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Current time:0:00Total duration:14:43

Let's say we had the reaction
molecule A plus molecule B is in dynamic equilibrium with
molecules C plus D. Which just means that the rate
of the forward reaction is going at the same rate as the
backward reaction, or the reverse reaction. There will be some equilibrium
concentrations of A, B, C, and D, and we can figure out
what the equilibrium constant is if we want. And I'll say it again. I've said it like four
times so far. The fact that the forward
reaction rate is the same as the backward reaction rate
doesn't mean that all of the concentrations are the same. The concentrations themselves of
each of the molecules could be very different. They're just not changing
anymore because the forward and backward rates
are the same. Now, given this, given that
we're at equilibrium now, what's going to happen if I
add more A to the system? So remember, it was
in equilibrium. The concentrations
were constant. But now all of a sudden, I'm
adding more A to the system. So now, the odds of an A and a
B particle, even though I'm not adding any more B molecule
to the system, the odds are slightly higher that an A and
a B are going to collide in just the right way, so the
forward reaction is going to be more likely. So if we add more A to the
system, you're going to have more A's. They're going to bump with
more B's, so the B's are actually going to go down
a little bit, right? Because more B's are going
to be consumed. And even more important, the
C's and D's are going to definitely be increased. And the way it would really
happen, you would add more A. Those A's would bump into some
more B's, and so this forward reaction, all of a sudden, it's
rate would go faster than this backward reaction. So the reaction would go
in that direction. Then you would have more C's
and D's and maybe some of those are more likely to bump
and go back in this direction. Eventually, you would reach
a new equilibrium. But the bottom line is you'll be
left with more A, a little bit less B because you
didn't add more B. So more B's going to be used
to consume with those extra A's you just added. And then those are going to
produce more C's and D's in equilibrium. And you can imagine, if you
added more A and more B, let's say if you added more B as well,
then the reaction is going to go in the forward
direction even more. I don't think this is an amazing
insight of the world. I think this is kind of obvious,
that if you stress this reaction by adding more on
this side, that naturally it's going to move in
the direction that relieves the stress. So if you add more A, you're
going to have more A's bumping with B's and go in that
direction and maybe consume a little bit more B's. If you add more of both, the
whole thing's going to go in that direction. Likewise-- let me rewrite
the reaction. I'll do it in a different
color. A plus B, dynamic equilibrium,
C plus D. If I add more C-- I think you
get the point here-- what's going to happen? Well, that's going to drive A
and B up, and it's maybe going to consume a little
bit extra D. And then if you added more C and
D, then, of course, it's going to produce a
lot more A and B. And this idea, it seems pretty
common sense, but there's a fancy name for it, and it's
called Le-- let me put a capital L-- Chatelier's
principle. If you've watched enough of
these videos, you know I have to be careful with
my spelling. And all it says is that when you
stress a reaction that's in equilibrium, the reaction
will favor the side or one side of the reaction to
relieve that stress. When they say stress the
reaction, that's like adding more A, so the reaction's going
to move towards the forward direction to relieve
the stress-- the quote, unquote stress--
of that more A. I mean, that's not stress in
its traditional way of thinking about stress, but
that is a kind of stress. You're somehow changing
it relative to it. It was nice and comfortable
before in a nice, stable environment. So given Le Chatelier's
principle, let's think of some other situations. Let's say if I had A plus B
plus some heat, and that produces some C plus D. And maybe it produces
some E as well. So if I were to add
heat to this system, what would happen? So in order for the reaction
to progress in the forward direction, you need heat. The more heat you have, the more
likely you're going to progress in the forward
direction. So Le Chatelier's principle will
say we're stressing this reaction by adding heat, so the
reaction will favor the direction that relieves
that stressor. And so to relieve that stressor,
you have more of this input, so you're going
to consume more A. So the stable concentration of
A once we reach equilibrium will go down. B will go down because they're
going to be consumed more. The forward reaction
is happening more. And then C, D, and
E would go up. Now, if you did the opposite. Let me erase what I just did. Let's say instead of adding
heat, you were to take away heat. So let's say you were
to take away heat. Let me make sure my
cursor's right. So if you took heat away
from the reaction, what will be favored? Well, then you're going to be
favoring it in the other direction because there'll
be less heat here. I mean, all of this
is together. There'll be less heat for this
reaction to occur, so this rate will start dominating this
rate over here, right? If you take away heat, the rate
of this reaction will slow down, this one will be
bigger, and so you'll have more movement of concentration
in that direction, or the reverse reaction will
be favored. Now, let's think of another
stressor-- pressure. Now, imagine that we had-- we
mentioned the Haber process before, and this
is the reaction for the Haber process. Nitrogen gas plus 3 moles of
hydrogen gas in equilibrium with 2 moles of ammonia gas. Now, what's going to
happen if I apply pressure to this system? I'm going to apply pressure. So if you think about what
happens with pressure, everything all of a sudden is
getting squeezed, although the volume isn't necessarily
decreasing, but something is somehow making all the molecules
want to be or forcing them to be
closer together. Now, when things are getting
closer together, the stress of the pressure could be
relieved if we end up with fewer molecules. Think about it this way. PV is equal to nRT. We learned this multiple
times, right? And let's say we could write
P is equal to nRT/V. Now, if we increase
the pressure, how can we relieve that? Remember, Le Chatelier's
principle says that whatever's going to happen is going to
relieve the stressor. The reaction is going to
go in the direction that it relieves it. Well, if we lower the number of
molecules, then that will relieve the pressure, right? You'll have fewer things
bouncing against each other. So if we lower the number of
molecules where you can kind of view it-- I mean, I shouldn't
have written it this way, because it's not quite an
equation, but I want you to think of it that way. Let me erase this. This probably wasn't
the best intuition. If I have a container--
nope, too shocking. If I-- nope, same thing. If I have a container and I'm
applying pressure to it, and in one option I could have 2
molecules-- let's say I could have 4 molecules
in some volume. And in another situation, let's
say they get merged and I only have 2 molecules,
right? In either of these, the reaction
can go between these, these 4 could merge to
make 2 molecules. Actually, let me use this
example up here. Let's say this nitrogen
molecule is this blue one here. Actually, let me do it in
a more different color. This brown one right here, it
can merge with 3 hydrogen. It could produce this. So this is another way of
writing this reaction, maybe in a more visual way. Now, if I'm applying pressure,
if I'm applying pressure to this system, so pressure I just
imagine is kind of more force per area from every
direction, which of these situations is more likely to
relieve the situation? Well, the situation where we
have fewer molecules bumping around because it's easier to
kind of apply or I guess squeeze them together than when
you have more molecules bumping around. I'm doing this very hand
wavy, but I think it gives you the intuition. So if you apply pressure to the
system, if pressure goes up, you're applying-- this
doesn't mean the pressure goes down. This means pressure is applying
to the system. But the pressure is going up,
what side of the reaction is going to be favored? The reaction's going to be
favoring the side of that has fewer molecules. And this side has 2 molecules,
although they'll be bigger molecules obviously, because
it's not like we're losing mass in one direction or the
other, as opposed to this situation where we have
4 molecules, right? 1 mole of nitrogen gas and
3 moles of hydrogen. And just to bring this all back
to the whole idea that we saw earlier with the kinetic
equilibrium, let's just imagine a reaction like this. And to show that it works with
Le Chatelier's principle is consistent with everything
we've learned with equilibrium constants. So let's say we had the reaction
2 moles, or the coefficient of two, 2 A's in the
gaseous form plus B in the gaseous form is in equilibrium
with C in the gaseous form. And let's say initially where
our first equilibrium, our concentration of A is 2 molar,
or our molarity is 2, our concentration of B is 6 molar
concentration, and then our concentration of C is 8 molar. So what's the equilibrium
constant here? The equilibrium constant here is
the product, concentration of C, that's 8 molar divided by
2 molar squared, because of this, 2 squared times 6. Which is equal to 8/24,
which is equal to 1/3. Now, let's say we were to add
more A, and I'm not going to say exactly how much. We could actually get quite
complicated with the mathematics, but let's say after
adding more A, we have a new concentration. Now, let's say our concentration
of A is 3 molar. You might say, hey, Sal,
didn't you add 1 molar? No. I actually added probably
more than 1 molar. What happens is, whatever I
added, that'll push the reaction towards the right
direction, or towards the forward direction, and so some
of it will get consumed and go in that direction, but
whatever's left over is here. So I might have added more than
1 molar concentration to this system. But whatever was extra beyond
the 1 was consumed, and I'm just left with this equilibrium concentration of 3. So I didn't necessarily add 1. I could've added
more than that. And let's say that our new
equilibrium for C is 12 molar, which is consistent
with what we say. We should be producing-- if
we add some A, then our concentration of C should go up,
and the intuition is that the concentration of B should go
down a little bit, because a little bit more B is going to
be consumed, because it's going to be colliding with-- or
it's more likely to collide with-- more A particles
or A molecules. So let's see what B's new
concentration is. So remember, the equilibrium
constant stays constant. So our equilibrium constant is
now going to be equal to the concentration of C, right? That was the reaction. So it's 12 molar-- whoops-- I
don't have to write the units here-- divided by our new
concentration of A, that's 3. But remember the reaction. The coefficient on A is 2. So it's 3 squared times the new concentration for B, right? There's no coefficient here so
I don't have to worry about any exponents. And let's just solve this. So you get 1/3 is equal
to 12 over 9B. So if we just cross-multiply,
we get 9 times the concentration of B is equal to
3 times 12, which is 36. And so divide both sides by 9. The new concentration of
B is 4, or 4 molar. So B is equal to 4 molar. So that makes sense. We added some more A
to the reaction. So we started with 2 molar of A,
6 molar of B, 8 molar of C. When we added more A, at the
end, we added a bunch. It went in that direction, maybe
it went back and forth a little bit, but it stabilized
at 3 molar of A, 12 molar of C, so C went up, so that
definitely went up. And notice, our stable
equilibrium concentration of B actually went down, and this
is consistent with what we were saying, that the reaction
moves in that direction, more C gets produced, a little
bit of B gets consumed. So anyway, hopefully you're
fairly comfortable now with the whole notion of stressing
a reaction and Le Chatelier's principle.

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