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PROFESSOR: OK, it's time to get started.
Pay attention to the clicker questions.
All right, I'll give you 10 more seconds.
Not bad, we seem to be in the 70's.
All right.
So if you didn't have time to click in, let's consider
what's going on here.
What's true about the relationship of q and k here?
So is q less than or greater than k?
Less than.
And so you have to think about what that means in terms of
where we are now in the reaction and where the reaction
is at equilibrium, and I like to think about it in terms of
products -- are there more products now, or are there more
products at equilibrium.
So if q is less than k, then there are more products at
equilibrium, and so the reaction will shift
toward products.
So it's going to move in the direction to reach equilibrium,
since k is a bigger number than q, and both of these terms are
products over reactants, there'd mean there's more
product at equilibrium then there are now, and so the
reaction is going to shift toward products.
All right, so we're at 73%.
Let's see if by the end of today, we can get up to 90 on
these kinds of questions.
All right, so we're going to continue to consider external
effects on k in today's class.
I also wanted to mention that on each of the handouts for the
lectures, I have the corresponding reading material
listed on the handout, and I've listed it as section numbers in
the chapters rather than page numbers, so that's a little
bit of a difference.
And the reason why I've done that is that we've had many
different versions of the book in this class, and they don't
seem to change the section titles or the section numbers,
but they do change the page numbers associated with them.
So if I give you section numbers then you should be able
to use of whichever version of the book is available to you.
So that's how I have it listed now, and the reading assignment
is also listed on each problem-set as well.
So when you're going over the handouts later studying for the
exam, you can see where in the book this particular lecture,
the material is, so you can go and do that reading.
All right, so we talked at the end of the class last time
about the Le Chatelier's principle, and this is a
principle that you can use to predict the direction of change
of a reaction -- whether it'll shift to the right or
shift to the left.
And simply stated, that systems tend to respond in such a
way to minimize the stress.
So if something is added to the reaction, the reaction will
shift in a way to minimize that stress caused by the thing
that's added to the system.
So, we talked about adding reagents and removing products
last time, and now we're going to go on and talk about
what happens when you change the volume.
So if you have a system at equilibrium and you change the
volume of it, and here in particular we're talking about
a gaseous system, what's going to happen?
So what do we know about volumes when it comes to gases?
We say that a decrease in volume of the gaseous
system causes an increase in total pressure.
What equation leaps to your mind when I say those words?
Exactly, p v equals n r t.
Yes.
So there's a relationship between pressure and volume --
n is the number of moles, r is our gas constant, and
t is our temperature.
This is one of the things that usually sticks from high
school, and if you haven't seen it before, it's pretty easy to
get up to speed with this equation.
So Le Chatelier's principle then predicts that the system
will respond if possible in a way to reduce the
total pressure.
So if you decrease the volume and you cause an increase in
total pressure, Le Chatelier says wait a minute, the system
wants to respond to minimize that stress, so that it's going
to respond in such a way to reduce that total pressure.
So let's look at some examples of this.
So we have an example of a reaction where we have two
moles of p 2 gas going to one mole going to p 4 gas.
And I have a little cartoon here to show this.
So we have our p 4 gas up here, and our p 2 gas.
All right, so what's going to happen if we're going to change
the volume of this system?
So if we're going to decrease the volume of the system,
then the reaction should shift toward products.
So let's think about that.
So here in the middle we have our sort of system, and then on
one side what happens when you decrease the volume and
on the other side we're expanding the volume.
So first, let's consider our system when we're going to
compress that volume, decrease that volume.
So what would be true is that the system's going to respond
in such a way to minimize that stress that has been put
forward by decreasing the volume, and you're going to try
to compensate for that, and in this case you can compensate by
a shift toward products.
Why is that?
Well, it all has to do with the stoichiometry of
this equation up here.
So for every two moles of reactants, you get
one mole of products.
So if you shift from the two to the one, that will cause a
decrease in pressure, and so that will help compensate for
the increase in pressure caused by the switch in volume.
So you're going to respond in such a way to compensate
for that stress.
So a shift to the reaction to the right decreases
the total pressure.
So now let's think more about why this is true.
Let's think about it from a math perspective and also just
this qualitative perspective.
So let's consider it in terms of q and k.
So suppose our volume is decreased by a factor of 2,
let's just make it easy, and we have constant temperature here.
So this change will increase the partial pressure of both
gases, the reacting gas and the product gas, and it's going
to increase them both by the same amount, by 2 initially.
So let's look at our equation for q, the reaction quotient.
We have partial pressure of products over the partial
pressures of the reactants, a raise to the coefficient
to the equation.
So initially you're going to increase the pressure, partial
pressure of both gases by 2 -- then your product by 2, the
reactant by 2 -- but here it's 2 squared.
So overall, q becomes 1/2.
So q decreases by a factor of 2, and so q
becomes less then k.
And so, in the clicker question, we talked about what
happens when q becomes less than k, and what happens there
is you're going to shift toward products until q
equals k again.
So you can think about this in terms of q and k, and you can
also just think about it how many moles of gas are on one
side, how many mold of gas on the other side, and how are you
going to correspond, how are you going to decrease or
minimize the stress on the system.
All right, so let's think about what happens if we increase the
volume, so here is our system and now we're going
to expand it.
If we're going to increase the volume, what's going to happen
to the total pressure?
I'm hearing it, but I want everyone to answer.
Thank you.
All right, so it will decrease the pressure.
And what that's going to do is shift toward reactants, because
we want to compensate for that decrease in pressure, and to do
that, we can increase the pressure by switching or
shifting the reaction toward the reactants.
So again, we're going from one mole of product on one side
to two of the reactants.
So here, we see a shift toward the reactants.
And so, this shift to the left is going to increase the total
pressure to compensate for the decrease in pressure that was a
result of this applied force to the system.
And we could do this again in terms of q and k.
Now we're going to get a little trickier.
What happens if we add an inert gas to a system increasing
the total pressure at constant temperature?
So we're not adding one of the reactants or one of the
products, we're adding an inert gas, and we're increasing
the total pressure.
What's going to happen?
There's a couple of options, what do you think?
How about nothing.
Why would nothing happen?
Well, q depends on the partial pressure of the reactant
gas and the product gas.
And in this particular example, the partial pressures
are not changing.
We are changing the total pressure of the system
by adding an inert gas, but we're not changing
the partial pressure.
So let's take this opportunity to review partial pressures, or
if you haven't seen it before, I'm going to tell you
everything you need to know about partial pressures.
So here's a little review, or for the first time, description
of partial pressure.
So what is the definition?
The partial pressure is the pressure that each gas would
exert if it were present alone in the container.
So in this example, we have oxygen in a container, and
we have one atmosphere of pressure.
In the second one there's nitrogen in the container with
one atomosphere of pressure.
If you put this amount of oxygen and this amount of
nitrogen together in a container, then you would have
two atmospheres of total pressure, but you're still
going to have one atmosphere of partial pressure for oxygen,
and one atmosphere of partial pressure for nitrogen.
So it's as if the gas is alone in the container.
That's the definition of partial pressure.
So let's look at some equations.
So the partial pressure of a gas, p to sub a, is equal to
the number of moles of that gas, and we also have in this
equation, r, our gas constant, t, our temperature,
and v, our volume.
The total pressure on the system, which is what this two
atmospheres is, is equal to the partial pressure of gas a,
which would be the partial pressure of the oxygen at one
atmosphere, the partial pressure of the nitrogen at one
atmosphere, that's all we have here.
So the partial pressure of those two, of one plus one,
we have two for a total.
And that's equal to the total number of the moles.
So in these problems, we're going to be considering partial
pressure, and the question you're going to be asking
yourself in all of these different wordings of the
question is is the partial pressure of the gas changing?
That's a very important question to ask because
that'll help you answer the questions correctly.
So let's go back to the original question, what happens
if an inert gas is added to the container, increasing the total
pressure at constant temperature?
And the answer is nothing, because q is not affected,
because q depends on the partial pressures, and the
partial pressure isn't changing.
We aren't changing the number of moles of the gas in
question, we're just changing the total pressure of the
system, we're adding an inert gas.
And because partial pressures don't change, q doesn't change,
and if q doesn't change, there's no response
from the system.
So a lot of the questions in this unit, they're not actually
very difficult, but the wording can be tricky.
And so when you're doing these problems, you have to be
thinking about what is changing and what isn't changing to
be able to get these questions correct.
And if you do that, then these are some nice questions for you
to get right on the exams -- they're short, they're good
questions to go after.
All right.
Let's see how you're doing with this.
What happens if an inert gas is added to the container, but the
total pressure is now kept constant, temperature
is also constant.
So let's think about this.
OK, let's take 10 more seconds.
OK, we're going down into the 60's -- again, we're heading
toward the 90's by the end, but this is actually new
material, so that's fine.
All right, so let's think about it.
So the reaction's going to shift toward reactants.
Let's think about why, let's break it down.
And what was important is that hint that for the pressure to
be kept constant, the volume of the container must
have increased.
So let's take a look at that.
So if the total pressure was kept constant, if the total
pressure didn't change when you added an inert gas, the
volume must have changed.
Let's look at this.
Say oxygen is our gas of interest, and n
2 is an inert gas.
If we added an inert gas to our system, the total
pressure should go up.
If it doesn't, then something else must have changed, and
what must have changed is the volume.
Otherwise the total pressure would not have stayed the same.
So the volume of the container must have increased if
the pressure is the same.
So then we asked the question, if you increase the volume,
what happens to the reaction, how does it shift?
If you increase the volume, if you expand the volume, then
you're going to shift from, in this case where you have one
mole to two moles toward reactants, you will, as the
volume increases, then you're going to have a change in the
partial pressure of the gas -- all of a sudden the gas has a
lot more room, and its pressure is going to decrease, the
partial pressure will decrease.
It has a lot more space for itself -- it's like it's
in there all by itself.
Partial pressure, you can think of it as selfish gas molecules,
they don't care what else is there.
They're only thinking about how they're fairing
in this environment.
So then we want to shift in a way that compensates for that
stress that increases the pressure, so we move from one
molecule to two molecules.
So it's all about the partial pressure q and k.
So again in these questions you have to dissect
what's going on.
What has changed?
The questions are often phrased in such a way, something's kept
constant, but if something's kept constant, you have to ask
the question, to keep it constant, did something have to
change for that to be true.
So again, the wordings of these can be tricky.
Do you have a question?
STUDENT: [INAUDIBLE]
PROFESSOR: So, the question is is it necessary for it to
try to increase pressure.
This is just the predictive tool of Le Chatelier.
Le Chatelier would predict that if the system has changed in
such a way, such as volume increases so the partial
pressure decreases, that the system will shift in a way
to minimize that stress.
And to minimize that stress, you would increase the partial
pressure by doing the shift from one mole to two.
So again, this is Le Chatelier's predictive, how you
would predict the direction of the shift based on the simple
idea that a system where a stress is applied, the system
will respond in a way to minimize the stress.
And again, I talked about last time that this concept of
minimizing stress is difficult for some MIT students, but you
just sort of have to memorize that in this case we're
predicting how a system will shift to minimize that stress.
OK.
So if you need a review of partial pressure, go over this.
This part, it looks a little challenging at first, and some
of the questions you really have to sit down and dissect
what's going on, and for most people, once they get there,
that part, those are good questions, you look forward
to them on the exam.
So now, we've talked about adding reagents, we've talked
about removing reagents, removing products, adding
products, shifts in volume, and now we're going to talk about
changing the temperature.
So here, Le Chatelier's principle, it's a little
bit more fuzzy, but it still basically works.
So if you talk about raising the temperature of a mixture
at equilibrium, then Le Chatelier's principle would
suggest that the system is going to respond in such a
way to minimize that stress.
So if you're adding heat, the system wants to respond in a
way to absorb some of the heat, to counterbalance
that change to the system.
So let's take a look at this.
So let's think about raising the temperature of an
exothermic reaction.
So what's going to happen here?
Raising the temperature of an exothermic reaction, favors the
formation of reactants, would shift toward reactants.
Well, why would this be true.
Well, you can think about this very simplistically that in an
exothermic reaction, you're going to be producing heat in
the forward direction and absorbing heat in the
reverse direction.
So if it's an exothermic reaction, that means it's
exothermic in the forward direction, which would mean
it was endothermic in the reverse direction.
So if you raise the temperature of an exothermic reaction,
you're adding heat, the system wants to respond in such a way
to absorb that heat, so it would shift toward the
endothermic direction of the reaction, which is the reverse
direction toward reactants.
So add heat, system wants to shift to absorb
some of that heat.
So in this part you think about the direction of the reaction,
whether it's exothermic or endothermic, and then think
about compensating for that stress.
Endothermic reaction, again, they say it's an endothermic
reaction, it means it's endothermic in the
forward direction.
So heat is being absorbed in the forward direction -- again
this is just sort of a very simplistic way to
think about it.
So if you raise the temperature then, you want to absorb that
heat, according to Le Chatelier's principle, and
so that would favor a shift toward products.
So let's look some more at this.
The predictive tool here is delta h.
So delta h, whether the reaction is exothermic or
endothermic, is going to be a predictive tool in thinking
about the direction of change, when heat is added to a system
or heat is removed from a system.
So let's try this out.
Heat is added to a system.
You're given information about your predictive tool, delta h.
Which direction will the reaction go?
All right, 10 seconds.
Why don't you discuss with your friends and vote again.
All right, 10 more seconds.
Yes!
I knew we could get into the 90's.
All right, so what type of reaction is this,
endo or exothermic?
Exothermic.
And so we're adding heat to an exothermic reaction, so it
wants to shift to absorb that heat, so it wants to go in
endothermic direction, which is the reverse direction or
toward the reactants.
And that's how it's supposed to work too, more people
get the right answer after the group discusses.
All right.
So we've been talking about the equilibrium constant so far in
terms of it being a constant, it's called the equilibrium
constant, but it's only constant at a given
temperature.
So the equilibrium constant changes with temperature.
It's also true that rates of reaction change
with temperature.
And kinetics is our last unit in this course, and we'll be
discussing that quite a bit.
So how does the equilibrium constant change
with temperature?
Let's consider that.
So let's look at some of the equations that we know that
have delta h in them -- delta h, again, is our predictive
factor when we're talking about changes in temperature.
We know that delta g nought is minus r t natural log of k, so
the relationship with our standard free energy, our gas
constant temperature, and our equilibrium constant.
We also know that delta g nought equals delta h nought
minus t delta s nought.
If those equations don't look familiar, then you should go
back and review the material on thermodynamics because you're
going to need it in this unit, the next unit, the unit after
that, and then the next two units, and they we're
done with the class.
All right, so good to get familiar with this now.
All right, so we can rearrange these in terms of solving for
our equilibrium constant.
And so we see the natural log of the equilibrium constant
minus delta h nought over r t, plus delta s nought over r.
And so, it's reasonable to assume that for the things
we're talking about, that delta h nought and delta s nought are
pretty much independent of temperatures.
That would be true for pretty much any temperature we
would be talking about.
So that means that k, or the natural log of k, is going to
change depending on the temperature.
It won't change -- these other ones are constant at all the
temperatures, so there's a relationship between
equilibrium constant and temperature.
So you can consider an equilibrium constant at one
temperature, and an equilibrium constant at another
temperature.
So we can talk about the natural log of a k 2, of an
equilibrium constant 2, equal to minus delta h over r
t temperature 2, plus delta s nought over r.
We can do the same thing for another temperature, another
temperature 1, equilibrium 1, and if we subtract these two,
delta s is going to cancel out, and we're going to get this
equation, which you know is important because it has a
name, so this is the van't Hoff equation.
And at some point later in the course, I'll be asking you to
come up with this name, so I always get very excited when
people can tell me it's a van't Hoff equation later on, not
that that's on a test, but there are very few
equations that have names.
So here we go.
So what does this do?
Well, we can talk about the natural log of one equilibrium
constant over another, an equilibrium constant k 2, which
is the equilibrium at temperature 2, equilibrium
constant k 1, which is the equilibrium constant at
temperature 1, is going to be equal to minus delta h nought
over r, and then it depends on the temperature.
So you bracket 1 over temperature 2 minus 1
over temperature 1.
And this just comes from subtracting those 2 equations.
So again, you can think about how the equilibrium constant's
going to change with temperature -- that's what
this equation does for you.
So let's look at some of the things that fall out of
this particular equation.
Let's think first about a case where this delta h nought
is less than zero.
What type of reaction is that?
Exothermic.
All right, so now let's consider a case for this
exothermic reaction where we're going to increase the
temperature, and so that means that t 2 will be greater than t
1 -- we've increased the temperature.
So now let's think about what sign the natural log of k
2 over k 1 will have if these things are true.
So there's a minus in the equation, let's put
the minus down here.
Delta h nought is negative, it's exothermic reaction, so
that's another negative sign.
And if we're increasing the temperature, and t 2 is greater
than t 1, then this temperature term will also have
and negative sign.
So if you have negative times negative times negative, you're
going to get a negative for your net result over here.
And what that will mean in terms of k 2 and k 1, is that
k 1 will be greater than k 2.
So you can think about this mathematically by running
through this argument.
You can also sort of think about it in terms of what you
would expect if you increase the temperature in terms of
what's the relative ratio of products to reactants
at equilibrium for an exothermic reaction.
So if you're increasing the temperature of an exothermic
reaction, the reaction would want to shift in the direction
to absorb that heat in the endothermic direction.
So you would expect that the equilibrium constant that was
at the lower temperature would be larger than the higher
temperature one, that there'd be less products at this
new higher temperature.
So again, you can think about it in terms of Le Chatelier's
principle, or you can run through with this equation and
do the math and think about the relative size of k 1 and k 2.
So now let's think about what happens when you decrease the
temperature where t 2 is less than t 1.
So we can do the same thing, we have a negative sign here, we
also have a negative sign for delta h, it's an
exothermic reaction.
But now we have a positive sign for this temperature term.
So overall we're going to have a positive sign for the natural
log of k 2 over k 1, which mathematically is going to mean
that k 1 is less than k 2 or that you would expect more
products in the equilibrium constant at higher temperature.
All right, so that's an exothermic reaction.
So if you're ever having problems with Le Chatelier, you
can go and use the van't Hoff equation to think about what
you would expect in terms of this shift, what you would
expect if you change the temperature of an exothermic
reaction in terms of the magnitude of the new
equilibrium constants.
All right, now you can do the same for me in terms
of this delta h nought is greater than zero.
Which of the following are going to be true, and notice
there's a possibility that all of them are true or that just
some of them are true or that one of them is true.
All right, take 10 more seconds.
See, actually people did very well, because e is also true,
all of those are true, but d was also true.
So if we add the 30 and the 58 together, I'm very happy.
All right, so let's look at this -- we'll keep the
questions up and let's consider all of them in order.
So the first one is that the reaction was endothermic,
I think everybody saw that as being correct.
OK, the second one talked about the equilibrium constant is
higher at higher temperatures.
So let's look at the first one.
We increased the temperature here, and we see that if you
work out the math or just think about it, k 2 is larger than k
1, so the second equilibrium constant is greater than the
first when you increase the temperature.
So that would favor then, in an endothermic direction, if
you're increasing the temperature, it favors the
endothermic direction of the reaction, so you could think
about the ratio having more products at that new
equilibrium constant at that higher temperature.
So that one was also true.
Then let's think about when you decrease the temperature, when
temperature 1 is greater than temperature 2.
And so this would favor, then, the exothermic direction.
So you would expect less products at this
newer temperature.
So, this part asked about k 1 versus k 2, and we see
k 1 is greater than k 2.
And then d says there's fewer products and equilibrium when
the temperature is decreased, and that's just the word way
of saying the same thing.
So all of those are true.
But again, people did very, very well with this question.
All right.
So now we're going to combine what we learned on Friday with
what we learned today, and think about ways that
that can be applied.
So why should you care about Le Chatelier's principle?
Well, for two reasons -- one I think Le Chatelier had sort of
a good life plan that when stress is applied to the
system, one should respond in a way to minimize that stress.
I think that is an important life lesson.
But also, it's very useful in terms of thinking
about maximizing a yield of a reaction.
So if you were going to create an industrial process and make
lots of money and give some of it back to MIT to improve
things in terms of teaching chemistry at MIT, then you
would want to think about these principles, because you would
want to maximize your yield.
If you're going to be making a product, you want to make a lot
of that product, and so you want to be thinking
about these things.
So we've talked about the reaction with nitrogen
and hydrogen making ammonia last time.
And this is an exothermic reaction.
So, there are a lot of people who want to make ammonia --
it's for fertilizer, and as some of you have heard in terms
of terrorism, it also can be used as an explosive, so there
are a lot of people who want to make a lot of this.
So how do you maximize this yield?
It's an exothermic reaction.
So you can think about temperature.
So here, low temperature would favor product, which is good.
We haven't talked about kinetics yet, but low
temperature also slows the rate, which is bad.
Because you not only want to make a lot of your product, you
want to make it in a reasonable time frame and sell it and make
lots of money and retire early.
So you care about how fast the reaction is going as well.
So for an exothermic reaction then, you have to balance out
what's thermodynamically favorable and what's
kinetically favorable.
So thermodynamically, you would want a low temperature, but
kinetically in terms of the rate, that's not so good.
So they use a compromised temperature, which is 500
degrees c, so that's pretty high actually
for this reaction.
So what are other things that you could do to drive this
reaction toward products?
So yeah, what's something else you could do?
STUDENT: Increase the volume.
PROFESSOR: Right.
So you could think about volume here.
So here we have four molecules of gas on one side, and two
molecules of gas on the other side.
So you can think about changing that to favor your products.
What is something else you could do?
STUDENT: Enzymes.
PROFESSOR: Enzymes, we'll talk about that in a minute.
If you can't use the enzymes, what's something else
that you could do?
STUDENT: You could remove ammonia from the system.
PROFESSOR: Yup, so you could remove ammonia from the
system, which would also shift the direction.
So let's just put those two things down.
Remove products.
So if you remove ammonia from the system, and they actually
do this in the industrial process, so they'll stop --
they'll, at a certain time points, they will remove the
product and they'll say hey, let's shift the
reaction to make more.
You just changed q versus k, you removed your product, so
now we want to shift to minimize that stress
and make more product.
And the first response, compress the volume, so we go
from four molecules to two molecules, and that would be
favor -- again, the system would want to respond, you
compressed the volume, you increased the pressure, so you
want to respond in such a way that compensates for that
increased pressure, which is decrease the pressure by going
from the four molecules to two.
So both of those are great, and both of those
are, in fact, used.
Sometimes when you study chemistry you'll realize that
some of the -- people made a lot of money on this, and
that some of the principles are actually very simple.
You are learning principles in this class that, applied
correctly, could make you a lot of money.
Some of the scientific discoveries are actually
not all that complicated.
All right.
So someone said use enzymes.
I love that answer, I want to use enzymes to do a lot of
those things, and people are trying to do this.
Now, why would you want to use an enzyme?
Well, there is a lot of nitrogen in the air, but
there's not a lot of usable nitrogen, because n 2 gas is
in the air, but it's very difficult to break it
apart to form ammonia.
And when you think about bonding, you can probably tell
me why it's hard to break n 2 apart, and if you consider how
many bonds would be between those two nitrogens.
So it's there but it's hard to do.
So the industrial process that's used now pretty much is
the same industrial process that's been used for quite a
while, the Haber-Bosch process, which does not use enzymes.
And here are some pictures of these folks.
This resulted in two Noble prizes, the development of
the industrial process here, and it's still being used.
On a historic note, these were German scientists, and so,
working out this process was important to the Germans
during World War II.
It's still the process going on today.
So what about enzymes.
Maybe enzymes could do a bit better than this process.
It's not true yet.
Still the Haber-Bosch industrial process is the
one that's being used, but enzymes can do it.
So, some of the problems of a lot of the common industrial
reactions are that you use high temperatures -- remember we
said the compromised temperature was 500
degrees celsius.
We were talking about compressing the volume, so
you want to put in energy to compress that volume.
And so that is expensive.
And you might want to think about a way, could you do that
reaction at normal ambient temperature without applying
energy to the system.
Well, enzymes can do this reaction at ambient
temperatures and they don't have to have high pressure
associated with them, and the way the enzymes do this, and
here's a little cartoon of a space-filling model of
a particular enzyme called nitrogenase.
So for nitrogen, and the ase means it's an enzyme.
And the secret to this enzyme, how it can do what Haber and
Bosch got two Nobel prizes for -- this enzyme has received
no Nobel prizes, and it can do the same thing.
Its secret are metals.
It uses these clusters of metals in the context of the
protein environment to do this chemistry, and they have iron,
and they have also inorganic sulfide, and molybdenum, and so
these combinations of metals can do this chemistry.
And a number of scientists have been working for decades
now trying to understand how the enzyme works.
It's actually quite complicated and it's been pretty
controversial.
So let me just kind of show you at the heart of the enzyme,
here all the atoms of the enzyme, but if you rotate the
enzyme around and go in, the secret was that those
combinations of metals.
And one of the units that we're going to have second to last in
this courses is about transition metals, that metals
can do a lot of really important things in biological
systems, and also in industrial processes and other
things as well.
So I do also like to mention research is
going on here at MIT.
I've mentioned Dick Schrock before, he won a Nobel Prize in
chemistry a few years back, but not for his work on sort of
nitrogenase systems, but he was able to design a small catalyst
that can do this chemistry, that can catalyze this
reduction of nitrogen to ammonia at a defined
molybdenum center.
Again, we'll talk about transition metals
later in the course.
And so his laboratory is very interested in coming up with
better ways to split nitrogen than the industrial processes
that are currently used.
Some people are studying enzymes, other people are
trying to use some of the tools that come out the enzymes,
thinking about the enzymes and come up with other
catalysts as well.
So this work is going on here at MIT.
All right, so let's give one more example of Le Chatelier
from a biological perspective, and this has to do with
the following reaction.
The combination of oxygen with hemoglobin in your blood.
And the reaction here, we have hemoglobin plus oxygen has
oxyhemoglobin, so the oxygen is now bound to be hemoglobin.
This reaction is very important for you.
None of us would be alive if this was not happening.
We need to get the oxygen from our lungs to our cells.
So what happens if you decide to climb Mount
Everest, for example.
You may be thinking how hard can that be after a first
semester freshman year at MIT.
Really, I can think more broadly about it.
All right, how many of you have climbed a mountain so far?
Woah, a lot of people.
OK, something about going to MIT and climbing big mountains
seems to be connected.
How many people feel like after surviving the first year, why
not, go ahead and climb a mountain?
Any more people?
OK, no.
So you gotta -- you've been through the hard part now,
you don't want to risk anything anymore.
All right, so this slide seems to be cutting off some stuff,
so let's look over here.
So what happens to this reaction?
Well, as you go up higher and higher, the partial
pressure of oxygen changes.
So you're going to drop in partial pressure 0.2 to 0.14 .
So what happens to this reaction if this
value decreases?
What happens, what direction does it shift?
For products or reactants?
Right.
So, if this is going down, then some of your oxygenated
hemoglobin is going to a release its oxygen.
So it'll respond in such a way to compensate
for that added stress.
So what can you do to switch it back the other way?
Again, we want our blood to be oxygenated.
What happens to compensate for this if you're going
to climb a mountain?
What happens in the body?
Yeah.
STUDENT: The body can produce more hemoglobin.
PROFESSOR: The body can produce more hemoglobin.
So you often go to another altitude and hang out
for a while before you start going way up.
And what's happening during that time is your body realizes
there is a problem, it knows about the Le Chatelier's
principle, and so it's going to make more hemoglobin.
And if you add more hemoglobin to this reaction, what
direction does it go?
Toward products, right.
OK.
So I wanted to mention significant figures, and there
is -- I have to make a confession that when I first
started teaching this course, I had never really paid attention
to the rules of significant figures for logs.
In this unit, the next unit, the unit after, you get the
idea, there are going to be a lot of logs, and they
have special rules for significant figures.
So in the bottom of your handout, these rules
are explained, it's also in your book.
Pay attention to this.
It's going to help you on all the problem-sets
and in the course.
So I just wanted to point out special rules about
significant figures for logs.
All right, that's it for today, see you on Wednesday.