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PROFESSOR NELSON: So last time we just about finished up the
discussion of thermochemisty, and we went through a
description of calorimetry, how we actually make
measurements of thermodynamic a quantities.
I should mention, which I didn't last time, that you
know a calorimeter, apart from measuring heats of formation
also is used to measure standard thermodynamic
quantities like heat capacities.
You know how much heat does it take to raise the temperature
of something by a degree?
Well, you put the material inside the calorimeter, and
you have a heater and it's connected up to current
source, and you can accurately measure how much heat you're
putting in, because you can measure how much current is
going into it, and you separately are measuring the
temperature of the material, so it's a straightforward way
of measuring the heat capacities, thermodynamic data
of this sort.
So work on phase transitions, just ordinary
properties of material.
Chemical reactions, all the thermochemistry involved is
done using constant pressure or constant volume calorimetry
in a pretty routine way.
So we talked about how to use the results of measurements
like that to calculate reaction energetics,
enthalpies of reaction.
And I just want to end the unit on thermochemistry with a
real brief discussion of an approximate way of formulating
heats of reaction, and that is in terms of bond energies.
Now you know you're probably never going to sit down and do
a series of thermodynamic
calculations using bond energies.
You could always look up heats of formation of compounds and
from those determine very accurately heats of reaction,
by looking at the heats of formation of products and the
heats of formation of the reactants
and subtracting them.
And since that data is available for a really wide
range of materials, that's sort of the simplest approach
and the most accurate one to determine reaction energetics.
On the other hand, if you just want to think about molecules,
you know, you think about bonds.
You know if you say I've got methane or ethane, I'm going
to use a fuel and burn it and capture the energy.
How much energy do we have in there?
Well the way you want to be thinking about that is all
right, I'm going to take the starting material, the fuel,
I'm going to break bonds, and I'm going to form some final
product with new bonds.
Roughly what kinds of energetics
are we talking about?
In biological systems, it's also super important.
You know you think about ordinary biochemical
energetics, what's used is fuel in biology and what are
the products that are formed?
And again, it's just super important to be able to think
sort of semi-quantitatively in terms of bond energies,
because you don't want to always be going and looking up
everything in tables.
If you need to do an extensive series of calculations, it's
very useful to have the tables.
But if you just want to think about how something is
working, you know biological or system or an engine or
another application, it's very useful to just be able to
think in terms of bond energies.
So let's just go over bond energies.
And you've seen this in 5.111 and 2, so I'm not going to
belabor it but they're just a little bit of examples that I
want to go through, just so it's fresh on your mind.
So the idea, of course, is to be able to have an idea of how
much energy is involved in forming or
breaking a single bond.
So, if we just do this by example, we can take methane
so it's a real simple case because we know it's just got
four identical carbon hydrogen bonds, and we could say OK,
let's just break them.
So we'll start with methane gas and go to one atom of
carbon in the gas phase, plus four atoms of
hydrogen in the gas phase.
The energetics in this case aren't given by the usual
heats of formation.
You'd just say, if I want to know about methane, I can look
up its heat of formation from the elements in their
standards states. but the difference here is these
elements aren't in their standard states.
Carbon in its standard state at room temperature and
pressure is graphite, a solid, it's not carbon atoms in the
gas phase, and hydrogen in the standard state is hydrogen
molecules H2, not individual atoms.
All right, so we'd like to have this bond energy for
these things.
So let's just to make a cycle that will allow us to
determine this.
So this'll be our first step in that cycle, or one way of
getting there.
And this delta H of reaction is going to be by definition
four times our bond energy for a carbon hydrogen bond.
Now let's go to be elements that are in their standards
states at room temperature and pressure.
So let's go to carbon as graphite.
And let's go to two molecules of hydrogen gas, so this will
be step two and this will be step three.
And this will allow us to work because now we're starting in
standard states, so we can just use
regular heats of formation.
And that's the key.
So if we look at step one delta H1 there as we've
defined it, four times our bond energy.
Two, this is just negative heat of formation of methane,
which we can look up.
And three, is similar.
Essentially it's the heat of formation of carbon in the gas
phase and hydrogen atoms in the gas phase, commonly called
heats of atomization, but these are also tabulated just
like heats of formation of ordinary compounds.
In fact, in the appendix to your book that has
thermodynamic data, you can find carbon gas phase atoms
including heat of formation, and same with individual
hydrogen atom as opposed to hydrogen
molecules in the gas phase.
So those data are available.
So we can write delta Hc from atomization, and two times
delta H of H2.
This is the way this is tabulated and this is molar.
And of course now we've written a cycle.
So we know that to get from here to here is the same as to
go through these two steps.
In other words delta H1 is the sum of delta H2 and delta H3.
And so this says that our four times the carbon hydrogen bond
energy is just given by minus the heat of formation of
methane plus the heat of atomization of carbon, plus
the two times the heat of atomization
for hydrogen molecules.
And if we look up the numbers that are tabulated, what we
discover is that our carbon hydrogen bond energy is 416.2
kilojoules per mole.
Very useful to have numbers like that stored in your head.
To have some basic idea of just ordinary chemical
energetics in terms of bond energies.
Hydrogen bonds, you know roughly 20 kilojoules per
mole, and so forth.
Just having those kinds of energetic enables you to think
in a qualitative way about what's likely to happen in an
ordinary chemical or biochemical situation.
OK, so this is in a sense the simplest case, because we've
talked about the dissociation of a molecule that consists of
all identical bonds.
But it's straightforward to go from there to
additional bond energies.
So from this we've got a value for a sort of typical carbon
hydrogen bond, right.
We can do the same thing now for ethane, and that's got a
single carbon carbon bond, as well as a bunch of carbon
hydrogen bonds, but now the we know the carbon hydrogen bond
energy, we can use that number plus the relevant heats of
formation to determine the carbon carbon bond energy.
And we can keep going and extend this to essentially
tabulate bond energies for a very large number of different
sorts of bonds.
So just as an example, ethane, so now we're going to have --
this I'll write it all out just some we can keep proper
track of all the bonds that we're going to be breaking.
So C2H6 goes to two carbon gas phase atoms, and
six hydrogen atoms.
So in this case, our, I can write this again as one delta
H1 is six times the carbon hydrogen bond energy plus the
carbon carbon bond energy that we'd like to determine.
And again we can go through now a similar cycle.
So again this now can be equilibrated with carbon
graphite plus three molecules of hydrogen in the gas phase,
just like before.
And so from that we can see that we'll end up with an
equality between these, and negative delta H naught
formation for C2H6 or ethane, plus two delta H of carbon
atomization, plus three delta H for
atomization hydrogen molecule.
So we've already seen we could look up these numbers and
this, and we've already determined a carbon hydrogen
bond energy.
So again, if we find all the relevant values, we can
determine that our carbon carbon bond energy is 342
kilojoules.
And so on.
So we can keep going and again build up a bigger and bigger
inventory of bond energies just to provide us with the
kind of intuition that we'd like to have to be able to
think about ordinary molecular energetics.
Any questions about bond energies?
All right.
Of course, if we know bond energies, we also could make
estimates of heats of formation.
So if we have some new compound, we don't know its
heat of formation, we know its structure.
We know what bonds are involved.
We could get an estimate for what we might expect for its
heat of formation if we know those, and I think I'll just
write this out on the board by example, but not work through
the numbers explicitly.
You know, let's say we were going to make, you know, n
pentane C5H12.
And of course again keeping proper track of all the carbon
carbon and carbon hydrogen bonds.
But now in principle we know the carbon carbon bond
energies and the carbon hydrogen bond energies.
So we should be able to figure out the heat of formation of a
compound like this, even if we don't know it from a table.
And it's again through a similar sort of cycle so if we
start with the atoms in the gas
phase, and go to n pentane.
So this is now the bond energies which
we basically know.
And again make our cycle of involving the elements in
their stable form at standard temperature and pressure.
Then again, here's our one, two, three.
Here's our minus delta H of atomization, and here's our
heat of formation.
And that's the point, is that now we can use what we've
already elaborated for the bond energies and the known
enthalpies of atomization, and we could
determine the heat of formation.
If we do that, we wind up with a value of minus 152.6
kilojoules.
If we measure it.
If we look up in a table what the actual value is, we
discover it's minus 146.4 kilojoules.
So it's a few percent off.
Why is it off?
Why isn't it right?
STUDENT: [UNINTELLIGIBLE]
PROFESSOR NELSON: Yes, exactly.
You know you can't use the C-H bonds from methane for every
C-H bond, or the C-C bond in ethane for every carbon carbon
single bond.
There'll be variations, usually of modest magnitude,
in the values.
So this isn't going to be an exact calculation, but
certainly this is close enough for allowing us to think
qualitatively and semi-quantitatively about
ordinary energetics.
Another illustration of it is if we look at neopentane -- so
same chemical formula.
Same number of carbon carbon and carbon hydrogen bonds.
So according to the way we've done this calculation, we
should get exactly the same number, because all we're
doing is adding up the total number of bonds,
and it's the same.
But what we discover is that for neopentane, delta H of
formation, neopentane is minus 166.1 kilojoules.
So again it's different in this case
in the other direction.
But certainly close enough that allows us to think
reasonably about what might happen in ordinary situations.
Any questions about bond energies?
OK, now we're going to go to a new topic, and let me say, the
topic we're going to start now, the second law, is really
in a sense, this is the start of the whole rest of the term.
So far, what we've covered essentially is conservation of
energy, how you think about energetics and what
contributes to them.
That is heat and work.
How to account for them all properly.
What happens under constant volume or constant pressure
conditions that led us to the definition of enthalpy right
and so forth.
And once you get your arms around these ideas, and it
takes some work to do that, but once you do, the actual
execution of things can be pretty straightforward, and
some of this can seem like just you know a lot of
accounting.
It's not.
It takes some effort to become accustomed to it and familiar
with how to properly think about heat and work and how
they contribute to energy and so forth.
But given that amount of effort, a lot of this can be
reduced to fairly straightforward
calculational stuff.
What we're about to start is a little different from that in
some sense, and really in some ways more difficult because
what we're going to try to understand starting with this
next set of topics is what determines whether something
happens spontaneously?
We haven't really talked about that in any of this.
We've talked about whether reactions are endothermic or
exothermic.
But as you're going to see that doesn't determine what
happens spontaneously.
It certainly is one of the factors, but by no
means the only one.
What determines what happens just by itself?
Or another way of saying that is, if I have some sort of
system, it could be an engine.
It could be a chemical reaction.
It could be a biochemical system or some part of it.
What determines where the equilibrium is and the
direction that things would have to go in order to reach
or approach the equilibrium?
What is the equilibrium state of something?
Actually, what we've discussed so far really doesn't allow us
to determine that, and there are a few easy ways to sort of
bring that out.
One is you can imagine constructing some kind of
cyclic process that you know it could remove heat like a
refrigerator does from a cool region and export the heat to
a hotter region, like the room, or outside a window if
it's an air conditioner or something like that.
Just move the heat.
That's it.
Why not?
I mean you could do that and not necessarily
violate the first law.
Or what about, here's a simpler case.
What if I just have a chamber with a divider, and I pump out
half of it, and I put gas in this half, so there's gas and
there's vacuum, and then I remove the divider.
So I certainly expect that what happens is the gas fills
the volume.
Why?
The first law didn't tell me that that needs to happen.
Why doesn't it go back the other way, anyway, even
without the barrier?
Why doesn't the gas just by itself all wind up over here?
Let's say it's a pretty dilute gas, so even if it does all
end up in this half, you know the energy of any sort of
unfavorable repulsions between molecules is negligible.
Maybe they even have weak attractions?
Why doesn't that happened?
It's a good thing for us that it doesn't happen.
Why in this room doesn't the air all just sort of collect
over there somewhere?
Or just the oxygen molecules in the air and
suffocate all of us.
The first law doesn't have anything to
say about that actually.
Whatever the reason it doesn't happen, we're not going to
find the answer to that in the first law.
So that's what the second law of thermodynamics and the
development of entropy is all about.
It's to help us understand what is the direction of
spontaneous change?
What governs that, and what's the equilibrium state that is
reached when the spontaneous change occurs?
OK?
So the second law is going to give us basically a principle
that'll tell us the direction of spontaneous change.
One way of thinking about that is it's going to allow it to
sort of determine the direction of time.
You know if we see initial or if we see one state and we see
another state, we'll be able or know, time must have been
going this way, right because the spontaneous change would
happen in this direction.
So, I'm going to put up a couple of just statements of
the second law of thermodynamics.
I'll start with kind of an easy one.
So maybe I'll start with that here.
Here's a statement by Clausius.
The first law says that the energy of
the universe is constant.
We're always conserving in one way or another.
And the second law is that the entropy of the universe is
increasing.
Now we're going to define entropy presently.
But I'm writing this up just to give you some foreshadowing
of how we're going to try to guide our thinking about
spontaneous processes.
This tells us about conservation of energy.
This tells us about which direction things move in
spontaneously, namely the direction that leads to an
increase in this soon to be defined quantity entropy.
It's going to turn out to be a state function, and it'll help
us understand the direction of spontaneous change.
You know, like the first law the second law too was
developed before there was a very well defined
understanding of the molecular nature of matter.
We're going to talk about the second law and entropy in
macroscopic terms.
Later on in the semester we'll also discuss them in
microscopic terms.
We'll do a little bit of what's called statistical
mechanics, of microscopic, basically discussing the
microscopic origins of entropy in terms
of disorder of molecules.
But really, the thermodynamics of entropy and the second law
evolved because people were trying to build things.
It was the Industrial Revolution. people were trying
to build engines, steam engines, and other kinds of
devices, refrigerators other things.
And they couldn't understand -- first of all, why couldn't
you make it perfectly efficient?
You know, you'd have a steam engine.
There'd be a source of heat.
You could extract work.
Never seemed like you could get it all and
turn it all into work.
Why not?
And if not, what were the limitations on it?
What was the best efficiency you could achieve?
So in a very practical way people were motivated to think
very hard about these kinds of questions.
So let's just look at some of the things that brought out
the need for this kind of consideration.
So let me start by drawing one of the impossibilities.
We could try to build a heat engine.
Where we've got some temperature T1 which is hot,
and some amount of heat, q1 goes into our system, and our
system runs as a cycle.
It goes around and around, because in any practical
situation we're going to need to keep going and repeating
the process.
Some amount of work comes out.
We'll call it minus w, because work is always defined as the
work that's done by the environment on the system.
So if I'm drawing the arrow this way, then I want to write
it as minus w.
Now this would be just swell.
This is it.
That's the whole heat engine.
This is running in a cycle.
So we know that, delta u is zero.
So that would mean that, you know, the amount of work out
would just equal the heat in every time around.
Never happens.
Can't make it happen.
The only way it works is if instead there's
an additional part.
Somewhere heat is also lost to a cold reservoir.
In other words, you know I burned some fuel.
I consumed something to produce my hot source, and I'm
going to then try to turn it into work, which I can, but
only partly.
There's going to be heat loss somewhere to a colder region,
and I can't avoid it.
And it was discovered that this was always the case.
So this can be built where, just to be clear on these. q1
is a positive number, because again that's heat going from
the environment to the system.
This is our system running in a cycle.
So q1 is positive, minus w is positive, that is work is
being done on the environment by the engine.
And minus q2 is positive.
And T1 is greater than T2.
That I can build.
Here's another thing we could try to do.
It's sort of the opposite.
We could try to produce a refrigerator but not have to
do any work.
So let's go the other way, in other words, let's go from T2
which is cold we're going to go in this direction now,
which means q2 will be a positive number and here's our
system running in a cycle.
And now here's minus q1 now is going to be our positive
number, and we're going to remove heat from q2 and move
it, or from T2 a cold region and move it to some place
that's hotter.
Also a pretty nice idea.
And also obviously impossible.
It only works if I put work in there to make it happen, then
yes I can take heat out of a cold body and move it up into
a hotter reservoir of some sort.
So here's a refrigerator.
Again T1 is greater than T2, and in this case q2 is
greater than zero.
I'm removing heat, minus q1 is greater than zero, that is
heat is going up this way.
Work is greater than zero.
Has to be.
Of course the thing is running in a cycle. delta u is zero.
So that means that work must be minus q1 plus q2, right?
And work is greater than zero, so that's saying minus q1
right that's a positive number, must
be bigger than q2.
In other words, I am taking some heat away
from the cold region.
The heat I'm dumping into the hotter region is bigger than
the amount that I taking away here, because it's also got
this amount.
That's why, you know, if it's a humid or warm summer
afternoon and you think oh what I a great idea, I've got
this little fridge in my room.
I'm going to open it up and cool the room off.
Turns out not to work too well.
Because of course what the fridge is
doing, it's cold inside.
So it's a machine that's removing heat from the inside
and dumping it outside.
There are coils in the back of the fridge, and if you touch
them they're a little bit warm and so on.
The heat is being lost to the room.
How much heat?
Well, the work that goes in is also added.
So the amount of heat that goes out into the room is
bigger than the amount of heat that you're taking
away from the space.
That's bad.
So if you try opening the refrigerator door to cool off
your room at the same time as you're dumping heat into your
room, all you discover is the net effect is to
heat up your room.
Because you know the amount of heat that the thing is
removing from somewhere on the inside there and struggling to
do that because you've left the door to the fridge open,
right, is not as big as the amount out heat that's coming
out in those coils on the back.
It's not as big because that heat has not only the heat
that you're removing from the inside, but also
this amount of work.
It's all conserved because delta u is zero.
The thing is running in a cycle.
Try it sometime.
You can verify this experimentally in a very
simple way.
It's not a very effective way to work, and it wastes a
little bit of energy, but could be worth it in the
interest of knowledge and discovery.
Let me write an alternate statement of the
second law of Clausius.
All spontaneous processes are irreversible.
When I open that panel in the container, I've got gas on one
side and vacuum in the other and it expands in there, it
happens spontaneously, and it's irreversible.
It's not just going to go back.
And now let me write a mathematical statement of the
second law.
And we'll celebrate it's importance
by using this color.
Going around in a cycle, if I integrate the quantity dq
reversible over T, I get zero.
Remember this, this is our special function.
Let me just keep going a little more though.
Let me also write another version.
If I write dq for an irreversible path over T,
right, less than zero.
It's not the same.
This is a state function.
This isn't.
Remember of course that the heat that's exchange between
the system and the environment depends on the path.
If there is a reversible path, and you see how much heat was
exchange with the environment, and you look at that divided
by temperature integrate over the whole thing, you discover
that as long as the path is reversible, it only depends on
the states at the beginning and the end of the path.
In other words if you have various ways you can go from
state one to two that are reversible,
it's always the same.
We saw that for just a couple of examples, a
couple lectures back.
Remember, we looked at these thermodynamic cycles and just
said, well let's just calculate those.
And those were, those all involved reversible processes.
So, in effect, we looked at different paths to go from one
state to another.
We had a cycle, and one path went to here, and the other
path went back.
We discovered that, in fact, this one zero around the cycle
or the way we expressed it at the time is the value of this
was the same regardless of the way we got from one state to
the other, whether we took a straightforward route or a
more circuitous route, as long as they were reversible.
We're going to define dS as dq reversible over T, and since
we're dealing with a state function delta S going from
state one to two of dq reversible over T is S2 minus
S1 state function.
And S as state function is the entropy, and of course the us,
it will always be special.
All right, now, the second law, you know it's difficult
to understand and to get your arms around, because you know,
do I really calculate this?
I actually have to find a reversible path in order to do
the calculation.
Normally, you know, when you calculate delta u and delta h
and so forth, you have various ways to calculate them that
can be pretty straightforward.
But here, you actually have to determine the heat exchanged
between the system and the environment, and you know that
does, in general, depend on the path.
But it turns out that this quantity, you know, the
differential of that heat over the temperature, also depends
on path unless it's a reversible pass.
Any reversible path will lead to the same results.
It doesn't depend on path among reversible paths.
What that means in practice is if we say OK, start here and
end here and calculate the entropy, you actually have to
sort of figure out a reversible way of getting from
here to here so that you can do that calculation.
So it's really very different from ordinary functions that
you've been dealing with.
A lot of people, you know, because the second law is
difficult to understand, you know you can easily go online
and see thousands of violations of the second law.
It's still easy to look, to go find claims of perpetual
motion machines.
Some of them cost a lot of money.
You can buy them, see how they work.
But and also, all sorts of schemes for producing energy.
Go online and see, right.
You're just, they're just crazy ideas about getting
energy from nothing.
Getting energy from the heat that's just already
present in the air.
I mean there's heat in the air, all this motion, it must
have energy.
We should be able to extract it and use it.
Turns out we can't do that.
It has to look something like this.
OK.
I want to look at some more statements of the second law.
To do that I just want to indicate one definition which
is just to be clear what these things mean.
These are what I've called heat reservoirs
or hot or cold bodies.
Here's what I mean to be more specific.
A heat reservoir is basically a very large thermal mass at
the same temperature.
And the real idea is that you can bring heat out of it as
much as you want and the temperature won't change.
It's an idealization. right If you have something that's big
enough, and it's at a constant temperature, of course you can
take heat away, but it's just so enormous that you won't
measurably change it's temperature.
In practice you might be able or approach that limit.
OK, now let's look at some more definitions or statements
of the second law.
This is from Kelvin of Kelvin temperature scale fame.
It's impossible to convert heat into work with a system
that operates in a cycle.
I'm going to write this all out.
I know you've got it in your notes too,
but it's super important.
To convert heat into work with a system that operates in a
cycle, without at the same time transferring some heat to
a colder reservoir.
OK, a couple of things to notice.
One is, you know, in some sense the second law is kind
of negative, right?
Must not have been done in the new age.
It tells us what we can't do.
And that really is in a sense what it's all about.
It's telling us what are our limitations of what can be
accomplished.
Very important part of this statement -- in a cycle.
Of course any practical engine has to keep going, so there's
going to be a cycle in some way or another, but it's
important because, you know, there are some -- if I'm not
operating in a cycle, I can convert heat
into work just fine.
You know, I could have a heat source, got a candle, and I've
got a piston, maybe I've got a weight on it.
I've got something stopping it.
And then I remove the stoppers and let it go, and because of
the heat, there's expansion.
It pushes up, right, and it'll be up higher somewhere.
Worked!
Heat turned into work.
I don't have a cold body somewhere.
But I only did this once.
There was just one stroke of the piston.
If I want to continue it and run in a cycle, somehow I've
got to have a place where the heat goes.
So this sort of diagram describes effectively the way
the Kelvin statement works.
This works fine and like we illustrated before, it doesn't
work fine if I don't have this part of it.
I won't be able to just take all the heat and convert it to
work, like I, in principle, could do, if I don't need to
run it in a cycle.
Here is yet another statement by Clausius And we'll
end with that one.
Maybe I'll just read this one, so we can
not go over the time.
So, it's impossible for any system to operate in a cycle
that takes heat from a colder reservoir, transfers it to a
hotter reservoir, without at the same time converting some
work into heat.
In other words, it's a statement kind of similar to
the Kelvin statement, but applying to the case of moving
heat from a colder to a hotter region.
Can't do it, unless some work is also converted to the heat
that gets released into the hot region.
I think we'll stop there.
Next time what we'll do is lay out a very well-defined kind
of engine called Carnot cycle that allows us to very
explicitly go through and say OK, let's just work out the
steps of isothermal and adiabatic expansions and
contractions.
In other words, how does the engine really work?
And just calculate it going around in a cycle and see
exactly how all this is born out.
And we'll see that next time.