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Lecture 23: Lindemann-Hinshelwood Part I

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OK are you guys I'm not really have a good weekend no
but wire you guys wearing ties you have to why you have to do the pledges quadriplegic OK when you look fantastic all right so we are up when I say we I mean Stephen Indian market a grating the mid-term I think it's about done and they're
going to post a last later on today there a key that keeps getting updated the current version of
that's on the Web already so it's been updated a few times so Our rumor has it that you did better than on mid-term 1 but I haven't seen the results yet so don't put too much weight on that and I'm hopeful but that so
yes yes he's posted scores are going to be posted today exams are going to be returned as PDF the a rapid return and I think you know that that's a
misnomer very rapid that
dictates for 5 days or so
OK but that will get a return as fast as we can I'm opposed to call my doing scored a day-old drop your loss to quizzes and let you know how
you doing in the class stability that score see what category you have gone into the final exam OK there's 1 quiz left Valdez of walkways Friday and it's really going to be on the stuff that we talk about today and also the stuff we talked about way a last Monday steady state approximation all the kinetic stuff that we've talked about really is covered by the squares to carry Freddie had 1 quiz that had kinetic stuff on
OK so that the very last quit as some of you probably have done well enough so you don't even need to take it you can drop to quizzes so if you've got 5 really
good ones already take the 1st 15 minutes off tomorrow from them in
the final exam is going to be like this it's comprehensive but it's going to emphasize the kinetic stuff that we're doing here at the end of the quarter because the kinetics that hasn't been on a midterm exam yet so half of the final exam is going to be kinetics 25 per cent thermal 25 % Stepanek but I'll break
it down for you problem by problems arise next week so don't worry about that but I just wanted to say that the stuff that we're
talking about now is going to be worth sort of 100 points on the final exams it's rather important
OK so we're going to review the steady-state approximation because I know that's not foremost in your minds anymore you've been studying for midterm too was all about thermal where do an example talk about the Lindeman Hinshelwood mechanism we started to talk about this last
Monday but I'm sure that that sort of a vague memory for you at this point we'll go back and look at it carefully so the whole
idea in the steady state approximation is that we
want a simplified bigger mathematical expressions for consecutive reactions rate reactions where here is a
reaction here's another reaction that follows that 1 there could be another 1 that follows this 1 right and in general the reactions are to be more complicated than their social Yoon example later
on all right but the basic
idea is that the mechanism for sequential reactions like this the mathematical expressions for the integrated realize gets exponentially more
mathematically complex as the mechanism gets they don't price increase in linearly with the size of the mechanism they increase exponentially so we need a mathematical tool that allows us to simplify what these expressions look like and that tool is
the steady state approximation we're going to use this simplified
several different types of reaction mechanisms including the Lindeman Hinshelwood mechanism and enzyme kinetic expressions right which were about to start talking about on Friday so the basic idea
is that we're going to set the time rate of change for all intermediates to 0 1 Watson intermediate while intermediate is something that shows up in a sequential reaction but it is not a product and it's not a reactive rights on this case the intermediate is obviously been organist at the time rate of change in beta 0 In solve the simplified kinetic expressions that result from
making this simplification
so for example we we've set the time rate of change of people the 0 but what is that while there's a rate of formation for B because B formed at a rate equal to K 1 time saying and there's a rate of consumption of the because turned into C at a rate times of K 2 times being so that there could be a minus K 2 times be an plus K 1 times a right and that difference as the equals 0 so if that's the case then
obviously that has to equal that that's were showing right here and so I commence offer beat the should say B steady state should the city-state subscript here the steady-state concentration of musical the K 1 or K 2 times that now 1 of the things that we
showed where the last a week ago Monday is that the steady
approximation is really only gonna work when K-1 is much much less than K 2 like
that if you plot with the concentration of the intermediate is doing you can convince yourself that only in this
limit is the concentration
of the intermediate going be Kwan's I constant right even in this limit it's not perfectly constant but it's Kwan's I constant bright and so
consequently if we make a here were saying the time rate of change of 0 right in other words bees not
changing its concentration is not changing at all as a function of time that's what we're assuming right in order for this
expression to approximate this expression K 1 has to be much much less than K 2
can everyone see that K 1 has to be he was advocate to has approximating 0
OK in order for this expression makes sense keep in mind that day is always going to go from its initial concentration to 0
according to this mechanism here great a is going to change a lot right so in order for me
not to change the
1 over K 2 was gotta be small very small robot
with me on that all right that's the assumption that were always making somewhere in the steady-state approximations
OK so here's the case here's the mechanism here the rigorous equations that describe the concentrations of a years with having a just like I said it starts up some initial concentration and then goes to 0 years see building up and use the concentration of the intermediate it's cause like constant y because K 1 potatoes just .
0 2 it's tiny this is
the land where the steady-state approximations can work pretty darn well for us notice all the simplified equations that results so we said the studies concentration of these key 1 overcame 2 times a day and so we can then plug that into the rate at which sees produced so here's the rate of the reaction In
terms of it's K 2 times
the R and I can just plug this expression from being bow and the cages are cancel and so the rate of this reaction just to be Bullock a one-time sale but
the and we know what that integrated rate expression is for a price just a first-order reaction with Ray constant K 1 and this is what it's equal to my work it out and so if I want to know what the rate at which the product is produced but obviously this is a decaying exponential it's
doing that it's doing
that right that's the integrated rail offer a and now and then integrate that To get their concentration C is a function of time and this is what that Interbrew gives me right here OK so c is a function of time is going to be given by that equation we get really simple find integrated rate last from the state approximation compared to the rigorous expression and the thing to keep in mind is this is
the best case scenario this is the simplest possible sequential reaction mechanism write to first-order reactions in sequence where there's only 1 reactant 1 product in each right and I think you can see there's
already a pretty big difference in complexity between these equations and these equations especially look at seat right that mathematical expression is a lot simpler than this 1 and as we make this
mechanism more and more complex these
equations blow up
exponentially OK
that's why we need a steady state approximation how well this work here K 1 over use tiny again look what works the -dash line is a steady state approximation is solid line is the perfect for the rigorous
equations for the concentration look at
a Likud CDC's what we really care about rights he's telling us what the rate of the reaction is steady state the approximations also predicting be right and you can see being is "quotation mark steady state concentration
doesn't change that much is changing but it doesn't change that much if I make
a one-over K 2 bigger I think should get worse you can see sword and see that it's getting worse but if I make a bigger yet now 2 . 9 you can't really see what's going on in I blow this out but when I blow up you can see that it's not doing a very good job of predicting be anymore but here's what the steady state approximation is doing steady-state approximation says b some fraction of a and a is changing a lot and so be is really not doing what the steady-state approximation is assuming now look at the difference between see here's the steady-state approximation of actual insurgents starting
to break down we expected to break down here it's not surprising that it does that and if we make is bigger yet we get a complete train wreck here's the steady approximation for this is the best 1 hears the solid line by big difference between the steady-state approximation of what the
concentration of C actually years OK so we wanna keep in mind the steady-state approximation doesn't always work right this is even
worse this is even worse than that that's the steady-state approximation that's what's actually happening no sorry
that idea to use OK so this
everyone understands steady-state approximations for students right here is a real world example for the state approximation here's a reaction
mechanism right 3 reaction if I rolled down the
actual kinetic rate expressions for this mechanism it would take 3 screens right it's a nightmare of enormous proportions the other let's see if
we can use the steady state approximation to simplify what's going on here 1st of all what we gotta do we have to find intermediates if
there are no intermediates there's no point using the steady-state approximation but the steady-state approximation only assumes that the time rate of change of intermediates is 0 identified in
hysterectomy mutual reproach presence of salt look at this guy right here in case he's produced by this reaction consumed by this reaction he does not appear as a product he is by definition an intermediate right what about this
guy that's another intermediate so
there's 2 intermediates in this case wait a 2nd there's
bromine following showing up here in is bromine and intermediate no b is a
catalyst right it's consumed that drives the reaction forward but is regenerated as a product without the
this reaction without the AA-minus this reaction grinds to a halt with that the reaction occurs at whatever rate is characteristic of the reaction alright but be
AA-minus is not consumed by the reaction because it's regenerated as a product even tho it's consumed by the 2nd step of the reaction so be AA-minus
is a catalyst you have to be able to recognize that it's not an intermediate and you can't apply the steady approximation to its concentration OK
what about their minds and altogether since the get itself applied city-state approximation following mechanism the weight of the formation of the product this is the product we care about right here by the way right its right of formation Schembechler K 3 times the concentration of this stuff times the concentration of this stuff everyone
agree with that OK so to
apply the steady-state approximation of the 1st thing we do is we write down this way right here we rate of the reaction is the way to the last step then we look at these 2 concentrations we ask ourselves Is that Intermediate Court is at an intermediate that's not an enemy that's cost or either the species intermediates yeah yes that guy we agree that guy it is an intermediate barriers and areas so he's getting generated by this step and consumed by the step so is definitely an intermediate let's apply the steady to its concentration steady-state approximation To do that we set the time change of the of that species 2 0 then rewrite a rate expression for but we look at the
mechanism here the mechanism for the reactions and we ask ourselves How is 0 in the are generated worse on the hour on beer is generated from this reaction right here so at that age 2 and 0 2 that concentration tends be AA-minus with a recurrence of take K 2 is generating all and the ah boat that's what this term is right here the rate at which
this guy it's getting generated by what by this step right here what it What's this guy this is the rate at which the on the odds getting consumed by this reaction right here OK so that rate is the steady the concentration of on the yacht comes a concentration of this stuff times case 3 so a minus sign in front of this guy and a plus sign in front of the sky that's the generation rate that's the consumption rate everyone see that OK we're
going to said that and that have to be equal 1 another if the steady-state approximation is valid OK so they can understand all for all and the artist and the concentration of going beyond our from this expression right here all right I can move the sky over the left hand side abiding by K 3 and the concentration of this stuff at the kinetic
expression for my steady concentration of all and the and
then I just played better and we're ready said the rate of the reaction is equal to this expression right here I can plug that into this expression and I've got a steady state approximation expression for the rate of the reaction so I just took that whole thing they're stuck in here and so not got K 3 that's the case we at the time this whole mess that's right here times the concentration of our product that's this bird that's a steady
state approximation integrated rate different or a long sought now
I'm looking at this and I'm asking myself can I simplified
further we will apply the steady approximation once looking
at this guy Are there any intermediates among these species here because could canceled that this sank all right is beyond intermediate know we agreed it was a
catalyst this age 2
and all last an intermediate all accounts with How could who focus now are either 1 of these 2 guys intermediates this guy could be right where is he he's right there and he's right there so he has an intermediate
so we can apply the steady-state approximation to his concentration to we're ready applied it to this guy now
we can apply it to this guy but the OK good so let's ask again are there any in the U.S. yes yeah so we said it's time rate of change to 0 this is the rate at which it's generated view there are 2 processors that consume it let's look back and OK it's generated by the reaction of protons with some NH 0 2 With the rate constant K 1 that what I did when sorry protons NH 0 2 with 3 constantly 1 this is the rate at which this stuff is generated and then there's true prophecies that remove it From the system White 1 of them is the back reaction here right the rate of that is just given by age 2 N O 2 plus tends to cater the minus 1 that's this guy right here and the other 1 Is this forward reaction here H 2 N O 2 plus reacting with minus the rate constant K 2 and that's this guy right here OK so the user consumption rates the services the generation rate for this species right
and then will reduce redness all for this guy right end if you do that this is the expression that you get and then what we do is we like this In 2 part original steady-state expression that we got after replied the steady-state
approximation the 1st time now we've applied at the 2nd time from point this mess
In for species here OK there it is and if there's some consolation no that's it
that's our steady state great loss it's a differential rate
right now there's going to be too limiting experimentally observed weight loss looking at the sky but it's possible
that this reaction will appear to have 2 different weight loss associated with that
because Of this denominator here all right if this term dominates organism 1 thing that this term dominates relative to their supporters see something else
let's think about that for a 2nd consider 1st the case where this guy K 2 times AA-minus is much larger than cater minus 1 all right if that's true then we can neglect cater minus 1 in this denominator because it there
an addition here part of 1 term is much much larger than the other then this will I have an
insignificant effect on the total size of the denominator so what we're looking for are edition
operations within the rate loss but there's 1 right
there that tells us that there are it's there is the potential that too distinguishable rate behaviors right if K 2 br is much much bigger than cater minus 1 then this guy simplifies to VS but we just leave out the key to the minus 1 Take-Two's canceled the beyond minus cancels and we end up with a rate but that's the right right there equal the K 1 times the proton concentration at times NHL that's the 1st thing
that can happen we would predict that this would be observed for example at high ball concentrations of
amicable my concentration higher and higher and higher at some point the reaction should stop depending on bromide should see any dependence at all anymore and that limit this
weight loss should apply to the reactions the other possibility Is that K to is much much less than the the minus 1 if that's true then I leave this guy out and I only keep became the minus 1 OK and in this limit the reaction does depend on B AA-minus and I get an effective rate constant that bundles together these 3 rate constants right here in in this case affected would be K 1 time state too divided by Kato
minus 1 everyone see that so I would expect to see a
transition from this behavior at low BER concentrations To this behavior that Heidi concentrations we should see the
reaction rate stop depending on B AA-minus
right in the land where investors Ray law applies Muncie that this will always happen when there is an additional operations somewhere in your weight loss there
will always be the possibility for limiting behaviors OK now we're
looking at different applications Of the steady-state approximation we're
going to talk about the mechanism of this reaction right here you molecular reaction of some
species :colon now if you look at this reaction right here in past the gives you products there is it's completely intuitive how this
reaction might occur the imagine for example India gas the species you've got a zooming around in the gas phase you've got be zooming around in the gas phase be collide generates and activated complex and then formed products right there's a
collision of a with the perform some activated complex that rates up to produce products by molecular reactions have an obvious mechanism in the gas phase a collided the generator that activate a complex of a and B and then this breaks up to give products that have some day-care appearance and the character
associated with but this
collision between a and B generates an energized transition state for now these are just words were
inmates were gonna talk about transition state theory later on it will be very important but right now it's just conceptual OK now the basic
idea is that this transition state here is located on this reaction coordinate here between some substrate which are these guys and product which are these guys it's located halfway in between is located in fact might at the peak of his energy level diagram but
once again we're gonna make this more qualitative later right now just it's just conception all
right that's held this happens it makes perfect sense to us they collides with
BT give products How does
that happen right there isn't any there's only a what kind of
reactions conform to this reaction mechanism what
kind of reactions do this they just react while this to remain tight you molecular reactions are you there I summarization spheres resveratrol remember that the stuff that's found in red wine it's so
delicious impedes the growth of cancer in your body and has all kinds of other benefits for you right you in it if you ingest kilograms of it right and decomposition
reactions is a it falls apart into a B
and C. right ages falls
apart at the decomposition reaction that's unique molecular right the reaction happens just to this route molecule by itself it falls apart this molecule here oscillates between 2 different chemical states because of isomerization around a double bond right there
right these are the 2 primary types of you know molecular reactions the how they
happen ,comma they occur although we understand what's going on we need a reaction mechanism in these 2 guys proposed 1 in the early 19 hundreds Linda men was an
English guy a physicist who did a lot of stuff for the British government during World War Two he was the science adviser to Winston Churchill and he was a really arrogant guy guys who most people hated but 1
of the things that he did is he worked out this reaction mechanism and then somebody who we didn't even know Hinshelwood now the
English guy came along and worked out of a mathematics and made a quantitative women cooked up the mechanism Hinshelwood refined its intial would ended up getting a Nobel Prize not only for this mean
Hinshelwood other stuff too but this is
1 of the things OK so for this reaction in America has so the Lindeman Hinshelwood mechanisms
basically postulates that this reaction occurs in 3 steps 1 a
collides with itself when that happens you generate activated a N ground state aid not activated this reaction can go back and that's what this is right here is activated a can collide again with another and so here's a colliding with a produce activated a activated a can collide with a again to become deactivated that's witnesses right here's a collision between activated a ground state a I get to ground state aid there's no more activated and finally activated a can fall apart and to be what can turn into
being this looks like a decomposition mechanism
phase decomposing needed to be OK so we can apply the steady state approximation to this guy the basic idea and the most important assumptions
Of the Lindeman insured mechanism and its assumption that turns out to be wrong all right the assumption is
that a star possesses an internal energy that exceed the activation In other words if there's a bond that
has the break In a deformed to be a star has enough energy within it to break that bond that turns out to be about assumptions because not all collision we are going to generate activated aid that has enough energy to break up necessarily right you can imagine there's going to be a wide distribution of collision energies because they can hit itself at a small angle and transfer very little energy to itself right that's 1 way in which they a stock and have a look a smaller amount of energy than it needs to react this is called the strong collision assumption of Lindeman insured theory theories the strong collision assumption is that these collisions that
occurred here right generated activated a that has enough internal energy to go on and do the reaction that you care about it decomposition on summarization as of now can we apply the steady approximation to this mechanism right here yes we can because there's an intermediate right there might be activated a is an
intermediate right here's a
reactants here's our products is formed and then consumed so only exist train ceiling within the mechanism that's the definition of an intermediate OK so what our rate expression is going to be like while the rate at which a stars formed here's the generation rate and these 2 processors that consume at the generation rate of course is that right 80 squared times K 1 and then how is a star consumed there's 2 things that consumer there's this reaction here a times a withering constantly minus 1 that's just the reverse of this of course and and a star can react with the rate constant K 2 right and so both of these 2 processes use up a stock and so they have a
minus sign in front of -minus
I'm here because a star used up by that guy minus sign here because a star's used up by that guy plus sign here because a star is generated by this guy OK
so you've got to be able to write these great expressions by looking at the mechanism then all
we do is set the sequel to 0 all of these other rate expressions for the other species a and B. all right and then we set the sky equal to 0 . that's what the steady-state
approximations receptors
sky-blue 0 and we solve for a star so if this is equal to 0 that I can set all the positive terms equal for all the negative terms right now and be more careful and I'm I'm putting this little steady-state subscript on me a star concentration because now
we're talking instead particularly about the steady-state concentration of a star and I can just offer a
star in this expression here to get it this guy right here OK and so once I've got that then I can just plug in a stock into that expression for the rate at which B is produced OK so I just plugged this hole in for a stock boom there it is all right and that's mine great
loss for the liniment insured mechanism what is the project
so look at this guy all right you see others of addition operation and the denominator that means there's
going to be too limiting behaviors depending on which 1 those 2 terms is bigger or smaller but that's the 1st thing to notice so
organ talk about those 2 limits if they've been in other words the minus 1 times a is much larger than K 2 right this would be in in the
limit of high pressures of at high pressure remember there's only 1 reacted in this thing you've got a vessel with a unit in the gas phase if you put a lot of pressure that's the limit
that you're in this guy right here this guy is much much bigger than K 2 in that limit and I that 1 of these aides canceled with 1 of these days I'm left with 1 A in the numerator but no ways in the denominator right and if you look at this point it's just I can cluster together these rate constants in 2 together effective rate constant times OK so the rate of this reaction if it conforms
to the 1 the 1 hand Lindenwood Lindeman Henschel would mechanism is developed like
this it's going to be apparently first-order well that's what you would
expect is looking at the mechanism of the yuan molecular reaction you'd expect it to be naively you expected to be first-order and it is at high
a pressure no what is this mechanism the mechanistic late well what is it mean for came eyes 180 much much bigger than K 2 it means that this reaction is fast compared to this reaction here all right the rate of deactivation at high pressures Nova to create
an excited a bicycle collides right away with a ground state aid to the deactivated
right in that limit you did a reaction that is first-order an now that's true for large A. This is what we what happens the various small what happens it's
small pressures low pressures for well
obviously this guy's going to be much smaller than that so I can get rid of him where is he under steady to In the denominator I don't have any a in the denominator cancels out still 80 squared so this K to hear cancels Becky to write their summit to K 1 0 times a squared it's going to act like a 2nd order reaction but at low pressures of similar a 2nd reaction this but it's just
a falling apart into products or summarizing right but that reactions can act like a 2nd order reactions where but what is new mechanistic
Lee Welch yeah that is much less than that right that means that this reaction is fast this reaction this
slow the rate of deactivation at low pressures right the rate at which if you form excited ate it doesn't give deactivated there it's not very likely to get deactivated B there aren't very many collisions to deactivate it is more likely to fall
apart once you form this guy it's more likely to undergo this reaction and it is the deactivation reaction at low pressures that's what it means so
this plot should make sense to us but here's what the Letterman Hinshelwood mechanism predicts let's look at this and make sure that we understand this because this is really everything in a nutshell here right
what's on its axis this is the pressure of 8 a lot pressure right Watson is that this is the lot of the rate constant the
effective first-order rate but what I mean by
that but if I pretend that the reaction has this by then this way constant here if I applied to this case right here it's going to be 2 times K-1 times a day is going to be politically effective get everyone see
that In other words if I
assume that the reaction always has this form the reactionary always has this form that I could write an expression
for K factor I became effective in this case is going to be able to 2 times the 1 times because that's what it would take that I would have to do it came what the effect would have to be equal to that in order for the effective finds give me this OK so there Is there a lot of the rate constant this is long the pressure of a and what this plot says is at low pressures abate the effective rate constants
depends on a it goes up as a as a as a function of the pressure of a entail it gets to a
point where there's no dependents of the effective rate constant on a at all and we just arrived these 2 cases right here the effective rate constant is 2 times 1 times a it does depend on aid goes up linearly with a that's officials right here right and this rate constant here doesn't depend on any and all it's
completely a independent now of course the rate of the reaction does depend on a up here because the rate is K 1 times a right but if I take the rate
constant right and I plot I'm only plotting the rate constant here right and uploading
the toll rate right here the rate constant becomes constant it's equal to this collection of rate constants from those individual steps here the rate constant depends on it OK so this is
classical behavior that's modeled by the Lindeman Henschel would mechanism at low pressure CZ the reaction rate
constant the effective rate constant depends on a linearly and a high pressures it does not
right here the reaction is acting up like a first-order reaction here the reaction is acting like a second-order reaction the London insured
mechanism explains that doesn't get it exactly right but it comes close OK there will get
that no and then there's a show would
break here's the expression that we'd arrived earlier let's recast this equation
right if I assume the rate looks like this in other words I defining an effective first-order rate constant
right just like I did for this
exactly the same way I did for this plot right now I can then say that the effective if this is true and this is true then effect is going to be cold all this nonsense including 1 of these days resale that's a square I'm a take 1 of those days the other 1 is right here so I take this if this is the effective and multiplied by a I get back OK so
this is my effective Hinshelwood first-order rate constant and I can just take the reciprocal of this so I can dig 1 overcame effective I move these 2 guys into the numerator there they are but I can't put this into 2 terms I could put the case minus 1 here and if I do that the EIS canceling I can put the catered to here and if I do that the kid to cancel right so this is my simplified expression if I take the reciprocal of this guy
that so this now
is my Lindeman potential would the equation if
you will what it says is a
pipeline at 1 overcame K effective versus 1 over a year the
concentration of a or the pressure of a partial pressure of a I should get a
straight line right and plotting that 1st 1 overriding a straight line with the slope 1 over 2 times K 1 and a positive intercept
everyone see that so that's how you tell whether you're reaction is conforming to the Lindeman Henschel would mechanism you make
that plot what is it 1 overcame a purses won over a 1 over 10
a versus won over a this is 1 over the partial pressure right How well does it work Welch not that great I mean OK it's working pretty well at low pressure by this is 1 overpressure so it's working well here that's low
pressure right but the Letterman
into a theory predicts a rate that is too here's what we project like this is 1 over the rate aerated predicts a rate that is too low that high a This is high acres is 1 over it is
very confusing but this is high
a lull a higher rate low rates because it's all reciprocal to told totally reciprocal plot by itself the lantern-jawed enjoyed theory predicts a rebound is too that's by there's a
good reason for that right will talk about it later on but where we expected to work Is Alloway all right we expect a
positive intercept see them and we are looking for a linear behavior down here all right we expected to roll off here with a manager would theory has a well-known
defect that we're going to understand in detail later but for now I expect this guide
a roll off like that at high eh where low 1 over
McCain but so all of this
is based on our application of the steady
approximation to the Lindeman Hinshelwood mechanism right actually the Letterman show would mechanism uses the steady approximation whether we want to or not OK but it's deepened
confusing so that's what this is
that the we're going talk about
1 more case right so what what were we just talking about we were talking about you know molecular reaction mechanism all right how does so how are you the molecular reactions occur you got a molecule it's just falling apart right what's the mechanism for that it's this Lindeman would mechanism we can understand that mechanism in terms of the steady approximation derive equations what other reactions we are going to be useful to look at what the steady-state approximation it turns out that enzyme reactions on another case where that's true end that's
why the reactions of this former importance right reactions where a priest
equilibrium is established within the reaction mechanism
pre-Columbian 1 might talk about look at this they reacts with BT give some complex of a and B that's going to be the
enzyme substrate complex Amy can fall
apart to give AMD again right so a before they react the uniform complex AB complex they be falls apart to give
a and B separately again all right there could be an equilibrium that
involves this forward step and this reverse steps that's the Prix equilibrium then maybe can fall apart give products this looks like an insane
enzyme reaction doesn't right
substrate reacts with enzyme to form and subject complex and sent substrate complex falls apart to give the enzyme and substrate or a reaction occurs in generates product
but there's lots of other reaction mechanisms also adhere to the screen equilibrium model but so that's
like yeah this is the end this is
a problem we're supposed to plot the data and find out whether the reaction conforms to the Lindeman Hinshelwood mechanism right so
what is this this is about pressure of some of their ass
call it and this is the effective
rate constants that we're measuring further the the fact of first-order rate constant that were measuring for the reaction of this gas to form products To
be an isomerization reaction could be decomposition reaction some intermolecular reaction so the question is
does this data conformed to the Lindeman Hinshelwood mechanism Hardy attack well
you have to to make a plot of 1 overcame effective verses 1 over partial pressure and see if there's any linearity in that plot at low pressures remember and so you
take 1 over P. you take 1 overcame effective right to take each 1 of these guys take the reciprocal and now you plant them and here are these these data points are the plot does this look like Letterman Hinshelwood mechanistic behavior sort of I mean it's
really ugly and it's sort
of linear at low ebb right and has this deviation that were expecting are right we would hate to see this on a test because it's sort of
nebulous whether this is really conforming to the liniment Intuit mechanism or not it's not very linear right I should really concoct a data set that looks better than but that's what you
would do right the point is that you make this plot to take these reciprocal plot this 1st as this and you look for a linear behavior over here right you know getting
really good when your behavior in this case this is actually this is real data actually certain part of the reason 1 thing you'll learn if you ever have to do experiments is that experiments never or very rarely and here perfectly to the theory that you're trying to apply to them this is a case of that right now so
getting back to the Prix equilibrium we can use the steady-state approximation again
and we'll do that more on Friday for that Saturday about steady state approximation
Pigmentdispergierender Faktor
Stromschnelle
Körpergewicht
Enzymkinetik
Pigmentdispergierender Faktor
Hope <Diamant>
Computeranimation
Wassertropfen
Stromschnelle
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Genexpression
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Phthise
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Oktanzahl
Fluoxetin
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Genort
Sense
Reaktionsmechanismus
Fluoxetin
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Konzentrat
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Funktionelle Gruppe
Genexpression
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Kartoffelstärke
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Methadon
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Bukett <Wein>
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Gen
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Kochsalz
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Hydroxybuttersäure <gamma->
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RNS
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Konzentrat
Sterblichkeit
Druckausgleich
Computeranimation
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Mühle
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Vorlesung/Konferenz
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Reaktionsmechanismus
Molekül
Reaktionsmechanismus
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Homogenes System
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Enzym
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Enzym
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Cyclopropan
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Druckausgleich
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Computeranimation
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Vorlesung/Konferenz
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Single electron transfer
Reaktionsmechanismus
Vorlesung/Konferenz
Reaktionsmechanismus
Computeranimation
miRNS
Vorlesung/Konferenz
Reaktionsmechanismus

Metadaten

Formale Metadaten

Titel Lecture 23: Lindemann-Hinshelwood Part I
Serientitel Chemistry 131C: Thermodynamics and Chemical Dynamics
Teil 23
Anzahl der Teile 27
Autor Penner, Reginald
Lizenz CC-Namensnennung - Weitergabe unter gleichen Bedingungen 3.0 Unported:
Sie dürfen das Werk bzw. den Inhalt zu jedem legalen und nicht-kommerziellen Zweck nutzen, verändern und in unveränderter oder veränderter Form vervielfältigen, verbreiten und öffentlich zugänglich machen, sofern Sie den Namen des Autors/Rechteinhabers in der von ihm festgelegten Weise nennen und das Werk bzw. diesen Inhalt auch in veränderter Form nur unter den Bedingungen dieser Lizenz weitergeben.
DOI 10.5446/18961
Herausgeber University of California Irvine (UCI)
Erscheinungsjahr 2012
Sprache Englisch

Inhaltliche Metadaten

Fachgebiet Chemie
Abstract UCI Chem 131C Thermodynamics and Chemical Dynamics (Spring 2012) Lec 23. Thermodynamics and Chemical Dynamics -- Lindemann-Hinshelwood Part I -- Instructor: Reginald Penner, Ph.D. Description: In Chemistry 131C, students will study how to calculate macroscopic chemical properties of systems. This course will build on the microscopic understanding (Chemical Physics) to reinforce and expand your understanding of the basic thermo-chemistry concepts from General Chemistry (Physical Chemistry.) We then go on to study how chemical reaction rates are measured and calculated from molecular properties. Topics covered include: Energy, entropy, and the thermodynamic potentials; Chemical equilibrium; and Chemical kinetics. Index of Topics: 0:00:06 Lindermann-Hinshelwood 0:03:46 The Steady-State Approximation 0:21:26 Two Limiting Experimentally Observed Rate Laws 0:24:40 Elementary Reactions 0:31:10 The Strong Collision Assumption 0:38:04 The Kinetics of Pressure-Dependent Reactions 0:45:44 Reactions Where a Pre-Equilibrium is Established

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