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Out tutoring objective deals with reaction mechanisms.
We are going to learn how to take reaction mechanisms
and examine them to find what is called the rate determining step
and from that predict the rate law of a reaction.
So this is another way to obtain a rate law
that we have not seen at this point.
We start with reactants that turn into products
but it doesn't happen usually in one fell swoop.
In a one step process. but it doesn't happen usually in one fell swoop.
In a one step process.
It takes a series of steps.
These steps are called elementary steps.
So they are simple reactions, what is colliding with what
to produce a intermediate?
Those thing then react and run into each other
collide to produce something else, on the way a reaction.
So reaction mechanism is the sum
of these elementary steps it is
that whole sequences of steps that come together
that take you from reactants to products.
Lets look at this reaction.
We have 2NO and O2 producing 2NO2.
This is an observable fact.
We you are monitoring that reaction
you find that some N2O2 exists
but just for a brief time.
It is not a reactant, it is not a product.
So what would that mean?
Take cannot mean that
2 molecules of NO collide with
one molecule of 02 and out pops
some NO2. It means that
on a molecular level that the two
molecules are doing something else.
They do not collide to produce it in a one step process.
Here is what is proposed, and this would account for the
observance of the N202.
An NO collides with a NO that
produces N2O2.
Then the N202 can collide with
an O2 to make the final product.
These are the elementary steps.
Those are your elementary steps of this mechanism. These are the elementary steps.
Those are your elementary steps of this mechanism.
They have to add up to give you the overall reaction.
So these N2O2 will be able to
cancel out when you add these reactions together.
They are intermediates so they appear in the mechanism
for a brief time, but they are not in the overall reaction.
They get produced in an early elementary step
and then consumed in a later step.
So we see here, it is getting produced in this step
and it is getting consumed in this step.
so we see they would cancel out
as you add those together and give you the overall reaction.
Your elementary steps must add up to give you
an overall reactant that you are looking for
over wise it is certainly not a good mechanism.
You use experimental evidence to help you
support the steps that you choose for a mechanism.
You are not going to be given a reaction and asked
to determine the mechanism.
But you ought to be able look at a mechanism
and see if it is a realistic or plausible mechanism for a reaction.
[Here is] some terminology that goes along with this concept.
An elementary step is uni-molecular
if it only has one reactant.
Its bi-molecular if it has two molecules.
Two molecules have to collide in order to produce a bi-molecular
That is by far the most common elementary step.
A Tri-molecular, which is very rare, would be three molecules
colliding into single space in order for a reaction to take place.
It is rare because it is difficult for three molecules to actually collide
if they are moving about randomly, into a single space and time.
Here is a question for you to consider and answer.
We have that mechanism
what are the elementary steps in this mechanism.
Well if you choose bi-molecular, the you are correct.
Now when you look at, elementary steps
and you want to come up with the rate laws
elementary steps, you actually can
and I will put a star by that. You actually can
use the coefficients
to determine the orders of those elementary steps.
Now you could not do this for an over all
reaction, but you can for a elementary step.
If this were an elementary step reaction, but you can for a elementary step.
If this were an elementary step
you would see a one coefficient
for the reactant A we can raise it to
the first power here.
And you know we do not write the first power down.
A plus B going to products
we have the reactants A and B
both raised to the first power.
If and elementary step looks like this,
A has to collide with an A to produce products
then that would be the same as
2 A, but we usually do not write it that way
simply because
we are showing the collision of what collides
with what when we write our elementary steps.
But that would give us a second power for A.
Now remember, and this is important to note,
because students want to go back
to and overall reaction and use the coefficients.
This does not apply for an overall reaction.
How do you determine for an overall reaction?
Experimentally, you have to determine it experimentally.
Once you determine it experimentally,
we are going to see that our rate law
needs to support that experimental information. we are going to see that our rate law
needs to support that experimental information.
One of the steps in our
mechanism is going to be determined
is called a rate determining step, and it is the slowest step.
It is what is going to determine how fast the overall reaction is.
It can only be as fast as the slowest step.
This elementary step, in which we determine to be the
rate determining step is
whatever rate it goes at, whatever
speed that slowest step proceeds at
is the speed or the rate of the overall reaction.
Now to give you a good feel for this
we are going to look at a specific
example. So you are going to see me constructing these little booklets
and from that process be able to
determine, your rate determining step, and see how
that actually determines the over rate of the reaction. determine, your rate determining step, and see how
that actually determines the over rate of the reaction.
We are going to simulate a mechanism
and talk about a rate determining step through the simulation of
this mechanism. It is going to have three steps
our final products is going to be our little booklet.
So we are going to staple this booklet together.
The Booklet is made up of an orange sheet and a blue sheet.
There are going to be three steps to this mechanism.
One person is going to collate the sheets together.
One person, that is going to be me,
is going to fold them into three parts
and hand it off to the next person, who is going to
staple the booklet, to give us this final product.
So what we just demonstrated
was a process that represents
a chemical reaction taking place. was a process that represents
a chemical reaction taking place.
It is representing a three step mechanism. We have the
collating of the pages, the folding of the pages,
and we have the stapling of the pages.
If we are going to look at the over rate of the reaction
we do not look at from the beginning to the end and say 'How Long did that take?
We look at it coming off the assembly line
and say 'How many can we produce
in a certain amount of time?'
and monitor that. So that is what you first need to learn about rates.
Its not start to finish, but coming off
the assembly line. So we produced six booklets Its not start to finish, but coming off
the assembly line. So we produced six booklets
in 45 seconds
so if we were to find the rate we do 6
divided by 45, and that would be booklets per second.
Now we want to think about what is the rate determining step
for that process. We have
the process of collating
the putting two pieces of paper together.
And if we speed up that process
it would not come off the assembly line any faster
We also had the stapling end of the process which
if he could just staple just lightning fast
and staple as fast as he could
it still would not come off the assembly line any faster.
But my step, which is to meticulously fold it
into a tri-fold, to put them together and toggle them,
and fold it, that was a slow step. Now if I could
figure out a way to do my step faster
that would speed up the overall process. So we
call this the rate determining step.
Now that you have an understanding
of the rate determining step
and how it drives the reaction
lets examine another mechanism.
We see the overall reaction above.
We see that someone went
can experimentally determined this to be the rate law
for the overall reaction.
Here is the proposed mechanism
it is a three step mechanism.
We have got some thing here to observe.
First thing we want to figure out in
which step would be the rate determining step. First thing we want to figure out in
which step would be the rate determining step.
Well they way that you would figure this out,
is to come up here and write
the rate laws for each of these elementary steps.
For this elementary step, rate
is equal too, and we are going to call the rate constant
K1 because it is just step 1,
times H2O2
raised to the first power. And I minus [I-] raised to the first power.
For this reaction, rate is equal to K, raised to the first power. And I minus [I-] raised to the first power.
For this reaction, rate is equal to K,
and I am going to give it a K2, for it is a different rate constant.
times OH minus [OH-]
times H plus [H+], both raised to the first power. times OH minus [OH-]
times H plus [H+], both raised to the first power.
This one is rate is equal to, and I will name it K3,
HOI times H plus [H+] times I minus [I-]
We have to look and say
which one of these rate laws
match the over all rate law?
What that would be the first one here.
It matches that
and that will tell me, that the first step
is the rate determining step.
If none of those matched that overall rate law
something else is going on we have not discussed
or you have a bad mechanism.
We see underneath here, I have written
that this one would be the slow step
because it was the rate determining step.
Its rate is exactly equal to the rate of the reaction.
That is the slowest step, and as fast as the reaction can proceed. Its rate is exactly equal to the rate of the reaction.
That is the slowest step, and as fast as the reaction can proceed.
What you need to understand, that it is not
the sum of these rate. We do not add them up to determine the rate of the reaction
it is how fast it is coming off the assembly line.
It is going to come off the assembly line as fast it is how fast it is coming off the assembly line.
It is going to come off the assembly line as fast
as we can do this reaction here. It is going to come off the assembly line as fast
as we can do this reaction here.
We see down at the bottom of the page, the mechanism
that was on the previous page, that we determined the slow step for.
I want you to look at the mechanism
and determine which substances are the intermediates.
Remember, the intermediates are the substances
that disappear in the overall reaction.
They would get canceled out. that disappear in the overall reaction.
They would get canceled out.
The get produced in an early step
and re-consumed in a later step.
OK, if you chose number 4 then you are
absolutely correct. Those are our intermediates. So we have
this substance, being produced
and then consumed, so it is not in the overall reaction.
We have OH minus [OH-] being produced and consumed. and then consumed, so it is not in the overall reaction.
We have OH minus [OH-] being produced and consumed.
Everything else is part of the overall reaction.
OK so this is end of out learning objective 10
where we were determining a mechanism OK so this is end of out learning objective 10
where we were determining a mechanism
looking at mechanism, and seeing how we can predict
rate law from those mechanism
and learning about the
slow step, and the rate determining step.