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CATHERINE DRENNAN: All right.
Let's just do 10 more seconds.
All right.
Does someone want to tell me up
here-- hello everyone up here--
how they got the right answer?
Over-- everyone
over there, I guess.
OK, I'm coming over there.
So for the answer we have
an MIT chemistry bag.
That's quite special.
AUDIENCE: Well, since
your delta H value
acts as the y-intercept,
you know that it's negative
because it's negatively.
And then you also see that
you're ascending positively,
so that you know your
slope is positive.
And so delta S also has to be
negative to make it positive.
CATHERINE DRENNAN: Interesting.
That was a different answer
than I would have given,
but that works
really well, which
is always great to ask people.
So you could also think about
this in terms of your going
from a situation with a negative
delta G to a positive value.
So you know that
temperature is going
to make a difference there.
And so that would also tell
you how the temperature changes
is also going to tell
you the answer to that.
So that's great.
All right.
So a couple of announcements.
As you can see, see what
I am wearing on my legs.
So I was pleased with the
performance on exam 2.
The average was 84.7,
not quite as celebratory
as 87.4, which was
the average on exam 1.
Yeah, those are real numbers.
That's a little weird.
But anyway, I was still very
happy with the performance
on the exam.
I would like for exam
3 for the average
to be back at 87, though.
I like it to be very close
to 90 because I want everyone
to have an average
around 90 in the class
to demonstrate excellent
knowledge of chemistry.
So I try to write
an exam where I
feel like if you have excellent
knowledge of chemistry,
you can get a 90% or above.
So this is my goal.
I all want you to have
this fundamental knowledge
of chemistry that you can go out
and solve the energy problem,
that you can go out
and be ready to tackle
the next great
challenge in health.
There's so many
challenges facing us,
and we don't even know
what they are right now.
So the Ebola situation, I
think, is a real case in point,
that we don't know what the
next challenge facing us
is going to be.
So we need to be ready with
our chemistry knowledge.
The only thing I can tell you
about the challenges facing us
is that knowing
chemistry is going
to be really important in
tackling those problems.
So everyone needs to have
an excellent understanding
of chemistry to go on
and do well and solve
the problems of the world.
So 84.7, really good.
You're getting there.
Awesome.
So celebratory
tights, or leggings.
Not really sure the difference
between tights and leggings.
Anyway, I was very pleased.
But next unit, next exam,
has thermodynamics, chemical
equilibrium, and acid-base.
We're already done with
thermodynamics and moving on.
So it's just for the
rest of the semester,
we're about halfway there,
the rate seems to accelerate.
So problem set 5,
thermodynamics.
It's a little shorter because
you have fewer days to do it.
So we wanted to
make sure that you
would be able to get it
done by Friday at noon.
But that's when it's due.
It's all thermodynamics.
We've already covered all the
material on the problem set.
You can do it
already right away.
All right.
So also in my wardrobe-- I'm
commenting a lot on my wardrobe
today-- you'll notice maybe
that I've worn this shirt once
before, and my goal was really
to have a different shirt
every time.
Now, this shirt is appropriate
for today's lecture.
But I just wanted
to say that I did
try to get a different shirt.
And I ordered new shirts on
thermodynamics and chemical
equilibrium this year,
and one of the shirts
had to do with entropy.
And I just thought I would
share with you briefly,
because it's a good
review of thermodynamics,
what happens when you do
things related to entropy.
So just to sort of
show you what happened.
In September, I was
super organized.
I ordered my shirts.
And on October 3, they shipped.
They went through
Indianapolis to Massachusetts
and to Newton Highlands
about somewhere
around October 8 or 9.
That's where I live,
so that's really good.
But the shirts
were about entropy,
and then look what happened.
Wait-- rerouted for
delivery to a new address.
Jersey City, Cincinnati, Ohio,
Warrendale, Pennsylvania.
And then yesterday they were
in Springfield, Massachusetts,
to which I say, entropy!
This does not happen when you
order T-shirts about kinetics.
That's all I'm saying.
Those arrive record time.
Thermodynamics is not
always your friend,
but you can embrace it.
And I'm going to try
to calm myself down
because we should be
at equilibrium today.
Today, the topic is
chemical equilibrium.
It's the calm part
of thermodynamics.
We'll get back to entropy.
There will be some
mention of entropy.
But I'm going to
try to calm down.
I'm going to
remember my reaction
quotients and my
equilibrium constants
and think about
chemical equilibrium
and embrace the
T-shirt I am wearing.
Because I feel powerful in this
T-shirt, like I can control
thermodynamics with the
power of the Green Lantern,
with chemistry knowledge
at my fingertips,
and I will move forward.
So chemical
equilibrium is calming,
but it's also a dynamic process.
We can't forget that the
reactions have not stopped.
It's just the rate of
the forward reaction
is equal to the rate of
the reverse reaction.
So there's no net change,
but the dynamic process
is still happening.
So my goal in life is to
have equilibrium in my life.
The amount of work-- the rate at
which work comes into my office
equals the rate at which
I complete the work
and it leaves my office.
Most of you should
also have this goal.
The rate at which
problem sets come in
equals the rate at which you
complete the problem sets
and the problem
sets get turned in.
You don't want to get
yourself in a situation
where that equilibrium
is messed up,
where the rate of
things coming in
is just nowhere at all equal
to the rate at which things
are completed and going out.
So goal-- chemical equilibrium.
We want to be at this calm place
where we're still working hard.
It's still dynamic.
But the net change is good.
It's in our favor.
The amount of work we're doing
equals the work coming in.
So I like chemical equilibrium.
I feel like it brings
hope to thermodynamics.
So let's look at an
example of a reaction,
and we're going to talk a lot
about this reaction in chemical
equilibrium.
So we have N2, nitrogen,
H2, hydrogen coming together
to form ammonia, and
it has a delta G0
of minus 32.9
kilojoules per mole.
So let's think about
what's happening
in the beginning of this
reaction, when we're just
starting this reaction.
So we can plot
concentration on one axis,
and we can have time
on our other axis,
so time increasing over here.
And we can think about
what happens when
we're starting our reaction.
We can have hydrogen
starting out.
We have some hydrogen in
a certain concentration,
and it's going to come
down and level out.
So that's our H2.
We'll also have
nitrogen as a reactant.
It comes down, levels out.
And we're starting
with our product.
We have no product.
We're just starting the
reaction with the reactants.
So product increases
and then levels out.
So we're decreasing
our reactants,
increasing our product,
but the lines go flat.
The reaction is still happening.
The reaction in the forward
direction is happening,
and then in the reverse
direction is happening.
But there's no net
change at equilibrium
because it's reached
a state where
the rate of the forward
reaction equals the rate
of the reverse reaction.
And let me also just
point out some arrows
and some nomenclature here.
So we have this
double arrow, and you
can see this a couple
of different ways.
But this indicates the
reaction's going forward,
and the reaction is going back.
And that's necessary if
you're at equilibrium.
The definition of
equilibrium-- rate
forward equals rate
back at equilibrium.
So you'll be seeing
those arrows a lot.
All right.
So let's think about
delta G now in terms
of what's happening at different
times in this reaction.
So if we're just starting
out this reaction,
and you don't have
a lot of products,
so you're in this case
of pure reactants.
At this point, you're
going to be moving
in the forward direction.
You need to make products.
And so at this point, the
forward direction is negative.
It's spontaneous.
You're spontaneously
making your products.
You don't have
any at this point.
And so your delta G is
going to be negative.
So up here with pure reactants,
your delta G is negative.
Now, if you started
with pure products--
you could do that as well-- the
NH3 will dissociate and form
the reactants.
And so that would be over here.
And so when you have
excess products,
now the forward
direction is positive.
It's not spontaneous because
it's the reverse direction that
is spontaneous.
So over in this part
of the graph over here,
our delta G is positive.
So if you have pure
reactants, you spontaneously
go to products.
If you have pure products,
you're going toward reactants.
And when you have the
right amount of reactants
to products, the right
ratio for equilibrium,
then you get to equilibrium and
delta G equals 0 at that point.
So delta G is going
to change depending
on your ratio of
products to reactants.
So delta G changes
as the proportion
of the reactants
and the products
change, and there's an
equation to describe that.
So we're back now
to having a few more
equations in this unit.
We noted a lot of people forgot
to bring calculators to exam 2.
There weren't so
many calculations.
We're back to more
calculations again.
So delta G equals delta G0
plus RT natural log of Q.
So let's think
about these terms.
So delta G is Gibbs free energy
at some particular point,
at some ratio of
product to reactant.
Delta G0 is Gibb's free energy
under standard conditions,
in their standard state.
Q is our reaction quotient,
products over reactants.
And R is the universal gas
constant, and T is temperature.
So this equation
describes how delta G
will compare to delta G0
at a particular condition,
a particular ratio, of
products and reactants.
So let's talk
about this reaction
quotient Q. So let's look
at a made-up reaction--
A plus B goes to C plus D.
And think about that reaction
in the gas phase.
When we think about that
reaction in the gas phase,
we're going to be talking
about partial pressures.
And so Q over here is
products over reactants,
but here we're writing it out
in a more complicated long way.
So let's go through this.
So we have this P sub
C. Well, what is that?
That is partial pressure.
So partial pressure
of gas X, and this
is partial pressure of
gas C, and C is a product
and D is a product.
So we have products
over reactants.
Partial pressure of
gas C over P reference,
partial pressure
reference, which is 1 bar.
The reference is 1 bar, so
we're dividing that by 1.
And this is raised
to the little c,
which has to do with the
coefficients of the reaction.
Then we have partial
pressure of gas D
over our reference raised
to the small d-- those
are our products,
products are C and D--
over reactants--
partial pressure
of gas A over reference
raised to little a,
partial pressure of gas B over
reference raised to little b.
But 1, the number 1,
the reference is 1.
So most of the time you'll see
the following expression-- Q,
our reaction quotient,
partial pressure of gas
C to the stoichiometry little
c partial pressure of gas
D raised to the stoichiometry
d, over reactants,
partial pressure of A little a
partial pressure of B little b.
And you will note that
because we had 1 bar
and these are in bars, our
units are going to cancel here.
So Q does not have units.
I knew this should be
very exciting for you.
You won't be losing one point
for lack of units with Q.
There are no units.
So this is very exciting.
So there's the expression
that you'll mostly see for Q.
And now we can kind of forget
that the reference is there.
We can ignore the reference.
You just don't freak out
later if you have no units.
It's OK.
All right.
So we also could be
talking about solution,
and here we're going to
talk about concentrations.
In a lot of these
problems, they're
talking about things
in the gas phase,
and they're giving
you concentrations.
Don't worry about it.
It's all OK.
So here our concentration
reference, C for concentration,
is 1 molar.
And you'll see this
term-- C in brackets.
That means concentration of.
And so if you see
that like this, you
would express this in words as
the concentration of C-- again,
that's a product-- over the
reference-- we'll get rid
of the reference in a
minute, but we'll keep it
for now-- raised to the little
c, concentration of product D
over this concentration
reference of 1 molar
to the little d,
concentration of reactant A
over this
concentration reference
raised to the little a,
concentration of other reactant
B over our concentration
reference of 1 molar raised
to the little b.
And then, again, this
is concentration.
And we can get this
expression, which
is usually the one you see, of
just products over reactants.
But I will make one
point here, that you
need to know how to
balance equations
to be able to do these.
So in the beginning
of the book there
was something about
balancing, limiting reagents,
stuff like that.
If you feel like you
did not master that,
you want to go over it.
You need to be able to balance.
And in a lot of
these problems you
may be thinking about
limiting reagents again.
All right.
So that's Q. You can see
it as partial pressure--
and I'm going to do a
little partial pressure gas
review on Friday--
or concentrations
with these brackets.
So at equilibrium,
delta G equals 0.
And it's a dynamic
process-- rate
of the forward reaction equals
rate of the reverse reaction.
Q now is the
equilibrium constant,
because the ratio of products
to reactants at equilibrium
is the definition of the
equilibrium constant.
So when delta G
equals 0, Q equals K.
So we now can go back to
this expression that we had
and rewrite this
for the situation
at equilibrium, when
delta G equals 0.
So when delta G
equals 0, this is 0
and Q is K. K is products
over reactants at equilibrium.
And we can rewrite this
or rearrange it now--
delta G0 equals minus
RT natural log of K.
And that's a very important
equation that you'll
be using a lot in these units.
So it relates delta G0 with
our equilibrium constant K,
and it depends on temperature.
All right.
So K-- same expression as
Q, except for something
very important, which is
that the concentrations are
the concentrations
at equilibrium.
So it has the same
form as Q, but it's
only the amount of products
and reactants at equilibrium.
So in the gas phase,
we would write
K in terms of our
partial pressures.
And again, this is this
little symbol to remind you
these are the concentrations
at equilibrium.
And in solution it
would be expressed
in molar or something else.
So K is, again,
products over reactants,
but only those concentrations
at equilibrium.
Whereas Q is at any
point, any concentrations
of products over
reactants for Q. For K,
it's those concentrations
at equilibrium.
All right.
So now let's think
about Q and K together.
So we can rewrite
this expression again.
We just derived a new
expression for delta G0,
and that was minus
RT natural log of K.
So we can say delta
G equals, and now
have this other
expression, minus RT
natural log of K plus
RT again-- gas constant
and temperature--
natural log of Q.
And we can simplify
it, bring out the RTs,
and now we have delta G
equals RT natural log of Q/K.
And this is, again, a
very important expression
that you'll use a lot.
Because it tells
you about delta G,
whether the reaction is
going to be spontaneous
in the forward direction
or the reverse direction,
depending on the
ratio of Q and K.
This is a very
important equation
for chemical equilibrium.
So let's think about what this
means, what comes out of this.
So the relationship between
Q and K, if Q is less than K,
what is the sign of delta G?
You can just yell it out.
AUDIENCE: Negative.
CATHERINE DRENNAN: Yep.
So it's going to be negative.
Just mathematically, you
can look at that expression.
So it'll be negative.
And again, we give
you all the equations
on the equation sheet.
That will be there.
So you just need to know
how to think about it.
And so the forward
reaction will occur.
And so if we think
about this, it's
going to mean that
at equilibrium there
are more products than there
are at this time for Q.
There's less products
in the Q expression.
K is greater than Q. So we
need to make more products.
So delta G will be negative,
and the forward reaction
will occur.
So when Q is greater than
K, delta G is positive.
And so when Q is
greater than K, that
means there's more
products now in Q
than there are at equilibrium,
too many products.
So we need to shift it
in the reverse direction.
So delta G will be positive.
So again, thinking about
the ratio of Q and K
tells you about what direction
is going to be spontaneous.
Is it spontaneous in the forward
direction or in the reverse?
OK.
So let's continue
doing a little example.
We can do a little
calculation here
on the board for
the same reaction.
We're given a
value of K. And now
we're told some partial
pressures of these gases
and asked which direction
the reaction will go.
So let's write the
expression for Q.
So Q, again, is going
to be equal to products
over reactants, and our
product here is our NH3.
So we're going to be talking
about the partial pressure
because it's a gas of NH3.
And am I done with the top
part of this expression?
AUDIENCE: No.
CATHERINE DRENNAN: No.
What do I need?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yep.
Again, you need to remember the
stoichiometry of the reaction.
So now we have the partial
pressure of N2 on the bottom,
and I'm good, and the
partial pressure of H2.
And again, I have to
remember that there are three
H2s in this balanced reaction.
So we have the partial
pressure of H2 to the third.
I can put in my numbers.
1.1 squared over 5.5 over 2.2--
these numbers may be made up,
that's OK-- equals 2.1
times 10 to the minus 2.
And we're back to thinking
about significant figures
a lot again in this unit.
So I have two there, so I'm
going to have these two here,
and I'm good.
But now this is the
value of Q. And I
want you to tell me with
a clicker question, what
direction is this
reaction going to go?
All right.
Let's just take 10 more seconds.
All right?
So recognized,
for the most part,
that yeah, most
of these numbers,
that Q is a bigger
number than K.
And so then you have to
think about what that means.
And so when Q is greater
than K, then you're
going to go toward reactants.
And so you're going to
dissociate the product
until you achieve
equilibrium again.
And so this means when
you have a big Q number,
you have too many products
compared to the equilibrium
state, and you want to
dissociate your products.
So you're going to
go spontaneously
in the reverse direction.
Now, I just want to
mention one point.
For the first part
of the course,
I only taught that
material once before.
But the second part
I've taught a lot.
So I have lots of
years of experience
of what people do wrong
on exams on this part,
and I will share one thing.
A lot of students--
and faculty too,
I'm this way--
right-left challenged.
So they write left
when they mean right.
They write right
when they mean left.
And so for those of you who are
like me and have this issue,
draw an arrow.
When people draw an arrow, they
always draw it in the direction
that they mean, or they
say toward reactants
or toward products.
And so I've seen so
many times on the test
they'd explain the answer
beautifully, and then
write the wrong direction down.
So if you are
right-left challenged,
try toward reactants or toward
products or draw an arrow.
So that is my suggestion
to you, as someone
who's also very bad with
saying the right direction,
left or right, when I
mean that direction.
OK.
So what does K tell us?
So K tells us about
the ratio of products
to reactants at equilibrium.
And if you have a
very big number for K,
it's going to tell you
something about the ratio
of those products
to the reactants,
and there'll be a lot of
products for reactants.
So let's just think
about this for a minute.
So when K is greater
than 1-- K is
products over reactants
at equilibrium--
that's going to
mean high products.
So you'll have a
higher number there.
But if you have a
small value for K,
then you're going to have
low products at equilibrium.
So this is a good thing just to
think about and check yourself
when you're doing
these problems.
Does my answer
actually make sense?
And if you see a big value
for K, you're like, great.
If I want products, I want a
big value for the equilibrium
constant, means
that this reaction
is going to give me
a lot of what I want,
a lot of my products.
Because again, K is products
over reactants at equilibrium.
All right.
So let's do an example here.
So here is an example
where K is greater than 1.
It's actually 6.84
at room temperature.
We have a delta G0 of minus
4.76 kilojoules per mole.
Two NO2 molecules going to N2O4.
So we're going to start
with 1 bar of our reactants
and no products.
So we can look at
what that's going
to look like in our plot of
concentration versus time.
And here the
concentration is indicated
as partial pressure, which
is one type of concentration.
So we're starting only
with our reactants.
It will go down.
We'll have no products
in the beginning.
At time 0, no products,
and that will grow up.
The curves will level off as
you reach the equilibrium.
Again, the reaction's
still going
in the forward and
reverse directions,
but the rates are
equal so there's
no net change in the
concentrations then.
So now we can
actually do some math
and figure out how this changes.
And so this is one way
that you can set up
these kinds of problems.
And we will be doing
these types of problems
in chemical equilibrium,
acid-base equilibrium.
There'll be many
examples, so it's
good to become
familiar with them
if you haven't seen it before.
So we can calculate
the partial pressures
at equilibrium using
this information.
So when we started, we
only had our reactant.
We had 1 bar, 1.000
bar, to be specific,
and we had no product.
As the reaction goes,
we will form product,
so it's plus X. Some amount of
product is going to be formed.
What is the change
in partial pressure
as you go to equilibrium?
What happens to this?
What changes?
What do I put in this area?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: What?
AUDIENCE: [INAUDIBLE]
CATHERINE DRENNAN:
I heard minus--
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: So I heard
minus X. I heard somebody say
[INAUDIBLE].
It's minus 2X because you have
to remember the stoichiometry
of the reaction.
And so we have two of these
going to one of those,
so it's minus 2X.
And then at equilibrium,
we have 1.000 minus 2X is
our concentration
of our reactant,
and our concentration
of product is plus X.
So now we're going
to solve for this,
and we need to write an
expression for the equilibrium
constant.
So why don't you
write that for me.
All right.
10 more seconds.
I'd like 90%.
Oh, so close.
OK.
So again, the trick here, you
have products over reactants,
and you have to remember
the stoichiometry.
So we can go back over here.
So we have partial
pressure of our product
over partial pressure
of our reactant.
And we have the
stoichiometry there,
and now we can
continue to plug in.
So the concentration,
the partial pressure,
of our product is
X at equilibrium.
The partial pressure of
our reactant is 1 minus 2X,
and that whole term
is raised to the 2.
So it's all that
whole term is squared.
And if you solve for X,
you will get 0.38 bar.
So you definitely want to
remember calculators and things
on this exam.
And that value,
then, is our product.
So that's the answer to the
partial pressure of the product
because that's what X is.
Another thing that
I've seen on exams
is that people solve for X.
They're happy to solve for X.
But then they don't
remember what X was.
So keep track of what
things belong to what.
So then we want to also
find the partial pressure
of our reactant.
That was 1 minus 2X.
So here we have 1
minus 2 times X.
And you can see the
significant figure fun
that one can have in this
because we have multiplication,
division, subtraction, and
pretty soon we're going to have
log significant figure rules.
So there's going
to be a lot of fun.
And of course, then this
is our reactant over here.
So let's just go back
to the diagram, which
is up above in your notes, on
the same page in your notes,
and plug these in and think
about what this means.
So we've done the math.
And we see that our reactant
at equilibrium is 0.238 bar,
and our product at
equilibrium is 0.381 bar.
So K is greater than
1, more products,
and you see that that works out.
So if you were asked just to
explain what you expected,
you could say K
is greater than 1.
I expect more products.
But if you do the math, and
you'll often do the math,
you can calculate what
the partial pressures
are at equilibrium of the
reactants and of the products.
All right.
So now let's think about
the relationship, again,
between delta G0
and K. So here is
our expression we saw before.
We can also rewrite
this to solve for K.
So sometimes you will be given
information about delta G0
and asked to calculate a K
at a particular temperature.
But we can also think
about what we would expect.
So if K is large, what is going
to be true about delta G0?
And so why don't you
tell me what you think.
All right.
Let's just do 10 more seconds.
So it would be a
large negative number.
And we can think about
this, that if you really
are lying on the side of
products in your reaction,
then that would be
consistent with a bigger
value and a negative value.
So you can think
about-- again, we
talked about in
terms of formation,
is the thing that's being formed
more or less stable compared
to its elements, the
same sort of idea.
You can think about
the relative stability
and whether you'd expect more
products or more reactants
at equilibrium based
on these values.
So let's do an example and
prove that this is true.
So let's consider
baking soda again.
Baking soda works really well
for this unit on thermodynamics
and chemical equilibrium.
So again, we have
our baking soda
going to CO2, which is very
important, the CO2 gas, which
helps our bread rise.
And we calculated last
week that the delta G0
at room temperature
for this process
was plus 36 kilojoules per mole.
And that would
mean that it's not
spontaneous in the
forward direction, which
is really bad for our bread,
because we need the CO2
gas to cause it to rise.
But the good news was
that if we remember
to turn on the oven
in baking bread
and put it at a normal
temperature in the oven,
that the delta G0 is minus
15 kilojoules per mole.
So then it becomes a
spontaneous reaction.
Now we can think
about this in terms
of our equilibrium constant
K. So if we do the math here,
at room temperature K would be
4.9 times 10 to the minus 7.
That is a small number.
That means that there is very
little product at equilibrium
at this temperature.
Very little CO2 gas to be able
to be used to make our bread
rise.
But now if we calculate
K at this value,
at a negative number for
delta G, K is now 55.
So we have quite
a lot of product
then to be used to
make the bread rise.
So you can think
about things in terms
of delta G's and
whether something's
going to be spontaneous and
give you the product you want.
But the equilibrium
constant gives you
that information as well.
If it's a very small number,
you have very little product.
And if you want
product, if you're
trying to industrially
make something,
that's a very bad thing.
But if you have a big value for
K and this negative delta G0
value, then that's
good if you want a lot
of the product at equilibrium.
So equilibrium
doesn't just matter
for things like baking soda or
forming ammonia from nitrogen
and hydrogen.
Chemical equilibrium
applies to large
molecules as well,
such as enzymes in your body
that are catalyzing reactions.
So I'm going to share with
you an In Their Own Words.
And this is Nozomi
Ando, who I will just
mention is an MIT
undergraduate-- was
a graduate of MIT, was an
undergraduate here in physics,
majored in physics.
And she is now a professor
of-- do you want to guess?
AUDIENCE: Chemistry.
CATHERINE DRENNAN:
Chemistry, that's right.
She's a professor of chemistry
at Princeton University.
So here is a good example
of what can happen.
And she was very
happy-- actually,
I think 5.111 didn't exist then.
I think it was 5.11
when she was here.
But nonetheless, so
here in her words
about chemical equilibrium
and the proteins she studies.
[VIDEO PLAYBACK]
-My name is Nozomi
Ando, and I study
a protein called Ribonucleotide
Reductase, or RNR for short.
It catalyzes the reaction of
converting ribonucleotides,
or the building blocks of RNA,
into deoxyribonucleotides,
or the building blocks of DNA.
It's the only means of
getting those letters for DNA,
so it's important for
DNA synthesis and repair
and replication.
It knows, for
example, when there's
an imbalance in the pools
of the letters for DNA
or if there's a lot
of letters of the RNA.
And this controls the sort
of state that RNR is in.
And RNR can be in an equilibrium
of active and inactive states
that are sort of regulated by
the alphabet soup in the cell.
When it's active,
it's very compact,
but then it has to make a really
dramatic structural change
to go into an inactive state.
I have this imagery
of Transformers
because it's just so dramatic.
So when it's active,
it's compact, like when
a Transformer's a car.
And when it's inactive and it
makes a sound, che, che, che,
che, and then it
expands into a robot.
The letter A, or adenosine,
pushes this equilibrium
from the active
to inactive state.
And it tells RNR, OK, we have
enough of the DNA letters,
so stop.
For humans, it's really
important to study RNR
because it's the protein that
is essential for making letters
of DNA.
So it's essential for
DNA replication, which is
essential for cells to divide.
And we want RNR to function
normally for our health.
But in cells that are
dividing too quickly,
such as tumor cells, we
want to slow it down.
So actually, RNR is a
really important target
for anti-cancer drugs.
But also, because RNR
exists in every organism,
we can start looking
at differences
between different species.
So for example, it could be
anti-bacterial and not just
anti-cancer.
[END PLAYBACK]
CATHERINE DRENNAN:
So that's an example
of how you have a
shift between two
states, an inactive
and active state,
that it's just a
chemical equilibrium.
And binding one thing shifts
the equilibrium one way.
Binding something
else shifts it back.
So chemical equilibrium--
a lot of nature
works by just suddenly
shifting the equilibrium
between different states.
So understanding
chemical equilibrium
is pretty important.
All right.
So now we're going to apply
stress to our chemical
equilibrium, and we're going to
talk about the principle of Le
Chatelier.
So here, a system in equilibrium
that's subject to stress
will react in a way that
tends to minimize that stress.
And whenever I talk to
MIT students about this,
I feel like I need to
really emphasize this point.
MIT students experience
a lot of stress,
but often do not tend
to react in a way
to minimize that stress.
They say, all right, I am just
struggling with this double
major.
I don't know what to do.
It's just so much work.
So maybe I should
triple major instead.
No.
[LAUGHTER]
Ask yourself, what
would Le Chatelier do?
Minimize.
Double major, single major.
So this is a principle that
can apply to your life.
And I recently saw a chemistry
major, and we had lunch.
He was a former MIT
student, and now he's
a CEO of a company, small
company, in Cambridge
that's designing computer
software platforms.
He was a chemistry major, and
he's doing software design
and building like
little computer tablet
things for restaurants.
So I said, would you use
your chemistry at all?
And he goes, oh, I use some
things all the time, especially
Le Chatelier's principle.
I'm all about
minimizing the stress.
So that was one
thing that he really
grabbed onto in chemistry.
So again, if you
think about this,
about minimizing the
stress, you can predict
the direction of the reaction.
And in nature, this
really works pretty well.
So Le Chatelier's
principle gives us
a way to predict the
direction the reaction will
go if you ask, which direction
will minimize the stress?
So let's look at an example.
We're back to N2 and
H2 making ammonia.
So here's a slightly different
plot than I drew over here.
Now, this is a reaction
sensitive to temperature,
so the equilibrium
constant's going
to change with temperature.
So every plot of this may look
a little different depending
on what temperature it's at.
But some things are the same.
If you start with
hydrogen and nitrogen,
you'll have some of
those to begin with.
They will be above 0.
But ammonia will start--
if you had no product,
you'll have ammonia at
0 and it'll rise in.
The other thing that
should look similar
is that as the reaction
runs for long enough it'll
reach equilibrium, and you'll
have the lines level out.
You'll reach an
equilibrium state.
Now what happens if you
stress that equilibrium state?
So say you add hydrogen. And
so this line adds the hydrogen,
then it goes down.
Then you're also going to use
up some nitrogen as you're
using up the
hydrogen. You're going
to shift it to
make more product.
Now, say, you make product.
You add product-- sorry,
you're adding product.
It's going to shift to minimize
the stress, use up the product,
have the product dissociate,
and make more hydrogen and more
nitrogen.
So here's the plot.
Now, let's just think about
what's happening at each step.
So if we're adding
more reactant,
you have more reactant.
You have too much reactant.
Shift to minimize the
stress, and you will shift
the reaction toward product.
Get rid of the reactant,
use up the reactant.
Let's get back to the
equilibrium condition,
minimize the stress.
Now, we can think about this
in terms of Q and K again.
So when you have
reactants added,
then Q is going to fall
below K momentarily.
And so that means-- if you think
about our equation over here,
recall this equation,
this important equation.
So with Q less than K, you
get a negative delta G,
and that's going to be
spontaneous toward the right,
toward products.
So again, you respond
by making more products.
You shift to the right.
And you saw that over here.
It's shifting toward products.
You're using up the hydrogen.
You're using up the nitrogen.
You're trying to return to
an equilibrium condition.
Now what happens if
we add more product?
And so when you're adding more
product, Q is greater than K
momentarily.
You have too many products
now compared to equilibrium.
And when Q is greater than K,
you get a positive delta G.
And that means that
it's spontaneous
in the reverse direction,
or non-spontaneous
in the forward direction.
So you shift toward reactants.
You shift to the left.
So again, you can
use this equation
to think about what direction
is now going to be favorable.
And so here you added product.
The product gets used up.
It's shifting toward
reactants, and you're
making more reactants until
you reach equilibrium again.
And we have one minute left
for a last clicker question.
What happens when
you remove products?
All right.
So let's just take
10 more seconds.
All right.
So we'll just put
it up over here.
So that means that
delta G is going
to be negative in that case
because you remove products.
So Q is less than K,
and so the reaction
is going to be spontaneous
in the forward direction.
Delta G will be
negative, and you'll
move to make more products.
Again, minimize the stress.
You took away products.
You need to make more.
Minimize the stress.
Do problem set 5 and
minimize your stress.
