The following content is provided under a Creative
Commons license.
Your support will help MIT OpenCourseWare continue to
offer high-quality educational resources for free.
To make a donation or view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: So, let's finish up new material today, new
material Monday, and then we're done with the kinetics unit.
So, we're going to talk about temperature, collision theory,
and activated complex theory today.
So on the first day, we talked about kinetics, we talked about
factors affecting the rates, and this was in your handout on
that first lecture, but just to kind of review for a minute
what we have talked about so far and where we're going.
So these are some of the things we came up with are factors
affecting the rates of reaction.
People said mechanism affects the rate of reaction, so
we talked about mechanism.
Concentration of material, nature of material -- that
pretty much fits in in every lecture.
We've been talking some about that, say, if it's first order
or second order, if you have a concentration term in there, or
if you don't have a concentration term.
The nature of the material, we're going to talk
more about that today.
Temperature is going to be the number one topic for today, and
on Monday the number one topic is the use of catalysts
to speed up reactions.
So, by Monday we'll be done with this list.
So, what about temperature.
So temperature, when I said what affects the rate of
reaction, temperature was one of the things that you yelled
out at me, and that's very much true, it has one of the biggest
effects on the rate of reactions.
So when we were talking about the gas phase, there was an
observation made that the reaction rates tend to increase
as the temperature increases.
And most of you can think about this and are aware of it, and
today we're going to talk more about the quantitative affect
of this, how much does the rate of a reaction increase when the
temperature increases, how do you know what
equation do you use.
So, this idea has been around for awhile.
So in 1889, Arrhenius plotted rate constants versus
temperature, and he found that if he plotted the natural log
of those rate constants versus inverse temperature, then he
would get a straight line.
So let's look at the plot that he had.
So, he found that if he plotted natural log of the rate
constants versus 1 over temperature, so units or kelvin
to the minus 1, that he got a straight line.
Which means that the rate constants are varying
exponentially with inverse temperature.
So you have to use a natural log to get a straight line,
if you don't use natural log, you wouldn't get
a straight line here.
So, let's look at some of these terms that we have here, and
here's our equation for the straight line, the natural log
of k equals minus e a -- e a is activation energy, which we're
going to talk a lot about today -- over r t, r is our friend
the gas constant, and t is temperature, plus
natural log of a.
So what is a?
A is called factor a or sometimes it's called
pre-exponential factor, and had has the same units as k, k
being the rate constant.
So, let's think about this equation and this plot, and
about factor A and this other term, activation energy.
So, factor A and activation energy depend on the
reaction being studied.
So, it depends on the nature of the materials involved.
So I said we'd talk again about nature of the material.
So one kind of reaction's going to have one activation energy,
another one will have a different one.
So, let's think about this term, factor A.
What is it exactly?
Do you think it would be temperature dependent?
What do you think?
How many people think yes, it would be temperature dependent?
How many people think no?
Some people are not going to commit.
The answer is no.
So, let's think about what factor A is.
So, factor A is the rate constant at some really,
really large temperature.
So, if we look at this plot, the natural log of k equals the
natural log of A, or k equals A if you're along this axis here.
If you're along that axis here, that's at zero, so that's when
1 over temperature equals zero, then the rate constant
equals factor A.
What would be true about the temperature for 1 over
temperature to equal zero?
Very big.
So, at the biggest, hugest temperature you can imagine,
the rate constant's going to equal A.
And so, factor A depends on the nature of the material being
studied and you would have to determine what that value is.
But it's definitely not temperature dependent.
It's the value of the rate constant at an
enormous temperature.
So, what about activation energy?
Do you think that's temperature dependent?
Why don't you tell me what you think.
All right, let's take 10 more seconds.
All right, so I'd like you to discuss this amongst yourselves
and see whether you're happy with this answer.
All right, go ahead and vote again.
All right, 10 more seconds.
And the correct answer is?
So, I think most of you will remember this answer.
So the activation energy is not temperature dependent.
You can calculate what the activation energy is by
plotting the rate constants versus 1 over temperature,
and then you get it from the slope of the line.
So, activation energy depends on the nature of the
material, but it isn't temperature dependent.
So, we're going to talk a lot more about that today.
Before we leave this, I just want to say one thing
about a clicker question.
So, at the end of last time, we had our first repeat winner
for a clicker competition.
So, we thought that we might have an opportunity for
recitations to see if they can -- if some other recitation
can get a second win.
So on Monday we're going to have our final clicker
competition, and if somebody else does tie, one section,
that we'll have a final one-time clicker question
that'll be a run-off.
And whatever recitation wins, we have special prizes for the
members of that recitation.
So that's on Monday.
So you may want to review catalysis on Monday, if you
feel that you're in the running for the grand champion
clicker recitation.
All right, so let's go back to this.
So we found out that no, activation energy is not
temperature dependent.
All right, so what are you going to see on your equation
sheet on the final, you may see a couple of different forms
that are all equivalent to each other.
So here is the plot of a straight line.
It will often be written with just natural log of a and e
a over r t terms reversed.
So that's as a straight line, and this is often what you see
for the Araneus equation.
You can also get rid of the natural logs and have the term
here exponential, so you have k equals factor A times e
to the minus, activation energy divided by r t.
So, those are the equations, they're all equivalent to each
other that you might see on your equation sheet.
So, it's also true that non-gases can exhibit this kind
of behavior, and let me just give you one example of a
non-gas that exhibits this kind of behavior and you
tell me what this is.
What is that?
Crickets.
So crickets exhibit Arrhenius behavior.
They will chirp faster as the temperature gets hotter.
So if you're out camping, sometimes it can be in the
summer, it can be actually be quite deafening, and you were
very happy to go back to the city where you only have
ambulances and cars going, and you don't have this
incredible racket at night.
You can actually calculate what the temperature is by counting
the number of chirps of the crickets and using
a little equation.
I think it's you count for 14 seconds and add 40 and that's
the temperature in fahrenheit or something like that.
So, not only gases do this kind of behavior.
All right, so activation energy, we're going to be
talking a lot about this today.
So, what is it exactly.
So let's think about two molecules coming together.
So when two molecules come together, you have this
bimolecular process going on, but every time two molecules
come together, they're not going to go on and
form a product.
You're only going to form a product when those
molecules have a critical amount of energy.
When they have the energy which allows them to react,
that activation energy.
If they have enough energy when they come together, they will
go on and form products.
So, that's what activation energy is, it's this critical
amount of energy that they need to react with each other.
All right, so let's just think about what affects that
critical amount of energy, and of course, temperature is
going to be involved.
So let's think about that.
So if we have fraction of molecules on one side, and we
have kinetic energy down here, let's think about
how temperature is involved in this.
So, at a low temperature, the fraction of molecules that are
going to have enough energy to react is going to be less.
And let's think about at a higher temperature, go like
this, so this is high temperature, and we have
low temperature up here.
And then over here we would have our activation energy, the
energy needed for a reaction.
And you see, if you're at low temperature, only a small
number of molecules are going to have enough energy to react,
but if you're at higher temperature, a large number of
molecules are going to have that critical energy to react.
So, at low temperature, not many can react, at higher
temperature, many more will have that energy -- will be
able to overcome that activation energy, will have
it and they can react.
So, temperature plays a big role here.
So we can use this idea of activation energy to
predict a rate constant.
Let's look at an example.
So, often people have sucrose in their diet, and the
hydrolysis of sucrose will form glucose and fructose as part of
this digestive process.
And we can think about the rates at which that would
happen in the body.
So normally, we're at 37 degrees, that's normal body
temperature, and we have the observed rate constant
for that is 1 .
0 times 10 to the minus 3 per molar per second.
But what happens if your body temperature is lowered, what
happens if it was at 35 degrees, and to be able to
answer that question, you need to know what the activation
energy is for this process, and here it's 108
kilojoules per mole.
So, we want to ask the question, what is the new
rate constant at 35 degrees?
So, we can take our Arrhenius equations, and we can put a
1 by the rate constant for temperature 1, and a 2 for the
rate constant at temperature 2.
Now we can combine these equations.
Natural log of a cancels out, it's not temperature dependent
so it doesn't stay in here.
And so we can solve, we can subtract these two, or that's
equivalent to dividing them, and so natural log of rate
constant 2 over rate constant 1 equals minus the activation
energy over the gas constant times this temperature term.
Does this look at all familiar?
Does this look like some other equation you
saw once upon a time?
Do you remember what that equation was called?
Van't Hoff equation, right.
And there we were comparing what instead of rate constants?
Equilibrium constants, and instead of e a, what
term did we have here?
Delta h, right.
And it's good you remember that because we're going to
come back to that at the end of today's class.
So, we can use this equation and plug in the values.
We put in our activation energy, and remember to pay
attention to units, because if you're going to cancel your
joules with a gas constant, you want to make sure you've
converted your kilojoules to joules, and then we can plug
everything in and solve for k 2.
So, here k 2 is 7 .
6 times 10 to the minus 4 per molar per second,
so it's slower.
And this is one reason why it's really nice to have your body
temperature stay the normal temperature.
If you get too cold, your body processes, your enzymes are not
functioning, everything is slowed down.
And if you get too hot, that's not good either.
So you really want to maintain it, and so for some of you who
come to MIT from warmer climates, let me introduce
you to the L.L.
Bean catalog, they sell coats, they sell boots,
and all sorts of things.
So, this winter, when you come back for IAP, or for the term,
you're well-prepared, and you don't have to prove that it's
not good if your body temperature goes
below 37 degrees.
All right, let's think about what else this equation tells
us, and the other thing it tells us is that if you have a
very large activation energy, if this e a is a very, very big
number, that means that your rate constants will be very
sensitive to temperature.
So if this is really, really big, there's going to be a big
change in your rate constants as temperature changes.
And keep that in mind, we're going to come
back to that later.
So, what do you think happens to the rate of an enzymatic
reaction at liquid nitrogen temperatures?
We looked at going from 37 degrees to 35, liquid
nitrogen is pretty cold.
So, not a whole lot happens at liquid nitrogen temperatures as
far as enzymes go, and I'll just mention, so it slows way
down, that this is a trick that I use in my research.
So we have crystals of enzymes, and we can try to get
structures in particular states by taking the crystals that
have enzyme in it, and starting a reaction, and then dunking
the crystals in liquid nitrogen to kind of stop it at a
particular stage, and then you look at what the
structure looks like.
So that's one use.
So, we're going to also take a look at other reactions, we
don't have any enzymes here, but some other things, and look
at what happens when we get things to be very cold.
[EXPERIMENTING]
So, Dr. Taylor is pouring out some liquid nitrogen.
PROFESSOR: All right, so what we're going to look at is a
reaction that we can see pretty easily here.
Have any of you used glow sticks before, maybe trick or
treating or some other point.
So basically, you may or may not know how they work.
There's two compartments in glow sticks that have two
different chemicals in them, and they're trade secrets so we
can't put them on the board.
But basically, what we have here is a reaction.
A lot of reactions we know give off heat, or they
give infrared light.
Here we have a reaction that gives off energy
as visible light.
So would you call this an endothermic or an
exothermic reaction here?
Yeah, so this is an exothermic reaction.
So just as Professor Drennen was talking about with slowing
down enzymatic reactions we can think about if we can
slow down this reaction.
So, we're just going to put it in the liquid nitrogen -- so,
keep an eye on this, we'll do several controls of
different colors here.
So, see if an orange glow stick works the same.
What we're looking to see is if we can slow down this reaction.
What would we see if we slowed it down or even
if we stopped it?
Yeah, we're not going to see anymore color.
So are you starting to see color loss in this
first reaction here?
So, we'll try green, and yellow.
So, it looks like this first one might have stopped
already, you see there's no color anymore here.
What would you expect if this heated back up
to room temperature?
Yeah, so hopefully, if you just keep your eye on this, we'll
continue on with the lecturing, because I don't know how long
it will take for it to warm back up the room temperature.
But keep an eye and we'll see if we get the temperature back
high enough to see the glow again.
And since we do have a liquid nitrogen here, it's too hard
to resist freezing a flower.
This has nothing to do with kinetics -- we can't even
try that connection.
But we will be freeze a flower.
OK, she will try making -- good catharsis pre-exam.
PROFESSOR: So, has anyone had liquid nitrogen
applied to them?
It's used in doctor's offices, if you want a little something
removed perhaps, put a little liquid nitrogen on
and burn it off.
Good premed training for you.
I think we're pretty frozen here.
So it looks the same, but as you can see -- [INAUDIBLE]
-- So I think that's all we can do [LAUGHTER]
[APPLAUSE]
PROFESSOR: So I actually heard something interesting on NPR
about liquid nitrogen removing warts and things like that, and
they were talking about how there was something that was
actually better than liquid nitrogen for doing that.
Did anyone hear the story about what the thing was that was
better than liquid nitrogen?
It was duct tape.
And so some scientists had looked at what all the
uses of duct tape are.
And since most of you are planning on being scientists
or engineers, duct tape will probably be an important part
of your life in the future.
And so, duct tape worked better than liquid nitrogen
for removing warts.
And they found that duct tape worked really well
for a lot of things.
There was only one thing they tried that it did
not work well for.
Anyone want to guess what that was?
Repairing ducts, yes.
Not really good -- there were many, many, many better ways to
repair ducts than with duct tape, but I guess they decided
that calling it wart removal tape was just not quite as
catchy as duct tape, so it's still referred to as duct tape.
OK, so when you lower the temperature, things
tend to slow down.
But if molecules are going to react, they need to have enough
energy, they need to have a high enough temperature
to overcome this activation energy.
So that critical amount.
So again, when the molecules come together, and let's just
look at these for a minute, when they're coming together to
react, and if they're going to react, there needs to be some
energy associated with this, because you're probably going
to have to break a bond, and that's going to take something,
and then you may have to form a new bond, for example.
So there needs to be a critical amount of energy, you need to
be able to overcome that activation energy barrier
for the molecules to react.
So it always takes some energy for things to react and so you
need to have enough energy.
And so, if those molecules have that energy, they'll come
together, react, and you'll form products.
If they don't have that energy, they're just going to go
back to what they were, unchanged reactants.
So you need to have sufficient energy to overcome that
activation energy barrier.
So, this is just a little movie that shows two molecules coming
together, and if they have enough energy to react, you
will see a spark, and then the molecules will react.
So, here we go.
Molecule in red, in green, they're checking
each other out.
Do they have enough -- they had enough energy,
and they reacted, and went on to product.
All right, so let's talk about this activation energy barrier
and these activated complexes.
OK, so in this example, you have n o 2 plus c o, and
they can come together and form n o plus c o 2.
And that's going to take some amount of energy to react.
And so, I'm drawing something that's called an activated
energy diagram, and we have potential energy on one axis,
and on the other we have what's called a reaction coordinate.
And so, the reactants are going to have a certain amount of
energy, so our reactants are going to have some amount of
energy, and our products will have some amount of energy.
But even though in this case the change, the products are
lower in energy, and you have a delta e for the difference
between the reactants and the products, they can't go
directly to products.
They have to overcome an activation energy barrier.
So they have to overcome some kind of barrier
before they can react.
So, only ones that can overcome that barrier, that have enough
energy to overcome this activation energy barrier, so
the activation energy for the forward reaction, only those
will be able to react.
There's also an activation energy barrier for the reverse
reaction on this side.
So, if you go from products to reactants, you have to overcome
that activation energy barrier.
And up here, this is called the activated complex, so you have
some kind of activated complex or transition state, so the
molecules will come together, they'll form some kind of
transition state up here, and then go down into products.
OK, so most of you are sort of familiar with the concept, I
think, of this activation energy, we've been talking
about it, but the idea of an activation energy barrier, I
think is something that probably all of you can
personally connect with.
So, for me, one of the things that I find really hard to do
is get started writing a long National Institutes
of Health grant.
They're about 25 pages long, they're single spaced, font 11,
and they have point 5 margins on every side of the page, and
it's really dense, and it takes a long time to sort of
get going on that.
And so, there is deadlines, and MIT is very particular, you
need to have it to the Office of Sponsored Research five full
business days before it's due at the National Institutes of
Health, and then the Department needs to sign off on it.
And I'll be looking at my calendar and I have those dates
marked, and checking how many days I have left, and
eventually, just like getting started, it's like oh, there's
so much to do, I have to read the literature, the new stuff
that's come out on my topic.
And I have to think about what projects I'm going to do in the
future, and I have to write about the progress that I've
made so far, which is not really what I want it to be.
And so, I think a lot about how -- it's just overwhelming.
But then eventually something happens.
Either it's tremendous fear that there's so few days left
and you just have to do it.
Sometimes it's going and getting an enormous cup of
coffee and sitting down.
You know, people have been known to sort of chain
themselves to their desk, like they're not going to get to get
up until they've written the introduction to the grant.
So, a lot you can connect with this.
That any new thing you start, there's some barrier that
you have to overcome to get started with it.
And often once you're started it's not that bad, and some of
you may be thinking, exam 1 was a long time ago, I recall
there was a lot of material on exam 1.
And it seems really scary.
But then the fact that I mention the final exam over and
over in class is helping you get that energy that you need
to overcome that activation energy barrier and start
studying, because once you start studying you go, Oh
yeah, I remember this, this wasn't so bad.
So you just need to get over that activation energy
barrier and you're all set.
So, molecules have to do the same thing, and the ones that
have higher temperature have an easier time getting
over that barrier.
So, here we can talk about this general process.
We can look at the individual numbers involved.
So in this particular case, there's an activation energy
for the forward reaction of 132 kilojoules per mole, and there
is an activation energy for the reverse reaction of 358
kilojoules per mole.
And there's also a delta e for the reaction, which is this
line, from reactions to products, which in this case is
minus 226 kilojoules per mole.
So do you think this reaction is endothermic or exothermic?
What do you think, exothermic or endothermic?
It's exothermic, and if you look back in your notes, we
talked a little bit about the relationship between
delta h and delta e.
And they're actually pretty similar.
A delta h usually equals delta e plus a change in p v.
So, for gases it's about 1% or 2% difference, and for solids,
there's really negligible difference between
delta e and delta h.
So, they're pretty similar types of values.
So, we can we can think about what delta e really is, and so
delta e in terms of activation energy is going to be equal to
the activation energy for the forward reaction minus the
activation energy for the reverse reaction.
And in this particular case, we have 226 kilojoules per mole is
our delta e, and for our forward reaction, we have 132
kilojoules per mole, and for the reverse reaction, the
activation energy for the reverse reaction is 358
kilojoules per mole, and so these should all equal
up to each other.
And so, if you know two of these values, you can
calculate the third.
And this is one of the equations that you have to
memorize for the final, because it has, its sort of a
conceptual thing that you need to understand what this diagram
says, that this plus that is equal to that, that these all
add up to each other.
And if you have a negative value here, that means it's
an exothermic reaction.
So, this delta e is a change in internal energy of the system,
and you can determine that value experimentally, say, with
a calorimetry experiment.
OK, so let's keep this in mind and go on and take a look at
how this connects back with some other things that we have
already talked about in this course.
So, for an elementary reaction, and I think for all of us,
there's always some activation energy barrier to overcome.
There's always some positive activation energy to overcome.
And because there's always this activation energy to overcome,
increasing the temperature is always going to increase the
rate of an elementary reaction.
It's always going to make it easier to get over that
activation energy barrier.
But for an overall reaction, increasing the temperature
may not increase the rate of the reaction.
So let's consider why that would be true.
So, here is a reaction that we've talked about before, we
talked about this proposed mechanism where we have a fast
reversible step and a slow second step.
So we learned last time that we can write the rate of product
formation from the second step, there are two molecules of n o
2 being formed, so we have 2 times k 2, the concentration
of n 2 o 2, and the concentration of o 2.
But this is an intermediate, so we need to solve
for that intermediate.
So, in this case, we have a fast reversible first reaction
and a slow second reaction.
So this intermediate is going to build up, and it's going to
be more or less an equilibrium with the reactants, because
this is very fast, and only a little bit of this is siphoned
off to make product, and so this creates an equilibrium
type situation.
So why don't you solve this intermediate for me, this is a
review from the last class.
OK, let's just take 10 more seconds.
Very good.
So, we can solve for this, equilibrium constant for the
first step, products over reactants, and then if you
solved for this, the intermediate here, which is the
product in the first step, then it would be equal to k
1 times n o squared.
We can take that term and we can we can plug it in, so over
here, we can substitute it into this equation, and so we have 2
k 2, big K 1 times n o squared times o 2.
All right, so here is our rate then.
And if you missed some of this, this was in the
notes from before.
So now let's think about the effect of temperature.
So, k 2 is an elementary rate constant, and so its
temperature -- if you increase the temperature, its
rate will increase.
So here again is our equation, the activation energy is always
positive, there's always positive, there's always
some barrier to overcome.
So if you increase the temperature, you're
always going to increase the elementary rate.
Well, what about equilibrium constant?
So we've talked about this back in chemical equilibrium that
the effect of temperature on the equilibrium constant
depends on whether the reaction is endothermic or exothermic.
And you told me before, the equation, and that's the Van't
Hoff equation shown here.
And so look how similar those equations are.
So for an elementary rate constant, we had e a, and for
equilibrium constant, we're talking about delta h.
So, if you have, here the reaction is exothermic, and if
you increase the temperature of an exothermic reaction,
what happens to k?
It decreases.
So, again, it would shift, then, to the endothermic
direction, decreasing k.
So let's look at this then.
So we have in this k obs term, we have an elementary rate
constant and an equilibrium constant, so if you if you
increase the temperature, the rate constant increases, but
the equilibrium constant is going to decrease.
So, the magnitude of the increase or decrease depends on
the size of the activation energy or the size of delta h.
So for this particular example, there's no way that you would
know this so I'm telling you, that the activation energy is a
small number, or you might be able to look it up in your
book, and delta h is a very big number and it's negative,
it's an exothermic reaction.
So if you have a very small number for e a, that means that
the rate constant will increase only a little bit, it's not
that sensitive to a change in temperature, because e
a's a very small number.
But if delta h is a big number, then the equilibrium constant
would decrease a lot with temperature because this
is a big number here.
So in this particular example, increasing the temperature
actually decreases the observed rate, because delta h is, in
this particular example, a much bigger thing.
So if you were given either numbers or some information
like that, you should be able to rationalize what might be
true about the rate of the reaction.
so, a large activation energy means that the rate constant
is very sensitive to changes in temperature.
A large delta h means equilibrium constant
is very sensitive to changes in temperature.
And, as we've talked about, e a is always positive, so the
elementary rates always increase with temperature,
whereas delta h can be positive or negative, so equilibrium
constants can increase or decrease with temperature.
And here, the magnitude of delta h indicates the magnitude
of the change, how much k will change, will it be a big change
or a small change, whereas the sign of delta h indicates the
direction of the change.
So, just want to review one thing and then
we'll stop for the day.
So when a stress is applied to a system, an equilibrium, the
system tends to try to minimize that stress.
So we're back to LeChatelier's principle.
And so, just one more clicker question and
we'll stop for the day.
So increasing the temperature is going to do what again?
Again, thinking back to LeChatelier.
And just 10 more seconds.
Very good.
So, we're going to finish up these notes on Monday.
We're going to think about LeChatelier in a new way, we're
going to think about it in terms of activation energy,
which is really fun, because we tie back what we learned in the
middle of course to what we're seeing now in
the end of the course.
All right, have a great weekend, everybody.