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Now, one of the most important things to affect the rate

of a reaction is the temperature at which the reaction occurs.

So, generally speaking, as the temperature

increases, the reaction rate will generally increase.

And there's an expression which relates this change in

rate with temperature, which is called the Arrhenius equation.

And for this we need to consider a process in a slightly

different way, to the way that we've been considering it up until now.

So if we've got a reactant A going to B, we need to consider this as an equilibrium

process where one is balanced against the other, so

we now have a forward reaction and a back reaction.

And they will both have individual rates.

So the rate for the forward reaction, the rate constant, we'll call K forward, Kf.

And the rate for the back reaction, we'll call Kb for back reaction.

1:12

Now as this is an equilibrium process as it proceeds, it will proceed towards

equilibrium, and the position of the equilibrium

will be defined by an equilibrium constant.

The equilibrium constant is given the symbol capital K, so you

need to be distinguish capital Ks from small ks for rate.

This is an equilibrium constant.

And that is simply defined as the rate of the forward

reaction divided by the rate of the back reaction.

1:47

Now from thermodynamics, we know how to relate the

equilibrium constant to the temperature T is given

by something called the van't Hoff isochore,

and that tells us that the rate of change of the

natural logarithm of the equilibrium constant

as a function of temperature is equal to the enthalpy change for the

reaction, divided by the gas constant, times the temperature squared.

2:29

So if we substitute this equation here for the equilibrium

constant into the van't Hoff isochore, we now have that the rate of

change of the natural law of the forward rate, over the backward rate.

The rate of change with temperature is equal to the

enthalpy of the reaction divided by RT squared.

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So in order to simplify matters, what we do here is

we treat the forward and the back reaction separately from one another.

So If we do that, then we've got the rate of change of the log

rate of forward reaction with respect to temperature.

4:16

So we can we can integrate this expression.

First, we'll separate the variables as we've done in the past.

So on the left-hand side, we're going to have, the change of log Kf.

And on the right-hand side, we're going

to have the activation energy divided by RT

squared times the change in temperature.

So now we're in a position to integrate both sides of this equation.

Re-integrate the left-hand side, we're just going to

get the natural log of Kf and we can include also on this side

the constant of integration because if we had an upper and a lower

value for Kf, or log Kf, then we would have minus log Kf.

So that can actually be included as just dividing that by a constant A.

6:05

So we can now rearrange this equation if we take the inverse log.

This will become an exponential on the

right-hand side, and we will end up finally

with Kf equal to A times the exponential

of minus Ea over RT, and this expression

here is known as the Arrhenius Equation.

So, this is a very, very important equation.

It's relating the rate of the forward reaction to both the

temperature and the activation energy for that reaction.

7:06

Ea is the energy required for the forward

reactions, so let's just think about what that means.

So if we have an energy scaled on the vertical

axis, and we have reactants sitting at some energy here.

So this is the energy level of our reactants.

7:28

And let's suppose that this is an exothermic process, so the

products are going to be sitting down here at some lower energy, so

as the reaction proceeds, there is going to be an amount of energy

given out by this reaction, which is going to be the enthalpy,

overall enthalpy for the reaction.

7:58

Now for this reaction to proceed going from reactions

to products, they don't just go downhill in energy.

They have to actually first all go uphill in energy, some maximum energy

and then turn and come downhill, and they go down to the products.

8:18

So, this point here at the top, where you've got maximum

energy there, that is the energy required for the forward process.

In other words, that is the activation energy for the process.

As you go up from the reactants to this state at the top, this highest

energy state is something known as the transition state of the process.

9:44

But what is this transition state?

Well, let's just consider a very simple example.

Transition state you got a reaction

occurring, bonds breaking and bonds forming.

So this transition state, intermediate state might involve the lengthening of

a bond or change in bond angles, and anything like that would require energy.

10:09

So here's just a very simple process; let's suppose we got atomic

hydrogen reacting with bromine, to give you hydrogen bromide and atomic bromine.

So on our energy diagram here, we will have reactants and products.

And we will have our process, which is going to go up

over a hill, through a transition state and down to the products.

So our reactants are hydrogen and

bromine, the products are hydrogen

bromide and atomic bromine.

So the state in the middle, the transition state, the high energy state,

will probably be some combination were you got bromine bonding to bromine.

You got the hydrogen coming in and probably forming a bond with the bromine.

And this bromine-bromine bond is going to be

different to the bromine-bromine bond just in bromine.

11:38

The Arrhenius Equation is very powerful and gives us a

way actually, to determine the activation energy for a process.

We can see how to do this.

If we take the Arrhenius Equation and just rearrange this,

put it back into a form of logarithm, so we would

now have the logarithm of the rate of reaction is equal

to the logarithm of the pre-exponential factor minus Ea over RT.

We now have the equation for a straight line, so on the Y

axis, we plot log of the rate of reaction, and on the x axis,

we plot one over the temperature, so what we need to do is we need to record

our rate of reaction, our process, a variety of temperatures.

And monitor the, the rate of that process.

And if we do that, the slope of this straight line plot

is going to be equal to minus the activation energy, divided by R.

At all times, you need to remember that the temperature

here, T is in degrees Kelvin, absolute temperature.

The intercept here, will be the, the term natural log of A.

So from that we can get this, pre ep,

pre-exponential factor, and in a moment, I'll tell you

a little bit more about the form of this

pre-exponential factor, and what's its physical meaning actually is.

13:19

We can understand the former theory in the equation if we just consider

a very simple process, so let's take a example of one of those

A and B reacting to give molecule C and D, and we just

consider this; this is a very

simple bimolecular process, two molecules coming together.

So here's molecules A and B; they're the reactants, and they've got to come

together, and they have to collide for the reaction to occur, so they collide.

They form some high energy transition state.

13:58

Now, clearly, in order for these things to react, they have to collide.

So the, this rate of reaction should

in principle, be proportional to the collision frequency.

The higher the co, collision frequency, the higher the rate of the reaction.

The collision frequency is given the symbol Zed.

14:21

But if we think about this for just one moment, if we were at atmosphere

of pressure, normal pressure of gas and we were at room temperature,

then we can easily show that the number of collisions is actually about 10 to

the 28, an enormous number, every second in each centimeter cubed of gas.

Therefore, this pressure and temperature, we can work out

that if every single collision resulted in reaction, then

all gas phase reactions would be over in, 10

to the minus nine or thereabouts of a second.

15:19

And the reason that not all collisions result in reaction is because

the molecules must also possess sufficient energy to react.

If they don't possess the energy EA, the activation energy,

they simply can't react, and we can look at this in terms of what is

called a Boltzmann plot, where y axis here, we plot the number of

molecules, and for instance, the gas, which will have a particular energy E.

15:55

And it follows a distribution, so for example, if we were at low

temperature, then we would have lots of molecules but fairly low energy.

And less molecules but a fairly high energy.

But there will be a distribution; if we increase the absolute temperature,

then we will decrease the number of molecules in the lower energy range.

But we will increase the number of molecules at the high energy range.

16:26

So at some point here, there will be an

activation energy for the process, which will be some

energy, and our molecules, in order to react when

they collide, they must have an energy greater than Ea.

So if we are at, at the low temperature, the fraction of molecules

which can actually react as shown by the hatched area over here.

All the molecules in this region down here, simply

will not react; they'll collide, but they won't react.

16:58

If we increase the temperature, then we can get more molecules

reacting now because we have more molecules which have an

energy greater than Ea, and Boltzmann said that this

fraction here, which has an energy greater than Ea, is given

by its proportional to the exponential of minus Ea over RT.

Exactly the same expression that we see in our Arrhenius Equation.

17:29

So we can see here that the Arrhenius expression, which is

our rate of reaction, is equal to some pre-exponential

factor, which is now really just the collision rate.

So they have to collide.

But when they do collide, they must have sufficient energy,

and that's given by the fraction minus Ea over RT.

So we can now see the entire form of our Arrhenius expression.

First, they must collide, then they must

have also sufficient energy in order to react.

18:10

So let's look an example of the Arrhenius Equation in action.

And we're going to look at the rate of decomposition of N2O5.

So the first thing we do is we make some

measurements at a variety of temperatures, and this case we start

in degree centigrade, zero degree centigrade, 25, 45 and 65

degree centigrade, and then we measure the rate of the reaction.

In this case, in units of per minute so zero centigrade,

the rate is 4.7 times 10 to the minus 5 per minute.

And so on, across the table, we measure the rates at the different temperatures.

18:50

Then we need to process this information, so the first thing

we do is we take the natural log of the rate.

So this won't have any units at all, so natural log of 4.7 times

10 to the minus 5 minus 9.965 and so on across the table.

19:10

Then we need to convert the temperature into Kelvin.

So we have it in degree centigrade.

We are going to add on 273.

So zero becomes 273.

25 becomes 298.

So on, then finally we're going to take the inverse of the

absolutely temperature, which will be in units of Kelvin to the minus 1.

So the inverse of 273 is 3.66 times 10 to minus 3 and so on across the table.

20:03

Gives us a straight line plot.

The slope of the plot we can determine, and that comes out to minus

12516 Kelvin, and we know from the Arrhenius equation that

that's going to be equal to minus the activation energy over R, so now we

have an equality, whereby we can calculate the activation energy.

We know the value of the gas constant R is

8.314 Joules per mole, to degree Kelvin.

So if we multiply the slope by the gas

constant, we will therefore get the

activation energy Ea being equal to

104058 Joules per mole, which is

approximately 104 kilo joules per mole.

So we have now determined the activation energy of the process.

We can also get the exponential factor by

determining the intercept of this plot, and if

we do that, we find that the intercept is equal

to 35.84, which we know is equal to the

natural log of the pre-exponential factor A.

Taking an antilog of that gives you the pre-exponential

factor of 3.686 times 10 to the 15th, the

units here will be per minute because the initial

rates were measured per minute, and we want to get that

into per second, then we are going to have to divide

by 60.

So, we divide that number by 60.

We get 6.143 times 10 to the 13 per second.

We're happy with the units per second, because remember

the pre-exponential factor is essentially related to the collision frequency.

So, this should be units of frequency.

So here we've used the Arrhenius

expression to work out for the decomposition

of N2O5, both the activation energy of

the process and also the pre-exponential factor.

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