okay let's start
now suppose that the mechanism of this
overall reaction consists of these three
one-step reactions remember all
reactions in the mechanism has to be
elementary reactions elementary step
elementary one-step reactions clear and
so this is suppose the mechanism now
condition is for one condition is for to
be this to be a plausible acceptable
mechanism for this reaction when you add
up these elementary one-step reaction
side by side you should get the overall
reaction so in this case I try to make
it that way
actually so E is cancels out G cancels
out and H cancels out and I have a plus
B is equal to C plus D this is now to a
a to a and so on alright so it obeys
that now these products like GE and H
which do not appear in the main overall
reaction equations are called
intermediates so simply I should say
here e G and H are intermediates now the
intermediates usually have very very
short lifetime like 10 to the minus 10
10 to the minus 9 seconds and so on
so it is very difficult to determine the
presence of intermediate in a chemical
reaction
if you can determine the presence of
intermediate those intermediate in your
chemical reaction that means that will
be a very good evidence that this
mechanism is a correct mechanism yeah
you should remember again the mechanism
is sort of theoretical I simply consider
the steps which will give me the rate
law of the reaction now the question
here is remember again be I can write
down the rate law for this reaction
right and I can write down the rate law
for this reaction I can write down the
rate law for this reaction by looking at
this let me just write it down here rate
of this reaction rate of the first step
would be equals to now tell me okay
let's call that K 1 K 2 and K 3 K yes
times what yes a concentration molar
concentration of a how about the power 1
n times B and power 1 you see again if I
didn't know that this step is elementary
one-step reaction I can't write it but
if I'm told that that is an elementary
reactions write down the rate law I can
write it clear so the rate for this
second step here would be equal to let's
call that K to call that k1 times the
concentrations of G and times the
concentration of a and then this is
going to be again rate 3 equals 2 k3
times the concentrations of e times the
concentration molar concentrations of 8
now all exponents are one because
coefficients in the elementary
one step reactions are 1 1 1 so that's
why I write the rate low this fashion
alright now the question here is how I
can write down the rate law or derive
the rate law from the mechanism now the
rate of these 3 reaction elementary
reactions will not be equal to each
other right some will be a little faster
sample people slower and so on but
there's always going to be a slowest
step in the mechanism and the rate of
the overall reaction has to be equal to
the rate of the slowest step in the
mechanism rate of the slowest step in
the mechanism now for example suppose
that this is the slow step and we
indicated like this and this is let's
say fast and this is fast so if I knew
that this is the slow step then the rate
of the overall reaction will be equal to
the to this rate law k2 times G times a
but there is a problem here the problem
is G is an intermediate you never have
the intermediate in your in your rate
law so now what are we going to do by
using some equilibriums and other things
in fact by using these equations we can
convert the concentrations of G in terms
of concentrations of a and B put it
there and then obtain the rate law has
the concentrations of which will include
only concentrations of a and
concentrations of B which means the
reactants only the reactants in the
overall reaction reactants in the
overall reactions now so this this is
simply called the slow step this is
called rate determining step rate
determinant
step that means that determines rate
actually the rate of the overall
reaction is equal to that now can we
explain that it I mean how can we
visualize that rate determining step
this is again you see a highway and a
tall tall Plaza
you know what tall means toll plaza
means Turkish what we call them para the
demonstrators
yes yes exactly right so that tall means
Keesha you pay there and go there and so
on so here it says the depending on the
number of lines at the toll here that
determines the rate of how many cars
will pass through this highway like this
now war logical to me at least logical
analogy can be made as follow to
describe or to understand the rate
determining step
suppose that I have pipes of different
diameters connected to each other right
I can have pipes and then I suppose that
pump water from this side I pipe system
so the water will fill these pipes and
then will go and come out from here now
one of the pipe here will determine how
much water will flow per unit time right
which one is it the narrowest one the
one which the smallest diameter so that
actually corresponds to the slowest step
in the mechanism so always the rate of
the slow step is equal to the overall so
in this case that pipe whatever the
water passes through there per unit time
would be the same as the water coming
here it doesn't matter how much water
you pump in that determines the actually
rate of flow water through the system
through the system so that's very
similar you can think about the rate
determining step as this is a matter of
fact okay now now here again we talked
about this what is required for a
plausible mechanism the rate law must be
able to be devised from the rate
determinant stop that means I should be
able to determine the rate law if I know
the mechanism but if I know the correct
mechanism of my overall reaction correct
mechanism if I can't determine the rate
law from the given rate mechanism that
means that mechanism is not valid it's
no good for that reaction so what I do I
try to propose
another mechanism actually so that it
will give me the rate law or give me the
correct rate law so that is it says the
rate law must be able to be devised from
the rate determining step in other words
the rate though and the mechanism has to
be consistent right that's what it means
has to be consistent now the
stochiometry must be obtained when all
steps are added up this simply means if
I simply add these three steps side by
side I should get this that's what mr.
geometry means right
so geometry must be obtained at all
steps added up each step must balance
just like chemical equation they have to
be balanced like any equation all
intermediates are made and used up again
otherwise I can't really get the overall
reaction if all the intermediates formed
and not used up during the mechanism
during the chemical reaction right so
that's what it says any catalyze is used
and regenerated if there's a catalyzed
remember the catalyze does not is not
included in the chemical reaction but it
changes the speed of the reaction it is
not consumed so we usually don't have
the if the catalyzed enters there it
must be regenerated again used and
regenerated means that the
concentrations of the catalyze will not
should not change if I have the
catalyzed as a matter of fact okay here
is again the mechanism overall reaction
I have them overall reaction here like
this and rate law is given as K times
no.2 to the square like this now this is
experimentally given rate law remember
it is experimentally determined this is
my overall reaction by looking at this
actually I can easily understand that
this is not a one-step elementary
reaction for clear now it may be
possible some reactions will be one-step
elementary reactions in that case in
fact the rate law if this had been
Elementary one-step reaction the rate
law would be K is equal to n o - not one
here not to n times the concentrations
of C all right but it is not what does
that mean that means that there this
reaction have has a mechanism that means
series of reaction
now the again let's see how we approach
it if the first step is the rate
determining step the coefficients on the
reactant side are the same as order in
the rate law as I told you we just
discussed that so the first step of the
mechanism begins may begin like this it
says right you know let's write it down
here
so you know - usually it's good to at
least put the state of the reactants and
the products in the overall reaction the
main reaction because it's a gaseous so
we will know that its reaction is taking
place in gaseous state and see all gas
and going to a no gas and plus carbon
dioxide gas the rate law experimentally
determined rate law is given as K times
n all - to the square that means
second-order reaction with respect to
the concentrations of nitrogen dioxide
as a matter of fact so it says that
since there are two you know no2 to the
to the first that maybe like no.2 and
no2 colliding with each other now this
is the mechanism again I'm trying to
propose or find out a correct mechanism
which will fit to this rate law
mechanism and the first step we can say
no.2 plus no.2 going to let's say n all
plus no.3 this may be possible again
here I don't put the states remember
they're all in the gaseous state right
so nope
vier alright so no.2 no.2 going to n o
and no.3 now we do not see the no.3 in
the stochiometry so it is an
intermediate remember which needs to be
used in a faster next step because when
I add up the one step elementary
reactions of the mechanisms right the
English they must give me the overall
reaction so in all three which is an
intermediate should be used up during
the chemical reaction so in the second
step the no.2 no.3 I guess plus carbon
monoxide reacts to give you no.2 plus
carbon dioxide like this and that is B
the next step actually now this is this
is consistent I should say with this
overall reaction when we add them up
side by side we get this now if this is
the correct mechanism for this reaction
and this is a rate law which should be
the slowest step the first one this is
the slow we just put it slow like this
and then this one here again to show
that it's faster than it's not the
slowest step we just put it fast faster
than this that's what it means like this
so now from this I can sense the slow
step determines the rate of the overall
reaction rate of the overall reaction
will be equal to the rate of the slow
step so the rate would be equal to K
times that's pole that k1 and k2 here
like this those are two different
reactions rate is equal to k1 times the
concentrations of no.2 raised to the
power 2 and that becomes the rate law
now here k1 here is equal to K right K
is equal to k1 which means the rate
constant of the first step in the
mechanism rate constant of the first
step in the mechanisms so this is like
you know very simple but it simply says
that this mechanism is a plausible
mechanism right possible mechanism and
it gives me this rate law by using these
but do I know I mean can I be sure about
that
no not really
only if there is some way of determining
the presence of no.3 nitrogen trioxide
during the chemical reaction then I was
there huh that's it you know
I propose that the no.3 nitrogen
trioxide will be an intermediate and if
I determine that that would be a very
good evidence that this is a correct
mechanism but if I can't determine that
I can't be sure that this is but at
least that mechanism fits to the rate
law or it is at the or let's put it this
way consistent with the rate law I say
that okay until somebody finds better
mechanism and shows the evidence that
it's better than this it would be it so
what you have to remember that the
mechanism is sort of a theoretical work
right theoretical you just try to fit it
to the rate though as a matter of fact
but if you sure that the mechanism is
correct than by using that mechanism you
can derive the rate law for your overall
reaction okay
now since the first step is the slow
step says I guess still say that gives
the rate law if you add up all
individual steps - you get the
stochiometry I mean checking that these
are this is a plausible mechanism each
step balance this is a plausible
mechanism I mean that's all it says
plausible right I mean can be we can't
be sure now
this is another reaction and equation
for the reaction is you see nitrous
oxide nitrogen monoxide with the bromine
to form the nitrosyl bromide
this is nitrosyl bromide that compound
the rate law for this reaction is found
to be rate is equal to K times no.2
natsound so on
now because the term molecular process
are rare this rate low suggests a
multi-step mechanism what does this
means now actually what this simply
means try to tell you this by looking at
this rate law and this equation here you
can say oh that's elementary one-step
reaction right here right because the in
the rate law the exponent of the
constant concentrations the same as the
coefficients of those in balanced
equations 2 1 and so on
but here if this is elementary one-step
reaction the what will be the
molecularity of that reaction
termolecular means that requires
collusions of 3 molecules at the same
point at the same time and which is very
rare and very seldom takes place so it
says probably it is not one-step
reaction then there must be a mechanism
right mechanism of that and let's see
what it says here are the mechanisms for
the first step is 1/n Oh nitrogen
monoxide reacts with bromine nitrosyl
broma bromide actually bromine and then
in the second step this reacts with
another no2 get this like this now how
can we derive the rate law from that
reaction
okay so the reaction was no.2 nitrogen
monoxide plus gr - given - okay - nobr
nitrosyl bromide like this alright now
we go to the mechanism and the mechanism
the first step and the second step now
the first step here is the fast
equilibrium step so I should write down
here in front of this fast equilibrium
okay now this is the slow step like this
let's say if that is the slow step then
I will write down the rate of the
reaction would be equal to the rate of
the slow step the rate of that slow step
will be equals to K times let's call
that K to call that k2 k2 times a no
br-2 and concentrations of n Oh like
this okay but this can't be the rate law
for the overall reaction why because it
contains intermediate that means it
continues the species that does not
appear in the main equations so I should
be able to replace it with the
concentrations or the species of the
concentrations that appear in the
chemical reactions now to replace that I
take that first step fast step an
equilibrium step that means that the
reaction for rate let me write down the
rate forward reaction that means this
way since it's a what's elementary
reaction would be equal to let's say k1
times the concentrations of n Oh times
the concentration of br-2 okay now this
is the forward reaction
now how about the reverse reaction the
reverse reaction let's call that leaders
reaction that means consider the
reaction that way that will be equal to
K minus 1 this is the reverse rate
constant that means reaction for the
reverse rate constant and and times the
concentrations of nobr - you know dr -
like this
alright equilibrium means as we will see
in the next chapter the rate forward
becomes great reverse what does that
means that means that K 1 times and O
and P R 2 is equals to K reverse that's
this one here times the N o BR 2 like
this alright now what I'm looking for
here I'm looking for this one to replace
that because it is an intermediate now I
get it from there and the concentration
is n o gr 2 would be equals to K 1
divided by K minus 1 and up here I have
the concentrations of you know and the
concentrations of ER - right this is it
ok so I get it put it there the rate of
the reaction overall reaction will be
equal to K 2 times K 1 over K minus 1
and I have here and all concentrations
of molar concentrations of n over here
but there's another n o there that means
NO x n o square and then I also have BR
two concentrations of BR 2 like this so
now this is
consistent with my rate law how the K in
my rate low here I call that where I put
that anyhow the rate constant this is
all the overall rate constants so I can
simply write it down as follow K times
the concentrations of n o nitrogen
monoxide raised to the power 2 and the
bromine concentrations of bromine this
is my rate law you see again how I
derive the rate law from the mechanism
of a reaction so if I know the mechanism
I can derive the rate law vice versa if
I know the rate law I can propose a
mechanism so in this case the rate law
for that reaction is actually known so
it is this one here and I proposed the
mechanism from two steps these two steps
and assuming that this is no yes two
steps one is as fast equilibrium and the
other one where I guess I have right
down here slow step and derive the rate
law and the rate law is the same form as
this this K is equals to the K 2 K 1 and
K minus 1 again K 2 K 1 and K minus 1
have only meaning if I have the
mechanism elementary one-step reactions
the elementary one step reaction here
clear yes
all right the rate the overall reaction
depends upon the please are all the same
things I guess we said slow step again
the rate loaf for the step would be
equals to this we have done that but we
just replace that and rate forward rate
reverse and finally we get this rate law
which is the experimentally determine
rate law is the same things again all
right now catalyzed catalyzed are very
very important in chemical reactions as
we described the definer catalyzed
catalyzes the substances which is not
consumed during the chemical reaction
but it increases the rate of a chemical
reaction actually increases the speed of
a chemical reactions right so if there
is a very slow reaction if you would
like to increase the rate of that
reaction speed of that reaction we try
to find out a catalyzed what does the
catalyzed do the catalyzed changes the
mechanism of the chemical reaction
chemical reaction so when the mechanism
of the chemical reaction changes that
means that the slowest step which
determines the rate of the overall
reaction also changes and the slow step
would be catalyzed mechanism in fact is
always higher or faster than the slowest
step without the catalyzed mechanism so
this way in fact the presence of
catalyzed increases the rate of the
reaction increases the speed of the
reactions mfa so recap in this case here
is the example the decomposition of
hydrogen peroxide to the water and
oxygen without the catalyzed has this
reaction coordinate this is the reaction
coordinate again you see again this is
the activation energy right for that
path without the catalyze the activation
energy here from here to there right now
remember
the activation energy the higher the
activation energy the lower will be the
rate constant from the Arrhenius
equation remember there are any
equations relate those the higher the
activation energy lower the rate
constant so in the second reaction we
add hydrogen bromide and the hydrogen
bromide in active solutions actually
hydrobromic acid you can look it that
way
increases the rate of the decomposition
reactions of peroxide how change in the
mechanism now with with the presence of
the catalyze the mechanism is this what
is the activation energy here can you
tell me the activation energy for
overall reaction okay this one here or
this one bigger one why because if these
two maximum corresponds to two step
reactions elementary reactions the one
which has higher activation energy will
be slower right higher activation means
it's slower the lower the activation
energy the faster the reaction is rate
constant because the rate constant will
be bigger and again when the presence of
the catalyzed we have a faster reaction
faster decompositions of hydrogen
peroxide now we can classify to catalyze
as homogeneous catalyzed heterogeneous
catalyzed enzymes homogeneous catalyzed
means that the catalyzed and the
reactants are all in the same phase
right for example in solutions if you
have the ends the cattle are dissolved
in solutions and my reactants are also
dissolved in solution in that case it
will be a homogeneous catalyzed
heterogeneous catalyzed usually
catalyzed phase is different than the
reactant phase how
for example the catalyzed are solid but
the reactants may be gas so that will be
heterogeneous catalysis that erosion is
catalyzed now enzymes as I said you
there big molecules proteins actually
that catalyzes the
biological reactions in the living
organisms right now let's look at these
quickly in fact the reactions and the
catalyze are in the same phase this is
the emergence catalysis many times
reactions and catalyze are dissolved in
the same solid as seen below like fizz
now when we have heterogeneous catalyzed
reaction usually take place on the
surface of the catalyst and the catalyst
is solid then the reactants are gas and
all the reactants are adsorbed on the
catalyzed surface and the reaction is
taking place at the surface actually in
that case it's the heterogeneous
catalyzed one example is the catalyzed
that we use in our exhaust pipes
automobile exhaust pipes and they
convert the nitrous oxides to nitrogen
and carbon monoxide to carbon dioxide
that comes out from the exhaust pipe of
automobiles right
and those catalyzed are solids and
they're usually platinum or iridium
podium containing solid catalyzed and
the gas which comes out from the engine
goes over those solid catalyzed and on
the solid catalyzed in fact the carbon
monoxide is converted to carbon dioxide
nitrous oxide nitrogen dioxide are all
converted to nitrogen so this way the
gas that comes out of the exhaust pipe
becomes clean right that is the
heterogeneous catalyze the catalyzed we
have in our exhaust pipes in automobile
exhaust exhaust pipes now enzymes again
as I said biological catalyze they have
a region where the reactants attach this
is important again they are very
specific
they only increases the rate of certain
reactions and to do that they have to
really have a very special part of the
catalyzed where the reactants depending
on the shape of the reactants fits to
that place in the catalyzed and then
proper reaction that take place
proper products in fact are also taking
place that region is called the active
side you see again where the biological
molecules or reactant molecules attach
there it's called the active side the
reactants are referred to as substrates
so the substrates goes binds to the
active sides and the chemical reactions
are taking place now let's look at this
is again a simple approach to explain
how the substrates and the enzyme active
sites fit to each other right this is
for example like the shape of a molecule
substrate molecule what does that mean
that means that reactant molecule right
biological reactions and it comes and
fit to this place like this then the
reaction takes place they'd simply
divides it into whatever the reaction is
so you see again the enzyme or the
catalyze are very specifically designed
to accept that substrate there so that
the proper reaction biological reaction
will take place and the proper product
will forms otherwise there would be
nothing actually okay now let's go to
the
chapter 15
these topics chemical kinetics or
chemical equilibrium are easier to
understand actually than the others
now chemical equilibrium what is its
chemical equilibrium actually all
chemical reactions are reversible now
what does that means for example a plus
B going to C plus D just an example if
this reaction is reversible that means
that when you start with a and B you get
C and D now if you start with C and D
only you will get a and B so these kind
of reactions are called
reversible reactions right so this is
called reversible reaction reversible
reaction now what does this means that
means that at the beginning for example
you can start from here at the beginning
the rate of the forward reaction will be
simply high and the rate of the reverse
reaction will be
what suppose that I have started with
this only in a closed container the rate
of the forward reaction will be certain
values the rate of the reverse reaction
will be zero right why because at the
beginning I will have no C and D I
started with only a and B but as the
time goes on because in creations of the
C and D will increase and as the
concentration increases what does that
means that
that means that the rate of the reverse
reaction that way will start increasing
right so now at the point let's call
that K forward k reverse at the point
where the rate of the forward reaction
equals to the rate of reverse reaction
what does that means that means that the
you can take any species here in
reactant or the product the amount of
that species formed in one unit time is
equal to the amount the same amount of
species is consumed in the same time
right that means again that the
concentration when we reach the rate
forward equal to the rate reverse let's
write it down k forward let's a and B
like this equals to K reverse and C and
D here like this now when these are
equal that means that we reach to the
equilibrium why from that point on the
concentrations of any species a B or C
or D will never change does that mean
the reaction stopped no the reaction
continued reactions in fact forward and
reverse reaction continues but with the
same rate that means that the amount of
any species produced and consumed per
unit time is the same that means the
concentrations are not changing another
way to describe that we say the
equilibrium is dynamic equilibrium
dynamic equilibrium so this is the
dynamic equilibrium that means reactions
did not stop but the rates are equal in
both directions dynamic equilibrium and
actually almost all reactions are
reversible and and almost all reactions
have equilibrium constant which is
actually this the if I simply rearrange
that K reverse divided by K forward yes
no just the opposite here if I take this
over here that over there
I will be again as follow K forward
divided by K reverse will be equals to
concentrations of C times the
concentrations of D and divided by
concentrations of a and concentrations
of B so this is simply as long as the
temperature is constant KF and kr will
be the same right they will not change
what does that mean that means the ratio
which is I described here as K sub C
that means the equilibrium constant in
terms of concentration molar
concentrations will never change so
every reaction will have an equilibrium
constant actually every reaction is
reversible but sometimes we just say I
reacted a and B and I get product C and
D it's complete reaction no the reaction
is never complete but if the when we
wish to the equilibrium if the
concentrations of the reactants are much
much smaller than the concentrations of
the product what does that means that
means this is very very small this is
very high what does that means KC will
be very large
going to infinite for example
right if 99.99% of the reactants are
converted to the products KC would be
very large so for those reaction we say
the reaction is taking place in one
direction it's a complete reaction but
it is not on the other side if the
equilibrium constant is very very small
10 to the minus 20 10 to the minus 30
what does that means that means very
very small amount of a and B reactants
is converted to the product very small
let's say 10 to the minus 11 percent
whatever now those reaction when we deal
with reactions like that we just say the
reaction is not taking place right
because the amount of the product that
formed when we reach to the equilibrium
is so small even sometimes we can't
collect it we cannot even detect it
right but we will always remember
reactions are reversible every reactions
has an equilibrium constant but for
those reactions that have very very
large equilibrium constant we just say
that the reaction is taking place in one
direction for those reactions which have
a very very small equilibrium constant
we say the reaction is not taking place
right but in reality it's not that way I
mean okay let's look at the example here
the same things again chemical Goutham
occurs when a reaction and its reverse
reaction proceeds at the same rate as we
said over here in the figure above the
equilibrium is finally reach in the
third picture this is the decompositions
of n2o4 - 2 no2 molecule and actually
the colors are different n2o4 molecule
colorless gas and the others are brown
color and you can see that now this is
again shows how the concentration of the
reactants and the products changes as
the reaction goes on and also the rate
of the reaction you see again at the
beginning for example if we started with
n 2 o 4
come on come on it's shame I PR be like
Oh true shake it up category since I put
the Black Madonna
maurella meatballs might not about me
I'm not poms around
luckily someone bigger than okay so the
concentration at the beginning if you
start with n 2 O 4 just like same as
here highest here and decreases at the
time goes on concentrations of no.2
increases as time goes on the rate
forward and the rate reverse rate
forward decreases as the rate reverse
increases they will simply cross in one
point that is where we have the
equilibrium right that's where we have
the equilibrium and we can write down
the equilibrium concentration
equilibrium concentrations and
equilibrium expressions I should
probably say and finally we get the
constant and that is called the
equilibrium constant okay I guess we
will stop here