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In the previous lecture, we'd begun our study

of reaction of rates by defining what we mean

by a reaction rate, and then trying to

relate those reaction rates to the concentrations of reactants.

You'll recall we had data like the following, in which we would measure,

for example, say, the concentration of a particular reactant as a function of time.

And we could observe that the typical sort

of curve look something like this, as the reactant

disappears as a function of time. We then define the rate of the reaction

at any particular time as being the slope of the graph at that particular time.

Which meant that the rate, if we write it in

calculus terms, is the derivative of the concentration as a

function of time, or it could be approximated by the

amount by which the concentration is changing as a function

of time.

We then noticed or observed that the rate of the

reaction could be related to the concentration of the reactants.

And at that the particular case of the C60O3 composition reaction, we

found that the rate was actually

proportional to the concentration of the materials.

We also noted that experimentally as observed, that the rates of

the reaction are commonly functions of the concentrations of two different reactants.

So if we have a reaction taking place which is the reaction of A with B,

then the rate depends upon the concentration of

A, and depends upon the concentration of B.

And we need to find these integers n and m because those

are going to be the functional dependencies

of the reaction rate on the concentrations.

So, what do we need to do?

We need to find a way, if we're going to understand the

relationship of rate to concentration, to find these integers n and m.

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One way we can do this is with a method

that is referred to as the method of initial rates.

Which says that we could actually look at the concentrations as

a function of time right at the beginning of a reaction.

So imagine we've got some reaction of A reacting

with B to go on and become products of some sort or another.

And we can set up an initial system in

which we have, say, an initial concentration of A

and some initial concentration of B, and we watch

the concentration of A disappear as a function of time.

The initial rate is simply

the slope of this curve, right at the outset.

Which could be measured.

We could measure the A as a function of time, and

then figure out what the slope is back here at zero.

And that would then give us a means by

which we would measure the initial rate as a

function of the initial concentration of each of the

reactants for this particular prototypical reaction A plus B.

Let's imagine that we've done that experimentally here, and here's

what a set of experimental data might wind up looking like.

For example, perhaps we start with an initial concentration of reactant A, which

is 0.1 molar, and an initial concentration of reactant B, which is 0.1 molar.

And let's imagine hypothetically then that what

we have observed is the experimental reaction rate.

Turns out to be one mole per liter per second the rate of change

of the concentration as a function of time.

Then what we could do is change the concentration

of A while not changing the concentration of B.

And when we do that we might observe hypothetically.

That reaction the reaction rate changes from one

to two moles per liter per second when

that change takes place.

Then we could revert back to the original concentration of A,

but this time double the initial concentration of B, and again measure.

How much does this slope change when we have changed the initial concentrations.

And this is just sort of a prototypical reaction, but what we've noticed is

in this set of hypothetical data, the reaction rate has gone from one to

four by doubling the concentration of B. These data should be sufficient now to

tell us what the values of n and m are in the rate law expression.

Let's see how that would work out. Here's how we could proceed, alright.

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If I take the ratio between these two equations, I can see that I double

the rate by doubling the concentration of A and holding the concentration

of B constant.

In essence I can can-, cancel out the

concentration of B and the rate constant, and

the doubling of the reaction rate is due

entirely to the doubling of the concentration of A.

That tells me immediately that N is actually equal

to one, because the, the amount by which I've

increased the concentration of A originally, doubling it, has

gone straight over and into doubling the reaction rate.

Let's contrast that to the rate for experiment three.

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Now the rate is 4.0 moles per liter per second.

That's K multiplied by the initial concentration of A in this particular

case, raised to the nth, which is of course now we know one.

But that value is 0.1 Times the concentration of B

initially raised to the m, and that value is now 0.2.

Let's compare now the first experiment and the last experiment,

and we see that when we hold the concentration of A fixed and double the

concentration of B, the rate goes up by four, not two.

That means in taking the ratio of the concentration of B, I must have squared

it, to take that factor of two, and turn it into a factor of four.

So immediately I can tell that M is equal to four.

From this, for this particular reaction, we can tell then,

that our rate law, is simply the rate constant multiplied

by the concentration of A raised to the first power,

times the concentration of B raised to the second power.

And we have successfully determined now a reaction rate law.

In general what

we'll discover is that when we are working through, reaction rate

laws, we'll find a variety of different things of this type.

In some cases, what we may discover is

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Now in the case of, for example, A and B, it might be possible

that the reactant, reaction rate in fact probably will depend on both A and B.

We have a different type of second order reaction now that we see here.

Second order.

And it's second order because we in fact have not one but two reactants here.

Each one of which is raised to the first power.

Overall, then, the sum of the powers, one plus one is

two, and we'll refer to that as a second order reaction.

These are some of the more common rate laws which are observed experimentally.

Let's spend a minute looking at those rate laws and thinking about what happens.

Let's go back and talk now about the first order reaction.

Remember the equation for a first order reaction

is this one here from the previous slide.

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Minus, oops, I've got minus on both sides

of the equation, let's move it to one side.

We'll have, if we move the A to the other side of the equation,

one over the concentration of ada, is equal to minus kdt.

And that equation can in fact be integrated to give

us the logarithm of the concentration of A, minus the logarithm

of the concentration of A0, is equal to minus kt.

Or in the alternative we can rewrite this equation as, the concentration of

A is equal to the initial concentration of A multiplied by e to the minus kt.

This is all true for a first order reaction.

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And it's interesting to look at

this particular equation because, we could actually

plot this equation. This is just an exponential function.

If we draw the equation for an exponential function,

the concentration of A is a function of time.

When T is equal zero, we have the concentration of A0.

And we get a normal kind of exponential decay

curve, which asymptotically approaches the the, the, the t axis.

That's exactly the

kind of thing that we observed in the case of the C60O3d composition.

We refer to this particular form of the equation as the

integrated rate law, as you can see in this table here.

These two equations are in fact completely equivalent to each other.

I can get one from the other, they're just different ways of

writing the rate law, and either one of them might be useful.

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As we saw in a previous

slide, it's actually possible to have a second order reaction as well.

This is a first order reaction.

If we go to the second order reaction, the rate law depends upon the concentration

of the reactant squared, instead of the

concentration of the reactant to the first power.

We can similarly integrate that rate law as we did

in the example given just a minute ago by writing again.

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If DADT, the change in the concentration of the reactant A

is given by minus K, times the concentration of the reactant squared.

I could rewrite this as one over A squared DA is equal to minus kt.

And now when I integrate this, I wind up with one over

the concentration of A, minus one over the concentration of

A at t equal to zero, is equal to kt.

Or the alternative, I can rewrite this now

as one over A is equal to one over A0 plus KT.

What that suggests is if I were to plot

one over the concentration of A, as a function

of time, I'd actually get a straight line. For a second order reaction, one

over the concentration of A as a function of time is simply a straight line.

And you can tell actually that one over A increases linearly with time.

So, that's an example of an integrated rate law for

a second order equation, and you can see we've included that

in our table as well.

We could actually keep this up in a couple of other ways, but one of the things we'll

actually considered is in fact, what these equations predict

for what is called the half life of a reaction.

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where we have A is equal to A0, E to the minus KT.

Let's let the time at which half of the material has gone away be the half life.

We going to call that how long does it take for that to happen?

At that point, the concentration of A is

equal to half the concentration we started off with.

We can plug this back into the above equation

and say, the concentration of A is equal to

one half of the initial value, which is equal

to the initial value times E to the minus KT.

We can cancel both sides here, maybe take the logarithm of both sides.

We have the logarithm of one half is equal to

the logarithm of e to the minus kt is minus kt.

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If I put the negative sign into the logrithm,

I've got a logrithm of two is equal to kt.

I should note this t one half right?

Because this only is true at that time where we fit the half life.

So, the half life for a first order reaction is simply equal to the logarithm

of two, divided by k. Notice that this is independent of

the initial concentration of A.

It does not matter how much A I start off with.

The amount of time that it takes for half of that to go

away is a constant that depends only upon the rate constant of the reaction.

And this logarithm two is just a proportionality here.

That's a fascinating result.

Half lives of first order reactions are independent

of the amounts of material which are present.

We can get lots and lots of

different kinds of experimentally observed reaction rates using the

methods that we've talked about in this particular lecture.

Here are some examples of them listed for a variety of different reactions.

And there are some interesting results here when we study these.

Because one of the things we notice is that for some

of the reactants, if we look at say this one in particular.

There are two nitric oxides and one oxygen, and

the rate law corresponding to that, we'll notice is

second order in nitric oxide and first order in oxygen.

That makes it look a little bit like the

stoichiometric coefficient on the nitric oxide which is two.

Matches up to the exponent on the nitric oxide.

And the stoichiometric coefficient on the oxygen, which is one, matches

up to the exponent on the oxygen here, which is also one.

That might seem like a general result, but we can immediately

dash our hopes about that by looking at the second illustration.

Here we have two nitric oxides reacting with two hydrogens.

Again, what we observed is that the

stoichiometric coefficient on the nitric oxide lines up,

but the stoichiometric coefficient on the hydrogen does

not, and in fact that's generally the case.

Look again, icl reacting with

h2, we don't get the expected exponents back over here from

stoichiometric coefficient. Sometimes we do, sometimes we don't.

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Let's look at the last one down here.

This is a particularly interesting reaction

in which ozone reacts with chlorine,

chlorine atoms, and notice that the rate law does seem to line up.

Where each of the reactants

appears to the first order.

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But some of these can even be more complicated.

Look at this particular reaction, here, hydrogen and bromine reacting together.

It turns out, in this case, we wind up with a rate law in which the

coefficient, I'm sorry, the exponent on the bromine is not even an integer.

It comes out to be a one half.

So rate laws can actually be quite complicated.

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Part of our task is going to be to understand why is it that in some cases

the rate laws seem to line up with

the stoichiometrical coefficients, and in others they don't.

And in general, what is it that determines what the values of n and m are?

Those powers, those exponents, those orders

on the concentrations of the reactants.

And that's what we're going to to take up in the next two lectures.