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

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

Â So, generally speaking, as the temperature

Â increases, the reaction rate will generally increase.

Â And there's an expression which relates this change in

Â rate with temperature, which is called the Arrhenius equation.

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

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

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

Â process where one is balanced against the other, so

Â we now have a forward reaction and a back reaction.

Â And they will both have individual rates.

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

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

Â 1:12

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

Â equilibrium, and the position of the equilibrium

Â will be defined by an equilibrium constant.

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

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

Â This is an equilibrium constant.

Â And that is simply defined as the rate of the forward

Â reaction divided by the rate of the back reaction.

Â 1:47

Now from thermodynamics, we know how to relate the

Â equilibrium constant to the temperature T is given

Â by something called the van't Hoff isochore,

Â and that tells us that the rate of change of the

Â natural logarithm of the equilibrium constant

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

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

Â 2:29

So if we substitute this equation here for the equilibrium

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

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

Â The rate of change with temperature is equal to the

Â enthalpy of the reaction divided by RT squared.

Â [BLANK_AUDIO].

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

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

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

Â rate of forward reaction with respect to temperature.

Â 4:16

So we can we can integrate this expression.

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

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

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

Â to have the activation energy divided by RT

Â squared times the change in temperature.

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

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

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

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

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

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

Â 6:05

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

Â This will become an exponential on the

Â right-hand side, and we will end up finally

Â with Kf equal to A times the exponential

Â of minus Ea over RT, and this expression

Â here is known as the Arrhenius Equation.

Â So, this is a very, very important equation.

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

Â temperature and the activation energy for that reaction.

Â 7:06

Ea is the energy required for the forward

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

Â So if we have an energy scaled on the vertical

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

Â So this is the energy level of our reactants.

Â 7:28

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

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

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

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

Â overall enthalpy for the reaction.

Â 7:58

Now for this reaction to proceed going from reactions

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

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

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

Â 8:18

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

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

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

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

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

Â 9:44

But what is this transition state?

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

Â Transition state you got a reaction

Â occurring, bonds breaking and bonds forming.

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

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

Â 10:09

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

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

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

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

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

Â So our reactants are hydrogen and

Â bromine, the products are hydrogen

Â bromide and atomic bromine.

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

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

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

Â And this bromine-bromine bond is going to be

Â different to the bromine-bromine bond just in bromine.

Â 11:38

The Arrhenius Equation is very powerful and gives us a

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

Â We can see how to do this.

Â If we take the Arrhenius Equation and just rearrange this,

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

Â now have the logarithm of the rate of reaction is equal

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

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

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

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

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

Â And monitor the, the rate of that process.

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

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

Â At all times, you need to remember that the temperature

Â here, T is in degrees Kelvin, absolute temperature.

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

Â So from that we can get this, pre ep,

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

Â a little bit more about the form of this

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

Â 13:19

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

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

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

Â consider this; this is a very

Â simple bimolecular process, two molecules coming together.

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

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

Â They form some high energy transition state.

Â 13:58

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

Â So the, this rate of reaction should

Â in principle, be proportional to the collision frequency.

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

Â The collision frequency is given the symbol Zed.

Â 14:21

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

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

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

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

Â Therefore, this pressure and temperature, we can work out

Â that if every single collision resulted in reaction, then

Â all gas phase reactions would be over in, 10

Â to the minus nine or thereabouts of a second.

Â 15:19

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

Â the molecules must also possess sufficient energy to react.

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

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

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

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

Â 15:55

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

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

Â And less molecules but a fairly high energy.

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

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

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

Â 16:26

So at some point here, there will be an

Â activation energy for the process, which will be some

Â energy, and our molecules, in order to react when

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

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

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

Â All the molecules in this region down here, simply

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

Â 16:58

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

Â reacting now because we have more molecules which have an

Â energy greater than Ea, and Boltzmann said that this

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

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

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

Â 17:29

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

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

Â factor, which is now really just the collision rate.

Â So they have to collide.

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

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

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

Â First, they must collide, then they must

Â have also sufficient energy in order to react.

Â 18:10

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

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

Â So the first thing we do is we make some

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

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

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

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

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

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

Â 18:50

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

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

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

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

Â 19:10

Then we need to convert the temperature into Kelvin.

Â So we have it in degree centigrade.

Â We are going to add on 273.

Â So zero becomes 273.

Â 25 becomes 298.

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

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

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

Â 20:03

Gives us a straight line plot.

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

Â 12516 Kelvin, and we know from the Arrhenius equation that

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

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

Â We know the value of the gas constant R is

Â 8.314 Joules per mole, to degree Kelvin.

Â So if we multiply the slope by the gas

Â constant, we will therefore get the

Â activation energy Ea being equal to

Â 104058 Joules per mole, which is

Â approximately 104 kilo joules per mole.

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

Â We can also get the exponential factor by

Â determining the intercept of this plot, and if

Â we do that, we find that the intercept is equal

Â to 35.84, which we know is equal to the

Â natural log of the pre-exponential factor A.

Â Taking an antilog of that gives you the pre-exponential

Â factor of 3.686 times 10 to the 15th, the

Â units here will be per minute because the initial

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

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

Â by 60.

Â So, we divide that number by 60.

Â We get 6.143 times 10 to the 13 per second.

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

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

Â So, this should be units of frequency.

Â So here we've used the Arrhenius

Â expression to work out for the decomposition

Â of N2O5, both the activation energy of

Â the process and also the pre-exponential factor.

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Â