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PROFESSOR: We're pretty much ready to get started.
Let's settle down and take a look at the clicker question.
So take 10 more seconds.
Very good.
Most people had this right, and the trick here is just to
think about the sign of delta g, whether it's negative and
positive, and think about the equation and the influence of
temperature on that equation.
So whether it's going to make delta s a bigger factor or a
smaller factor and how that will play out.
So you can look at the equation and figure out what
the signs have to be and then the influence of temperature.
All right, so let's move to the first slide that I have
today, which is welcome to the middle of the semester.
So I'm Kathy Drennan and this is lecture 19 of 36, which
means that you are halfway through the course.
And so what you heard on Monday before the exam was
about thermodynamics, and if you don't have all your delta
g's and delta h's and if you don't have entropy full in
mind, that's OK, we are not really leaving thermodynamics,
we're going into chemical equilibrium, which is all
about thermodynamics.
So you're going to hear a lot more about delta g, delta h,
temperature, and about our friend, entropy as we go along
in the second half.
So it's the middle of the semester.
That means that you've had two of the sort of four hour exams
-- remember the fourth hour exam is actually combined with
the final exam.
So the final exam is 200 points of cumulative material,
and 100 points of new material, so you've done two
now of those four sort of hour exams.
Just to remind you of some of the topics you have seen and
the topics coming ahead, this is on the syllabus, and we're
actually where we're supposed to be on the syllabus, so if
you want to take a look at that ever to see where we're
going, we're actually on track.
So the first half of the course are a lot of basic
principles, and we're moving today into chemical
equilibrium, so a lot more delta g's,
delta h's coming on.
And then we're going to move into acid base, which is acid
base equilibrium.
So we're not going to be leaving equilibrium or
thermodynamics.
Then we're going to go into oxidation reduction, which is
also about equilibrium, and then transition metals, and
end with kenetics.
And so, these altogether represent the fundamentals
needed for the study of biochemistry, organic
chemistry, any kind of chemistry, biology, the life
sciences, many things -- so you're getting all the
fundamental principles of chemistry in this class.
So along those lines, I just thought I would share
something that happened to me on Wednesday when I was riding
the T, the Silver Line, in particular.
So I don't know if any of you have had this kind of
experience yet where you're on public transportation and
someone next to you says, "So, are you a student?" Maybe you
have a notebook out or a book out or something.
And you say, "Yes," and they said, "Oh, what are you
studying?" And then you say "Well, you know, chemistry,
math, physics." And they're like "Oh." And then they go
busy themselves doing something else.
Or they tell you "I didn't really like those subjects in
school." And then they stop talking to you.
So how many people have had that experience so far, that
you scare the person next to you by what you study?
OK, a few people.
If I ask that question four years from now, I think it'll
be like half the class, eight years from
now it'll be everybody.
So this is kind of common.
But what happened to me on the Silver Line was actually
pretty exciting, because I've had that other
experience many times.
So, I was wearing my keys around my neck, because
women's professional clothes, no pockets, with a little MIT
cord, and the person next to be on the Silver Line said
"So, do you go to the Georgia Tech of the east?" And I said
"Well, I am a professor at MIT." And he said "Well, I'm a
mechanical engineer and I went to Georgia Tech." And so he
said, "So, what do you teach?" I said, "Chemistry." And he
said, "Chemistry --
I really wish I had paid more attention in chemistry." I
said, "Well, what do you do now?" And he said "Well, I
work for the army, and I'm part of a team that has
mechanical engineers, electrical engineers, and
chemists, and we're trying to figure out ways to detect
explosives." So I said, "Oh, well have you heard of the
head of the Chemistry Department at MIT, Tim Swager
who works in the area, and he said, "Yes," he knew the name.
And so he said that one of his porblems on the team, and in
talking to the chemist, his chemistry language is not
really good enough, and so he was really struggling, and
that the best results of this team effort would be if
everyone could really talk to each other.
So he wished that he had paid more attention in his
chemistry class.
And so I, of course, my connection with chemistry was
by wanting to understand biology, but for everyone it
might be a little bit different.
So, as I mentioned in the beginning of the class, one
challenge that you have this semester is figure out what
your connection to chemistry is.
What are you going to use chemistry for?
How does this fit into what you want to do?
And maybe for some of you, you're not going
to know that yet.
Hopefully it won't take till you have a job doing something
to realize that you're lacking something in your education.
And so, probably if you stay in science and engineering,
need chemistry, good to learn it all now.
And from exam 2, we see that you are learning it now and
that's really great.
So, some day maybe someone in this room will be involved in
national security, and if you're not a chemist, you'll
be able to talk to the chemists and make really good
progress toward that work.
So, chemistry -- it's important in medicine,
national security, the economy, energy initiatives, a
lot of the big things going on now.
It's a fundamental, and you will get in this course, the
fundamentals.
So if you have any good tea or bus or airplane conversations,
let me know --
I'll catalog those for future reference.
It's always a touch of how what we do here connects with
the real world.
All right, so we're not going far from thermodynamics, we're
going into chemical equilibrium.
And we're going to be talking a lot about delta g.
So if delta g is not your friend yet, don't worry, you
can still bond with delta g.
All right, so chemical reactions, chemical reactions
can go into a state of equilibrium.
And it's a dynamic equilibrium, the reaction is
still happening, but if a reaction is an equilibrium,
the rate of the forward reaction will equal the rate
of the reverse reaction, so there'll be no net change in
composition.
So let's take a look at an example.
So let's look at a reaction in which we have nitrogen gas,
and we have hydrogen gas, and they are
reacting to form ammonia.
Suppose we just start with nitrogen gas and hydrogen gas
and we don't have any ammonia left.
So if we consider concentrations versus time.
So say we start with, we have some nitrogen gas, some
concentration of nitrogen gas.
As it reacts with hydrogen, the concentration will
decrease and then level off.
We'll start with some amount of hydrogen gas, and its
concentration will also decrease and level off.
And in the beginning we won't have any of the product, any
of the ammonia, so its concentration will increase
and then level off.
So when these reactions level off, you're reaching
equilibrium.
The reaction is still happening, but the rate of the
forward reaction is equal to the rate of the reverse
reaction, so there's no net change.
So the concentrations are staying the same, but there's
still -- the reaction is still going forward.
So let's think about the case when we have pure reactants
when we haven't formed enough products yet to reach
equilibrium.
So if we have pure reactants, the reaction is going to be
spontaneous in the forward direction.
So we'll have a reaction that is spontaneous
in the forward direction.
And what will that mean about our friend delta g?
Is it going to be greater or less than zero?
So it'll be less than zero -- so delta g or the forward
reaction will be less than zero.
So when delta g is negative, the reaction is spontaneous in
the forward direction.
So what about when you have pure products, then the
reaction should be spontaneous in the reverse direction.
So you have spontaneous in the reverse direction, and what
does that mean about the sign of delta g?
Greater or less than zero?
Greater, so it'll be positive.
So, we can think about that in terms of a plot.
We can think about free energy versus the
progress of the reaction.
So, the progress of the reaction is going this
direction as you go along.
So in the beginning if you have pure reactants, delta g
is going to be less than zero.
And you're going to proceed along in the forward direction
spontaneously.
If, on the other hand you have pure products, delta g will be
positive, so you'll be spontaneous in the reverse
direction, and the reaction will go in the reverse
direction until what happens?
Till you reach equilibrium.
And what would delta g be at equilibrium?
Zero.
So you see, there is a great relationship between delta g
and equilibrium, so we have not left delta g behind.
So, delta g is going to change as the components of the
reaction change.
As you have more products or more reactants, you're going
to have a different delta g.
So let's look at some equations.
So delta g is the change in free energy, the difference in
free energy, at some point in the reaction at any time with
amount of composition.
We also have delta g nought, which talks about delta g
under particular conditions, so it's sort of the standard
free energy.
We have a term called q, which is the reaction quotient,
which tells you about products and reactants.
We have our friend r, which is the gas constant, and this
depends on temperature.
So temperature is a term involved here.
So the delta g at any point in the reaction is going to
depend on the delta g nought for that reaction, and the
reaction quotient, products and reactants, and we'll
define this in a minute, and then the temperature as well.
So we need to know more about what q is, what this reaction
quotient is.
So this slide looks a little bit scary, but it's going to
be fine, because a lot of the terms are going to cancel out.
So we're going to talk about q, this reaction quotient, and
we're going to have different types of problems -- some
where we're talking about gases, and others we're
talking about solutions.
So you can see two different kinds of q's, one that depends
on partial pressure of the gas, and another that depends
on concentration.
So here is our equation again that we just saw -- delta g
equals delta g nought plus r t natural log of q, and here
we've expanded the term for q.
So p to the sub x is the partial pressure of a
particular gas, and so in this equation, we have a plus b
going to c plus d, the top of the line or the products.
So we have the partial pressure of gas c, and this is
over a reference, partial pressure, raised to the power
c, the coefficient, and then also we have d, the partial
pressure of gas d, over a reference raised to the
small letter d.
On the bottom of the terms for the reactants, partial
pressure of gas a over a reference raised to the
coefficient a, partial pressure of gas b over the
reference raised to the coefficient b.
Now what's great is that the partial pressure is one bar,
and so that basically cancels out as far as we're concerned.
And so the term for q is much simplified.
You won't see problems that have the reference in them, so
you can just think about q in terms of
products over reactants.
And you just have to remember that the coefficients in the
reactions do matter.
If you're talking about solutions, the only difference
is that we'll be talking about molar, so we have one molar,
so the reference term here also cancels out.
When you see something in brackets, like c in brackets,
here that telling you it's a concentration term.
So here, q is the concentration of c raised to
its coefficiency.
The concentration of d raised to its
coefficient d over reactants.
Concentration of a to a, concentration
of b raised to b.
So the thing in x indicates it's a concentration term.
So q is just products over reactants, considering the
stoichiometry of the particular reaction.
So what about the equilibrium constant k?
So at equilibrium you told me that delta g equals zero, and
at equilibrium q, the reaction quotient equals k, the
equilibrium constant.
So we can consider that in terms of this expression, if
we're talking about this expression at equilibrium,
delta g is going to equal zero.
And so, we just set this whole term equal to zero.
We can rearrange the equation bringing delta g nought to the
other side.
And so we have delta g at nought equals minus r t
natural log of K, because at equilibrium, k is equal to q.
So now we have another term to solve for delta g, and an
equation that relates delta g nought to be equilibrium
constant k.
So k, the equilibrium's constant has the same form as
q, but we're only talking about the concentrations or
the partial pressures of things at equilibrium.
So it's the same, it's product over reactants, same
expressions as q, but in the corner you say at equilibrium.
So you're only talking about the concentrations of things
at equilibrium if you're solving for k.
If you're solving for q, it's the concentration or the
partial pressure at any time in that particular reaction.
Important thing, products over reactants.
All right, so, we can rewrite the equation one more way.
So I just told you that delta g nought is equal to minus r t
natural log of K. So we can substitute into the expression
minus r t natural log of k, and now we can
rearrange this equation.
And so, if we rearrange it, we have delta g, the delta g at
any particular point in the reaction equals r t natural
log of q over k.
And this equation is helpful for people if they're thinking
about what is the equilibrium constant, what concentrations
do I have now -- we're not at equilibrium, what
concentrations do I have now, what is q now in the reaction,
how does that compare to k?
And when you know what those values are, you'll know
something about the direction of the reaction because you'll
know if delta g is positive or negative.
It'll to be spontaneous in the forward direction or in the
reverse direction.
So this is a handy equation for thinking about the
relationship q k delta g.
So let's think about that relationship for a minute.
If q is less than k, what is the sign of delta g?
So it would be negative, which means the forward direction of
the reaction will occur.
So if you think about it, you can think about in terms of
products and reactants.
So at equilibrium then, if q is less than k at equilibrium,
there are more products than there are right now, so you
need to make more products.
So you would have delta g would be negative, you'll see
that mathematically, and you can think about it in terms of
whether you're going to make more
products or less products.
If q is greater than k, what is delta g?
So it would be positive and the reverse
direction would occur.
So you think that at q, if it's larger than k, it has
more products in its terms, and at equilibrium there are
less products, so you need to go in a direction that will
get rid of some of those products so you'll reach
equilibrium again.
So this equation is very helpful in thinking about the
direction of the reaction -- which direction will it be
spontaneous.
So let's look at an example.
K is given in this example, and then we have a bunch of
partial pressures.
And we're asked which direction the
reaction will go.
So what do I need to do to answer this question, what do
I need to calculate?
So if I'm given k, and a bunch of partial pressures, what do
I first need to calculate?
Q, right.
So let's calculate q.
So q, we're going to talk about products over reactants,
so we're going to talk about be partial pressure of the
ammonia, and there are two of them being formed.
Our reactants, we want to talk about the partial pressure of
the nitrogen, and the partial pressure of the hydrogen gas,
and again include the stoichiometry, so we have
three there.
So now I can plug in my values, so I'm told I have 1 .
1 bar, and down here I have 5 .
5 and 2 .
2 to the 3, again, considering the stoichiometry.
And if you do the math, you get 2 .
1 times 10 to the minus 2.
That's your q value.
Now given your k value, which the problem states, and this q
value, let's do a clicker question, tell me which
direction the reaction will go.
All right, 10 seconds.
So, 77%, pretty good.
So, q is greater than k here.
And if q is greater than k, what will be
true about delta g?
I have another clicker question for that -- yell out
the answer.
What is it?
Positive, right.
So you're going to shift towards reactants, so you're
going to go in the reverse direction.
So you think about whether there are more products or
less products at equilibrium, and so there are more
products, so we have to think about which direction it'll
go, and here, ammonium will dissociate until equilibrium
is reached again.
OK.
So let's think more about what k is going to tell us.
So k tells us about the mixture of products and
reactants at equilibrium, whether we can expect low or
high concentration of reactants at equilibrium.
So let's look at another example.
So when you have k that's greater than one, so more
products than reactants at equilibrium, you can think
about this in terms of higher products at equilibrium.
When you have k less than one, we're going
to have lower products.
So again, think about k in terms of products over
reactants at equilibrium.
So if k is greater than one, there are more
products than reactants.
If it's less than one, there will be less
reactants than products.
So let's look at an example of that.
Let's look at when k is greater than one.
And I have the equation up there and I'll
write it here as well.
So we have 2 n o 2, and two double arrows, and n 2 o 4.
So we have a k value here of 6 .
84, so that's greater than one value.
So let's think about this reaction.
So over here instead of concentration, we're going to
talk about partial pressure because we're talking about
gas, and we have time.
So initially we have a reactant.
And the reactant starts at some concentration and
decreases and then reaches a straight line, so reaches
equilibrium.
So we have our reactant here.
Originally we have no product, and so product is going to go
up and be formed and then it's going to level off as you
reach equilibrium.
So initially, what is true about q and k?
So with no products, what is true about q and k?
Q is less than k.
And so what's true about delta g?
Less than zero, it'll be negative, so it'll be
spontaneous in the forward direction.
So you're going to be spontaneous in the forward
direction and you're going to make your product.
So now let's calculate what the concentrations are going
to be at equilibrium.
So initially, so you have your initial pressure for the
reaction, 2 n o 2 going to n 2 o 2, and our initial
concentrations are given as one bar,
and we have no product.
Now we talk about the change as we go toward equilibrium,
how much does the reactant change?
What do I write here?
What change?
Minus x minus something x?
Minus 2 x.
So again, we're considering the stoichiometry,
and what's over here?
So just plus x, and then at equilibrium we now have 1
minus 2 x and x.
So we're talking about equilibrium concentrations, so
we're talking about k, so k equals 6 .
84, which is going to be equal to the partial pressure of the
product over the partial pressure of
the reactant squared.
So it's going to be equal to x over 1 minus 2 x squared.
So x, if you calculate it out, should equal point 381 bar.
And then if we do 1 minus 2 times 0.381 bar, we get 0 .
238 bar.
So if we go back over here, our products at equilibrium --
oh, I guess I should write what -- so x is our product
and this is our reactant.
So the product we have 0 .
381, and our reactant at equilibrium is
going to be 0.238 bar.
So we have more product than reactant at equilibrium, which
is consistent with the value of k being greater than one.
So, when you know something about the equilibrium
constant, you know something about the reaction and whether
you would expect more products or reactants at equilibrium.
So again, you can think about this in terms of q and k.
All right, so now we are going to go toward our
next clicker question.
So if we can rewrite the expression for delta g equals
minus r t natural log of k, and express it in terms of k,
and now we can think about then what is that
relationship.
If you have a large value for k, what do you expect to be
true about delta g nought?
All right, let's give that 10 seconds.
OK, why don't you discuss for a minute with your friends
whether you agree with that 78% or not.
All right, now we're going to re-poll, so click in again.
Now give the right answer.
Interesting.
It actually usually goes the other direction, that after
there's a discussion, more people come to the same
conclusion.
So I guess it's a matter of if the professor asks that, you
assume that that must have been the wrong answer.
Was that people's logic?
So what are the points of doing this, and we're going to
actually do this kind of thing a few more times in class, is
that collectively, and I was actually just at an education
meeting about science down at the Howard Hughes Medical
Institute, that statistics show that if you have a group
where everyone in the group has the wrong answer and
they're allowed to discuss it, there's a good chance they'll
come up with the right answer.
So that it's not just about one person in the group having
the right answer, convincing everyone else that they're
right, that the act of discussing often
leads to new answers.
So now that you know that it's not a trick on my part to tell
you you got it wrong, we'll see next time whether this, in
fact, holds, that the act of discussing helps give the
right answer.
Anyway, so if k is large, it is true that delta g nought
would tend to be negative and more -- it would be negative
and more on the large side.
So one can think about if there's more products over
reactants, that's going to indicate something about the
delta g nought for the reaction, and the k -- the k,
if it's greater than 1, if it's a big number, then there
are more products than reactants, and delta g nought
would be negative and would tend to be a large number.
All right, so this is more actually kind of simple
bookkeeping involved that if you know steps in the
reactions and you know equilibrium constants, you can
calculate an overall equilibrium
constant for that reaction.
So you can write a reaction as the sum of different
components.
And so up here, we are going to try to add these first two
equations to get this net equation.
So we have 2 gas p 3 c l 2 going to 2 p l c 3, and that's
equilibrium 1.
And so then in the next reaction, some of that is
being consumed reacting with another c l 2, giving you p c
l 5, and the net reaction we're interested in has 2 p 5
c l 2's going to 2 p c l 5.
All right, what do I have to do before I can add these
together effectively and have things cancel out?
I need to multiply what?
Second equation by 2, yeah.
So to get that to work I would need to have 2's there, then
this is going to cancel out and these will add up, so we
have 5 and then we have two of the main products.
So if I do that, and I have the equilibrium constant for
one and for two, how am I going to get equilibrium
constant for three?
I'm going to multiply k 1 by k 2 and k 2.
So I'm going to have to multiply k 2 in there twice,
because there's two of those -- we'd multiply that up.
So if you have different parts of reactions and you can sum
them together, then you can multiply out the individual
k's to get the new value of k.
So that's something that's just useful that you'll run
into in doing these types of problems.
All right.
Now we're going to think about how
equilibriums respond to stress.
And I always feel like I need to pause.
MIT students, you guys are some of the smartest, most
talented scientists in the world.
I don't know if you fully appreciate how
smart you all are.
But this concept is tough for MIT students.
So Le Chatelier's principle says that a system in
equilibrium that is subjected to with stress tends to react
in such a way to minimize that stress.
I have advisees coming to my office saying, "I'm
double-majoring in this and that and I'm taking five
classes and I have UROP and I have this lab exercise --
I don't know what's going on, so I've been thinking a lot
about it and I think I should add a third major." Le
Chatelier would be very unhappy with that.
Le Chatelier would say that the appropriate response is to
drop one of the majors, to minimize the stress.
So in doing these problems, what I want to encourage you
to do is think the opposite of what you would do.
Minimize the stress.
If you think about that you'll be all set.
All right.
So Le Chatelier's principle actually is very useful.
If you think about minimizing the stress, you'll be able to
predict the direction that the reaction will shift.
So the reaction's going to shift in a way to minimize the
stress, so you can predict it.
If you're thinking along these lines, you say, oh, that
system was stressed, and then you can think about how that
reaction or that system is going to respond.
So let's give some examples.
So, we have a system in equilibrium -- we started out,
we had our nitrogen and our hydrogen
and we had no product.
We reacted the hydrogen and the nitrogen, their forming
products, and eventually they reach equilibrium, their delta
g equals zero there's still the reaction going on, but
there's no net change.
All right, now the system is going to be stressed.
So we're going to add more of a reactant.
How will the system react to minimize that stress?
What is it going to do?
It now has too much of one of the reactants.
You'll form more products, and that's going to use up some of
your other reactant, which will go down, and you're going
to form more product until equilibrium is reached again.
And one thing that I'll mention that the ratio has to
be the same.
The equilibrium constant is a constant given the same
temperature, but you're not necessarily always going to
have the same concentrations, but you should have the same
ratios, the same value of k.
All right, so what about if you add more product, what's
going to happen?
What direction will the reaction shift?
Right, it's going to shift toward the reactants, so
you're going to make more of each of the two reactants
until again, you find your equilibrium again.
And then the reaction's still going but there's no net
change anymore.
All right, so let's think about this -- let's think
about this in terms of the math as well.
If you want to stick with the math, that's OK, you can think
about calculating delta g's here.
So if you're a system in equilibrium and you add more
hydrogen, the system will respond to
minimize that increase.
So it's going to make more product that will minimize the
increase, it shifts to the right.
And this can be explained in terms of q and k.
So, you can think about, again, do you have more
products now or did you have more products at equilibrium,
and if you're adding more reactants, momentarily q will
become less than k.
And so, you would get a negative delta g, which would
make it spontaneous in the forward direction.
So some people like to think about this in terms of delta g
equals r t natural log of q over k.
So you think about if q is less than k, what sign you
have for delta g, and that's tells you whether the reaction
is spontaneous in the forward or the reverse direction.
And let me just give you one hint for taking
exams in this unit.
That use of the arrow is really good, or saying "toward
product" or "reactant." I can't tell you how many people
write -- know what the answer is and write left when mean
right, or write right when they mean left.
But if you draw an arrow you never really get it wrong, and
when you say products or reactants it's a lot harder to
make that mistake.
So if you're not good with right or left, which let me
tell you a large fraction of people are not, hedge your
bets, you can write everything, more products,
arrows, and then write if you want, then you're pretty sure
that you have it all in there.
OK.
So again, we can explain this in terms of q or k.
All right.
So let's just quickly talk about adding more products.
We did this already.
If more products are added. then q is going to be greater
than k momentarily, and you would shift toward reactants,
shift toward the left, and again, we saw that down there.
So tell me a final clicker question and then we're done.
What happens if you remove products and why?
OK, 10 seconds.
See if we can get in the 90's.
No, no 90's today, we'll have to do it next time,
but we got 73 right.
So we're going to make more products as well.