We saw in the last video that if we defined enthalpy, H, as
being equal to the internal energy of a system plus the
pressure of the system times the volume of the system-- and
this is an almost arbitrary definition.
But we know that this is a valid state variable.
That no matter what you do in terms of how you get there,
you're always going to have the same value.
Because it's the sum and product of
other valid state variables.
But this by itself isn't that useful or that intuitive.
But we saw in the last video that if you assume constant
pressure-- and that's a big assumption, but it's not an
unreasonable assumption for most chemical reactions.
Because most chemical reactions, you know, we're
sitting at the beach with our beakers, and they're exposed
to just standard temperature and pressure, or at least some
pressure, that's not changing as the reaction occurs.
If we assume constant pressure, we saw that the
change in enthalpy becomes the heat added to the system at
that constant pressure.
That P there is just to show you that hey, this is assuming
that we're dealing with heat being added
at a constant pressure.
Fair enough.
So how can we use these concepts in any useful way?
Let's say that I had some carbon in its elemental form
as graphite, and I add to that-- say I
have a mole of carbon.
And I add to that 2 moles of hydrogen in
its elemental form.
It's going to be a gas, and it's going to be as a
molecule, right?
If I just have a bunch of hydrogen in its gaseous state,
let's say, in a balloon, I'm not going to have individual
atoms of hydrogen.
They're going to bond and form these diatomic molecules.
And if I react them, I'm going to produce a mole of methane.
CH4.
But that's not all I'm going to produce.
I'm also going to produce some heat.
I'm going to produce 74 kilojoules of heat-- plus 74
kilojoules of heat-- when I produce that one mole.
I'll do a lower case k for the kilo.
When I produce that one mole of methane.
So what's happening here?
So first of all, how much heat is being added to the system?
And let's assume that this heat just gets released from
the system.
That this isn't an adiabatic process.
I haven't insulated the system from anything.
That this just gets released.
It just goes away.
It gets released.
So my question is, how much-- you know, I started off with
this container, I guess we could call it, that's at a
standard, fixed pressure.
And maybe I had a bunch of-- well, I
wanted to do the carbon.
I'll do it in gray.
I have a bunch of solid carbon lying around.
Maybe some type of dust. And then I have some molecular
hydrogen gas.
Each of those dots is actually two hydrogen atoms. And then,
I don't know, maybe I shake it up or something to make them
react, and then I get a bunch of methane gas.
I'll do that in green.
So now I just have a bunch of methane gas, and I released 74
kilojoules.
So how much heat was added to the system?
Well, we released heat from the system.
We released 74 kilojoules.
So the heat added to the system was minus 74
kilojoules.
Right?
If I asked you the heat released, then I
would have said 74.
But remember, we care about the heat added to the system,
is 74 kilojoules.
And I just showed you that that's the exact same thing as
the change in enthalpy.
So how can we think about this?
What is the enthalpy of this system
relative to this system?
Well, it's going to be lower, right?
Because if you take enthalpy-- so the change in enthalpy is
the enthalpy of your final system, minus the enthalpy of
your initial system.
And we got a negative number.
We got minus 74 kilojoules.
So this has to be lower than this by 74 kilojoules.
So this enthalpy right here is less than this
enthalpy right here.
So if we were to actually draw it on a diagram, if I were to
draw the reaction-- let's say that this is just time or
something, as the reaction proceeds, that axis.
And on the y-axis, I'll draw enthalpy.
So the reaction starts off at your initial enthalpy, Hi and
that's this state right here.
So you start there.
I'll do it in the yellow of that container.
You start there.
And then, I don't know, you shake it up, or I'm not going
to go into the activation energies, so it might have
little hump and all of that, but who knows.
But then we end up at our final enthalpy.
We have this final enthalpy, right here, after the reaction
has occured.
That's this state right here.
This is H final.
So you can see, you've had this drop off in enthalpy.
And what's interesting here is, is not so much what the
absolute value of this enthalpy is here, or what the
absolute value of this enthalpy here is.
But now that we have enthalpy, we can kind of have a
framework for thinking about how much heat energy is in
this system relative to this system.
And given that there's less heat energy in this system
than that energy system, we must have released energy.
And you know, to some degree, I told you that from the
beginning, right?
I told you that energy is released.
And the word for this we use is exothermic.
Now, if you want to go the other way-- let's say we
wanted to go from methane and go back, you'd have to add
heat into the reaction.
If you wanted to go backwards through this reaction, go
upwards, you would have to add that heat content to get that
positive delta H, and then you would have
an endothermic reaction.
So if a reaction releases energy, exothermic.
If a reaction needs energy to occur, it's endothermic.
Now you might be asking, Sal, where'd that energy come from?
So I started at this enthalpy here.
And enthalpy has this weird definition right here, and
then'll ended up as that other enthalpy here.
And as you see, enthalpy, the pressure
we're assuming is constant.
Let's say the volume isn't changing much in this
situation, or maybe doesn't change at all.
So most of the change is going to come from the change in
internal energy, right?
There's some higher internal energy here, and some lower
internal energy here that's causing the
main drop in enthalpy.
And that change in internal energy is really a conversion
from some potential energy, up here, into
the heat that's released.
So there was some heat that was released, 74 kilojoules,
and so our internal energy dropped.
And what all of this does is, it gives us a framework so
that if we know how much heat it takes to form or not form
certain products, then we can kind of predict how much heat
will either be released, or how much heat will be absorbed
by different reactions.
And so here I'm going to touch on another notion.
The notion of heat of formation, or sometimes it's
change in enthalpy of formation.
So the way they talk about it is, the change
in enthalpy of formation.
And it's normally given at some standard
temperature and pressure.
So you put a little, usually it's a naught, sometimes it's
just a circle in there.
And what that is is, what is the change in enthalpy to get
to some molecule from its elemental form?
So for example, if we want it for methane-- if we have
methane there, and we want to figure out its heat of
formation, we say, look, if we form methane from its
elemental forms, what is the delta H of that reaction?
Well, we just learned what the delta H of that reaction was.
It was minus 74 kilojoules.
Which means that if you form methane from its elemental, I
guess, building blocks, you're going to release 74
kilojoules of energy.
That this is an exothermic reaction.
Because you've released heat, you can say that the methane
is in a lower energy state, or it has a lower potential
energy, than these guys did.
And because it has lower potential
energy, it's more stable.
I mean, one way to think of it is, if you have a guy, you
have a mountain here, and down here, you have a ball.
And this isn't, you know, a complete, direct analogy.
But the analogy to potential energy is, look.
When you're down at a lower potential energy state, you
tend to be more stable.
And so, in the everyday world, if you have a bunch of methane
sitting around, the fact that it has a negative heat of
formation, or a standard heat of formation, because I have
that naught here, or a negative standard change in
enthalpy of formation-- those are all the same things--
tells me that methane is stable relative to its
constituent compounds.
And actually, you can look these things up.
You don't have to memorize them, but it's good to know
what they are.
And I copied all of this stuff-- actually, let me get
the actual tables from Wikipedia down here.
I copied all of these directly from Wikipedia.
These give you the standard heat of formation
of a bunch of things.
And if you look down here, for-- let's see if they have
methane-- right there, this is what we were dealing with.
They're telling us essentially the delta H of the reaction
that forms methane.
This point table right there tells us that if we start off
with some carbon in a solid state, plus two moles of
hydrogen in a gaseous state, and we form one mole of
methane, that if you take the enthalpy here minus the
enthalpy here-- so the change in enthalpy for this
reaction-- at standard temperature and pressure, is
going to be equal to minus 74 kilojoules.
Per mole.
And this is all given per mole.
So if you have a mole of this, two moles of this, to form one
mole of methane, you're going to release 74
kilojoules of heat.
So this is a stable reaction.
Now there's a couple of interesting things here, and
we'll keep using this table over the next few videos.
You see here, monoatomic oxygen has a positive standard
heat of formation.
Which means it takes energy to form it.
Right?
That if you have a reaction, let me just say, the reaction,
I'll write it this way.
One half of molecular oxygen as a gas to go to one mole of
oxygen in its gaseous state.
This tells us that this state has more
potential than this state.
And in order for this reaction to occur, you have to add
energy to it.
You have to put the energy on the other side.
So you'd have to say, plus 249 joules.
So you might say, hey, Sal, that doesn't make sense.
Oxygen is just oxygen.
Why is there a heat of formation of oxygen?
And that's because you always use the elemental form as your
reference point.
So oxygen, if you just had a bunch of oxygen sitting
around, it's going to be in the O2 form.
If you have a bunch of hydrogen, it's going to be H2.
If you have a bunch of nitrogen, it's going to be N2.
Carbon, on the other hand, is just C, and it tends to be in
its solid form as graphite.
So all heats of formation are relative to the form that you
find that element when you have a pure version of it.
Not necessarily its atomic form, although sometimes it is
its atomic form.
Now, in the next video, we're going to use this table, which
is a very handy table-- I cut and pasted parts of it-- to
actually solve problems. In this last video, I gave you
the heat of formation, and we just thought
about it a little bit.
In the next few videos, we're going to use this table that
gives us standard heats of formation to actually figure
out whether reactions are endothermic-- meaning they
absorb energy-- or exothermic-- meaning they
release energy-- and we'll figure out how much.