00:12:13.200
Now why did I do all of this?

00:12:13.200 --> 00:12:15.450
Because I wanted to do
an experiment with an

00:12:15.450 --> 00:12:18.810
irreversible system, and this
was a very simple experiment

00:12:18.810 --> 00:12:20.580
with an irreversible system.

00:12:20.580 --> 00:12:25.580
Now, we said that delta S,
which a long time ago I

00:12:25.580 --> 00:12:28.090
defined as Q divided by T-- and
in this video, I said it

00:12:28.090 --> 00:12:29.380
had to be reversible.

00:12:29.380 --> 00:12:33.000
And I wanted to show you right
now that what if I didn't make

00:12:33.000 --> 00:12:36.830
the constraint that this
has to be reversible?

00:12:36.830 --> 00:12:39.000
Because if this doesn't have to
be reversible, and I just

00:12:39.000 --> 00:12:43.640
use this definition right here,
you'll see that your

00:12:43.640 --> 00:12:52.200
delta S here would be-- you just
divide everything by T--

00:12:52.200 --> 00:12:54.870
because our temperature was
constant the entire time, we

00:12:54.870 --> 00:12:58.410
were just on a reservoir--
you'll see that this is going

00:12:58.410 --> 00:13:00.260
to be your delta s.

00:13:00.260 --> 00:13:03.240
This is your total change in
the, I guess you could say,

00:13:03.240 --> 00:13:05.950
your net heat added
to the system.

00:13:05.950 --> 00:13:09.090
So this is, let me say,
this is the heat

00:13:09.090 --> 00:13:11.012
added to the system.

00:13:11.012 --> 00:13:13.290
Let me do it this way.

00:13:13.290 --> 00:13:19.200
Heat added to the system,
divided by the temperature at

00:13:19.200 --> 00:13:20.655
which it was added.

00:13:24.840 --> 00:13:26.290
Which is a positive number.

00:13:26.290 --> 00:13:28.410
Even though we got to
the exact same place

00:13:28.410 --> 00:13:30.060
on this date diagram.

00:13:30.060 --> 00:13:32.780
So in an reversible system,
this wouldn't be

00:13:32.780 --> 00:13:34.220
a valid state variable.

00:13:34.220 --> 00:13:37.830
So it's only a valid state
variable if it's reversible.

00:13:37.830 --> 00:13:40.670
Now, does that mean that you
can only talk about entropy

00:13:40.670 --> 00:13:43.950
for reversible reactions?

00:13:43.950 --> 00:13:44.740
No.

00:13:44.740 --> 00:13:46.300
You can talk about entropy
for anything.

00:13:46.300 --> 00:13:50.050
But what you do is-- and this
is another important point.

00:13:50.050 --> 00:13:57.630
So let's say that I have some
irreversible reaction that

00:13:57.630 --> 00:14:00.730
goes from here to here.

00:14:00.730 --> 00:14:03.110
And I want to figure out
its change in entropy.

00:14:03.110 --> 00:14:03.370
Right?

00:14:03.370 --> 00:14:05.410
And it might have done all
sorts of crazy things.

00:14:05.410 --> 00:14:08.000
It's an irreversible reaction,
its path might

00:14:08.000 --> 00:14:08.630
have gone like that.

00:14:08.630 --> 00:14:11.410
That's assuming it's
quasistatic, if we can even

00:14:11.410 --> 00:14:13.020
look at its path like that.

00:14:13.020 --> 00:14:15.700
If we wanted to figure out its
change in entropy, though, we

00:14:15.700 --> 00:14:18.470
wouldn't worry about the heat
that was added to it and the

00:14:18.470 --> 00:14:20.130
different temperatures at
which it was added.

00:14:20.130 --> 00:14:21.300
We wouldn't worry about that.

00:14:21.300 --> 00:14:22.380
We would just say, OK.

00:14:22.380 --> 00:14:26.900
What would it have taken for a
reversible system to go from

00:14:26.900 --> 00:14:29.100
this state to this state?

00:14:29.100 --> 00:14:31.290
And then maybe a reversible
system would have done

00:14:31.290 --> 00:14:32.490
something like this.

00:14:32.490 --> 00:14:34.710
Sorry, I want to make
it a smooth curve.

00:14:34.710 --> 00:14:37.060
Maybe a reversible system
might have done

00:14:37.060 --> 00:14:38.180
something like that.

00:14:38.180 --> 00:14:43.540
And this change, this heat added
by the reversible system

00:14:43.540 --> 00:14:47.160
divided by the temperature for
the reversible system, would

00:14:47.160 --> 00:14:49.090
be the change in entropy.

00:14:49.090 --> 00:14:56.280
And this change in entropy--
we could call this S final,

00:14:56.280 --> 00:15:00.540
and this is S initial, it's
going to be the same for both

00:15:00.540 --> 00:15:04.060
systems. It's just, we don't use
the irreversible system to

00:15:04.060 --> 00:15:05.310
figure out our entropy.

00:15:05.310 --> 00:15:08.750
We would use the reversible heat
and temperature to figure

00:15:08.750 --> 00:15:10.460
out the actual change.

00:15:10.460 --> 00:15:12.300
Hopefully that clarifies
something else.

00:15:12.300 --> 00:15:15.370
It's, on some level, a subtle
point, but on some level it's

00:15:15.370 --> 00:15:16.140
super important.

00:15:16.140 --> 00:15:18.160
Because you can't-- the
thermodynamic definition of

00:15:18.160 --> 00:15:20.360
entropy has to be this.

00:15:20.360 --> 00:15:23.560
It has to be heat added to a
reversible system divided by

00:15:23.560 --> 00:15:24.820
the temperature that
was added.

00:15:24.820 --> 00:15:26.750
Not just heat to any system.

00:15:26.750 --> 00:15:28.870
It just happened to work when
I did it, and I should have

00:15:28.870 --> 00:15:32.540
been clearer about it when I
first explained it, that it

00:15:32.540 --> 00:15:35.790
worked only because it was
a Carnot cycle, which is

00:15:35.790 --> 00:15:37.040
reversible.