Lecture 21. The Steady State Approximation.

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Lecture 21. The Steady State Approximation.
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UCI Chem 131C Thermodynamics and Chemical Dynamics (Spring 2012) Lec 21. Thermodynamics and Chemical Dynamics -- The Steady State Approximation -- 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: 0:03:36 Svante Arrhenius 0:05:59 Arrhenius Equation 0:10:25 Consecutive Reactions 0:20:23 Steady State Approximation 0:33:58 Irvine Langmuir and the Lightbulb 0:42:20 Lindeman-Hinshelwood Mechanism 0:48:18 Applying the Steady State Approximation
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OK so we posted the results request 6 this morning you guys did well the meeting was a little lower than last rate pretty high phase B
lower grades OK mid-term
is Friday and we're going to have 1 more quick as a week from Friday with 7 them were done were winding down what we'll do is wearing a review in in detail of what's going to be on the mid-term on Wednesday a
walking through a problem by problem in review that material for you what will not
be honored for share these 2 topics from Chapter 19 right at the end right polarization Fuller chemistry were not going to get to those 2 topics but I think everything
else in Chapter 19 will be included if where we're going to get to really get right up
to section 19 . 8 by the end of a lecture today OK so we're in the final stretch is just 7 lectures including this 1 laughter and we've just got 1 quiz laughter but
most of the points in this class are still to be decided to two-under .period finally we have 100 .period midterm it was 1 quiz left from this is the core these next 2 weeks would be good weeks to focus on doing well the class because it could make a huge difference to you me so far we've basically decided for most of you what your quiz Graves going to be for the quarter just go 1 quiz laughed a ride and since you can drop to he pretty much you come close to locking in your quiz Grady you can still affected with Chris 7 a little bit and is taking mid-term 1 right but would get 300 points and exams slept in the costs and so it's tempting to think that
were coasting into the summer vacation period but in reality now is the time to do well all right so going certainly do a good job of reviewing this mid-term 2 for you on Friday CEO
exactly what's going to be on
so what we're gonna do today the talk a little bit about a radius Arrhenius developed a
phenomenal logical equation but we find very useful in chemistry it's used virtually all the time by people doing all kinds of chemistry and so it's very important to us but it's really phenomenal logical equation hopefully by the end of the
quarter will understand where it comes from there were going to talk a little bit about consecutive reactions in the steady-state approximation this is a very important this is right in Chapter 19
and there will definitely be a problem on returned to that involves this topic because it's 1 of my favorite topic ss
and never talk about the Langmuir Hinshelwood mechanism which is a complicated mechanism for you know molecular reactions that exploits the steady-state approximation the Langmuir Hinshelwood mechanism OK so a
radius is a Swedish guy and we
already know he created definition of acids and bases for us that involves Proton donation and
I hydroxide accepting the
he but for the purpose of of kinetics he recognize the relationship between temperature in reaction rate he recognized a trend in the scientific
literature for the temperature dependence of reaction rates that other people haven't noticed and so
will say more about that in the 2nd but he also discovered the greenhouse effect most people don't appreciate that there's the Arrhenius equation that relates that rate constant for a reaction to the temperature but he also figured out the greenhouse effect in 1896 and this is the paper that he wrote a great deal has been written on the influence of the absorption of the atmosphere upon the decline so he
understood climate change in
1896 Tindall particulars .period of enormous importance of this question to him was transferred to a world Trojans it is the mean temperature of the ground in any way influenced by the presence of heat absorbing gasses in the
atmosphere but the 1st thing
about this is actually 1948
yes that 48 but
but in this paper actually a radius works out the expected temperature
change for the No 1 concentrations of things like C O 2 in the atmosphere and gets a decent answer for that did the amount of warming that should occur and it's just really amazing that you
can even make that calculation in 1896 the Carbide on the influence of carbonic acid in the air upon the temperature of the ground
amazing all right that doesn't really have anything to do with kinetics but this
equation the right this is the famous Arrhenius equation and so what we have here this CD a year that's something called the activation energy
we haven't talked about that yet but we will before the end of the quarter In case is the a pre
exponential factor obviously it's before the exponential right but it's also called the frequency factor it's units have to match those of K oracle what this equation predicts 1st of all this activation energy the summit products some reactants Statement between the product react state there's an energy barrier that has to be surmounted that the top years something called the transition state it's the state of
the reactants when they're halfway to becoming products right that's the so-called transition statements on higher energy
state for the system that gets the highest energy state of the system it as it
progresses from reactants the product you have to get over this he at the very top of the speakers the transition state by definition and so that the height all of this energy barrier here with respect to the reactants state that the activation energy and we're not
really more about it right now but I just wanted you know where that comes
but so if you go to the lab and you carry out a measurement of a reaction rate as a function of
temperature would you notice and everybody in this room already knows this is the rate of most chemical reaction goes up as you should increase the temperature so
here's what that data might look like this is supposed to be the rate constant on this vertical axis here this is the temperature on this horizontal axis and as we increase the temperature it turns out that we're talking about this reaction right here right the dissociation of N O 2 but the reactionary goes on and and on up and on all right and what our readers noticed is that if you replant this data as Of the rate constant verses and 1 over temperatures these data file almost on a straight line and
he knows that this was true for all wide variety of different types of reactions reactions that occur in solution reactions that occur in the gas phase this is a very general observation that if you
recast rate versus temperature data there's a lot of great verses 1 of attempted 80 get a straight line in this equation explains why that would be the case right of
chemical reactions adhere to this functional form because I
just take the log of all sides of this equation I get lot equals logged a minus the IEEE over T and so you would expect a plot of like Avis is 1 of a the straight line with the negative slope in the slope is going to be CA overall are and intercepts to be logged a and so but
this would explain phenomena logically only why the data would adhere to this functional form he didn't explain what this activation energy
was in detail and he didn't explain what it was in details we have to
understand those things yet and we're going to talk about something called transition state theory that allows us to understand what these
parameters are in more detail but suffice it to say this is a phenomenal logical reaction to the equation rather that is very important to us so canister
constantly bearing the temperature the reaction measuring the reaction rate as a function of temperature calculating an activation energy
using that equation right there write all kinds of campus
for all kinds of reactions
but OK so we're not going to see more about that for now will only pointing out that there
is this parameterization of them the rate constant with
temperature from which we can extract some parameters pay any mind as he said at an end he said rather right that
have relevance in terms the kinetic mechanism but we haven't said what the relevance is yet willing to come back to it OK so
consecutive reactions are sequences of reactions which prodding raises wins so far many times in chemistry
what we observe His reaction occurs in the product of the reaction serves as the reactants for following reaction in the product that following accesses reactors for another reactions he hit sequences of reactions occurring right now the single reaction in isolation of others which is what we always talk about and kinetics turns out that some relatively unusual what we often see a cascades of reactions occurred this is an
example of a a case where there's true reactions that occurred in sequence right we've got this methyl bromide reacting with whatever the heck this thing again right now ethyl ester all right in any intermediate is formed by that reaction that this guy right here which is Esther onto which we've appended but a methyl group and then and hydrolysis occurs right now we end up with this guy all right where this Apple group is small enough I should actually have a plus Apple group here this melodic acid misses the synthesis more water gas and from this medical ethyl ester so the point is it's important to understand consecutive reactions because consecutive
reactions are sort of the rule not the exception right reactions are usually consecutive there's usually more than 1 thing happening in sequence rather than just having something happening in isolation of other chemical reaction all right because they teach how
the constitution many reasons yes so what to talk about 1st is the
simplest consecutive reaction right 1 where we've
got irreversible conversion of ADB and irreversible conversion of Beta seat with these 2 rate concentrate here no reverse ability in either 1 of these 2 reactions all right the simplest
possible case of 2 consecutive reactions the minimum number of number of react react some
products now without going through and deriving these equations are derived for you in Chapter 19 this is equation 19 . 2 7 B. This is equation 19 . 2 7 C this equation here we didn't arise all right this is just the 1st order the reaction of a going to be we derided into greater rate was like 3 times OK so these are just the integrated rate lost for a B and C. for this reaction mechanism right here that's not
really what we're interested in right as I'm going to show you what we
wanna find our simpler expressions for these integrated rate right because in general the consecutive reactions that were really care about are going to be much more
complex than this right and I
think you'll agree that these equations are already getting pretty complex even for the simplest case where we stripped down the number of reactants and products to the absolute minimum were already getting pretty complex mathematical expressions here imagine if this was a plus B goes to see blood the policy goes to AFP was G was
and opposed choosing the metrical efficiency not all 1 I think you can see this is going to start to be a nightmare scenario for us so we would like to
do is look at these complex equations see if there's a way to strip down the complexity but and generate a
simpler way to think about these reactions that with the steady state approximation does for us but allows us to simplify
these integrated lots for reactions that consecutive OK so let's 1st of all look at
what these rigorous equations predicted and see if we can see something generalisable from what they predict that will help us In terms of figuring out what this steady
approximation of what these equations predicted about the versus time Beasley intermediate here write an intermediate it is not a
reactant and not a problem right by definition it's something that is formed to try
and least in the middle of this reactions what we may see in the lab as we put a into a beaker see comes out but the was there transient wing
but was formed and then it was consumed it was formed
by this reaction was consumed by this reaction but what we want now is and what is what happens to its concentration is a function of time but let's consider the case 1st work K 1 is faster than K 2 that fast that's a lot right qualitatively what we expect to happen if this fast and this is slow
these gonna build up right because
where else Formby rapidly but that's only the dictator see slowly and so what we're going to expect qualitatively is the larger the disparities between K 1 and K 2 in this direction where this is fast and this the slow the higher the
concentration of B is going to be so I calculated the concentrations of a B and C
vs time right for the case for cable 1 overcame a 2 is 5 the Member K 1 is faster than K 2 so the ratio between K 1 and K 2 is going to be bigger than 1 right all right so let's say we make it 5 if that's the case in the world where the start off with a equals 1 Moeller OK on this access and this is just a time axis here so you can see a just a case like it would in any
you know molecular reaction just an exponential decay of a not very interesting
the look happens to be b increases reaches a peak and then decreases as a function of time so there is a big spike in me as a consequence of the
fact that the 1 is faster reform and it's slowly reacts to give us
right we reform be rapidly and then slowly reacts to give us see and you can say here is the concentration of C is a function of time that makes sense now if
we change this ratio to make it better not make a 1 faster relative to Kato to you see the good bigger here's where was whose word is right the
concentration of the intermediate builds up more if I make a 1 versus case fast
and K 1 or K 2 equals 20 bigger yet here's 10 was 20 OK case qualitatively these equations tell us exactly what we expect intuitively we expect as we make a
1 faster versus K 2 weeks that the build of this intermediate to be more more prominent we expect the peak concentration of the intermediate to go up and up and up as we make that number
bigger but that's 20 that's 50th think you get the idea now let's look at the other possibilities right if we make a 1 slower than K 2 what's going to happen but we
slowly Formby be reacts quickly right under those conditions we don't expect the EU to build up very much like we slowly formed the and as formed the reacts faster that's what that limit
means right there right K 2 is faster than K 1 bite so this is slow this is fast and if we look at it with we make that ratio 1 to 10 in other words K 1 of the key to his 0 . 1 look the forms but it doesn't we never get very much of the Internet leaks away White but the concentration of the never gets very large compare that to the case for K 1 advocate who was 10 potluck hoping
enormous difference OK and if
we increases now I made the ratio 1 to 20 in other words . 0 5 we made the K-1 even slower compared to take to be gets smaller there's the wet . 1 B would . 0 5 the woods . 0 2 smaller
yet by so what's interesting about this is if you look at this now it looks like these becoming clause I constant as a function of time all right it's concentrations hardly changing but the concentration is not quite 0 all right it's becoming almost invariant is a function of time that's the key
thing is that we want a focus on right so there's an assumption that we can use that we can bring to bear on reactions like this that allow us to simplify the mathematics enormously and that's the
steady-state approximation basically what we're going to do is soon for all of the
intermediates in the reaction but the time rate of
change of those intermediate time rate of change of the concentration is going to be
0 so here are the
top rate of change in the would be 0 that's the steady-state approximation what is that do for us now it's going to allow us to
do I simplified equations for the rate at which the product appears mainly that we care about how fast is the product appears a function of time what's the integrated weight loss for the product and that's what we really care about here so what we wanted you supply the steady approximation derive new integrated rail offer the product that is simpler than what it would be if we had to work out all the complex mathematics so when
all simplified equation that result so for this reaction mechanism right here the time rate of change of the going to be K 1 times because at the rate of these forms be plus the one-timed a and B is consumed by this reactors will be minus K 2 times be so that's the rate would be consumed that's the rate at which abuse formed and what we're saying in a steady-state approximation is that since that time reader changes 0 those 2 things happen equal 1 another the rate at which
its formed the rate at which it's consumed are going to be equal to 1 another and if that's true
then indeed saw steady state right in the special case that we're talking about the steady approximation B is going to be able to carry 1 over K 2 times a very
simple much simpler the whole truth about the reaction the rigorous equation that we talk about on an earlier slide also to
again in the 2nd we always have to keep 1 thing in mind are right it's tempting to
mindlessly applied a steady state approximation to everyday reaction mechanism that we come in contact with but it's important to understand that this reaction this assumption
write the steady-state approximation is
an approximation and this approximation fails badly 15 1 it is faster than K 2 right if K-1 is faster than K 2 we're going to give you build up a B it's concentration is not going to be constant is a function of time not even close right what we've learned by looking back In calculating concentration of B is the steady state approximation can be expected to work when K 2 is faster than K 1 right when that when the intermediate is used up rapidly in subsequent chemical steps but for 1 yes In other words this is the the case with
K 1 overcame 2 is small K 2 is fast and he won a slow so to reconcile this equation with this equation right here we insist that K 1 be much much less than K 2 right if that ratio approaches 0 0 these 2 results can be reconciled
with 1 another right this 1 says the tiring change of
a 0 this 1 says the Thai rate of change in the matches the time rate of change of ages racial by the 1 Kate to write but if K whatever K 2 is their the
concentration of these are going to be very low at all times right that's really what we're assuming in the approximation we deviate from that assumption the
steady-state approximation is not going to work very well let me
show you that both of presented to
guests just pointed that out of sir some this suggested Expedia method for dealing such reactions becomes disabled so solution yet OK so here's the integrated weight loss for seed all right
if if this is the concentration of the right then we know statistical the K 2 times b right and so consequently obliges just plug-in for me from the steady-state
approximation solution really Bibi said state should really be beasts of steady state and so when I plugged in for OK to cancel a distinctly one-time that
Is the differential rate laughter DC for the 1st see the words the CVT is the 1 time so I couldn't solve that despite integrating in the normal way to derive an integrated rail office see and when I worked through the integral I find that the concentration of C is equal to the initial concentration of 8 times won over this yet the term here but it depends on K it doesn't even
depend on K 2 all right so
this is new integrated rate subject to the assumption that we are applying the steady-state approximation right so here are the integrated laws that we have from the state approximation this is the 1 that we derive directly by thinking about the steady state of B
by setting DVD to equal the 0 we derived this
equation directly and then this 1 has never changed as the first-order integrated rate laugh or a you know molecular reaction of any kind that
was always the same same as it is here since that is here this 1 we derived by just making a substitution of the into the differential rate law offices which is what we did on the previous life I submit to you that these equations are a lot simpler than these equations but that degree
of simplification here is tiny compared to what it normally right because we're talking about the most elemental sequential reaction mechanism possible right now right as you make that bagels to be goes to see if you may as you make that incrementally more complex these
mathematics get exponentially more complex right so the degree of simplification that you get becomes very significant which is why we use the steady-state
approximation so much in chemistry to use a lot it's overused How good is it let's look at some data how well that the steady approximation work let's 1st examine a case for we expect it will work well right what cases that K 1 of the key to small right in other words were forming the
intermediate ends reacting fast right we look at this data
before the -dash lines are the steady-state approximation is solid lines are they full and correct
answers write for the concentration is a function of time like this is
just the concentration of these 3 guys as a function of China's before right check this out it's -dash lying here almost falls on top of that solid line would get excellent agreement is green date every year by the way is the 1 that we care about malls we wanna know
what the product is doing is a function of time but the product concentrations were use the steady-state approximation get at that and here we're
going to do a great job but we got an excellent Of the steady-state approximation to the true answer look at what they're steady-state approximation predicts for the intermediate it's it's killing this right on top of it right and of course is doing a a good job for the reactant as well rights of the steady-state
approximation works great here the world's great because
that's much less than 1 now let's make it works right let's make it a little bit bigger now we're not killing quite as well it's a little bit hard to see a still looks great seeing love for pretty good said but not quite as good as it did here is kind of hard to see that there is a difference but it's still
doing a pretty good job don't you think right now we care mostly
about this green curve and the dashed line here is almost right on top of that Green Line not quite now you can see a little bit of
daylight there but not much
of the steady-state approximation still working pretty well here now what if we make this 10 In other words we're gonna
start forming the intermediate faster than it gets consumed but this is a case where we expect the steady-state approximation to fail all right and we're going to remember that even tho nobody else does even tho other people use the steady-state approximation indiscriminately we will not do that we will always remember that this is a case where the steady-state approximation
really should work very well doesn't work very well what am I talking about check this out here is the -dash line from the steady approximation for the product here's what the product concentrations really doing that's a pretty big dealt right that doesn't look like if it's that very well at all right in the
intermediate but terrible
except except and you're actually not not quite as terrible as you might think but but it's still a distilled a pretty good job I took it's doing a poor job in K 1 over Katie was 10 by making whatever 250 it gets
downright nasty right look at this year's the -dash streamlined here the solid green line there's miles between these 2 curves here right and many
people will apply the steady approximation without any qualms at all 2
cases that conform to this problem this he said of of rate constants right here all right so you
have to be careful that the intermediate is not building up because in that case he doesn't make any sense to apply the steady state approximation
parts of the steady approximation of what we do
but if we want to apply the steady-state approximation to the Connecticut Protestant kinetic mechanism my 1st
sequential reaction 1st of all we right a the process look at the end I'm talking about the process that generates product usually
Susan what we wanna know the words in this
case the CDT is K 2 times right that's the rate at which sees formed then what we do His we ask Father intermediates in this weight loss identifying them and then apply the steady-state approximation
to all of them here we can
find an intermediate it's right there right and it shows up in the differential rate laughter the formation of product as well perfect
this is a great candidate for the application of steady state approximation but then what we
this at the time change for these intermediates 2 0 or in this case DVT is equal to 0 and then we write equations for the creation the intermediate quite reasons although the states the concentration of the intermediate that's what we did here all right so I said at the time rate of change of the intermediate 2 0 and I said what that means for this reaction mechanism is at the rate at which the EU's formed has the equal the rate at which it's consumed by these 2 rates have to be equal 1 another now I just the road for the concentration of B but you saw
the concentration of those of the intermediate alright and
inevitably the concentration of the intermediate appears in
the rate expression for the concentration of the products
and so the next thing to do is just plug the sand To the differential rate offer the product right substitute suppression the riddle of the product the words DCT K 2 times Beach right now instead of being paid 2 times the it's going to be TK 2 times be steady state and we just plug in what we said is the steady-state concentration abuse K 1 overcame 2 In this case the 2 Katie's cancel and that then becomes art differential rate laughter C and we can integrate that an
integrated reel off we want to but
remember this works if and only if the 1 is less than the 2 they say the ready I know it OK now and take a little
conceptual breakthrough all that information is sinking deeper into work or Texas a retired what Irving
Langmuir Irving Langmuir was 1 of the 1st Great American chemist end of it we Lewis he invented Lewis .period structures
Lewis called them Languedoc structures that's not true I just made that is the father of surface
chemistry revolutionize the light bulb he discovered how to manipulate modern molecular what layers on water who cares what that turns out to be rather interesting to do turns out you revolutionize the light all yeah but the
incandescent light bulb mean there's been several revolutions yet in the meantime that this is a
page from and Keisha interested Jean-Louis notebook right he actually made Lewis .period structures that's why they call it was .period structures like animal as a notebook or a cocktail napkin but pretty cool and he
wasn't very mean the see but so
Langmuir figured out how to make the light bulb work properly lots of people had worked on the light Edison's famous for the light right but he wasn't the first one to make an incandescent light bulb turns the other light bulbs for his his
work better right when you think about it this is enormously unlikely way to make light right what is this that is a tungsten tungsten-filament that is almost sold almost so
tiny in diameter need microscope the see and tungsten what we know about Constance tungsten a very stable metal How does it compare to other plant not as stable as planned it's less noble than platinum right right does a react Indiana we buy a reading Wayne wedding ring do you want a tungsten wedding ring it also said he would be impressed by that the
problem with tungsten is it reacts rapidly Van Exel Thurmond with oxygen
deformed tungsten and so you know if you're going to choose a metal to make life with you might not choose tungsten in
effect Edison didn't you Stockton chose carbon
by the original light bulbs or carbon the only lasted 40 hours because carbon filaments felt like oxygen needed by then a guy
named here on Webster figured out how to make a really really good carbon filaments at General Electric beckon like 1897 and he made carbon
filament light bulbs that lasted 100 hours these were the case of carbon filament light bulbs but the only lasted 100 hours and
100 hours really wasn't good enough I minnows like a couple months at most In your light bulb had to be replaced and at that time these
were not cheap but until 1906 or labeled a carbon filaments so turns out lying there was not I figured out that we should use tungsten for this development there was another guy the General Electric in 1908 who did it 1st he made
tungsten-filament why tungsten especially in view of
this 1 makes tungsten special anybody now Thompson
turns out is the best at something no he capacity that's a good a like that answer but that's not the correct what problem you going to have when
think about this and put 100 volts across this
filamentous that With of a human hair right what probably going to have a film it's going to get hot right melting did you say nothing melting Wells is going to happen
what happens to solids when you hear them up and get really hot even before they melt but say that it doesn't decomposed somehow His yes that the person you'd expect but a dozen decompose it could still fall apart house but about sublimation what
sublimation direct transition from a solid of a gas phase right metal sublime right so it turns out Thompson does have the highest melting point of every man in the periodic table all right and it has been extremely low rate of sublimation all right and so the problem with tungsten was that when you turn on the light bulb as soon as you turn on the light while the tungsten Woodstock a sublime quantities glass surfaces here and return is liable black red the daily by the slide Bob
screwed and it's clear 2 weeks later starting to turn gray 2 weeks after that it's getting darker and so the model like becomes of your light bulbs decays exponentially as
the inside surfaces of his like all where their act right because what happened is that Thompson would evaporate under the light bulb and then this tungsten-filament this tungsten film would react with trace oxygen
and water are right and would coat the inside his libel turned black
Selangor solved that problem right he sold the light bulb
turns black problem and he's all the time so when he saw the light bulb turns black problem he created
another problem right this Fellman said last very long and yet has yet to solve that problem too so
just to make a long story short what he did as he failed the inside is involved with nitrogen and it turns out but the reason these tungsten
light bulbs worth of work were were
turning black it's because of this reaction here but the reason that the Thompson was blind under the surface of this glass was
because it was reacting with trace waters that was inside the light bulb all right and that former the tungsten
oxide and it's actually the tungsten oxide that's blind on the surface of this glass right now the tungsten directly the tungsten
oxide actually got sublime metal oxides it turns out to have much higher vapor pressures than Parra metals the metal oxides you wouldn't necessarily expect that so he felt find
out into the field this whole thing with nitrogen got rid of all the water that not only did the tungsten-filament last longer than that surfaces of his glass and didn't turn black anymore yet a
patent for that so the way that
he made light bulbs is the same way the say whether they're made today by the essentially
solved the incandescent light
bulb problem of course is completely obsolete today look at me those opposed to rest rest
your thought process so we can go back to
thinking about kinetics we got
8 minutes let me just type but the
figured out something about how really simple reactions occur so even a reaction that looks like this can be geometric In
other words not elementary
even reaction at the symbol of this as this can have multiple steps but
how is that possible well
but as an example if there
is a barrier for the reaction to occur right if in the process of going from a to B. they have to surmount an energetic barrier right it can be necessary for there to be an activation of a that has to occur before the reaction can happen how could
that happen let's say a is in the gas phase aching collide with itself and if that collision is energetic enough you could generate an excited a that has
a lot of kinetic energy a lot internal energy that and
then that activated a could give rise to products right so there's several steps here it reacts with itself together and activated a that's his orange thing then of course they can
deactivate as well right so there's a bad reaction this is
just the Midas was so this is just this going back to the messages written all the reaction for a year there were activated a collides with another AM loses its energy
but it also happened and finally this activated a
could give rise to be that is the Langmuir Hinshelwood mechanism for you know molecular reactions turns out that
lots of reactions
that look as simple as this occur by this mechanism right here all right this activation of them reacted by itself but it's it's
activating itself passed a worker before the reaction can happen right the Hinshelwood mechanism
now were a possesses internal wonderful or to give BES said that but why do we I bring this up
right after the steady approximation
I think you know the answer we're going to apply the steady-state approximation OK so let's think about what happens to the time rate of change of the intermediate here does everyone agree that the intermediate is a start How do I know a star's new media I don't see a star here but it's not a product and not a reacted and its formed during the reaction and then it's consumed In the course of forming the product business all
right so the 1st thing we have to be able to do is spot the intermediate in the reaction but can be nontrivial depending on how complex the reaction mechanism OK then what we will
what we want to do is write a differential rate while forward but with the time change of any looking at this reaction right here I can see a is formed by 8 2nd reaction of a with itself with the recounts the tale 1 inside the right K-1 1 times a squared that's the rate at which it is formed now can anyone see the the film's story based camels won a stock as a consequence of this reaction
what else can happen when a stock can react with a and it can get consumed by this reaction right here right so I have to write a term for that some nifty minus 1 minus sign 1st book is this makes a star smaller not bigger this 1 makes the Questar are bigger and what what is that right it's a star times a withering constantly minus 1 and finally a targets consumed by this reaction right here right inside that K 2 times a star and that's a minus sign to that someone site is a star gets smaller that the miners signed as a started smaller and that the plus side as they started right so
this is my differential rate about people like that for the intermediate because the next thing I would do the steady-state
approximation studied equal to 0 so this is just analogous rate Lofa AT is consumed by this reaction formed by this reaction those are the only 2 that matters Ciarlini 2 terms
why is this minus Q 1 8 squared and not to k 180 squared because there pertains To this reaction right here
think about that and if the answer does
arrive but In the short answer is
you only consuming 1 a in this industry Woodside to ways getting consumed forming 1 his own reality only losing 1 that's why it's of 1 another to OK and the different rail of abuse just as everyone can see that yes yes yeah now what we're going to do with the steady-state approximation said this guy equal the 0 out here is not 0 period
and can work out what the kinetics of insured reaction OK so what I said this guy with the 0 still agree with that now I'm just rearrange this equation and remove them negative terms to left-hand side in the positives the right hand side and set them equal to 1 another and analysts offer a stark and then applied a star into the expression for DVD tea and
this is going to give me an integrated rate lockable Lindeman Hinshelwood reaction In a mid-table about that on on Wednesday


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