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

Â to have some types of comparisons between different reactions.

Â And one way to do this is to

Â measure something called the half-life of a reaction.

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

Â 0:33

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

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

Â it's A, to drop from it's

Â initial concentration to half it's initial concentration.

Â 0:51

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

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

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

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

Â So, for example, natural log of A divided

Â by A0 is equal to minus k1 times the time.

Â That will do.

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

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

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

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

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

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

Â This expression can now be rearranged.

Â The minus sign can be lost by recognizing that we can

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

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

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

Â So, quite a simple expression.

Â 2:22

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

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

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

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

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

Â 3:02

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

Â seconds, which is approximately 32 minutes.

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

Â 3:26

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

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

Â But there are alternative ways to do this,

Â there's easier ways that are less time consuming.

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

Â assuming that we have a first order reaction,

Â then we use the first order integrated rate expression.

Â 3:57

Let's supposed that we start with an

Â initial concentration of A0, which we call a.

Â And then allow the reaction to proceed so that sometime

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

Â So therefore, the concentration at any given

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

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

Â So 2 times t then at some initial time t1,

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

Â And at some later time, t2, the concentration

Â of a is going to change to a minus x2.

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

Â Therefore, you're going to have the initial

Â time logarithm of the concentration at time t.

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

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

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

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

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

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

Â And then we'll end up with an expression

Â which is the log of, a minus x1, divided

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

Â And on the right-hand side, we will

Â end up with k1 into t2, minus t1.

Â This expression here is called the interval formula as it

Â corresponds to the interval between times t2 and t1.

Â 6:46

So, now by knowing a, we can determine

Â x at two different times and that is sufficient

Â information to determine the rate constant k1.

Â We don't need to determine the concentration at

Â many, many times in order to do this.

Â But, we would need to know that the reaction corresponded

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

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

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

Â And by this we mean that the reaction is

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

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

Â So if we take two molecules of NOBr that decomposes

Â into two molecules of NO and one molecule off bromine.

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

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

Â k2, times the concentration of the reactant NOBr squared.

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

Â One half for the stoichiometry, times the rate of change

Â of the concentration of NOBr, over function of time.

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

Â 8:34

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

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

Â This is also second order.

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

Â order rate constant k2, times the concentration of hydrogen to

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

Â 9:04

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

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

Â So, for example, that would equal to minus the

Â change in concentration of hydrogen as a function of time.

Â 9:25

So, two second order processes.

Â Slightly different, ways of writing the equations.

Â We can also see here the units of k2

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

Â We've got k2 here.

Â We've got a concentration squared.

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

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

Â 9:55

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

Â So the units of the concentration are going to

Â be the inverse units of concentration, that is

Â moles to the minus 1 times, decimeters cubed,

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

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

Â 10:37

Okay, so if we have a general reaction, A

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

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

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

Â So we proceed as we did before.

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

Â We need to get everything with A onto one side of

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

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

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

Â 11:28

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

Â we're going to have minus k2 times dt.

Â Okay, so this is now something whi, which we

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

Â And we can integrate the right-hand side.

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

Â 12:07

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

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

Â integrate 1 over A squared, with respect to A.

Â And the integral of 1 over A

Â squared is minus 1 over A.

Â So that would be our upper integral for when we substitute

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

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

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

Â right-hand side is straightforward as we had before.

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

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

Â So if we do that, this would just

Â change into 1 over A, minus

Â 1 over A0, equals k2t.

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

Â 13:49

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

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

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

Â 14:23

And this time the intercept here will be minus

Â 1 over the initial concentration of A [SOUND].

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

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

Â 15:40

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

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

Â divided by k2 times the initial concentration of A.

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

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

Â 16:07

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

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

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

Â What would happen if we went to

Â higher order reactions, third order reactions, for example?

Â Well, these are rather uncommon because a

Â third order process would require a minimum

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

Â It doesn't happen very often.

Â There are examples.

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

Â combining with oxygen to give you two molecules of NO2.

Â This is indeed a third order process.

Â And the rate expression can be written as a third order

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

Â 17:05

We could go further, however, this will

Â give you a very complicated integrated rate expression.

Â So for our purposes, as they are very

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

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Â