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## Equilibrium constant

Current time:0:00Total duration:12:52

# Keq intuition

## Video transcript

Let's see if we can develop some
intuition as to why the equilibrium constant equation
looks the way it does. Just as review, this is it:
equilibrium constant. It would be the concentration
of our molecule Y raised to its coefficient power or, if
we're thinking in moles, raised to the number of moles. If we think of these as kind
of the mole ratios, or the molar ratios-- or we could
just view them as the molecular ratios, either way--
times the concentration of our molecule Z. Now, we're not doing
some calculus here. d is just the number of moles we
need of Z for every c moles of Y, b moles of X,
and a moles of V. So it's Z to the d power divided
by the concentration of V to the a power and
X to the b power. So it's a nice, little, clean
equation, but why does it look this way? And I actually made a video
earlier today where I started exploring this with
natural logs. And I think I got someplace,
but that one started to break down. And I think I've come up with a
much simpler reason why this looks this way. So I've deleted that video, and
I think I've come up with a much more intuitive one that
explains more of why this works and actually some of the
other things we're going to learn about the equilibrium
constants in future videos. So what makes a reaction
happen? Or what does equilibrium mean? It means the rate at which the
forward reaction is happening. So that means that the rate of
this happening, of V plus X turning into Y plus Z--
I can't forget the coefficients-- is going to be
equal to the reverse reaction, is equal to the rate of
the reverse reaction. So our c moles of Y plus d moles
of Z turning and going the other way, turning
into the V and the X with certain ratios. It doesn't mean necessarily that
the concentrations are the same, because we could
have one where we end up heavily favoring the
forward reaction. Where we end up with much higher
concentrations of Y and Z, or we might heavily favor the
backwards reactions where we have more V and X. But when we're in equilibrium,
we're saying that our concentrations have reached a
stability point, which implies that the rate of going in this
direction is equal to the rate going into that direction. So let's just think a little
bit about what drives these rates, what drives these
rates of reactions. In order for this forward
reaction to happen that I drew in purple, what needs
to happen? We have to have a molecules
of V roughly. And let's say in any volume of
space, we have to have some V molecules, and preferably a
V molecules, being in the vicinity of b X molecules. So there's got to be b of these
X molecules, and they have to be in the right
configuration and in the right place and kind of close
enough in order for the reaction to happen. So the reaction is really going
to be driven by, if you think about it, the probability
of finding a V molecules and b molecules all
within close enough confines that they can actually react. So you could say that the rate
is going to be driven by-- maybe it's going to
be proportional. Let's say it's just equal to--
let's say some constant that takes into account things like
temperature and how the molecules are actually
configured. Because it's not dependent
just on them being there. You have to have worry about
their kinetic energies. You have to worry about their
shape, because some shapes are going to be more conducive
to reaction than others. So let's just let that be taken
into account with a K. And we're talking about the
forward reaction, right? So in order for the forward
reaction to happen, let's call that K plus for the
forward reaction. We have to have a molecules of
V there and b molecules of X. So what's the probability of
having a molecules of X? Or what's a rough approximation
of the probability? Well, the concentration. Let's think about
this a second. When we write the concentration
of the molecule V, which I think when I did this
was the blue one right here, what is that given in? That is given in moles
per liter. Moles is just a number, so this
tells us, look, in any given volume, roughly how
many of the molecules do you expect to find? That's what concentration is. So if I wanted to figure out the
probability of finding a of these molecules, because
that's how many I need, I need to multiply this by
itself a times, because I need a of them. The probability of having just
one molecule in just some small fraction, you would just
use the concentration once. But you're going to use it a
times, because you want a of those molecules there, right? You could look at it like what's
the probability of having five heads? Well you would multiply
the probability of one head five times. So the forward reaction
probability is going to be the concentration of V to the a
power, and, of course, that's not enough to have the
reaction happen. You also need to have b of
the X molecules there. So you have the concentration
of X to the b power. And I want to make sure
you understand this. My claim is that this is
approximation-- or actually it's a pretty good way of
calculating-- the probability. So let me write it this way. The rate is equal to some
constant that takes into account the temperature and the
molecular configurations times the probability of having
a V molecules and b X molecules in a sufficiently
small area all at the same time. And the best way to approximate
that is with their concentration. Obviously, the higher the
concentration, the higher the moles per liter, the more likely
you're going to find that many of molecules in kind
of that little small space that you care about, and
the temperature and the configuration are going
to matter more. But if you use the concentration
as the probability of a-- let
me switch colors. If the probability of having a V
molecule in some volume-- if we assume that the solution
is homogeneous, that the V molecules are roughly evenly
distributed, it's going to be-- this isn't even
an approximation. It's going to be the
concentration of the V molecules times the volume under
which we care about. If we want the probability of
a, where a is a number, it could be five V molecules, a V's
in some volume, it's the probability of finding
this a times. So it's going to be equal to--
and this is just from the probability concepts that
we learned in the whole probability playlist. So if you want to have five
heads in a row, it's 1/2 to the fifth power. If you want to have V molecules
there, five of them at the same time in some volume,
or a of them, it's going to be V to the a power
times the volume. If you also care about the
probability so you want all of that, so a V's and b X's in some
volume, then you're going to have to multiply all
of them together. So it's going to be equal to the
concentration of V to the a power times the concentration
of X to the b power times the volume. So the probability of finding
the right number of V particles and X particles in the
right place in some volume is going to be proportional
to exactly this. And we're saying that the
reaction rate, the forward reaction rate, is also
proportional to this thing. So that's where we get the
forward reaction rate. So the rate forward is equal to
the concentration of our V molecules to the a power times
the concentration of our X molecules to the b power. Now, if we want to find the
reverse rate, so this is the rate forward. If we want to find the rate of
the reverse reaction, let's say that that's equal to some
other constant-- let's call that K-minus-- the same
exact logic holds. We're just going in this
direction now. If we look at our original
one, we're going in that direction. So for this reaction, we
do the same thing. We literally just do different
letters, so the reverse reaction is just going to be
the concentration of the Y molecule to the c power, because
we need c of them there roughly at the same time,
times the concentration of the Z molecule
to the d power. Now, just at the beginning of
the video, we said that equilibrium is when these
rates equal each other. I wrote it down right here. So if the reverse rate is equal
to some constant times this, and the forward rate is
equal to some constant times that, then we reach equilibrium
when these two are equal to each other. Let me clear up some
space here. Let me clear this up, too. So when are they going to
be equal to each other? When the forward rate-- the
forward rate is this. That's our forward constant,
which took into account a whole bunch of temperature and
molecular structure and all of that-- times the concentration
of our V molecule to the a power. You can kind of view that as
what's the probability of finding in a certain volume--
and that certain volume can be factored into that K factor
as well-- but what's the probability of finding
V things, a V molecules in some volume. And it's the concentration
of V to the a power times concentration of X to the V
power-- that's the forward reaction-- and that has to equal
the reverse reactions. So K-minus times the
concentration of Y to the c power times the concentration
of Z to the d power. Now, if we divide both sides
by-- let me erase more space. Nope, not with that. All right. So let's divide both sides by
K-minus and both sides by this, so you get K-plus over
K-minus is equal to that, is equal to Y to the c
times Z to the d. All of that over that-- V to the
a times the concentration of X to the V. Let me put this in magenta just
so you know that this was this K-minus right here. And then, these are just two
arbitrary constants, so we could just replace them
and call them the equilibrium constant. And we're there where
we need to be. We're at the formula for the
equilibrium constant. Now, I know this was really hand
wavy, but I want you to at least get the sense that this
doesn't come from out of the blue, and there is-- at
least I think there is-- there's an intuition here. These are really calculating the
probabilities of finding-- this is the forward reaction
rate probabilities proportional to this. Because the more V concentration
you have, the more likely you're
able to find it. Although if you need more of
those particles around, you're going to have to multiply that
concentration by each other, because the probability's
going to get lower. Because you need more of them
together in order for the reaction to happen. Same thing for everything
there. But all this is derived from is
that the forward reaction should be equal to
some constant times the reverse reaction. Or actually, their rates should
be equal, but then when you actually calculate the
probability, you'll have a constant in there. Anyway, hopefully, I didn't
confuse you, but I just wanted to give you that this isn't
just some random equation. It really does, I think, come
from the reality that the higher the concentration you
have, the more the probability you have of the actual
molecules bumping into each other.