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I am going to continue our
blitz through a review of

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thermodynamics.
And we are going to start,

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right now, talking about
spontaneous change.

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And you all know what
spontaneous change means.

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It means some change that
occurs without intervention.

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Spontaneous change has a
directionality to it.

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For example,
if you put a rock at the top of

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a hill and you let go,
it rolls down,

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it doesn't roll up.
If you put a warm body next to

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a hot body, the heat flows from
the hot body to the cold body,

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and not the other way.
If you, for example,

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have a bulb of gas here and a
stop-cock and an empty bulb,

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you open the stop-cock,
that gas flows from the high

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pressure region to the low
pressure region,

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and it does not go the other
way.

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However, spontaneous change
does not have to be fast.

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You know that if you take a
bottle of ketchup,

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when it used to be a glass
bottle that you couldn't

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squeeze, and you turn it over,
that ketchup has a spontaneous

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tendency to flow out.
But it does not have to be

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fast.
What is, in chemical reactions,

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the key to spontaneity?
Is it the exothermicity?

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Well, certainly iron rusts.
You don't have to do anything

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to make it rust.
A very exothermic reaction.

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Delta H is about 824 kilojoules
per mole.

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If you have an acidic stomach
and you then ingest something

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that is a little bit more basic,
well, this reaction goes.

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It neutralizes that acid,
also an exothermic reaction.

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You did not have to do anything
other than drink the more basic

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solution.
This reaction,

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here, this hydrolysis of ATP to
ADP, adenosine diphosphate,

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--
-- proceeds in every cell of

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your body, and you don't have to
do anything to make that go.

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That is an exothermic reaction.
But what about an endothermic

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reaction?
How about this reaction?

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This is ammonium nitrate
dissolving in water to make

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ammonium ions and nitrate ions.
The delta H for this reaction

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is positive, plus 25.7
kilojoules per mole.

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It is an endothermic reaction.
Is this reaction spontaneous?

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Well, yes.
We are going to do an

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experiment anyway.
And the experiment we are going

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to do is, you are going to be
given one of these cold packs.

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And the TAs can start
distributing them.

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What you have to do is find the
little pouch inside the cold

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pack that contains the water.
You have to press on that,

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allowing the water to disperse
in the material,

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and you will see if the
reaction goes.

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If it goes, that cold pack
better get cold because this is

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an endothermic reaction.
It is going to remove some heat

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from the environment.
Some of you will get hot packs,

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and we will see if that is
spontaneous.

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Is an endothermic reaction
spontaneous?

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Yes.
Is an exothermic reaction

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spontaneous?
Yes, right.

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Is delta H the key to
spontaneity?

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No.
What is?

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Delta G, right.
So, delta G is the key to

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spontaneity.
You already know that when

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delta G is negative,
you have a spontaneous

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reaction.
When delta G is positive,

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you have a non-spontaneous
reaction.

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And very importantly,
when delta G is zero,

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that is when you are at
equilibrium.

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That is what we are going to be
talking about in just a few

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minutes, when delta G is equal
to zero.

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These are the conditions,
of course, for constant

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temperature and pressure.
Delta S, in this expression for

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delta G, is the change in the
entropy.

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It is a measure of disorder.
When delta S is positive,

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we have an increase in the
disorder.

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When delta S is negative,
we have a decrease in the

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disorder.
Where do these conditions come

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from?
You are going to talk about

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that in exquisite detail in
5.60, but I want to give you a

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physical understanding for what
delta G is.

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And that is this.
Suppose you have an exothermic

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reaction.
It turns out that not all of

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that exothermicity,
not all of that enthalpy,

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can be released to the outside
world to do useful work.

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Some of that enthalpy is
actually caught up in the

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product molecules,
or can be caught up in the

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product molecules.
Some of that enthalpy gets

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stuck in the product molecules.
It gets stuck in the vibrations

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and rotations of the product
molecules in the internal

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degrees of freedom.
It is that energy that is not

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available to do useful work.
For some exothermic reactions,

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here, the amount of energy that
is available to do useful work

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is something smaller than delta
H.

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It is delta H minus T delta S,
where T delta S,

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then, is a measure of how much
of that enthalpy is stuck in the

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internal degrees of freedom of
the molecule.

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And so, this delta G is called
the free energy because this is

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the amount of energy that is
free to do useful work.

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That is what delta G is.
For example,

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if you go home and turn on your
gas stove, oxidize some methane,

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burn some natural gas to heat
up the water for your pasta,

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that delta H is very negative.
But let's now calculate delta

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G.
Well, to do that,

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we need to know what delta S
is.

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Delta S, I will show you in
just a moment,

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is the change in the entropy.
We can calculate that.

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But knowing delta S then,
we can plug in the delta H and

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the temperature.
What we find is that delta G

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here is less negative than delta
H.

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Not all of this enthalpy,
not all of that energy is

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available to the outside world
to do useful work.

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Some of it, in this reaction,
gets trapped or caught in the

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internal degrees of freedom of a
molecule.

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00:08:04,000 --> 00:08:08,000
And it is not available to do
that useful work.

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Delta G, in this case,
is less negative than delta H.

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But let's take the eat,
breathe, exhale,

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pee reaction.
The oxidation of glucose,

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going on in every cell of your
body, this reaction also very

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exothermic.
Let's calculate delta G for

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this reaction.
Well, first of all,

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we need delta S.
From that, we can calculate

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delta G.
And, in this case,

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what do you see?
In this case,

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you see that delta G is more
negative than delta H.

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In this exothermic reaction,
we are getting back all of the

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enthalpy and then some.
We are getting back even more.

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And that extra energy is coming
from the energy that is tied up

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in the internal degrees of
freedom of the reactants.

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In this reaction,
here, delta S is positive.

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We are increasing the disorder.
If we increase the disorder,

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there are fewer ways to tie up
energy in the internal degrees

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of freedom of the molecule.
When we increase the disorder,

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that is what happens.
And so, your body has figured

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out how to get back all of that
delta H, and then some.

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We are getting that energy out
that is locked up in the

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reactants, in the internal
degrees of freedom of the

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reactants.
That is a physical feeling,

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there, for what delta G is.
Entropy for a reaction.

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The entropy is calculated from
the absolute entropies of the

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products and the reactants.
And so, what you need are the

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absolute entropies for each one
of the products multiplied by

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the appropriate stoichiometric
number.

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You sum them all up.
And you do the same then for

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the reactants,
the absolute entropy of each

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one of the reactants multiplied
by that stoichiometric number.

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Sum them up.
Subtract it.

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That is the entropy change for
the reaction.

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Now, one thing I do want to
point out is that in the case of

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entropies, here,
these are absolute entropies.

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That is rather unusual in
thermodynamics.

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In thermodynamics usually we
look at changes in energies,

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but here we are talking about
absolute entropies.

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That is a consequence of the
third law of thermodynamics.

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You will see how that arises
again in 5.60,

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but right now we are going to
use it to calculate the entropy

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for a reaction.
And some reactions here are

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completely entropy-driven.
For example,

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if you melt ice,
this is an endothermic

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reaction, 6.95 kilojoules per
mole.

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In going from solid water to
liquid water,

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you have to put energy into the
system to do that.

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But you know that this reaction
occurs spontaneously.

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If I had a cube of ice here,
it would melt.

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This reaction is
entropy-driven.

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Why?
Let's look at it.

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Let's calculate delta S for
that reaction.

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You need the absolutely entropy
for water in the liquid phase,

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that is the product,
minus the absolute entropy for

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water in the solid phase.
The difference between the two

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is positive.
We have increased the disorder.

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There are less ways to tie up
energy in liquid water than

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there are in solid water,
in ice.

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And that fact is going to make
this reaction then spontaneous,

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6.95 positive for delta H minus
T delta S.

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This term, T delta S,
now is larger in absolute

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magnitude.
Therefore, delta G is negative.

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This reaction is spontaneous.
It is certainly entropy-driven.

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The other concept I want to
talk about is the free energy of

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formation, which is analogous to
what we talked about last time,

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00:13:02,000 --> 00:13:09,000
the enthalpy of formation.
The free energy of formation is

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00:13:09,000 --> 00:13:14,000
the free energy for a reaction
that forms one mole of a

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compound from the elements in
their most stable form in the

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standard state.
And our standard state is one

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bar.
The free energy of formation,

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here, is tabulated like the
heats of formation.

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You can find tables of free
energy and enthalpy of formation

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for all molecules.
However, unlike the enthalpy of

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formation, the free energy of
formation can also be calculated

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easily by you,
which is the following,

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if you know the enthalpy of
formation for some reaction.

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00:14:00,000 --> 00:14:04,000
Or, for the reaction that
defines the enthalpy of

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00:14:04,000 --> 00:14:07,000
formation.
And if you know what the

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00:14:07,000 --> 00:14:12,000
entropy change is for that
reaction that defines the

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00:14:12,000 --> 00:14:17,000
enthalpy of formation,
then you can calculate the free

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00:14:17,000 --> 00:14:21,000
energy here of formation.
You have two choices in the

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00:14:21,000 --> 00:14:25,000
case of the free energy of
formation.

196
00:14:25,000 --> 00:14:30,000
You can look it up.
Or, if you know what delta H

197
00:14:30,000 --> 00:14:35,000
sub f and delta S
are for the reaction,

198
00:14:35,000 --> 00:14:41,000
that defines the free energy of
formation or the enthalpy of

199
00:14:41,000 --> 00:14:45,000
formation.
You can calculate the free

200
00:14:45,000 --> 00:14:47,000
energy of formation,
here.

201
00:14:47,000 --> 00:14:53,000
Here is a reaction that forms
one mole of CO two.

202
00:14:53,000 --> 00:14:58,000
And the delta G for this
reaction is minus 394 kilojoules

203
00:14:58,000 --> 00:15:03,000
per mole.
The delta G for this reaction

204
00:15:03,000 --> 00:15:09,000
is defined as the free energy of
formation for CO two.

205
00:15:09,000 --> 00:15:14,000
It is so because we are forming
one mole of this molecule from

206
00:15:14,000 --> 00:15:19,000
the elements in their most
stable form in the standard

207
00:15:19,000 --> 00:15:22,000
state.
The most stable form of the

208
00:15:22,000 --> 00:15:26,000
element carbon is graphite.
The most stable form of the

209
00:15:26,000 --> 00:15:31,000
element oxygen is molecular
oxygen.

210
00:15:31,000 --> 00:15:36,000
But what is important about the
free energy of formation,

211
00:15:36,000 --> 00:15:42,000
here, is that it is a measure
of a compound stability relative

212
00:15:42,000 --> 00:15:46,000
to decomposition to its
elements.

213
00:15:46,000 --> 00:15:50,000
For example,
the free energy of formation of

214
00:15:50,000 --> 00:15:55,000
carbon dioxide is negative.
What that means is CO two

215
00:15:55,000 --> 00:16:01,000
is thermodynamically
stable relative to decomposition

216
00:16:01,000 --> 00:16:07,000
into its elements.
Because delta G for this

217
00:16:07,000 --> 00:16:13,000
reaction is written as negative.
The reaction is spontaneous in

218
00:16:13,000 --> 00:16:18,000
this direction.
It is not spontaneous in the

219
00:16:18,000 --> 00:16:22,000
reverse direction.
That is, the decomposition of

220
00:16:22,000 --> 00:16:28,000
CO two to its elements
is not a spontaneous process.

221
00:16:28,000 --> 00:16:34,000
Therefore, we say that CO two
is thermodynamically stable

222
00:16:34,000 --> 00:16:40,000
relative to decomposition into
its elements.

223
00:16:40,000 --> 00:16:44,000
If you have a free energy of
formation for a molecule that is

224
00:16:44,000 --> 00:16:47,000
negative, it is
thermodynamically stable,

225
00:16:47,000 --> 00:16:52,000
but if you have a free energy
of formation that is positive,

226
00:16:52,000 --> 00:16:56,000
then that compound is not
thermodynamically stable

227
00:16:56,000 --> 00:17:00,000
relative to decomposition to its
elements.

228
00:17:00,000 --> 00:17:03,000
For example,
let's look at benzene.

229
00:17:03,000 --> 00:17:09,000
Here is a reaction that forms
one mole of liquid benzene from

230
00:17:09,000 --> 00:17:14,000
the elements in their most
stable form in the standard

231
00:17:14,000 --> 00:17:18,000
state, graphite and molecular
hydrogen.

232
00:17:18,000 --> 00:17:24,000
The free energy of formation
for this benzene is positive

233
00:17:24,000 --> 00:17:31,000
kilojoules per mole.
That means that this forward

234
00:17:31,000 --> 00:17:36,000
reaction, as written,
is not spontaneous.

235
00:17:36,000 --> 00:17:43,000
However, the reverse reaction
is spontaneous because delta G

236
00:17:43,000 --> 00:17:47,000
is minus 124 kilojoules per
mole.

237
00:17:47,000 --> 00:17:52,000
And so we say benzene,
here, is thermodynamically

238
00:17:52,000 --> 00:17:58,000
unstable relative to its
elements.

239
00:17:58,000 --> 00:18:02,000
But, even though you might have
a reaction, here,

240
00:18:02,000 --> 00:18:08,000
that is spontaneous,
that is the reverse reaction.

241
00:18:08,000 --> 00:18:12,000
You have a delta G formation
that is positive,

242
00:18:12,000 --> 00:18:17,000
meaning the reverse reaction is
spontaneous.

243
00:18:17,000 --> 00:18:22,000
Even though that reverse
reaction may be spontaneous,

244
00:18:22,000 --> 00:18:27,000
it also could be very,
very slow.

245
00:18:27,000 --> 00:18:32,000
When was the last time you saw
a pint of benzene decompose into

246
00:18:32,000 --> 00:18:37,000
graphite and hydrogen?
You haven't seen that.

247
00:18:37,000 --> 00:18:40,000
I haven't seen that and I am a
lot older.

248
00:18:40,000 --> 00:18:45,000
What this means is that even
though you might have a

249
00:18:45,000 --> 00:18:50,000
thermodynamic tendency to
compose, it does not mean that

250
00:18:50,000 --> 00:18:56,000
it is going to happen because
there is a question of the rate

251
00:18:56,000 --> 00:19:02,000
of the chemical reaction.
We have two different terms to

252
00:19:02,000 --> 00:19:08,000
describe the thermodynamic
tendency and then the kinetic

253
00:19:08,000 --> 00:19:12,000
tendency, and these terms are
the following.

254
00:19:12,000 --> 00:19:18,000
We call the molecules stable or
unstable when we are referring

255
00:19:18,000 --> 00:19:23,000
to the thermodynamic tendency.
That is when we are referring

256
00:19:23,000 --> 00:19:29,000
to delta G or the free energy of
formation.

257
00:19:29,000 --> 00:19:32,000
If it is negative,
the molecule is stable.

258
00:19:32,000 --> 00:19:36,000
If it is positive,
the molecule is unstable

259
00:19:36,000 --> 00:19:41,000
relative to decomposition.
However, we also have to

260
00:19:41,000 --> 00:19:45,000
consider the rates.
And we have another term to

261
00:19:45,000 --> 00:19:49,000
describe the rates.
And that is labile and

262
00:19:49,000 --> 00:19:52,000
non-labile.
If a molecule is non-labile,

263
00:19:52,000 --> 00:19:57,000
what that means is that the
rate, at which the thermodynamic

264
00:19:57,000 --> 00:20:03,000
tendency is realized,
is very, very slow.

265
00:20:03,000 --> 00:20:07,000
Benzene, for example,
is thermodynamically unstable

266
00:20:07,000 --> 00:20:11,000
relative to decomposition into
its elements,

267
00:20:11,000 --> 00:20:15,000
carbon and hydrogen,
but it is non-labile.

268
00:20:15,000 --> 00:20:21,000
The rate at which that reaction
happens is just too slow to be

269
00:20:21,000 --> 00:20:24,000
of any practical interest to
you.

270
00:20:24,000 --> 00:20:30,000
However, if this decomposition
reaction occurred very readily,

271
00:20:30,000 --> 00:20:35,000
then we would call that
molecule labile.

272
00:20:35,000 --> 00:20:41,000
That is, the rate of the
decomposition is something that

273
00:20:41,000 --> 00:20:48,000
is going to affect your ability
to handle that particular

274
00:20:48,000 --> 00:20:55,000
compound or that molecule.
Like the enthalpy of formation,

275
00:20:55,000 --> 00:21:02,000
the free energy of formation
can also be zero.

276
00:21:02,000 --> 00:21:07,000
Last time, we talked about how
the enthalpy of formation for

277
00:21:07,000 --> 00:21:10,000
hydrogen, oxygen,
chlorine, xenon,

278
00:21:10,000 --> 00:21:14,000
all of that,
had zero enthalpy of formation.

279
00:21:14,000 --> 00:21:18,000
Well, all of these molecules
have zero free energy of

280
00:21:18,000 --> 00:21:24,000
formation because they are all
in the most stable form already,

281
00:21:24,000 --> 00:21:29,000
in the standard state.
The free energy of formation of

282
00:21:29,000 --> 00:21:34,000
bromine liquid and
mercury liquid is

283
00:21:34,000 --> 00:21:38,000
zero.
Because they are in the most

284
00:21:38,000 --> 00:21:41,000
stable form at the standard
state.

285
00:21:41,000 --> 00:21:46,000
The free energy of formation of
carbon in the form of graphite,

286
00:21:46,000 --> 00:21:49,000
sodium solid,

287
00:21:49,000 --> 00:21:53,000
iron solid, iodine solid.

288
00:21:53,000 --> 00:21:57,000
Those free energies of
formation are all equal to zero,

289
00:21:57,000 --> 00:22:02,000
but the free energy of
formation here of bromine in the

290
00:22:02,000 --> 00:22:09,000
gas phase is not equal to zero.
Because bromine gas is not the

291
00:22:09,000 --> 00:22:12,000
most stable form in our standard
state.

292
00:22:12,000 --> 00:22:17,000
Liquid bromine is,
so it is going to take energy

293
00:22:17,000 --> 00:22:23,000
to produce bromine gas.
So Delta G of formation is not

294
00:22:23,000 --> 00:22:27,000
going to be zero.
Likewise, the free energy of

295
00:22:27,000 --> 00:22:33,000
formation of diamond is not
equal to zero.

296
00:22:33,000 --> 00:22:36,000
Because carbon,
in the form of diamond,

297
00:22:36,000 --> 00:22:40,000
is not the most stable form of
carbon.

298
00:22:40,000 --> 00:22:44,000
Graphite is.
This is going to take energy to

299
00:22:44,000 --> 00:22:47,000
form that configuration of
carbon.

300
00:22:47,000 --> 00:22:53,000
So free energies of formation
are very important in terms of

301
00:22:53,000 --> 00:23:00,000
understanding the thermodynamic
stability of molecules.

302
00:23:00,000 --> 00:23:01,000
Just a review,
here.

303
00:23:01,000 --> 00:23:05,000
If you want to calculate delta
G for a reaction,

304
00:23:05,000 --> 00:23:08,000
what can you do?
You can take the free energy of

305
00:23:08,000 --> 00:23:13,000
formation for each one of the
products, multiply it by the

306
00:23:13,000 --> 00:23:17,000
appropriate stoichiometric
number, since those free

307
00:23:17,000 --> 00:23:21,000
energies of formation are given
per mole, add them all up,

308
00:23:21,000 --> 00:23:24,000
and then do the same thing for
the reactants.

309
00:23:24,000 --> 00:23:30,000
Free energies of formation for
each of the reactants.

310
00:23:30,000 --> 00:23:32,000
Stoichiometric number.
Add them all up.

311
00:23:32,000 --> 00:23:36,000
Subtract the two.
That is the delta G for your

312
00:23:36,000 --> 00:23:40,000
chemical reaction.
Like delta H for the chemical

313
00:23:40,000 --> 00:23:44,000
reaction, which you calculated
from the differences in the

314
00:23:44,000 --> 00:23:48,000
heats of formation,
the products and reactants,

315
00:23:48,000 --> 00:23:51,000
note that this is products
minus reactants.

316
00:23:51,000 --> 00:23:55,000
This is different from when you
calculate delta H,

317
00:23:55,000 --> 00:23:59,000
for example,
from bond enthalpies.

318
00:23:59,000 --> 00:24:01,000
That was reactants minus
products.

319
00:24:01,000 --> 00:24:06,000
Something to remember.
But again, in the case of delta

320
00:24:06,000 --> 00:24:11,000
G, we can calculate it from
knowing the free energies of

321
00:24:11,000 --> 00:24:14,000
formation of the reactants and
the products.

322
00:24:14,000 --> 00:24:19,000
But we can also calculate it
from this expression.

323
00:24:19,000 --> 00:24:22,000
If, for example,
you know the enthalpy change

324
00:24:22,000 --> 00:24:28,000
for a reaction and you know the
entropy change for a reaction,

325
00:24:28,000 --> 00:24:32,000
you can calculate the free
energy change from these two

326
00:24:32,000 --> 00:24:37,000
quantities.
You have two choices to

327
00:24:37,000 --> 00:24:43,000
calculate the free energy change
of some chemical reaction,

328
00:24:43,000 --> 00:24:49,000
depending on what information
is known or is given to you.

329
00:24:49,000 --> 00:24:55,000
Now, I want to talk kind of
briefly about controlling the

330
00:24:55,000 --> 00:25:00,000
spontaneity of a chemical
reaction.

331
00:25:00,000 --> 00:25:04,000
And, to do so,
I am going to talk about this

332
00:25:04,000 --> 00:25:07,000
molecule, which is sodium
bicarbonate.

333
00:25:07,000 --> 00:25:12,000
Sodium bicarbonate is better
known as baking soda.

334
00:25:12,000 --> 00:25:17,000
It is what you put into some
batter of soft baked goods that

335
00:25:17,000 --> 00:25:22,000
you want to make.
What happens is that the sodium

336
00:25:22,000 --> 00:25:27,000
bicarbonate decomposes and two
of the products are gases,

337
00:25:27,000 --> 00:25:32,000
CO two and water.
And so, in your batter,

338
00:25:32,000 --> 00:25:37,000
what you get are these bubbles
of CO two and water.

339
00:25:37,000 --> 00:25:42,000
And what happens is that the
dough starts to harden around

340
00:25:42,000 --> 00:25:45,000
these bubbles.
And then, of course,

341
00:25:45,000 --> 00:25:49,000
eventually, the CO2 and water
leave, are completely driven

342
00:25:49,000 --> 00:25:54,000
off, but what that leaves behind
in the batter is a very porous

343
00:25:54,000 --> 00:25:59,000
structure, so porous that after
you remove it from the oven you

344
00:25:59,000 --> 00:26:04,000
can actually put your teeth into
it.

345
00:26:04,000 --> 00:26:08,000
Have you ever left out the
baking soda in some soft baked

346
00:26:08,000 --> 00:26:12,000
goods?
You cannot get your teeth into

347
00:26:12,000 --> 00:26:15,000
it, right?
So, this is an important

348
00:26:15,000 --> 00:26:18,000
reaction.
But this is a reaction,

349
00:26:18,000 --> 00:26:22,000
here, that is very endothermic.
It does increase in the

350
00:26:22,000 --> 00:26:27,000
entropy, but at room
temperature, this reaction has a

351
00:26:27,000 --> 00:26:33,000
positive delta G.
This reaction does not proceed.

352
00:26:33,000 --> 00:26:37,000
It is non-spontaneous in the
forward direction,

353
00:26:37,000 --> 00:26:41,000
as written.
And that is good because you do

354
00:26:41,000 --> 00:26:47,000
not want this reaction to start
when your batter is still on

355
00:26:47,000 --> 00:26:53,000
your kitchen counter.
You want it to start when it is

356
00:26:53,000 --> 00:26:57,000
in the oven.
What happens as we raise the

357
00:26:57,000 --> 00:27:02,000
temperature, here?
For the purposes of what we are

358
00:27:02,000 --> 00:27:07,000
going to do, we have the same
delta H and delta S.

359
00:27:07,000 --> 00:27:11,000
But now, when we calculate
delta G, the temperature,

360
00:27:11,000 --> 00:27:16,000
here, is going to be our baking
temperature of 350 degrees

361
00:27:16,000 --> 00:27:20,000
Fahrenheit.
And now, what you see is delta

362
00:27:20,000 --> 00:27:24,000
G is negative,
minus 14 kilojoules per mole.

363
00:27:24,000 --> 00:27:27,000
All of a sudden,
this reaction became

364
00:27:27,000 --> 00:27:33,000
spontaneous.
And we made it spontaneous

365
00:27:33,000 --> 00:27:39,000
simply by increasing the
absolute magnitude of T delta S.

366
00:27:39,000 --> 00:27:44,000
We increased it by increasing
the temperature,

367
00:27:44,000 --> 00:27:49,000
in this case.
We made it spontaneous by

368
00:27:49,000 --> 00:27:54,000
raising the temperature.
For some reactions,

369
00:27:54,000 --> 00:27:59,000
we can do that.
We can find a temperature at

370
00:27:59,000 --> 00:28:05,000
which the reaction will be
spontaneous.

371
00:28:05,000 --> 00:28:09,000
Let's look at the particular
conditions.

372
00:28:09,000 --> 00:28:14,000
Delta G is a linear function of
the temperature,

373
00:28:14,000 --> 00:28:19,000
so let me draw that for this
particular case,

374
00:28:19,000 --> 00:28:24,000
for the sodium bicarbonate
decomposition.

375
00:28:24,000 --> 00:28:30,000
The slope of this line for this
reaction is, of course,

376
00:28:30,000 --> 00:28:35,000
minus delta S.
The intercept is delta H.

377
00:28:35,000 --> 00:28:39,000
And now, let me just draw a
zero here.

378
00:28:39,000 --> 00:28:45,000
What I want you to notice is
that there is some temperature

379
00:28:45,000 --> 00:28:49,000
that I am going to call
T star.

380
00:28:49,000 --> 00:28:55,000
There is some temperature at
which the sign of delta G

381
00:28:55,000 --> 00:28:58,000
changes.
You see that below T star,

382
00:28:58,000 --> 00:29:04,000
delta G is positive.
Below those temperatures,

383
00:29:04,000 --> 00:29:08,000
such as room temperature for
the sodium bicarbonate

384
00:29:08,000 --> 00:29:11,000
decomposition,
delta G is positive.

385
00:29:11,000 --> 00:29:14,000
We have a non-spontaneous
reaction.

386
00:29:14,000 --> 00:29:18,000
Above some temperature delta G
has become negative.

387
00:29:18,000 --> 00:29:23,000
We have a spontaneous reaction.
We can calculate the value of

388
00:29:23,000 --> 00:29:26,000
this T star, where the
spontaneity switches,

389
00:29:26,000 --> 00:29:31,000
by setting delta G nought
equal to zero

390
00:29:31,000 --> 00:29:36,000
and then solving for T star.
Let's do that.

391
00:29:36,000 --> 00:29:41,000
I rearrange this to give me T
star is equal to delta H over

392
00:29:41,000 --> 00:29:43,000
delta S.
For this reaction,

393
00:29:43,000 --> 00:29:46,000
I know the values of delta H
and delta S.

394
00:29:46,000 --> 00:29:51,000
And for this reaction that
value of T star is 406 Kelvin.

395
00:29:51,000 --> 00:29:56,000
So, your oven has to be higher
in temperature than 406 Kelvin

396
00:29:56,000 --> 00:30:01,000
if you want to make some baked
goods that you can put your

397
00:30:01,000 --> 00:30:06,000
teeth into.
For some reaction,

398
00:30:06,000 --> 00:30:11,000
in this particular case here
that is endothermic,

399
00:30:11,000 --> 00:30:17,000
delta H greater than zero,
in which case we have an

400
00:30:17,000 --> 00:30:22,000
increase in entropy,
that reaction is spontaneous

401
00:30:22,000 --> 00:30:30,000
whenever the temperature is
above some temperature T star.

402
00:30:30,000 --> 00:30:35,000
But we also have reactions
which are exothermic,

403
00:30:35,000 --> 00:30:40,000
delta H less than zero,
and reactions in which the

404
00:30:40,000 --> 00:30:45,000
entropy decreases.
Delta S is less than zero.

405
00:30:45,000 --> 00:30:51,000
And I have plotted that line
here, delta G versus T.

406
00:30:51,000 --> 00:30:57,000
For the opposite condition,
we have the opposite slope

407
00:30:57,000 --> 00:31:02,000
because delta S has changed
sign.

408
00:31:02,000 --> 00:31:04,000
In this case,
the reaction will be

409
00:31:04,000 --> 00:31:08,000
spontaneous whenever the
temperature is below some

410
00:31:08,000 --> 00:31:12,000
temperature T star.
And you can see that by looking

411
00:31:12,000 --> 00:31:15,000
at the relative signs in this
expression.

412
00:31:15,000 --> 00:31:20,000
If delta H is less than zero up
here, then we have a negative

413
00:31:20,000 --> 00:31:24,000
quantity right here.
And, since delta S is less than

414
00:31:24,000 --> 00:31:30,000
zero, we have minus times a
minus, which is a plus.

415
00:31:30,000 --> 00:31:35,000
This term here is positive,
this term is negative,

416
00:31:35,000 --> 00:31:40,000
and so this positive term
better not be in absolute value

417
00:31:40,000 --> 00:31:44,000
very much larger than delta H
here.

418
00:31:44,000 --> 00:31:49,000
If it is, then we are going to
have a positive delta G,

419
00:31:49,000 --> 00:31:55,000
but if this term is small
compared to the absolute value

420
00:31:55,000 --> 00:32:01,000
of delta H, then we will have a
spontaneous reaction at low

421
00:32:01,000 --> 00:32:06,000
temperatures.
So, you want to keep this term

422
00:32:06,000 --> 00:32:09,000
small.
That is why it is spontaneous

423
00:32:09,000 --> 00:32:14,000
when T is less than T star.
But we can also have reactions

424
00:32:14,000 --> 00:32:18,000
that are exothermic and
reactions in which the entropy

425
00:32:18,000 --> 00:32:20,000
increases.
In this case,

426
00:32:20,000 --> 00:32:23,000
you have the best of all
worlds.

427
00:32:23,000 --> 00:32:26,000
In this case,
this reaction is spontaneous at

428
00:32:26,000 --> 00:32:30,000
all temperatures.
You cannot adjust the

429
00:32:30,000 --> 00:32:35,000
spontaneity of this reaction by
temperature.

430
00:32:35,000 --> 00:32:37,000
This is always going to be
negative.

431
00:32:37,000 --> 00:32:41,000
Delta H is negative here,
minus T delta S.

432
00:32:41,000 --> 00:32:44,000
This is going to be minus a
positive number.

433
00:32:44,000 --> 00:32:48,000
A negative minus a positive
number is always a negative

434
00:32:48,000 --> 00:32:52,000
number, always spontaneous.
And then, finally,

435
00:32:52,000 --> 00:32:57,000
we can have a condition where
we have an endothermic reaction,

436
00:32:57,000 --> 00:33:00,000
delta H is greater than zero,
and where the entropy

437
00:33:00,000 --> 00:33:05,000
decreases, delta S is less than
zero.

438
00:33:05,000 --> 00:33:07,000
In this case,
the reaction is never

439
00:33:07,000 --> 00:33:11,000
spontaneous, no matter what
temperature you do.

440
00:33:11,000 --> 00:33:14,000
You cannot adjust that
non-spontaneity with

441
00:33:14,000 --> 00:33:17,000
temperature.
Delta H, you can see that,

442
00:33:17,000 --> 00:33:21,000
is going to be positive.
And since delta S is negative,

443
00:33:21,000 --> 00:33:25,000
we have a minus times a minus,
that is positive,

444
00:33:25,000 --> 00:33:29,000
delta G is positive at all
temperatures.

445
00:33:29,000 --> 00:33:34,000
So we have failed to adjust the
spontaneity with temperature for

446
00:33:34,000 --> 00:33:38,000
some reactions that are
endothermic and in which the

447
00:33:38,000 --> 00:33:40,000
entropy decreases.

448
00:33:45,000 --> 00:33:56,000
What we have been talking about
are values of delta G that are

449
00:33:56,000 --> 00:34:02,000
standard state.
We have been talking about

450
00:34:02,000 --> 00:34:06,000
delta G nought,
one bar pressure.

451
00:34:06,000 --> 00:34:10,000
That is what we have been
talking about.

452
00:34:10,000 --> 00:34:17,000
What that means is that the
delta G nought for any reaction

453
00:34:17,000 --> 00:34:23,000
means that we have one
atmosphere each of the products

454
00:34:23,000 --> 00:34:28,000
and the reactants.
That is what delta G nought

455
00:34:28,000 --> 00:34:32,000
means.
That our famous reaction here,

456
00:34:32,000 --> 00:34:35,000
argon-boron,
argon plus boron going to

457
00:34:35,000 --> 00:34:39,000
carbon plus deuterium.
Let's take this reaction.

458
00:34:39,000 --> 00:34:44,000
When I give you a delta G
nought for this famous reaction,

459
00:34:44,000 --> 00:34:48,000
what that means is that that is
the delta G nought when the

460
00:34:48,000 --> 00:34:52,000
partial pressure of the reactant
A is one bar,

461
00:34:52,000 --> 00:34:57,000
and the partial pressure of
reactant B, is one bar and the

462
00:34:57,000 --> 00:35:02,000
partial pressure of product C is
one bar, and the partial

463
00:35:02,000 --> 00:35:06,000
pressure of product D is one
bar.

464
00:35:06,000 --> 00:35:11,000
That does not seem real useful,
because say you were going to

465
00:35:11,000 --> 00:35:17,000
start a reaction and you just
start it even with one bar of A

466
00:35:17,000 --> 00:35:21,000
and one bar of B,
this is not the delta G under

467
00:35:21,000 --> 00:35:27,000
those particular conditions.
We have to be able to calculate

468
00:35:27,000 --> 00:35:33,000
delta G under any conditions.
Not just our standard state

469
00:35:33,000 --> 00:35:36,000
conditions.
And that is what we are going

470
00:35:36,000 --> 00:35:41,000
to do with this expression.
We are going to be able to

471
00:35:41,000 --> 00:35:45,000
calculate a delta G at any time
during a reaction.

472
00:35:45,000 --> 00:35:50,000
We can stop a reaction,
and we can measure the partial

473
00:35:50,000 --> 00:35:54,000
pressures of the reactants and
the products present,

474
00:35:54,000 --> 00:35:59,000
and we can get from that,
then, a delta G for a reaction

475
00:35:59,000 --> 00:36:05,000
at that particular point.
That is what we are going to

476
00:36:05,000 --> 00:36:08,000
do, here.
What have I got written here?

477
00:36:08,000 --> 00:36:12,000
I have delta G is equal to this
delta G nought,

478
00:36:12,000 --> 00:36:16,000
the standard state delta G,
plus RT ln of the partial

479
00:36:16,000 --> 00:36:22,000
pressure of product C over some
reference pressure raised to the

480
00:36:22,000 --> 00:36:26,000
appropriate stoichiometric
number, times the partial

481
00:36:26,000 --> 00:36:31,000
pressure of the other reactant,
divided by some reference

482
00:36:31,000 --> 00:36:35,000
pressure raised to the
appropriate stoichiometric

483
00:36:35,000 --> 00:36:39,000
number.
And that is all over the

484
00:36:39,000 --> 00:36:44,000
product of the partial pressure
of A over that reference

485
00:36:44,000 --> 00:36:49,000
pressure, raised to the
appropriate stoichiometric

486
00:36:49,000 --> 00:36:53,000
number, times the same for
reactant B.

487
00:36:53,000 --> 00:36:58,000
What this ratio here of partial
pressures is,

488
00:36:58,000 --> 00:37:02,000
or what it is called,
is the reaction quotient,

489
00:37:02,000 --> 00:37:06,000
Q.
It is the ratio of the partial

490
00:37:06,000 --> 00:37:11,000
pressures present during any
time during the reaction.

491
00:37:11,000 --> 00:37:15,000
Not necessarily at equilibrium
-- at any time during the

492
00:37:15,000 --> 00:37:18,000
reaction.
That is what Q is here.

493
00:37:18,000 --> 00:37:21,000
And so in this way,
with this expression,

494
00:37:21,000 --> 00:37:26,000
we are going to be able to
calculate what delta G is for

495
00:37:26,000 --> 00:37:31,000
the reaction at any time outside
of this special standard state

496
00:37:31,000 --> 00:37:36,000
of one bar pressure for each one
of the reactants,

497
00:37:36,000 --> 00:37:42,000
one bar partial pressure for
each one of the products.

498
00:37:42,000 --> 00:37:46,000
Where this expression comes
from is, again,

499
00:37:46,000 --> 00:37:50,000
discussed in exquisite detail
in 5.60.

500
00:37:50,000 --> 00:37:56,000
You will see where it comes
from later, but right now we are

501
00:37:56,000 --> 00:38:02,000
going to use it to calculate
delta G at any time during the

502
00:38:02,000 --> 00:38:06,000
reaction.
But now let me point out that

503
00:38:06,000 --> 00:38:09,000
our reference pressure right
here is one bar.

504
00:38:09,000 --> 00:38:13,000
What I am going to do is
substitute a one in here for

505
00:38:13,000 --> 00:38:18,000
each one of the reference
pressures, so that my reaction

506
00:38:18,000 --> 00:38:20,000
quotient is going to look like
this.

507
00:38:20,000 --> 00:38:25,000
And I did not put in the bars
because what you are going to do

508
00:38:25,000 --> 00:38:31,000
is put in the pressures in bars.
And then, the reaction quotient

509
00:38:31,000 --> 00:38:35,000
will be just fine.
You have to use bar pressure

510
00:38:35,000 --> 00:38:37,000
here, since that is our standard
state.

511
00:38:37,000 --> 00:38:42,000
That is our reference pressure.
Now, keep this in mind.

512
00:38:42,000 --> 00:38:45,000
This is important.
We are going to use this again

513
00:38:45,000 --> 00:38:49,000
in just a moment.
Let's start talking about the

514
00:38:49,000 --> 00:38:53,000
equilibrium constant.
Here is our famous A plus B

515
00:38:53,000 --> 00:38:58,000
going to C plus D reaction.

516
00:38:58,000 --> 00:39:03,000
We already said if delta G,
for this reaction as written is

517
00:39:03,000 --> 00:39:07,000
negative, then that forward
reaction is spontaneous.

518
00:39:07,000 --> 00:39:13,000
If delta G for that reaction is
positive for the reaction as

519
00:39:13,000 --> 00:39:18,000
written, what that means is
then, the reverse reaction is

520
00:39:18,000 --> 00:39:21,000
spontaneous.
And now, very importantly,

521
00:39:21,000 --> 00:39:25,000
if delta G is zero,
then we are at chemical

522
00:39:25,000 --> 00:39:29,000
equilibrium.
Notice this is delta G equal to

523
00:39:29,000 --> 00:39:33,000
zero.
This is not delta G nought

524
00:39:33,000 --> 00:39:37,000
equal to zero.
A big difference here.

525
00:39:37,000 --> 00:39:43,000
This is delta G equal to zero.
Then, we are at equilibrium.

526
00:39:43,000 --> 00:39:48,000
What I am going to do is I am
going to bring back my

527
00:39:48,000 --> 00:39:53,000
expression that allowed me to
calculate delta G at any

528
00:39:53,000 --> 00:40:00,000
arbitrary pressures of the
reactants and the products.

529
00:40:00,000 --> 00:40:03,000
And I just said that
equilibrium, here,

530
00:40:03,000 --> 00:40:07,000
is defined by delta G being
equal to zero,

531
00:40:07,000 --> 00:40:11,000
so I am going to plug in a zero
right in there.

532
00:40:11,000 --> 00:40:15,000
If that is the case,
then I am going to rearrange

533
00:40:15,000 --> 00:40:19,000
this equation.
I am going to bring delta G

534
00:40:19,000 --> 00:40:23,000
over to the other side,
and that is what I do here on

535
00:40:23,000 --> 00:40:27,000
the next slide.
I have delta G nought is equal

536
00:40:27,000 --> 00:40:31,000
to minus RT ln of this reaction
quotient.

537
00:40:31,000 --> 00:40:37,000
Now, we have a special

538
00:40:37,000 --> 00:40:41,000
condition here.
And that is when delta G is

539
00:40:41,000 --> 00:40:45,000
equal to zero,
when we are at thermodynamic

540
00:40:45,000 --> 00:40:51,000
equilibrium this reaction
quotient has a special name.

541
00:40:51,000 --> 00:40:56,000
That special name is the
equilibrium constant,

542
00:40:56,000 --> 00:40:59,000
K.
This reaction quotient is equal

543
00:40:59,000 --> 00:41:05,000
to the thermodynamic equilibrium
constant.

544
00:41:05,000 --> 00:41:10,000
And it is so only when delta G
is equal to zero.

545
00:41:10,000 --> 00:41:13,000
Not delta G nought equal to
zero.

546
00:41:13,000 --> 00:41:18,000
We get the equation,
delta G nought is equal to

547
00:41:18,000 --> 00:41:23,000
minus RT ln of K.

548
00:41:23,000 --> 00:41:30,000
That is where this particular
expression comes from.

549
00:41:30,000 --> 00:41:34,000
So, the reaction quotient is
equal to K when we are at

550
00:41:34,000 --> 00:41:39,000
thermodynamic equilibrium.
Now what we are going to do is

551
00:41:39,000 --> 00:41:45,000
set up some equations here that
are going to allow us to quickly

552
00:41:45,000 --> 00:41:48,000
tell when our reaction is at
equilibrium.

553
00:41:48,000 --> 00:41:52,000
We are going to compare Q to K.
Let's do that.

554
00:41:52,000 --> 00:41:57,000
Here is that general expression
I had that allowed me to

555
00:41:57,000 --> 00:42:02,000
calculate delta G at any
arbitrary pressures of the

556
00:42:02,000 --> 00:42:08,000
reactants and the products.
Here is the expression delta G

557
00:42:08,000 --> 00:42:12,000
nought at equilibrium,
delta G nought equal minus RT

558
00:42:12,000 --> 00:42:15,000
ln of K.

559
00:42:15,000 --> 00:42:19,000
What I am going to do is
substitute this delta G nought

560
00:42:19,000 --> 00:42:22,000
into here right in there.
Let me do that.

561
00:42:22,000 --> 00:42:25,000
I am just going to plug it
right in.

562
00:42:25,000 --> 00:42:29,000
So, I have minus RT ln of K
plus RT ln of Q.

563
00:42:29,000 --> 00:42:34,000
Now I have the difference

564
00:42:34,000 --> 00:42:39,000
between two logs.
The difference between two logs

565
00:42:39,000 --> 00:42:45,000
is the log of the quotient of
those arguments of the logs.

566
00:42:45,000 --> 00:42:50,000
So, delta G then is RT times ln
of Q over K.

567
00:42:50,000 --> 00:42:55,000
Remember what Q is?

568
00:42:55,000 --> 00:43:02,000
It is this ratio of
instantaneous partial pressures.

569
00:43:02,000 --> 00:43:08,000
If we have a reaction running
and we somehow measure the

570
00:43:08,000 --> 00:43:15,000
partial pressures at some time,
and we calculate the reaction

571
00:43:15,000 --> 00:43:21,000
quotient from that and then we
find that Q is less than K,

572
00:43:21,000 --> 00:43:28,000
the equilibrium constant,
what that means is that we have

573
00:43:28,000 --> 00:43:34,000
fewer products present than
equilibrium says we should have

574
00:43:34,000 --> 00:43:38,000
--
-- because Q is less than K.

575
00:43:38,000 --> 00:43:43,000
If that is the case,
then what is going to happen is

576
00:43:43,000 --> 00:43:49,000
that the reaction is going to
proceed in the forward direction

577
00:43:49,000 --> 00:43:55,000
in order to make more products,
so that Q can get to the point

578
00:43:55,000 --> 00:44:01,000
of being equal to K.
You can also see that up here.

579
00:44:01,000 --> 00:44:06,000
When Q is less than K,
we had the log of a number that

580
00:44:06,000 --> 00:44:11,000
is going to be less than one.
The log of a number less than

581
00:44:11,000 --> 00:44:15,000
one is a negative number.
When Q is equal to K,

582
00:44:15,000 --> 00:44:19,000
delta G is going to be
negative, meaning,

583
00:44:19,000 --> 00:44:24,000
the reaction as written is
spontaneous in the forward

584
00:44:24,000 --> 00:44:28,000
direction.
That is what our equilibrium

585
00:44:28,000 --> 00:44:32,000
says.
It wants to make more products.

586
00:44:32,000 --> 00:44:36,000
On the other hand,
if we have Q greater than K,

587
00:44:36,000 --> 00:44:40,000
that says we have too many
products compared to what

588
00:44:40,000 --> 00:44:44,000
equilibrium says we should have.
And, of course,

589
00:44:44,000 --> 00:44:49,000
then the reaction is going to
run in the reverse direction.

590
00:44:49,000 --> 00:44:54,000
It is going to try to use up
those products and make more

591
00:44:54,000 --> 00:44:57,000
reactant.
You can also see that by this

592
00:44:57,000 --> 00:45:02,000
expression.
When Q is greater than K,

593
00:45:02,000 --> 00:45:07,000
we are going to have a log of a
number that is greater than one,

594
00:45:07,000 --> 00:45:13,000
that is going to be positive.
If delta G is positive in the

595
00:45:13,000 --> 00:45:17,000
forward direction for the
reaction as written,

596
00:45:17,000 --> 00:45:22,000
then it is the reverse reaction
that is going to be spontaneous.

597
00:45:22,000 --> 00:45:26,000
Another way to look at it is
shown here.

598
00:45:26,000 --> 00:45:32,000
Suppose we are in a situation
where Q is less than K.

599
00:45:32,000 --> 00:45:37,000
I am plotting here Q as a
function of the time in the

600
00:45:37,000 --> 00:45:44,000
reaction, and I am starting out
here at T equals zero and Q is

601
00:45:44,000 --> 00:45:47,000
less than K.
If Q is less than K,

602
00:45:47,000 --> 00:45:53,000
it means that we have fewer
products than what equilibrium

603
00:45:53,000 --> 00:45:58,000
tells us we should have.
The reaction is going to

604
00:45:58,000 --> 00:46:04,000
proceed in the forward
direction, so that Q will

605
00:46:04,000 --> 00:46:09,000
increase to the value of K,
so that we can attain

606
00:46:09,000 --> 00:46:13,000
equilibrium.
On the other hand,

607
00:46:13,000 --> 00:46:18,000
if we have a situation where we
start a reaction,

608
00:46:18,000 --> 00:46:21,000
but Q is greater than K,
in that case,

609
00:46:21,000 --> 00:46:26,000
we started the reaction by
putting all the products in

610
00:46:26,000 --> 00:46:32,000
instead of the reactants.
When Q is greater than K,

611
00:46:32,000 --> 00:46:38,000
it says we have more products
than equilibrium says we should

612
00:46:38,000 --> 00:46:41,000
have.
What is going to happen is that

613
00:46:41,000 --> 00:46:45,000
the reverse reaction is going to
proceed.

614
00:46:45,000 --> 00:46:49,000
It is going to try to use up
those products,

615
00:46:49,000 --> 00:46:54,000
so that we attain this ratio
given by the equilibrium

616
00:46:54,000 --> 00:46:58,000
constant.
That is how we are going to use

617
00:46:58,000 --> 00:47:04,000
Q, compare it to K to determine
whether or not we are at

618
00:47:04,000 --> 00:47:08,000
chemical equilibrium.
Professor Cummins will be here

619
00:47:08,000 --> 00:47:12,000
on Wednesday.
I will see you later in the

620
00:47:12,755 --> 00:47:15,000
week.