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CATHERINE DRENNAN: All right.
So 10 more seconds.
OK.
Let's quiet down.
So that is the-- you got the 70% with the right answer.
If people can just yell out what was wrong with number one.
AUDIENCE: Sig figs.
CATHERINE DRENNAN: Sig figs.
And what about numbers three and four?
What equation was that using?
AUDIENCE: Second order.
CATHERINE DRENNAN: Second order.
That's right.
So this is a good clicker question
for a problem that's going to be coming up on exam four
and also on the final exam to a larger degree.
On the final exam we have equation sheets
that have all the equations from the whole semester,
and so you need to figure out and remember which equation
goes with which problem.
It doesn't say, oh, here is the expression for first order.
Here's the equation for second order.
You need to look and remember which
equation goes with which thing.
But in terms of first order, what's
a way that you can remember what a first order
equation would be?
What's missing from first order?
Yeah, the concentration in the material.
So for first order it's independent
of the original concentration in the material, which
is why we can use first order equations
for nuclear chemistry.
Because the rate of decay of radioactive nuclei
are independent of the nuclei around them,
so it's a first order process.
So when you think a little bit about these equations,
you should be able to identify which equation
goes with which problem.
And so you can identify the type of first order problems also.
And for the second order it almost always says,
for this second order process, which
gives you a nice hint that that's a second order process.
And then you just have to identify the equation.
OK.
So today is more kinetics.
So we're in the kinetics unit.
We'll be in the kinetics unit for the rest of the semester.
It's our last unit.
So today I think is one of the most important lectures
in terms of the kinetics unit because we're
talking about reaction mechanisms,
and that is really an important part of kinetics.
So investigating reaction mechanisms.
So if you're going to describe how a reaction takes place--
often reactions don't occur in one step.
It's really uncommon for reactions to occur in one step.
So you want to describe the different steps.
And you describe those steps, which are also
called elementary reactions.
So you break down a complex reaction
into a series of steps, and then you
try to figure out if that mechanism,
if those series of steps, are consistent with the data
that you've collected on this particular reaction.
And over here I just show you some steps
in the natural biosynthesis of a vitamin biotin.
Biotin is important vitamin for us.
It's also used a lot in feedstock.
And so people buy a lot of biotin to put in feed stock,
and so there's a huge market of biotin.
Now right now it's made by what is estimated
to be a 13 step organic synthesis which produces--
that's a lot of steps.
It's really expensive, and there's
a huge amount of organic waste associated with making biotin.
And we're talking about in the tons level of waste.
So researchers have been trying to figure out
the mechanism by which nature makes biotin
because that would be a lot more environmentally friendly.
So when you're thinking about this,
what you want to say is, OK, if we write out a mechanism
is it consistent with the experimental data
and are there fast and slow steps?
Because if there's a step that's really slow maybe
you can do something about it.
Maybe if you're talking about an enzyme
you could re-engineer the enzyme so it would be a better enzyme.
Maybe natural selection didn't particularly work.
Maybe the cell doesn't really need as much biotin
as we do commercially, so there was no need to make it fast.
But now we have a need to make it fast,
so maybe we can do some evolution of this enzyme
and design something to be better.
So when you're talking about reaction mechanisms
you want to know what's fast, what's slow.
And if you want to use that product for something,
you want to figure out how you can change things
to have a better mechanism-- maybe
avoid some really slow steps so that you can do better.
Today is also World AIDS Day.
And understanding the mechanism of HIV protease
was really essential in designing inhibitors
against that enzyme.
And if you inhibit the enzyme you
stop the development of the disease.
So this was a very important thing,
and we have some pretty good molecules
to treat AIDS right now.
And some of the challenges have moved on to other things,
like less good health care in parts of the world where
this has affected.
So I feel like AIDS has kind of taken a backseat to hearing
about Ebola recently, but AIDS is still
a very important problem and one that smart people
like you could address.
So again, understanding reaction mechanisms
is very, very important.
All right.
So let's go to a simpler problem and a simpler reaction
mechanism.
We'll go to our friend over here where
we have two molecules of NO reacting
with one molecule of O2 going to two molecules of NO2.
So someone measured some rates for this
and came up with the following rate law
where you have a rate constant, which is just
called k obs for observed.
And we'll talk about that more in a little bit.
And they discovered it is second order with respect to NO
and first order with respect to O2.
So there's a couple of things right away
that we can ask about this.
One is, what is the overall order then of this reaction
from this experimental data?
What would that be?
AUDIENCE: Three.
CATHERINE DRENNAN: Three.
Again, if some of you missed it.
2 plus 1 is 3.
This is not where you want to lose your points on the exam.
There's going to be some tricky significant figures
in the kinetics units.
Save your points to lose there.
Count 2 plus 1 and get 3.
So is one step likely to have three different things come
together at the same time?
So how likely is it that all three things are
going to merge in one step?
No, it's not very likely, so it's no.
But if it did work that way, what would it be called?
What molecular reaction?
Do you remember?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: So if you have three things
it's a termolecular.
So they're rare.
So that's not how this works.
All right.
So let's look at some rate laws and try to write a rate
law for a reaction.
We're going to take our overall reaction,
we're going to divide it to two steps
and write a rate law for that and see if that's
consistent with the experiment.
So we finally got this up here.
Sorry the handwriting isn't perfect.
I got here early but then-- the first time I'm
going to use the boards a lot today.
Squeegee was gone.
There's always something.
So we're going to break this reaction down into two steps.
So in the first step we have our two molecules of NO
coming together to form an intermediate N2O2.
And this is a reversible step.
So in the second step of the reaction
we have our O2 molecule coming in,
reacting with our intermediate and forming
two molecules of NO2.
And so this is really pretty common
that when you have a multistep reaction that you
form an intermediate and your intermediate goes away.
So now we can think about how we would write the rate law
for this particular mechanism.
So starting up here.
This is being written as a series
of steps, which were also called elementary reactions.
And for an elementary reaction it occurs exactly as written.
That's its definition of an elementary reaction or a step.
So now we can write the rate law for the forward direction
of this exactly as it is written.
So that would be writing-- and these are my little K's
for rate constants.
So we have K1 times the concentration of NO to the 2.
So we're writing it exactly as written.
I said that for an overall reaction
you can't just look at the stoichiometry.
You have to think about experiment.
But for an elementary reaction you can write it just
from the stoichiometry, and so this
is how we would write it just from the stoichiometry.
So what would be the order of that reaction?
Just the forward reaction?
2.
And that makes it a what kind of molecular reaction?
AUDIENCE: Bimolecular.
CATHERINE DRENNAN: Bimolecular.
Bimolecular.
OK.
So let's write out the rate law for the reverse reaction now.
And again, exactly as written.
So we're going to have K minus 1 times the concentration
of our intermediate N2O2.
And that would be the rate law for the reverse reaction.
So what would be the order for this reaction now?
1.
Right.
We only have one thing in here.
And what do you call an x molecular
when there's one thing?
AUDIENCE: Uni.
CATHERINE DRENNAN: Uni.
Unimolecular.
For step two now you're good at this.
Let's do a clicker question.
All right.
10 more seconds.
Yup.
So we're just going to write this exactly as written.
It doesn't say it's a reversible reaction.
So it's K2-- and I'll put this down here.
K2-- our rate constant for the second step--
and then times the reactants.
So we have O2 and our intermediate N2O2.
So what would be the order of this reaction or this step?
Two?
We have two things in there.
And again, we call that bimolecular.
So we have uni-, bi-, and ter- molecular reactions for order
of 1, 2, and 3.
All right.
So we've written this out.
Now what we want to do is we want to write out
an overall rate law for this.
So we've written out the rate laws
for all the individual steps.
And we'll put that-- Yeah, I guess I can put it-- maybe
I'll try to write it down here.
I was going to have this all organized
but then the boards weren't cooperating today.
So I'm going to write the overall rate.
Let me try it here.
And this is for formation of N2O2.
And we're going to put a 2 in there,
and I'll explain that in a minute.
Then our K2-- so we're just writing this now
from the last step-- times O2 and our intermediate N2O2.
So the overall rate of forming NO2
is the 2 because 2 moles are formed.
So as these guys disappear-- as O2 and N2O2,
our intermediate, disappear-- and O's
going to form twice as fast because there
are two of them being formed.
So we put a 2 in there.
And then we just have a rate order or rate
law for the last step, K2 O2 intermediate.
So you can always write the rate for the overall reaction
from that last step.
Although sometimes we'll see if they're fast and slow steps
you can also write it from the rate determining step.
But we're not done here because we have an intermediate
in this expression.
And in a rate law you cannot have an intermediate in there.
You need to solve for the rate in terms of your rate
constants, your reactant concentration, and your product
concentration.
So we need to get rid of this.
We need to solve for this.
So how are we going to do that?
How are we going to solve for this?
So we want to think about now, what
is the net rate for this formation?
How is that intermediate being formed,
and how is it being consumed given those steps up here?
So it's only being formed in the forward direction
of the first step.
And let me just grab this.
So the forward direction of the first step, which is K1 times
NO to the 2.
And so this is where our intermediate is formed.
Our intermediate is going away in two different steps.
So it's going away in the reverse direction of step one.
So it's decomposing in that step.
And so it's decomposing at the rate
of K minus 1 times its concentration.
The intermediate is also being consumed in the second step.
It's reacting with oxygen and being consumed.
So here it decays and here it is consumed.
And it's being consumed by the reaction
K2 times the concentration of O2 times its concentration.
So that's the second step.
So now we just need to take that equation
and solve for the intermediate N2O2.
But we don't know net rate over there.
We have too many variables right now.
So at this point we have to use what's known as a steady state
approximation.
So steady state approximation and pretty much everything--
I'm talking about reaction mechanisms--
we're going to use the steady state approximation.
And that is that the rate of formation of intermediates
equals the rate at which they go away.
So the net rate is 0.
So we can set this whole thing now equal to 0.
So steady state approximation net rate
is 0, or the rate at which an intermediate forms equals
the rate at which that intermediate decays.
So I can rearrange this equation now,
and I'm going to bring the terms that have
the intermediate to one side.
So I'm going to put them over here.
So we have the term at which it decays--
K minus 1 times our concentration
of our intermediate.
And it had a negative but I brought it over here
to this side with a 0, so now it's positive.
And I'm going to bring the same for the rate law at which it's
consumed over.
So that's K2 times our intermediate concentration
and our oxygen concentration.
And now on the other side we'll have
the rate at which the intermediate is formed,
which is K1 No to the 2.
So this is another way to express the steady state
approximation.
The rate at which the intermediate goes away
equals the rate at which it's formed.
That's the steady state.
There's no sort of flux in the intermediate.
It's being formed and going away at equal rates.
So now we can use this to solve for the intermediate.
Now we're set.
Now we can solve for the intermediate.
And so let's do that over here.
So I'm going to pull out-- I had a straight line here.
This one's a little crooked.
I used to write a lot on the board
and used to evaluate professors by their good handwriting
and my ratings were always-- my overall rating was limited
by my handwriting to a large degree
and so I stopped writing on the board.
But now they've got rid of that as a criteria,
so I can write on the board again.
So I'm going to pull out the concentration
of the intermediate-- our N2O2.
and I'll pull it out of the expression leaving
K minus 1 and K2 times O2.
So I just pulled out the concentration
of the intermediate over here.
And then we leave the other side the same.
Our K1 times our NO squared.
So now I can solve for the concentration
of the intermediate.
N2O2 equals K1 times NO squared over K minus 1
plus K2 times O2.
Great.
So we've solved for the concentration
of the intermediate now.
Now we can take this and bring it back over here
and put that whole term into this.
And then we'll have a rate law that is expressed only
in terms of our rate constants and our reactants
and/or products.
So let's do more on the PowerPoint here.
So this is what we just came up with.
So the concentration of our intermediate K1 times
NO-- one of our reactants second order-- over K minus 1
plus K2 times the concentration of oxygen.
Now I'm going to plug that into this expression, which we just
got from writing the rate law for the last step using a 2
because we have two molecules of product being formed.
So we're going to plug it in there,
and that's going to give us this.
So we have our 2, we have a K1 from here, a K2 from here,
we have our oxygen concentration,
we have our NO overconcentration second order over K minus 1
plus K2 times concentration of oxygen.
So that might be the complete answer to some problems
but here we were given an experimental rate
law, which was second order in NO and first order in O2.
That does not match this.
Term has an O2 at the bottom.
This one doesn't have an O2 at the bottom.
These are not the same.
So that must mean that the mechanism has
fast and slow steps and that we need
to reconsider this expression in terms of fast and slow steps.
So let's go back now and think about our mechanism.
And so if we come over here-- let's
say that the first step-- let's bring this down again.
Let's say the first step is fast.
And we already said that it was reversible,
but we'll put that down too.
Is fast and reversible.
And step two is slow.
So I'm just going to propose that these are true,
and then we will recalculate what the rate law would
be if you have a fast reversible step followed by a slow step
and see if that agrees with experiment.
So to do that we have to consider
what it means if we have a fast step followed by a slow step.
So let's introduce a term-- very important term-- which
is the rate determining step.
Also known as the rate limiting step.
So the slow step of a reaction, if it's truly a very slow step,
is going to govern the overall rate of the reaction.
So let's think about this a minute.
So I told you that the extra problems for exam four
are long.
They're very long.
Sorry about that, but there were a lot of problems
and I wanted to get you ready for exam four.
You also have problem set nine due tomorrow.
So you've got a lot of problems to do.
So after class today I feel that many of you
are going to be really inspired to get
started on those problems.
And you may run out of here.
You might leap over the chair in front of you and race
out the door.
You may clear the table on the way out
because you're in a real hurry to start those extra problems.
You will run to the library to look for a table,
but all the tables will already be taken.
How did your classmates get out of class so fast
to get all of the tables in the library?
And they're already finishing problem set nine
and starting on the extra problems.
So you race back to your dorm, but all the tables
in the downstairs of the dorm are also
filled with 5.111 students.
So finally, on the fifth floor of one of the dorms
you find an empty table, and then you're really fast.
You got the problems out, you got your pencils out,
you got your calculator out, you've got old equation sheets
out, and ready to go.
It's like two seconds.
So it took you 40 minutes to find a free table
at MIT that didn't already have a 5.111 student sitting
and doing those extra problems and problem set nine.
So it was 40 minutes plus 2 seconds
to actually start doing the problems.
So the rate determining or rate limiting step
was finding your table.
40 minutes plus 2 seconds is pretty much 40 minutes,
and that's what happens in these reaction mechanisms.
If you have a really slow step that
governs the overall rate of the reaction.
Now a lot of you know also about rate determining steps
because some of you may be the rate determining step
in your group of friends.
The rate at which you get to dinner and eat
is determined by you being ready to go.
Some are in the room like me, yeah, that's me.
You know who you are.
So I'm saying that what you gotta
do to not let yourself be that person-- that rate determining
person, the RDS in your group of friends-- you need to get sleep
and you need to eat well and you need to make sure
you got your ATP.
And that means, of course, to get
enough sleep you gotta start problems sets early, especially
the extra problems because they're long.
Rate determining steps.
Very important.
So let's get back to our example.
And we made a proposal.
We made a proposal that step two was
going to be rate determining.
That was slow.
We made the proposal that step one was fast and reversible.
Step two was slow.
So what that's going to mean then
is that our rate law for the first step-- that's fast.
That's a big number.
The rate for the second step is slow.
Rate determining.
So that will mean then that K1 is going to be greater than K2
times O2, so we can drop out our concentrations here
to think about them.
They're going to be the same.
So what's left?
That means the rate constant for that reverse step is very fast.
It's a big number.
That's going to be big compared to K2 times O2.
So now we can go back and look at our expression
for the intermediate.
And we note that K minus 1 is in the bottom
as well as K2 times O2 is in the bottom of the expression.
So if K minus 1 now is really big compared to K2 times
O2-- again, that's the fast step; that's
the slow step-- then this term pretty much doesn't matter
and it can drop out.
Because this is really small compared to that.
And if we drop out this term we can rewrite the expression
like this.
The concentration of our intermediate rate constant K1
NO squared over K minus 1.
And now we can rearrange this equation,
bringing the concentrations to one side and our rate constants
to the other side.
So we have our intermediate N2O2 over NO squared
equals rate constant K1 over K minus 1.
What does rate constant K1 over rate constant K minus 1 equal?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: It equals the equilibrium constant K1
for the first step.
So if you have a fast reversible step followed by a slow step,
the first step is basically in equilibrium.
We can make that approximation that it is in equilibrium
and solve it by thinking about equilibrium expressions.
Fantastic.
We can be back to equilibrium expressions.
So let's think about this a little bit more.
Here I have a pretty picture for you.
So here we have our reactants forming our intermediates.
The intermediates are also going back and forming our reactants.
Reactants are forming the intermediates and then
back again.
Fast reversible.
Every once in a while an intermediate gets siphoned off
to products, but if this is a really, really slow step
it doesn't happen very often.
It doesn't really affect this very much.
And so basically, this is in equilibrium.
It's like this part doesn't even really matter.
It doesn't play in.
So when we have a fast reversible step followed
by a slow step, we can assume the first step is
in equilibrium and we can solve for our intermediate
using equilibrium expressions.
So let's do that.
Let's take our equilibrium expression for the first step
and now plug it in to our rate law.
So we can substitute this step now
or we can have rate constants or we can
have an equilibrium constant.
You can write either.
These are equivalent.
And we can put those back into this,
which was our original overall rate that we wrote.
We weren't done though because we had an intermediate.
So we can plug this now in for our concentration
of the intermediate.
And so now we get 2 times K1 K2 O2 NO squared over K minus 1.
Or we could just put that with the big equilibrium constant
and get rid of our little K1 over K minus 1.
Both of those are equivalent.
And now we can take all of our K terms and call them K obs.
So K obs is just the experimental rate constant.
It's the collection of rate constants that are measured.
And we often-- when we measure things,
we can't distinguish K1 from K2.
We sometimes try to do that, and that's
a little more complicated.
But in this case, all that was given
was an overall K observed.
And this was our experimental rate law K observed times O2
first order NO second order.
And now we see this expression agrees with this rate.
So the fact that we have a good agreement
means that a mechanism with this fast reversible step followed
by a slow step gives rise to a rate law that's
consistent with experiment.
It doesn't prove that's the right mechanism.
It's very hard to prove mechanisms are right,
but at least it's consistent.
So we can say this is a good guess, a good proposal,
for our mechanism.
So let's look at another example.
So in this example we have NO again.
We have two molecules of NO, and now we
have Br2 going to two molecules of NOBr.
And we're told that the experimental rate is
K obs times NO first order, Br2 first order
and asked, for this proposed mechanism,
which would be the slow step to give rise
to that experimental data?
So the first thing that we would want to do with all of these
is to write the rate laws for each individual step.
So for the forward step we have one molecule
of NO reacting with Br2 with rate constant K2.
So we get rate constant K1 times the concentration of NO times
the concentration of Br2.
Again, this is a step or an elementary reaction,
so we write the rate law just based on the stoichiometry
here.
Now we can do the same thing for the reverse rate.
So we have K minus 1 times the concentration
of our intermediate.
So in step two our intermediate, which is formed in step one,
is reacting with the second molecule NO,
forming our product.
And we can write the rate for this as well.
K2 times the concentration of our intermediate, NOBr2,
times the concentration of NO.
So again, these are steps.
They're elementary reactions so we
can write the rate law based on the stoichiometry
in that proposed step.
So now we can write the overall rate law
for the formation of NOBr, and we can just
write it from the second step like we did before.
Again, this is an example.
We're forming two molecules of product
so there's a two in there.
We have K2.
It's basically just this.
K2 times our intermediate, NOBr2,
times the concentration of NO.
But once again, we're not done because there
is an intermediate in the expression
and you can't have an intermediate in your rate law.
You need to solve for the rate law in terms of rate constants,
reactants, and products.
So we need to now solve for our intermediate
in terms of things that are allowed in the overall rate
law.
And so we want to think again about what
is the change in concentration of our intermediate
So we can do the same thing that we did before?
So the intermediate is being formed in the first step.
So we have the rate law for the first step.
K1 times NO times the concentration of Br2.
The intermediate is also decaying in the reverse part
of the first step.
So that's minus the reverse rate minus K minus 1 times
the intermediate here.
And then it's being consumed in the second step.
So it's going away by the rate K2 times the concentration
of the intermediate times the concentration of NO.
So again, this is exactly what we did with the first example.
We think about the change, how it's being formed,
and the two different ways that it's being consumed.
We can again use a steady state approximation
and set all of that equal to 0.
So let's do that and we'll solve again for the intermediate.
So on this next slide now-- I just put those things up there.
This is what you were just copying down.
If you didn't finish it's still here.
And here is the steady state approximation.
So again, the steady straight approximation
is the net rate of formation of your intermediate
equals the net rate of it's going away.
Net rate is 0.
So rearranging then we can bring the two terms that
involve the decay or the consumption
of our intermediate on one side and then set them equal
to the rate at which that intermediate is formed.
And then, we can pull out our terms for our intermediate.
So we pull out NoBr2, leaving K minus 1, leaving K2 and NO,
and set it equal to the rate law for the first step
in the forward direction.
We can solve for the intermediate.
Take this, divide it by this term.
So we have K1 times NO times Br2 over K
minus 1 plus K2 times NO.
Now we can take this.
We are done solving for our intermediate.
We have no more intermediates in there,
so we can now plug this back in to the expression we
had before.
So we can take this, plug it in here,
and that gives us this formation.
So we have 2 times K1, NO times NO-- NO squared-- Br2 on top,
K minus 1 on the bottom plus K2 times NO.
Now we were asked, what are the fast
and what are the slow steps?
So now we want to take this and think about,
if there's different fast and slow steps,
is it consistent then with the experiment?
So first, let's consider if the first step was slow
and the second step was fast-- or i.e.,
if we have K2 NO greater than K minus 1.
And this is a clicker question, so why don't you tell me how
this then, using this, changes?
OK.
10 seconds.
OK.
Let's now think about why that's true.
This involved doing a couple of steps in your head.
So if we have a first step that's slow
and a second step that's fast, the second step
involves the K2 times the concentration of NO.
That's the second step.
That's fast.
That's going to be a big number compared to K minus 1.
So if we look, both are on the bottom here.
And if this term, K2 NO, is much, much bigger
than K minus 1, then we can say K minus 1 goes way.
If we get rid of K minus 1, we can simplify
the expression even more.
We can get rid of our K2s and we can get rid
of one of our NOs, which gives us this.
So saying that the first step is slow
and the second step is fast gives us
a very different equation.
A lot of things cancel out.
So we can also write that expression as K
obs times NO times Br2.
And the overall order of that reaction would be what?
Yell it out.
AUDIENCE: Two.
CATHERINE DRENNAN: Yes.
So this is what we would get for a first step that's slow;
second step that's fast.
Now let's consider if the first step is fast
and the second step is slow.
So if the first step is fast that means K minus 1,
the rate constant for the reverse step,
is going to be a lot bigger than K2 times NO.
And so now we can look up at this expression
and say, OK, if this is much bigger than this
then that cancels out.
And then we're left with this expression,
which I can put down here.
So that leaves us-- we can't cancel any more at this point.
So that leaves us with 2 times K1 times
K2-- these-- NO squared Br2 over K minus 1.
So assuming different things about how fast and slow
the steps are gives you very different rate laws.
We can also write that to make it look a little simpler
as K obs, but you'll note that the overall order is
very different.
So what's the overall order here?
Three.
Right.
So let's remind ourselves what the experimental rate law was.
And it was NO first order Br2 first order.
So that means that this one would be consistent.
So the mechanism is likely to involve a slow first step
and a fast second step.
And so that's how you do a lot of these problems.
You think about what is going to change when you have
different fast and slow steps.
One of them will be more consistent
with the experimental data and one of them
will not be consistent.
OK Let's do one more fast example.
Here we have rate law for two molecules of ozone O3 going
to three molecules of O2.
And ozone has been in the news a lot recently.
So we want to keep our ozone layer.
We don't want it to go away.
So we have O3 going to O2 plus O,
and you're forming an intermediate O.
That intermediate is reacting with O3, forming
our two molecules of O2.
So let's just write out what our rate
is for the forward reaction.
So we have K1 times the concentration of O3.
For the reverse we have K minus 1 times concentration
of O2 times the concentration of our intermediate O.
For the next one we have K2 times our intermediate O
times the concentration of O3.
So now we're told that there's a fast reversible step
and a slow step.
So the rate will be determined by the slow step.
So we can write out the rate of formation of O2
based on the slow step, which happens
to be the second step, which is what we've done all along.
So there's not really a huge change right now.
So the formation-- again, two molecules of O2
were formed so we have a 2.
We have K2 times the concentration
of our intermediate O times the concentration of O3.
But again, O is an intermediate so we
need to solve for it in terms of our products
reactants of rate constants.
But now we're told something about fast and slow steps
right up front.
And so if we have a fast reversible step
followed by a slow step, how can we
solve for our concentration of our intermediate
in a simpler way than we've been doing?
What do we use?
Or what can we use?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: We can use the equilibrium expression.
So we can put that in.
We can say our equilibrium expression products
over reactants equals little K1 over K minus 1,
or equilibrium constant K1.
Solve for O and get either big equilibrium constant K1
or our little rate constant K1 over K minus 1,
and we have O3 over O2 here.
So this was a lot simpler than doing all of that.
So again, if you have a fast reversible step followed
by a slow step you can solve for the concentration
of your intermediate using an equilibrium expression which
you all know how to write.
So that makes your life easier.
Then we can substitute that back in
and we are able to put this back.
And we'll solve it for O and we'll plug in our K1
over K minus 1.
O3.
We had an O3.
So that's squared over O2.
Or we could write that in terms of K obs O3 concentration
to the 2 over O2.
So let's end with some fun, thinking about what
we would observe here.
First, the order and then what would
happen if we double things.
So what is the order with respect to O3?
You can yell that out.
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yup.
What is the order for O2?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Oh, sorry.
I had something here.
Double this what happens?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: The rate.
Well, four times.
Order here.
Some people yelled it out.
Oh, no.
Oh, man.
It's a clicker question.
I forgot about that.
Don't listen.
But luckily, everyone yelled out different answers.
All right.
10 more seconds.
OK.
Yup.
So it's minus 1.
And so if you double it, it will half.
And then finally, the overall order would be 1 because again,
the overall order is the sum.
So 2 minus 1 is 1.
And last clicker question.
All right.
10 more seconds.
I know you're really in a hurry to do those extra problems
for exam four.
I'm the rate determining step.
I am with my clicker question.
I admit it.
AUDIENCE: Yay.
CATHERINE DRENNAN: All right.
All right.
See you Wednesday.
Remember, final clicker competition
of the year before the finals.