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Friends, let us continue our discussion on complex reactions, and so far we looked at
several examples of this complex reactions, and simple schemes which are representative
of complex reactions. And we saw some characteristics of such reactions using this simple schemes.
We also saw two basic approximations that we use quasi steady state approximation or
quasi equilibrium approximation, which allows us to simplify the kinetics of the process.
Today’s session we will take a look at few examples, and see how we derive the kinetics
of these processes using the approximations that we saw in the last class. Now the first
question that we may ask ourselves is that how do we know that reaction is a complex
reaction? So the first indication that the reaction is a complex reaction comes from
the fact that the order and molecularity of the reaction do not match.
For example, if we look at oxidation of nitric oxide or other reduction of nitric oxide to
give rise to nitrogen and water. The order of this reaction is given or the kinetics
is given by… So, the molecularity here is four, two moles of N O and two moles of hydrogen,
so two plus two four, whereas the order is three, two, and one over here. So that is
a clear indication that this reaction is not occurring as a single step interaction between
two molecules of N O, and two molecules of hydrogen. But there is something else that
is that is happening, so that is one example. Another example, we can look at is sucrose
when it undergoes hydrolysis to give glucose and fructose. The rate of this reaction is
given by an expression, which looks something like this. It is not a order or power law
kinetics but it is a little more complicated reaction kinetics and which says that rate
is K some constant times C S divided by K n plus C S and it turns out that this constant
K is also not really a constant but it is a function of p H of the media. So in reality
the kinetics and or the order and the molecularity once again do not match. So, how do we now
get this kind of kinetic expressions when we have complex reactions?
Now before we start looking at few examples and deal with them. Let us look at how do
what is the approach for developing the kinetics of such reactions. So what we essentially
do is we write a mechanism of the reaction. So mechanism of the reaction is nothing but
the detailed description of the reaction as it takes place from reactant side to the product
side. So mechanism is the detailed description of the reaction and we write the mechanism
of a reaction with an understanding that each stage in this mechanism or each step in this
mechanism is a elementary reaction.
So we have a reaction, we have it does not occur as a single step process as we as I
said earlier but it occurs as a multi step process and we further make a assumption that
each step of this is an elementary reaction. If we take a example of sucrose so I am just
going to write this as S going to glucose and fructose, this example actually is a example
of a catalytic reaction and it involves an catalyst E, let us say we call that catalyst
E. So this is a overall reaction and we write the mechanism of this reaction as a detailed description and we will discuss
about this in detail little later on. But at this point let me just say that what happens
in this reaction is this sucrose binds to this enzyme E gives rise to some complex.
This complex interacts with water gives rise to some other complex, which call it as P
star then E P star gives us E plus G plus F. So, what we have done in short is we have
written the overall reaction in terms of it is constituent elementary reaction. Now there
are several things to be remembered when we write the mechanism. Mechanism of a reaction
must explain the conversion of a reactant to the product. For example, I cannot have
a mechanism of a reaction in which this fructose is not there. This is not a complete description
because the overall reaction over here has a product fructose. But if we omit fructose
from our mechanism then this is not a correct mechanism. So it has to explain the entire…
Now and each step of this mechanism we say that it is a elementary step and law of mass
action kinetics therefore, applies to such elementary steps. Then we can work either
quasi steady state approximation or quasi equilibrium approximation and derive the kinetics
of the process, which eventually leads us to the expression that we saw earlier. So
kinetics of complex process is consist of analysis consist of first of all the overall
reaction, the mechanism then some assumptions and the kinetic rate expression. So now this is of course, not always one way street we
are to remember that whatever mechanism we write and whatever kinetic rate expression
we get it must be experimentally verified. That is if I write a mechanism of this reaction
and if I cannot experimentally verify that this is the correct rate that is I experimentally
measure the rate of hydrolysis, I experimentally measure the sucrose concentration and see
whether this kind of expression applies to that experimental experimentally determined
rate. If it does not then I have made mistake either in writing the mechanism or writing
or making certain assumptions and therefore, I will have to go back to my mechanism step
and rework so that I get consistent kinetic rate expression. Now, in our discussions we
will be following mechanism assumptions and what is the final kinetic rate that we get.
We obviously cannot in these video tutorials verify them experimentally but whenever possible
I will try to show some data, which says that indeed, that is the case. Now before we go
on to examples one more point to be noted that when it comes to mechanisms, there are
two kinds of mechanisms. One what we call open sequences, this particular example or
this particular case is an example of an open sequence that is what we have is starting
with reactant we follow a series of steps, which end up in product.
So to be brief we have let us say reactant we have reactants, intermediates, products.
It need not be a linear sequence like this but that is a order in which information is
flowing and there is this is what we call open loop open loop sequences. This catalysis
is an example of an open loop open loop sequence.
The other type of mechanisms are what we call closed loop sequences closed loop sequences. Now in these kinds
of mechanisms, what we have is we start with reactants, they go to some intermediates some
more intermediates, which goes to products. However, unlike open loop so this is what
happens typically in an open loop sequence. But in a closed loop sequence the intermediates
let us say at the later stage go back to intermediates at an early stage in a way closing the loop
in a in a sequence of events. You probably are familiar with control terminology,
open loop control, closed loop control and so on. So in that sense, this is a closed
loop because you are going in one particular direction. But at some stage there is a feedback
into a previous stage and this is what makes these sequences to be called closed loop sequences.
So we will start with examples of closed loop sequences, two examples chain reactions actually
chain reactions and polymerization reaction and see what happens to kinetics of these
processes.
So, let us look at chain reactions to start our discussion we are going to look at complex
reaction, chain reactions as first example and there are several examples of chain reactions
few I mentioned earlier but let me reiterate once again. We have combustion reactions which
are chain reactions. We have decomposition reactions. We have auto oxidation reactions,
polymerization reactions and so on. All important reactions, combustion reaction of course,
as the name suggest any combustion process hydrogen and water combining to form hydrogen
and oxygen combining to form water or burning of hydrogen a combustion reaction.
Decomposition reaction, that is break down of complex molecule into simpler molecule
and we are going to see a example of that auto oxidation. Formation of peroxides that
is we looked at that example earlier that is you have a hydro carbon we imbibe oxygen
into it and it is a deadly combination because both fuel and oxygen are together and then
of course, the polymerization reaction.
So, let us start with an example of a chain reaction decomposition reaction, so decomposition
of acetaldehyde. Now what we have is acetaldehyde giving rise to methane and carbon monoxide.
The simple break down and if it was a simple reaction we would have expected the kinetics
to be k into concentration of acetaldehyde. Because the order is one, so first guess would
have been or first order reaction or rather the molecularity is one, so our first guess
would have been first order reaction. However in reality the experimental data suggest
that this particular rate actually is has a order 3 by 2. So it is r is k into concentration
of acetaldehyde raise to 3 by 2. A clear indication that it is not a complex reaction and in fact
this reaction actually takes place through several steps8 and let us looks at first the
mechanism and say example of closed loop sequence, so that will also tell us why we call this
as a closed loop sequence.
So, what happens in the first step that is the beginning of the reaction is acetaldehyde
breaks out to give radicals free radicals actually. It gives methyl radical and acetyl
radical C H 3 and C H O. So that is a first step I will go through this sequence and then
see the proper terminology for this. So, we have what we call first reaction as initiation
reaction, the reason for this will become clear as we move along. So, next step what
happens is this methyl radical extracts hydrogen from an acetaldehyde molecule and gives rise
to first product methane and another radical C H 3 C O. This C H 3 C O breaks down to give
methyl radical and the second product namely carbon monoxide.
So, once this third reaction takes place as you can see there is a feedback because the
product of the third reaction namely methyl radical, goes back as a reactant in the second
step further combining with acetaldehyde molecule and thereby closing the loop and these reaction
in other words can sustain on it is own, because moment we have this methyl radical that is
why this name initiation. So, once the reaction is initiated by formation of this methyl radical
it can be propagated indefinitely in principle. Why because if one dimethyl radical will give
this C H 3 C O will re again regenerate C H 3 and close the loop.
So these are what are called as propagation reactions, the name is self explanatory because
these are the set of reactions which are propagating by themselves. Now, if this last reaction
was not there then imagine what would happen just presence of one methyl radical is enough
to keep this process going cycles indefinitely and rate of such reaction will be extremely
fast or the reaction will end only when the entire acetaldehyde has been broken down.
That means always hundred percent conversions but the experience tells that that is not
the case. That is acetaldehyde decomposition does not
end only when acetaldehyde is exhausted and it does not happen at an infinitely fast rate,
which implies that there must be some other reaction, which is taking away the important
constituent, which keeps this reaction propagating namely these methyl radical. So what happens
as shown in the fourth reaction is this methyl radical combined with another methyl radical
to give rise to ethane. So, this now if there is a source of methyl radical in the initiation
step, the last step is a sink because in this methyl radicals are getting consumed or and
therefore, this reaction is referred to as termination reaction.
So this is a characteristic of a typical chain reaction. So what do we have here four steps
in general; initiation, propagation, termination. Propagation need not be only two step process
it could be several steps but the key point is that something at a later stage feedback
feeds back into an earlier step in a reaction. Now there are several things to be noted,
I mentioned earlier that in a reaction sequence all products have to be explained. So that
is that is clear that we have methane as one product, carbon monoxide as other product.
But, there is also a side step side product namely ethane and it does not show up as a
overall reaction and that is also an characteristic of a chain reaction that there will be some
products, which are insignificant and therefore, are not counted as major products of a reaction.
Why is it insignificant because had this reaction been significant then we would have formation
of ethyl radical and it is combination with another methyl radical and reaction terminating
instantaneously that means almost zero rate. But that is not that is not the case. So,
it is clear that termination reaction rate is fairly small compared to propagation reaction
step. So, now let us try to see how we go from this particular mechanism to deriving
the kinetics.
But before we do that let us rewrite this in a slightly different form, so that we see
the cyclic nature of this reaction. So, what I am going to do is I am going to write the
same reaction in a slightly different form. So, I start with acetaldehyde and I am going
to call it A. This acetaldehyde as so let me number these reactions as 1 2 3 and 4.
So the first reaction gives rise to a methyl radical, so that is my reaction one. This
methyl radical combines with another acetaldehyde and gives rise to C H 3 C O liberating the
first product methane, so that is my reaction two.
This radical C H 3 C O decomposes to regenerate C H 3 and the other product C O, so that is
my reaction three and two methyl radicals combined with each other giving rise to ethane,
that is my reaction four. Now you will appreciate the cyclic nature if you focus on once initiation
has happened, that step one has taken place the cycle can continue forever. Till such
time that termination takes place and that is the importance of termination reaction.
So even though we say that propagation reactions are faster but termination reaction is an
important role to play. So let us try to determine the kinetics of this particular process.
So the best way of doing this is let us write down the mass balances. So, assuming that
this reaction is taking place this reaction is taking place in a close system. I am going
to write mass balance of this particular system and to keep discussion little simple, I am
just going to label this species 1 2 3, this as A so A, this is already 2, this is 4, this
is product p 1. This is again species 4, so this is 2, this is 4, this is giving rise
to 2 and product p 2 and this is this is again 2 giving rise to some products. So I have
labeled my species 1 to 4, which are the key species in this reaction. So C 1, C 2, C 3,
C 4 is going to be the concentrations of these four species. So let us write down the balance
on acetaldehyde first and this reactions we are going to call 1 2 3 4, 1 2 1 2 3 4.
So, we will just simply write reactant rate constants as k 1, k 2, k 3, k 4 for these
four reactions or let us still better let us call this initiation rate constant k i
this propagation step one k p 1, this as k p 2 and this as k t. So let us write down
the mass balance equations as minus k i into C A or C 1 because we call this methanol as
acetaldehyde is one. So let us call this as species one to avoid any confusion. So, what
we are saying is that we will call the acetaldehyde species one to be consist so this C 1 is k i into C 1 minus k p 1 from
the second reaction into C 1 into C 2. Similarly, d dt of C 2 will be k i into C
1 coming from the first reaction getting consumed in the second reaction. Once again getting
consumed or getting formed in the third reaction and getting consumed in the fourth reaction
and for four reactions, since two molecules are involved we will write the rate as 2 k
t into C 2 square. Remember this is an elementary reaction so the order of this reaction is
same as it is molecularity. So we have C 2 square and we have two molecules two radicals
methyl radicals combining and hence we write the reaction as with the stoichiometric coefficient
minus 2. So what does that mean k t into C H C 2 square
is the rate of these reaction and therefore, we write this. Let us write the same thing
for C 4. Now it is getting formed in the second reaction so and getting consumed in the third
reaction, so we have C 4. Now, what is the rate of methane formation? As per this is
k p 1 C 1 C 2. What is the rate of C O formation? So, as per as per third reaction now if you look at the kinetics expressions,
it involves two species C 2 and C 4, which are radicals and their concentration is difficult
to determine. So what are we going to do? We are going to
use quasi steady state approximation. Why how do we justify quasi steady state? We say
that these radicals are highly active and so movement they are formed they participate
in the reaction in a fairly quickly and therefore, their concentrations are very small and dc
to dt or there dynamics is also very small and so we have quasi steady state. So before
we invoke quasi steady state approximation, let me recap we have the overall reaction
we have the overall reaction, which we wrote in terms of its constituent reactions.
So the mechanism of these reaction and we want to invoke certain approximations or make
certain assumptions. In this particular case, we make the assumption that these methyl radical
C H 3 and C H 3 C O radicals are highly reactive and therefore, there dynamics is very slow
or they form they get formed they get consume form consume, so very fast dynamics. So quasi
steady state is a reasonable assumption for the species so in order to invoke quasi steady
state we write the mass balance equations, which is what we have written over here.
The next step is to therefore, invoke now quasi steady state approximation. So we will
invoke it for species 2 and 4. So, let us start with species 4. So, there we see that
k p 1 C 1 C 2 that is if we said dc 4 dt is close to zero because quasi steady state k
p 1 C 1 C 2 is k p 2 C 4. So, that is the first result why are we doing this because
we have our rate expressions, we have our rate expressions, which are in terms of C
2 and C 4 concentrations and we do not know what these concentrations are so we want to
eliminate them. So that everything is in terms of C 1 because that is the only reactant over
here and p 1 and p 2 but they are not participating or there is no feedback, so in this particular
case not reversible reaction. So we want to eliminate C 2 and C 4 in terms of in terms
of … So dc 4 dt gave us gave us this particular.
Now what is the significance of this? If you remember for our overall reaction overall
reaction, we will have certain rate as we have been defining earlier and r C H 4 or
the rate of formation of methane must be the stoichoimetric coefficient times this r. So,
r C H 4 must be r similarly, r C O 2 must be also r, so this is r this also is r and
this approximation tells us that indeed that is the correct case that is k p 1 C 1 C 2
is same that is k p 1 C 1 C 2 is same as k p 2 C 4. That means both these are equal and
we call that as r. So, that is a good sign that we are not we our assumption is a good
assumption. Then we invoke the same assumption for dc to dt balance or balance for species
two and you will notice because of this quasi steady state approximation for species 4,
the second term and the fourth term or third term cancel each other and we are left with
k i C 1 is 2 k t into C 2 square from this rate expression.
What does this what does this mean? This means that I now have concentration of C 2 in terms
of C 1 and that is what we have been we have been looking for. So concentration of C 2
that is the radical in terms of species whose concentration we can measure. So, let us revisit
overall reaction mechanism, mass balances, quasi steady state approximation and the simplification
of that.
So, now let us look at what happens as a consequence of this and to do that I am going to this
is becoming too crowded. Now so let us remove some of the old part of the discussion.
And take it from the last point, and therefore then say that my rate of this reaction, which
is r C H 4 r C O over here, C 1 C 2, so that is k p 1 into k i by 2 k t raise to half C
1 raise to half into C 1. So, this is our substituted for C2 now, and the net result
is that we now have r as some constant k C 1 raise to 3 by 2, which is what experimentally
observed rate. So we now therefore, complete this discussion. So we have the desired rate,
this particular example illustrates several aspects. If you go back to our rate expression
we said that these first and third reactions are slower and so on.
So does it mean that their rates or rate constants do not matter? No, it does matter. In fact
if you look at the rate expression the kinetic constant k is actually has propagation constant k p 1 k
i and 2 k t. That is if we look at this particular expression. So rate constant of initiation
and termination do matter. So that is point one. Point two if we look at this approximation
for approximation for C 2 over here, we see that the initiation rate that is rate of this
first reaction equals the rate of termination reaction the fourth reaction and this also
is a typical of a chain reaction. That is reaction one is equal to reaction four, the
rates of these reactions. So that is one example.
I will quickly go through second example polymerization reaction and particularly chain polymerization,
because as you would know the polymerization reactions are of several types, condensation
polymerization, addition polymerization and so on. But we will look at only one such example
here that is chain polymerization, which is also a chain reaction mechanism. So in particular
we are looking at ethyl, the polymers whose monomer can be represented as C H double bond
C H X. So if X is hydrogen we have ethylene and polyethylene, if X is chlorine we have
vinyl chloride and polyvinyl chloride P V C, if it is styrene C 6 H 5, we have polystyrene.
So all these polyethylene, polyvinyl chloride, polystyrene and several others can be represented
polymerization reactions by looking at monomer C H 2 double bond C H X and initiator for
such polymerization reaction there are several it can be even light energy. But there can
be chemicals also like benzoyl peroxide benzoyl peroxide, which is shown over here. So the
first step is benzoyl peroxide gives rise to benzyl radicals and this benzyl radical
binds with R X, I am going to call this C H 2 C double bond C H as R and X as the appropriate
side group, there is H or chlorine or C 6 H 5 and so on. So we have phi joining with
R X giving rise to another radical phi R X dot then this phi R X dot in a propagation
reaction. So we have propagation now, phi R X dot binds
with R X gives rise to phi R R X radical and these reactions continue. So for example,
this particular radical addition of R X gives rise to phi R j R X. So, these are all propagation
reactions and then we have termination reaction phi R j minus 1 X combining with another radical
so having I monomeric units and j monomeric units combined to give phi polymer with I
plus j monomeric units.
Now the kinetics of this, we will write it as in terms of reaction scheme. So initiator
giving rise to initiator giving rise to free radicals, step one free radical benzoyl peroxide
combining with monomer giving rise to a live radical R 1. Then R propagation reactions
that are R 1 combining with one monomer giving rise to R 2, R 2 plus M R 3. So, in general
R j minus 1 plus M as R j, so these are all propagation reaction and the termination reaction
R j plus R i giving rise to P i plus j. So these are the typical rate constants of
these reactions initiation and termination being fairly small this should be minus sign
here fairly small compared to propagation reaction step. So how do we determine the
kinetics of this polymerization same strategy. We have a mechanism and we are going to invoke
quasi steady stet approximation for all these radicals R 1 R 2 R 3 R i R i R j R M. Remember,
this is a infinitely long process and therefore, we have polymer chain distribution and so
on.
So, let us how do we determine the kinetics in a simple manner. You can work out these
details by yourself, it is writing for a given species it is appearance and disappearance
rate. So, we have initiator of radical live polymer R 1 live polymer R j and the polymer
P j, which is in this particular case formed from R 2 live radicals R 1 and R 2 combining.
So, we now then invoke quasi steady state approximation for the species that means what
does it mean that means d phi dt. For example, it is appearance rate minus disappearance
rate that is true for any species rate the formation is a balance between appearance
and disappearance. So for quasi steady state approximation to be invoke this must be equal,
these two rates are equal and so on for all the species.
Now using some properties of infinite change because this is a polymer chain, we can actually
simplify that by in terms of initiation rate the total radicals that is R 1 R 2 R 3 R phi
R infinity is given in terms of these initiation rate r i and the initiation rate constant
k a. Then monomer consumption rate this is refer to as total radicals tau zero monomer
consumption rate. Radical concentration in a species R j in this particular form, which
can also be written as gamma times psi raise to j and the polymer generation rate which
is in terms of this R j j minus 1 into some constant.
Now, if we look at the structure of this we notice the following. The importance of initiator
for example, if initiator is not there I is zero then it turns out that all these rates
go to zero. That is if initiation rate is zero, then total radical concentration is
zero, monomer consumption is zero, individual radical concentration is zero, polymer generation
rate is zero. It also implies that if initiator concentration rate, why I am saying this because
in all these rate expressions r i is there r i r i r i r i and so on.
So if initiator concentration goes to zero, that means this goes to zero, r i will go
to zero and all these rates will go to zero, that is the importance of initiator. It is
importance can also be felt into the following manner expressed in the following manner.
If initiator concentration is small then i is small so r i is small and then all rates
will be small. In other words, we have to maintain initiator concentration at a high
value and furthermore it is often maintained at a constant value.
Now, if we look at these expressions once again we also notice the following these radical
concentrations for example, R j as a function of j. What is j? j is the number of monomeric
units one two three four and infinity it actually always falls down and that is to be expected
because here R j comes from j minus 1, j minus 1 comes from j minus 2 and so on. So j minus
2 concentrations is always higher than j minus 1 is always higher than j. So, it is monotonically
decreasing. Now if you look at the rate of polymer polymerization
we find that R j function is continuously decreasing as function of j, j minus 1 is
continuously increasing. Hence the product of the two, if you plot it as R P j as a function
of j it always goes through a maxima. And that is that distribution of the polymer that
we find.
The weight fraction of the jth component as a function of j or its molecular weight, we
always see that there is a unimodel distribution, and our kinetics can explain that.
So with this, we come to the end of this particular session, and here today we saw two examples
that of chain reactions and polymerization reaction, and how we use quasi steady state
approximation to determine the kinetics. In the next session, we are going to look at
catalytic reactions, and the kinetic rate expressions of catalytic reactions. Thank
you.