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When we consider how fast a reaction is going, it's sometimes useful

to have some types of comparisons between different reactions.

And one way to do this is to

measure something called the half-life of a reaction.

Now the half-life is, as its name suggests, it's a, it's a time.

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We give it the symbol tau, Greek letter tau, with a subscript a half.

And it's a time taken for the concentration of a reactant, let's suppose

it's A, to drop from it's

initial concentration to half it's initial concentration.

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And we can, very easily work out tau a half,

just by using our integrated rate expression, which we've already developed.

So, we can use any one if it's first

order, any one of the three which we developed before.

So, for example, natural log of A divided

by A0 is equal to minus k1 times the time.

That will do.

So, for the half-life time, the time T is going to be tau

a half so this is, here it's going to be tau a half.

Then the concentration A is going to be A0 over 2.

So if I put in A0 over 2 here, the A0s will cancel and

all you'll be left with will be logarithm of one half.

And that's going to be equal to minus k1 times the half-life time, tau a half.

This expression can now be rearranged.

The minus sign can be lost by recognizing that we can

just invert the log of one half to log of 2.

And bringing the tau one half to the left-hand side of the equation,

all you're left with is natural log of 2 divided by k1.

So, quite a simple expression.

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So, if we use the example that we had before for

the decomposition of azomethane for which we had a value for k1.

We can now work out the half-life time for that particular process.

So tau a half is going to be equal to the natural log of 2 divided by k1.

And we saw that that was 3.6 times 10 to the minus 4 per second.

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And if we do the maths, then that comes out to 1,925

seconds, which is approximately 32 minutes.

So that's the half-life time for that particular decomposition of azomethane.

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So if we want to follow the course of a reaction,

we could, in essence, measure the concentration at many, many time points.

But there are alternative ways to do this,

there's easier ways that are less time consuming.

So for example, if we just go back to our simple conversion of A into products and

assuming that we have a first order reaction,

then we use the first order integrated rate expression.

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Let's supposed that we start with an

initial concentration of A0, which we call a.

And then allow the reaction to proceed so that sometime

later, sometime t, then an amount x has decomposed.

So therefore, the concentration at any given

point in time A, is going to be equal to, a minus x.

Then, if we just consider two times in the reaction.

So 2 times t then at some initial time t1,

then a is going to be equal to, a minus x1.

And at some later time, t2, the concentration

of a is going to change to a minus x2.

[SOUND] So writing out the integrated rate expression at these two times.

Therefore, you're going to have the initial

time logarithm of the concentration at time t.

Which is a mi, t1, is a minus x1 divided by the initial concentration

a, is going to be equal to minus k1 times the initial time t1.

And then at the second time, we'll have the log of the concentration

which is, a minus x2, divided by the initial concentration.

And that's equal to minus k1 times time 2.

So, we can simplify these two equations by subtracting one from the other.

And then we'll end up with an expression

which is the log of, a minus x1, divided

by, a minus x2, on the left-hand side.

And on the right-hand side, we will

end up with k1 into t2, minus t1.

This expression here is called the interval formula as it

corresponds to the interval between times t2 and t1.

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So, now by knowing a, we can determine

x at two different times and that is sufficient

information to determine the rate constant k1.

We don't need to determine the concentration at

many, many times in order to do this.

But, we would need to know that the reaction corresponded

to first order kinetics for this to work [SOUND].

Okay, so now we've dealt with both zero order and

first order reactions, we can move onto second order reactions.

And by this we mean that the reaction is

second order overall, that is the order overall is two.

So, lets go back a couple of examples of such reactions.

So if we take two molecules of NOBr that decomposes

into two molecules of NO and one molecule off bromine.

And, the rate expression for that reaction is equal, rate

is equal to a second order rate constant, we'll call that

k2, times the concentration of the reactant NOBr squared.

As a differential equation, that will be equal to minus because it's being used up.

One half for the stoichiometry, times the rate of change

of the concentration of NOBr, over function of time.

So that would be one example of a second order process.

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Here's another example of a second order process, where we

take hydrogen and iodine to make two molecules of hydrogen iodide.

This is also second order.

But in this particular case, the rate is now equal to the second

order rate constant k2, times the concentration of hydrogen to

the power 1, times the concentration of iodine to the power 1.

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And so the overall order is two, one plus one, as a differential

equation that can be written in terms of either of the reactants.

So, for example, that would equal to minus the

change in concentration of hydrogen as a function of time.

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So, two second order processes.

Slightly different, ways of writing the equations.

We can also see here the units of k2

from this expression, because we take the upper example here.

We've got k2 here.

We've got a concentration squared.

On this side of the equation, we've got a concentration divided by a time.

So we bring the concentration squared down to the bottom here.

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All we'll be left with is one over time and one over concentration.

So the units of the concentration are going to

be the inverse units of concentration, that is

moles to the minus 1 times, decimeters cubed,

and the inverse of the time seconds to the minus 1.

So it's moles to the minus 1, decimetres cubed, seconds to the minus 1.

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Okay, so if we have a general reaction, A

going to products, a second order process, a simple process.

We would have to integrate the rate expression where we have the rate is

proportional to A squared equals some constant k2 times A squared.

So we proceed as we did before.

First of all, we need to separate the variables so the two variables, A and t.

We need to get everything with A onto one side of

the equation, everything with t onto the other side of the equation.

So, on the left-hand side, we will have dA on top, and on the

bottom, we are going to have concentration of A squared.

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And on the other side of the equa, the equation,

we're going to have minus k2 times dt.

Okay, so this is now something whi, which we

can integrate, so we can integrate the left-hand side.

And we can integrate the right-hand side.

We don't need to integrate the cons, constant, minus k2.

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And then at some time later, time t, the concentration will be A.

Okay, so on the left-hand side, we need to

integrate 1 over A squared, with respect to A.

And the integral of 1 over A

squared is minus 1 over A.

So that would be our upper integral for when we substitute

an A, then we need to subtract the lower integral.

So that will be minus, minus 1 over A0, which would be plus 1 over A0.

So that's the integral of the left-hand side, the

right-hand side is straightforward as we had before.

Don't integrate the k2, and integral of dt between 0 and t is just the time t.

We might want to just rearrange this to take the minus onto the other side.

So if we do that, this would just

change into 1 over A, minus

1 over A0, equals k2t.

And that would be our integrated rate expression for a second order process.

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[SOUND] Okay, so here is our second order integrated rate expression.

Okay, and it's a, the equation of a stra, straight line.

This time we will need to plot 1 over A on the y-axis against t.

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And this time the intercept here will be minus

1 over the initial concentration of A [SOUND].

Because we have a different integrated rate expression for second order than

for first order reactions, the half-life is also going to be different.

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This can be rearranged to make tau one half the subject.

And if we do that, we end up with tau one half equal to 1

divided by k2 times the initial concentration of A.

So we can see actually the half-life is going to

be different if we have different initial concentrations of A zero.

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So, for the second order reaction, these are actually the most common

types of reaction that require two species to interact [SOUND].

Okay, so we've dealt with zero, first, and second order reactions.

What would happen if we went to

higher order reactions, third order reactions, for example?

Well, these are rather uncommon because a

third order process would require a minimum

of three species all to collide at once, which is a fairly rare event.

It doesn't happen very often.

There are examples.

So, for example, if you take two molecules of NO

combining with oxygen to give you two molecules of NO2.

This is indeed a third order process.

And the rate expression can be written as a third order

rate constant times the concentration of NO squared, times concentration of oxygen.

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We could go further, however, this will

give you a very complicated integrated rate expression.

So for our purposes, as they are very

uncommon, we won't go to any higher orders.

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