# HeterogeneousÂ equilibrium

## Video transcript

Let's say we wanted to figure
out the equilibrium constant for the reaction boron
trifluoride in the gaseous plus 3-- so for every mole of
this, we're going to have 3 moles of H2O in the liquid
state-- and that's in equilibrium. It's going forward and backwards
with 3 moles of hydrofluoric acid, so it's
in the aqueous state. It's been dissolved
in the water. If it wasn't dissolved, if it
was in the solid state, you would call this hydrogen
fluoride. Once it's in water, you call
it hydrofluoric acid, and we'll talk more about naming
in the future, hopefully. Plus 1 mole of boric acid, also
in the aqueous state. It's dissolved in the water. H3BO3 in the aqueous state. So what would the expression for
the equilibrium constant look like in this situation? So you might be tempted say, OK,
that's easy enough, Sal. So the equilibrium constant,
you just take the right-hand side. That's just the convention. There's symmetry here. I could've rewritten it either
way, but let's just say you take the right-hand side and
say, OK, this is dependent on the concentration of the
hydrofluoric acid, the concentration of the HF, or the
molarity of the HF, to the third power, times the
concentration of the boric acid, H3BO3. And remember, this intuition of
why you're taking this to the third power is what's the
probability-- because in order for the reaction to go this
way, you need to have 3 molecules of hydrofluoric acid
being very close to 1 molecule of the boric acid. So if you watched the last video
I just made about the intuition behind the equilibrium
constant, this is indicative of the probability of
this reaction happening or the probability of finding
all of these molecules in the same place. Of course, you can adjust it
with a constant and that's essentially what that does. But that's on the product
side, or the reactant, depending on what direction
you're viewing this equation, divided by the molarity of the
boron trifluoride times-- and I'll do this in a different
color-- the molarity of the H2O to the third power. And that's, of course,
the H2O liquid. So there you go. We'll just figure this out. And my rebuttal to you is I
want you to figure out the molarity of the water. What is the concentration
of the water? Remember, the concentration is
moles per volume, but in this case, what's happening? I'm putting some boron
trifluoride gas essentially into some water, and it's
creating these aqueous acids. These other molecules
are dissolved completely in the water. So what's the solvent here? The solvent is H2O. This might be how the reaction
happens, but pretty much, there's water everywhere. The water is in surplus. So if you were to really figure
out the concentration of water, it's everywhere. I mean, you could say everything
but the boron trifluoride, but it's
a very high number. And if you think about it from
the probability point of view, if you say, OK, in order for
this reaction to happen forward, I need to figure out
the probability of finding a boron trifluoride atom or
molecule-- actually, molecule-- in a certain volume,
and it also needs 3 moles of water in that
certain volume. But you say, hey, there's
water everywhere. This is the solvent. There's water everywhere, so
I really just need to worry about the concentration of
the boron trifluoride. So you could say the forward
reaction rate, rate forward, is going to be dependent on some
forward constant times just the concentration of
the boron trifluoride. The water's everywhere, so you
don't have to multiply it times the concentration of
water, whatever that means, because the water's
everywhere. So the denominator here, you
do not put the solvent. So the correct answer for this
one is you only put whatever is actually dissolved
in the solution. Because frankly, the
concentration doesn't actually makes sense for everything else,
and if you think about it from the probability point
of view, that also makes sense, because there's
always water around. If you said, OK, what's the
probability of finding water at any small volume of our
fluid, it's going to be 1, so you could just multiply it by
a 1 there, but that doesn't make a difference. Now, what about the following
reaction? Any equilibrium where you have
different states of matter is called a heterogeneous
equilibrium. And so let me write another
heterogeneous equilibrium. So let's say I have H2O in the
gaseous state and that's essentially steam-- so it's not
going to be the solvent this time-- plus carbon
in the solid state. And let's say that that's an
equilibrium with hydrogen in the gas state plus carbon
dioxide in the gaseous state. This is a heterogeneous
equilibrium because you have things in the gaseous
and the solid state. And solid state, by definition,
it can't be dissolved either into the gas
or into the-- when we talk about solutions, we talked about
colloids and suspensions and mixtures before, but we're
talking about solutions. By definition, if this
is in the solid state, it's not dissolved. If this was dissolved, we
would write an aq here. It would be the aqueous state. So if you talk about the forward
reaction, what's the forward reaction going
to be dependent on? So the rate forward, well, the
solid, there's a big block of carbon sitting there. There's a big cube of carbon
there, and there's steam, there's water gas
all around it. So if you pick any volume,
especially if you pick some volume near the boundary of
the carbon, you're always going to have carbon around. It's just what matters
is the concentration of the water gas. That's what's going to drive
the forward rate, so the forward rate is going to be
dependent on some constant times the concentration
of the water gas. And, of course, the backwards
rate, so you need to get some H2, some molecules of-- let me
draw it like that, because it has 2 hydrogen molecules plus
a carbon dioxide, so maybe a carbon dioxide looks
like that. So the reverse reaction, so
rate, let's call that reverse, is going to be equal to some
constant times the probability of finding both of these
molecules in the same place. And, of course, the probability
is related to or it's on a first-level
approximation, depending on the concentration. So it's concentration of H2
times the concentration-- and to find both of them, you
multiply the probability, because you need this
and that-- times the concentration of CO. So when a reaction is in
equilibrium, these two equal each other-- this is an r right
here-- so this is going to be equal to the reverse
rate of reaction H2 times carbon dioxide. Divide both sides by the K's,
both sides by the H2O, and you get the forward coefficient or
constant or whatever you want to call that, divided by the
reverse constant-- I'm just dividing both sides by that-- is
equal to this-- let me just copy and paste that-- is equal
to that divided by this. You take that and you
divide it by that. And so if we call this the
equilibrium constant, because it's just two arbitrary
constants, so we can just call this the equilibrium constant,
you see that it actually makes a lot of sense to ignore
the solid state in your equilibrium reaction. So the two takeaways here
is when you're trying to calculate an equilibrium
constant, you should ignore-- especially when it's in a
heterogeneous equilibrium-- you should ignore the solution-- or not the solution. Ignore the solvent in that first
example, where I did it with boron trifluoride
with water. Water was the solvent,
so I ignored it. Because water is everywhere,
and you also ignore the solid state. Ignore the solid. Anyway, we'll probably use these
in future things where we actually calculate the
equilibrium constant. See you in the next video where
we'll learn about Le Chatelier's principle.