If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:12:35

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.