Boiling point elevation and freezing point depression | Chemistry | Khan Academy

Boiling point elevation and freezing point depression | Chemistry | Khan Academy

December 8, 2019 94 By Kody Olson


Let’s think about what might
happen to the boiling point or the freezing point of any
solution if we start adding particles, or we start
adding solute to it. For our visualization, let’s
just think about water again. It doesn’t have to be water. It can be any solvent, but let’s
just think about water in its liquid state. The particles are reasonably
disorganized because of their kinetic energy, but they still
have that hydrogen bonds that wants to make them be
near each other. So this is in the liquid
state, and they have a reasonable amount of
kinetic energy. You know, each of these
particles is moving in some direction, rubbing against each
other, bouncing off of each other. Now, to move it into the solid
state, or to freeze it, what has to happen? The ice has to enter kind of
a crystalline structure. It has to get pretty organized,
so let’s say it has to look something like this. The water molecules are going
to have a regular structure where the hydrogen bonds
dominate any kind of kinetic movement they want to do, and
all the kinetic movement, they’re just vibrating
in place. So you have to get a
little bit orderly right there, right? And then, obviously, this
lattice structure goes on and on with a gazillion
water molecules. But the interesting
thing is that this somehow has to get organized. And what happens if we start
introducing molecules into this water? Let’s say the example of
sodium– actually, I won’t do any example. Let’s just say some arbitrary
molecule, if I were to introduce it there, if I
were to put something– let me draw it again. So now I’ll just use that same–
I’ll introduce some molecules, and let’s say they’re
pretty large, so they push all of these water
molecules out of the way. So the water molecules are now
on the outside of that, and let’s have another one that’s
over here, some relatively large molecules of solute
relative to water, and this is because a water molecule
really isn’t that big. Now, do you think it’s going
to be easier or harder to freeze this? Are you going to have to remove
more or less energy to get to a frozen state? Well, because these molecules,
they’re not going to be part of this lattice structure
because frankly, they wouldn’t even fit into it. They’re actually going to make
it harder for these water molecules to get organized
because to get organized, they have to get at the right
distance for the hydrogen bonds to form. But in this case, even as you
start removing heat from the system, maybe the ones that
aren’t near the solute particles, they’ll start to
organize with each other. But then when you introduce a
solute particle, let’s say a solute particle is sitting
right here. It’s going to be very hard for
someone to organize with this guy, to get near enough for
the hydrogen bond to start taking hold. This distance would make
it very difficult. And so the way I think about
it is that these solute particles make the structure
irregular, or they add more disorder, and we’ll eventually
talk about entropy and all of that. But they make it more irregular,
and it’s making it harder to get into
a regular form. And so the intuition is is that
this should lower the boiling point or make
it– oh, sorry, lower the melting point. So solute particles make you
have a lower boiling point. Let’s say if we’re talking
about water at standard temperature and pressure or at
one atmosphere then instead of going to 0 degrees, you might
have to go to negative 1 or negative 2 degrees, and we’re
going to talk a little bit about what that is. Now, what’s the intuition of
what this will do when you want to go into a gaseous
state, when you want to boil it? So my initial gut was, hey, I’m
already in a disordered state, which is closer to what
a gas is, so wouldn’t that make it easier to boil? But it turns out it also makes
it harder to boil, and this is how I think about it. Remember, everything with
boiling deals with what’s happening at the surface, and
we talked about that in our vapor pressure. So at the surface, we said if
I have a bunch of water molecules in the liquid state,
we knew that although the average temperature might not
be high enough for the water molecules to evaporate, that
there’s a distribution of kinetic energies. And some of these water
molecules on the surface because the surface ones
might be going fast enough to escape. And when they escape into vapor,
then they create a vapor pressure above here. And if that vapor pressure is
high enough, you can almost view them as linemen blocking
the way for more molecules to kind of run behind them as they
block all of the other ambient air pressure
above them. So if there’s enough of them and
they have enough energy, they can start to push back or
to push outward is the way I think about it, so that more
guys can come in behind them. So I hope that lineman analogy
doesn’t completely lose you. Now, what happens if you were
to introduce solute into it? Some of the solute particle
might be down here. It probably doesn’t have much
of an effect down here, but some of it’s going to be
bouncing on the surface, so they’re going to be taking up
some of the surface area. And because, and this is at
least how I think of it, since they’re going to be taking up
some of the surface area, you’re going to have less
surface area exposed to the solvent particle or to the
solution or the stuff that’ll actually vaporize. You’re going to have a
lower vapor pressure. And remember, your boiling
point is when the vapor pressure, when you have enough
particles with enough kinetic energy out here to start
pushing against the atmospheric pressure, when the
vapor pressure is equal to the atmospheric pressure,
you start boiling. But because of these guys, I
have a lower vapor pressure. So I’m going to have to add even
more kinetic energy, more heat to the system in order to
get enough vapor pressure up here to start pushing back
the atmospheric pressure. So solute also raises
the boiling point. So the way that you can think
about it is solute, when you add something to a solution,
it’s going to make it want to be in the liquid state more. Whether you lower the
temperature, it’s going to want to stay in liquid as
opposed to ice, and if you raise the temperature, it’s
going to want to stay in liquid as opposed to gas. I found this neat– hopefully,
it shows up well on this video. I have to give due credit, this
is from chem.purdue.edu/ gchelp/solutions/eboil.html, but
I thought it was a pretty neat graphic, or at least
a visualization. This is just the surface of
water molecules, and it gives you a sense of just how things
vaporize as well. There’s some things on the
surface that just bounce off. And here’s an example where
they visualized sodium chloride at the surface. And because the sodium chloride
is kind of bouncing around on the surface with the
water molecules, fewer of those water molecules kind of
have the room to escape, so the boiling point
gets elevated. Now, the question is by how
much does it get elevated? And this is one of the neat
things in life is that the answer is actually
quite simple. The change in boiling or
freezing point, so the change in temperature of vaporization,
is equal to some constant times the number of
moles, or at least the mole concentration, the molality,
times the molality of the solute that you’re putting
into your solution. So, for example, let’s say I
have 1 kilogram of– so let’s say my solvent is water. I’ll switch colors. And I have 1 kilogram of water,
and let’s say we’re just at atmospheric pressure. And let’s say I have some
sodium chloride, NaCl. And let’s say I have
2 moles of NaCl. I’ll have 2 moles. The question is how much will
this raise the boiling point of this water? So first of all, you just have
to figure out the molality, which is just equal to the
number of moles of solute, this 2 moles, divided
by the number of kilograms of solvent. So let’s say we have 1
kilogram of solvent. This was, of course, moles. So our molality is 2
moles per kilogram. So we just have to figure out
what this constant is, and then we’ll know the temperature
elevation. And actually, that same
Purdue site, they gave a list of tables. I haven’t run the experiments
myself. They have some neat
charts here. But they say, OK water, normal
boiling point is 100 degrees Celsius at standard atmospheric
pressure. And then they say that the
constant is 0.512 Celsius degrees per mole. So let’s just say 0.5. So it equals 0.5. So k is equal to 0.5. And I want to be very clear here
because this is a very– I won’t say a subtle point, but
it’s an interesting point. So I said that there’s 2– the
molality of– I just realized I made a mistake. I said the molality of
sodium chloride is 2. 2 moles per kilograms. But
that would be if sodium chloride stayed in this
molecular state, if it stayed together, right? But what happens is that the
sodium chloride actually disassociates, and we learned
all about it in that previous video. Each molecule or each sodium
chloride pair disassociates into two molecules,
into a sodium ion and a chlorine anion. And because of that, because
this disassociates into two, the molality is actually going
to be two times the number of moles of sodium chloride I have.
So it’s going to be two times this. So my molality will
actually be 4. And this is an interesting
point. If I was dealing with–
and I wrote it here. So this right here is glucose,
and this is sodium chloride, or at least sodium chloride
in its crystal form. One molecule, I guess you can
view it, or one salt of it. I guess you could just view it
as one of these little pairs right here. But the interesting thing is
is you could have the same number of moles of sodium
chloride when you view it as a compound and glucose. But glucose, when it goes into
water, it just stays as one molecule of glucose. So a mole of glucose will
disassociate into a mole of glucose in water. Well, I guess it won’t
disassociate. It’ll just stay as one mole,
while a mole of sodium chloride will turn into
two moles because it disassociates. It turns into two separate
particles. So in my example, when I start
with a mole of this, I end up– actually, once I dissolve
it in water, I ended up with 4 moles per kilogram of molality,
because this turns into two particles. So given that the molality
is 4 moles. 2 moles of sodium, 2 moles
of chloride per kilogram. So I just use that constant that
I just got from Purdue. And I get the change in
temperature is equal to that constant, 0.5, times 4, which
is equal to 2 degrees. So my boiling point will be
elevated by 2 degrees. Now, if I had the same number of
moles, if I had 2 moles of glucose dissolved into my water,
I’d only get half as much, half as much
of an increase. Because the molality would
be half as much. Because it doesn’t turn
into two particles. In some textbooks, you’ll
actually see it written like this. You’ll actually see the same
formula written like change in boiling temperature, or vapor
temperature, or whatever you want to think, is equal to k
times m times i, where they’ll say this is the molality
of the compound you’re talking about. In this case, this number
would be 2, and i is the number of molecules or the
number of things that it disassociates into. So in this case, this
would have been 2. And that’s where we would have
gotten 4 times k, which is 0.5, which is 2. In the case of water, this would
be– oh, sorry, in the case of the glucose, this
would still be 2. But it only turns into one
particle when it goes in the water, so that would be 1. So you would only have a 1
degree increase in the boiling point of water. Now, freezing point
is the same thing. Change in freezing
point is also proportional to the molality. And you can either say the
molality of the original non-in-water compound times
the number of compounds it disassociates into, although
this k is going to be different for freezing than
it is for boiling. Of course, this k changes at
different pressures and for different elements. But the really big takeaway is
just to realize that even if you have a mole of this and a
mole of that, and they’re going to be dissolved into the
same amount of water, because this dissociates into two
particles and this disassociates into only one
for every– or this disassociates into two moles for
every mole of the crystal you have– this doesn’t
disassociate; it just stays as one– this’ll have twice as
large of an effect on the freezing point change or on the
boiling point elevation than the glucose will.