00:07

OK so mid-term too come right out where reviewed today

00:20

retired in some detail what's going to be honored now with

00:27

6 had an error on problem too I know several of you brought this to our attention regrade all of them are right and basically what

00:38

we're going to do is just credit everybody with problem too the

00:44

issue is this for particular reaction this was supposed to be the reaction gives the standard reaction gives energy with the superscript of 0 here you can see that there is no

00:56

superscript so in reality

00:59

the crew the correct answer would be about Katie equilibrium we can then conclude nothing that

01:07

would be the correct answer thorax and

01:10

is going to give you credit for that everyone give everyone credit for that

01:15

problem OK so we're going to be revising this Instagram I'm

01:19

assuming that it's going to shift to the right I guess it would have been working to give everyone credit

01:25

for problem too right Stevens can

01:29

the mid-term to review session today starting around 5 over and bring all

01:39

OK using just gonna work problems so you think that might be useful for you 10 that I think he

01:46

might even be planning to review sessions is that right 2 of the the

01:55

OK so the US State into the Facebook page and let you know when those are going to be but the 1 today is definitely

02:04

5 their right yeah so almost every video that we've generated has been

02:14

posted to YouTube only missing the last couple these videos have to be processed and edited and it's very time-consuming and so it's impossible to get these things posted immediately after the lecture may not be obvious to you but there's a lot of work that has to be done between the time the Rob video is taken on these 2 machines that here and it's put together in a form where the 2 video feeds are brought together in all of this takes time and so hopefully will have everything

02:51

up through lecture 20 posted by the end of the day or so right now lecture 19 is still missing but chance fixing up for

02:58

us and so maybe by the end of the day will be able to post that if those are interesting to you how many people I care about these

03:06

videos have used them there are

03:11

many more so if you wanna

03:16

find these there's 2 ways to do that you got you to search on my name and they all piled up in random order unfortunately all right in this list goes all the way down to all

03:31

includes all the matter post you 2 or you go to war

03:35

lecture page I tried post them manually and you know I have some degree of success in doing this but I'm usually by "quotation mark you 2 months and that always behind and so if you want the latest just go directly to YouTube in search

03:48

of my name you can find it

03:53

OK so what's mid-term to cover in principle it covers everything on this line all of Chapter 15 chapter 16 sections 3 4 and 5 Chapter 17 sections 1 2 4 and 5 of Chapter 19 1st 3 sections of that right all of this is this is the only kinetics that we talked

04:21

about so far at the beginning of chapter 19 and so as I thought about all

04:27

of this material in writing this exam From a practical perspective I

04:33

can't ask questions about all of this when was the

04:36

questions are just trivial on these

04:38

exams we don't want as trivial questions we want to ask more probing questions and so on what we

04:47

decided to do is actually

04:49

always ask about thermodynamics there will be any fanatics on the matter In

04:59

fact problem 1 is going to be to problems on the exam problem 1 it's going to relate entropy McConnell cycle and the lecture that he's on Madam vast his lecture 13

05:11

but if you look like 14 is there's lecture 12 were also impacting this concept all right but lecture 13 sort of the central lecture relating to this

05:22

concept and problem Tuesday to be about the Gibbs energy reaction equilibrium constant and things related to that and that's the central lecture there's 17 it turns out OK

05:35

recaps some highlights from from lectures 13 and 17 in this lecture try to emphasize a few

05:41

points that may be Warren emphasized enough the 1st time around all right now I already

05:47

made strong statements about what the extra credit problem was going to be In yesterday when I

05:52

started the work on the exam I started to look through I Googled and I started to find out what information you would be able to find about this

06:02

problem on the Internet and I basically concluded that it

06:08

would be hard for you to get the right answer on this extra credit problems it's I found some

06:14

really good papers about Lashell shall liaise principle and the Hebrew process but these papers

06:22

are mathematically intense I'm not sure you know the

06:27

last 8 pages of this paper but it's jammed full of equations in the talks about not ideology in the talks about concepts that we can really talk about In class

06:37

at all so I'd like in the next couple

06:42

of slides to explain to you

06:44

what I was thinking right now I'm not going to ask this question a midterm exam amassed a different question but I want to explain it you got up at the

06:57

end of what I'm explaining here I want you to tell me if this is what you're gonna write down if if it was a apologize but here's the

07:08

ammonia synthesis reaction right here its equilibrium constants if we double

07:17

hypothetically if we double the partial pressure of nitrogen In

07:23

order for this equilibrium constant to stay constant and independent of about the values of these partial pressures this partial pressure harmonious and happy increased by Rutshuru because the square that ammonia partial pressures got a double in order for this kind K-State constant right and so on Of

07:45

course this conforms to the exactly to the prediction of a shot liaise principal right if we increase and 2 we expect to increased the amount

07:53

of ammonia that produced right

07:55

if we increase and to hear we push this equilibrium to the right and generate more ammonia

08:04

that's exactly what we shall use principle critics all right so

08:09

why is the process the Hebrew botched process is different from that prediction wine why would we expected not to

08:18

adhere to their prediction the key

08:22

is in the catalyst particles what I'm trying to represent here is an Irish micro particles on some kind of indirect support right in the ammonia synthesis reaction In

08:34

the Haber Bosch processed iron is used as a catalyst what is the iron do but basically what the iron does

08:41

is it takes apart and generates but but also takes apart the H 2 takes apart the end-to-end produces activated

08:51

hydrogen and nitrogen on the iron surface no

08:54

this isn't actually literally correct way of dryness but all I need to represent by this is that we've created on the catalyst surface

09:04

activated nitrogen and activated hydrogen OK and you

09:10

can't think about that 1 way to think about that is that it's it pulls the nitrogen apartment produces a nitrogen species of bonded to the hydrogen it wouldn't happen this year this bond here is not meant to represent a single bond necessarily the presence of the iron catalyst activated Nigerian hygiene are produced on the island service ammonia is formed from these activated species so the reaction is occurring on the catalyst surface once you activate the nitrogen and hydrogen then they can

09:37

react to give your money under conditions that are much milder then would exist in the absence of the catalyst but that's the whole point of the catalyst is to read reduce the pressure and reduce the temperature necessary to drive this reaction forward

09:54

all right now once you understand this picture at once you agree that this picture is qualitatively correct I think you can see that if we increase the partial pressure of nitrogen and we get rid of most of the hydrogen on this surface because there's conservation of

10:11

catalyst areas In other words if we saturated catalyst with nitrogen and other Levering much room for hydrogen does it and so the

10:21

point is if the partial pressure of nitrogen is increased the part of becomes saturated with nitrogen at the expense of activated hydrogen

10:29

that's the key right if if I fill

10:32

up these binding sites of nitrogen and I get rid of most of the hydrogen is the reaction to be more efficient at that point or

10:38

less efficient while the

10:41

first-order this catalyst particles can operate at peak efficiency when the mole ratio between hydrogen and nitrogen is 3 2 1

10:51

because that's the racial we need to make ammonia right rights of the 0 water picture in your in your mind should be we want

10:58

3 times as much hydrogen on science

11:02

surface as there is nitrogen under those conditions we would

11:05

naively expect a reaction to the most efficient and generate the most ammonia but that won't

11:13

happen if you increase and to you will saturate these mining site of nitrogen and you will make a less ammonia In this case outcome would be shift equilibrium further toward the reactants state in defiance of the law shall use principle right this is

11:30

the difference right when you've got the catalyst particle you have to think about but on catalyst surface right if that if you can't generate a stoking metric mixture of the reactants on the catalyst surface so that they can react efficiently then you're going to favor the reactants relative to the products that makes sense how many people were about to say that on the matter well I

11:59

apologize to you then the fact that but I believe but we're going to do

12:10

instead is worded the whole problem directly off interim not a whole

12:18

problem because pretty long but part of the problem right now it's slacken be exactly the same so don't memorize the problem is I'm to change the numbers but it's going to be virtually identical to a problem was on mid-term 1 no mid-term wine most of you didn't dominate bitter 1 and so I think this is a great idea because this is going to force everybody wants to get these extra credit points to go back and look at midterm 1 make sure they understand how to do those problems because if you want these 10 extra credit points I might make it even 20 I haven't thought about it yet writer today

13:01

you're going to have to know undue do problems amid turmoil and I don't suspect most of you gone back in gone through and see what I do

13:11

wrong how could I have got network now you have to do that OK so I should have given more thought that they were botched problem for the extra credit but I think if we had assigned a problem it would adjust been a disaster for everyone except this young lady right over him OK so what we wanted

13:37

to in in what's left of this lecture is 0 4 this material and hit the high points and zoomed through this material and at the high point OK and then it's going to be here you don't have to

13:52

worry about genetics on Friday there hasn't been any kinetics on the midterm exam but you got a focus on these 2 topics which I think are really at the crux of the thermodynamics that we talk about since mid term 1 right you can focus all your attention on that you can work on nothing but those kinds of problems so problem

14:14

was going to be all about entropy and Connell cycle we've got 2 definitions frantically we've got a statistical definition for entropy Kailai W and we've got a thermodynamic definition for entropy This is what it looks like we talking about a reversible process IDS is DQ for reversible process divided by 2 we were talking about the car cycle he His flooding in

14:39

and out of this cycle right were converting heat into work but there is no mass fluxes OK so we're not

14:49

talking about an isolated system were not talking about an open system were talking about a closed system closed in the sense that there could be fluxes of heat and energy white but no

15:00

fluxes of material right

15:05

so the colonel cycles the most efficient exist existing cycle cables converting a given amount of thermal energy into work or by some

15:13

VS but that's what the sole

15:17

schematic diagram is supposed to indicate this is a hot the source whereas the war this is a cold weather wore white and were extracting he between hot reservoir on the holders of war in some of that he flexed gets converted into work by trying to represent you so the cardinal showed us is that we can think about a process that looks like this 1 are what are we looking at here this is a pressure volume plant this is an isaf amenity T-1 this is an ISO fermenting to this is the starting point for the cycle here at T 1 this is the higher temperature and so there are 4

16:03

prophecies that make up every Connell cycle right there's a

16:08

reversible Eisele thermal expansion the there's and Adia batik expansion volumes getting bigger here getting bigger here now there's a compression isothermal compression and an EU-backed compression those of the 4 steps but we spend so much time

16:25

talking about this ,comma cycle because every pressure volume

16:30

process can be carved up into little tiny cycles and if you think about this you what you've got this mosaic of cycles are right this downward process for this guy is the upper process for this guy and so these the lines that where you got adjacent Colonel cycles like this that line goes away because that I can't look back in various scandals of this end that can that so the

17:04

borders of these Colonel cycles that our mosaic together all disappear and were left

17:10

with a boundary that looks like it would be like a sock tooth but if you if you ImageNet there's a this 10

17:18

thousand of these Connell cycles it starts to smooth out anew as you increase the number of Colonel cycles you start to match up with this

17:26

purple border and consequently it's telling you

17:31

about the efficiency of this

17:33

random process even know this doesn't look anything like a Okano cycle the cardinal predictions applied to it that's why it's so

17:41

important now normally the colonel cycle we're talking about a reversible reversible process season we're talking about an ideal gas but it also places a rigid upward limit on the efficiency that you can achieve for a real process with real gasses that are not acting reversible but it places a rigid upper limit on the efficiency of those processes it's important for that reason to even know when I'm usually talking about ideal gasses reversible prophecies so

18:13

what we know for sure about this ,comma cycle well this is a nice of thermal expansion here the internal energy of an ideal gas only depends on its temperature the

18:26

internal energy of an ideal gas only depends on its temperature it doesn't depend on anything else and consequently this is a nice of

18:37

them so the internal energy of the gas is not changing anywhere along the size of the it's constant the internal

18:42

energy the gas is constant on the ice of because internal energy of an ideal gas only depends on its temperature but here the temperature is not changing OK so

18:55

that's true and this is an expansion so work is less than 0 that means that you have to be greater than 0 will cut because Hugh plus W past equals 0

19:09

right OK so we know she was

19:13

greater than 0 here in other words

19:17

we have flux of heat into the system right then down here same story but this is the other rights affirmed colder temperatures that the colder temperature than that and the internal energy of idle death doesn't only depends on temperature and temperatures constant here so that's not changing work is greater than 0 secures going be less than 0 so there's the flux of

19:41

heat out of the system OK and these 2 guys Arcadia bats so there's no fluxes of heat there all by definition because the radio that accuse Europe that's it we

19:53

know before we even start thinking too hard about this OK the

19:58

efficiency for Connell cycle is defined by this equation don't try to derive this equation it's just a

20:04

definition all right it's markets performed In was in response to the temperature gradient divided by the heat that's absorbed by the system just the absorb

20:17

heat that's only that high-temperature heat that the plots at the high temperature OK so

20:23

it's that's the work that's the heat work over heat he transferred from the hot reservoir to the system that number doesn't matter at all I would get 25 per cent for this process right how the Licano cycle we can write the efficiency in terms of the temperatures of those 2 reservoirs the hot-tempered part

20:50

reservoir on the cold reservoirs and we went through and derived

20:53

this equation laboriously all right we can write the work that happens in each 1 of those 4 steps this is the I sought formal expansion this is the 88 batik expansion this the isothermal compression this is the immediate that compression but at that is the work that happens and is about process right how do we know

21:21

that the internal energy of an ideal gas only depends on its the internal energy of an ideal gas only depends on its so consequently I can

21:35

calculate the change in internal energy and I know cues not changing because it's

21:40

really about process Q is 0 so that means the work

21:43

started the that because of the change

21:50

in the internal energy change in the internal energy equals work for needy that process by itself that's confusing about this is that the

21:59

constant volume the capacity and the volumes changing is that confusing

22:04

I think it's confusing right it's still correct the cost of buying the

22:10

capacity doesn't depend on the volume of its volume

22:14

independent for an ideal gasses 3 are over to .period right because

22:21

volume capacities 3 are overdue that doesn't depend on the OK so now just looking at this equation I can see right away but these 2 terms on the cancel because they're just going to be that says the opposite of that because all on changes as the direction of this integration right OK and so I interpret these 2 terms and I can simplify bellows these equations look like they can't possibly be right but I have triple

22:54

checked them and I am confident that they are exactly correct all right so

23:02

there's this relationship between this these 2 temperatures T 2 and T 1 right that's what this to teach you and won refers to this as the 1 in the 4th that's the 1 In the former OK so this expression applies to any 2 temperature volume .period

23:21

on right gamma is just

23:26

CPO receiving but CP what is it numerically for an ideal gas 5 have all right

23:40

over 3 have right for an ideal guest OK so we can make those substitutions and do some algebra divide substitute my goodness of we get this expression once we've done that right that's the total work and he transferred from the horrors of war

24:07

only now this is the total work for the whole kernels cycle here's the transfer from only from the hot reservoir

24:14

of course there's he transferred in the cold reservoirs well but we ignore that were calculating efficiency and so now we can

24:20

plug in these numbers into the plug that into that like that and that simplify it we get there now I'm not showing you this so that we can derive this expression for the efficiency of the carnal cycle I'm showing it to you

24:34

because in the process of deriving this equation we learn a lot about how to calculate entropy changes internal energy changes he work we can calculate that for each step of the condo cycle that be the thing that know how to do what is the

24:54

interview change for each of the 4 steps reversible Coronel cycle this is the expression that we have the firm anemic expression French this equation right here but so since the entropy is a state function we know that we can just a fad the entropy changes for each 1 of these 4 steps Bobo bloom bone not only that because it's a closed cycle because the carnal cycle ends

25:18

up where it started we knowledge allegedly changes gotta be 0 right it has to be the state functions so if I integrate over the whole cycle the matter what that cycle looks like as long as I end up where I started dealt S is going to be 0 that's

25:37

what that means writer steps 2 and 3 Arabia barracks accused 0 so that these 2 are easy to calculate by detonating a batik process that's needy batik across Q was 0 so the entropy changes 0 for the other 2 processes I can just

25:52

calculate what the heat fluxes from the work all right that those of the

25:57

temperatures and so this is the total we already know the total cycles gotta be 0 so that means that has gotten equal that right I feel in

26:10

terms of entropy nothing happened but for the adiabatic processes but there's no entropy

26:17

change for the better process is

26:19

just a vertical line here and here here's the entropy change we can calculate for the isothermal processes by its own terms of temperature and entropy we get a box that's what the colonel cycle looks like OK now the colonel cycle

26:37

applies to reversible process he's right but if the process is irreversible

26:44

efficiencies lower right it is bounded by the efficiency of the reversible process but it can be equal to or less than that of the reversible process and we went through and a derivation of an equation that looks just exactly but this equation right all right so this this is called the closets inequality says if we integrate the entropy changes over any

27:13

cycle process that does it work not every step is reversible right if there's 1 irreversible process embedded in some cyclical series of processes that end up at the same place by right this year ending up where you started right it's

27:33

going to be less than or equal to 0 yes that's Claudius OK and so this clause inequality leads directly to the 2nd laughter when an Amex which says that for all

27:47

prophecies reversible and irreversible Diaz is

27:51

greater than or equal to or greater than 4 irreversible processes and equal to 4 reversible process these DQ over to this is the most

28:02

rigorous state of 2nd laughter when anemic and we derive did by doing theirs and then talking about this and we talked also about this notice that and then we got to this but so this equation makes predictions about 3 types of processes there spontaneous and irreversible processes were that is an inequality by IDS is going to be greater than DQ over tea for reversible process is sequel the is that not spontaneous its last that's why the entropy is important it tells us about spontaneity OK some simple but important examples reversible phase transitions 4 face transitions there's a well defined and copy Of the phase transition but we can

28:56

look it up it's in there are tables of and copies of phase transitions and the change in

29:01

the centerpieces that Delta H a change in the end be divided by the temperature at which the phase transitions occurred were

29:08

talking here about formal phase transitions right face occurs we change the temperature like melting like boiling but if we plot the

29:20

entropy as a function of temperature but this equation tells us is that there are a stepwise changes in the entropy when we now when we bought right there stepwise changes in the interview that occur increases

29:35

in the case of these phase transitions that we're talking about right now right there is more and to

29:39

be in the liquid and solid want to be in the gas and liquids that's what these vertical lines are telling us

29:46

right there's a stepwise increasing the entropy as we undergo these phase transition

29:52

In terms of heating or cooling we can deduced that the change in the entropy is related to the heat capacity and with the right this expression for the change in the entropy where this sea it

30:08

depends on whether the process that we're talking about a cursor constant volume or constant pressure if

30:12

it's a constant pressure heating using Cesar that the constant volume meeting were using season the OK we use the see that's appropriate for the constraints that were posing on the system while we do the OK and so if we plot this is the change in the entropy divided by an hour but essentially were fortunate to the change in the entropy as a function of the temperature this is what the entropy looks like for a gas as a function of temperature as this curve shape it's increasing and we can talk about changes in volume the was my is sold for pressure volume work we to dealt S so consequently we can relate pressure and volume 2 Q right that allows us to write this expression from our normal general expression for the entropy and then by substituting from the ideal gas equation we can equate the change in entropy with this expression right here all right logged the final over the initial that tells us what's happening to the entropy is a function of volume and we can also by isolating the derived this expression for the change in the pressure notice here there is a minus sign here there is not but

31:36

entropy going down as we increase the pressure it's going out as we increase the volume OK

31:46

so if we want calculate the entropy for any kind of change we can now do that but I wouldn't calculated the function of temperature using this expression right here or a function of volume using this expression right here or the bunch departure using this expression right over here don't forget in the case of pressure only there's a minus sign right the final over initial final over initial final over initial

32:12

temperature volume question OK so these are pretty simple expressions we can even drive them if we have to

32:26

OK because this is a statement of salt if 2 things that happen

32:29

calculate the entropy change when gasses he even expanded when gas is pressurized and cool write any combination of things all right the nice thing about the entropy is that the state functions you can calculate 1 thing at a time you know you can get from point a to point B anyway you want to be in your should get the right answer if you're calculating each segment correctly pride

32:59

were a segment could be a change in temperature change in volume or change in pressure calculate any 1 of those things in any orders sequentially if

33:08

you do it correctly you should get the right dealt asked for the composite process that might involve several different variables right that's the beauty of this passes the state function

33:21

because vessel like Uganda I propaganda which argon gas is so this is straight out of a lecture

33:32

13 a couple the interchange

33:34

of organ gets 25 degrees of 180 a container volume 100 cubic centimeters compressed to 5 cubic centimeters signed heated so here changing the volume some Mason reversible processes and sepia constant write total change entropy is equal to the change in entropy for each 1 of these 2 processes done sequentially it shouldn't

33:59

matter which order you do them in either process 1

34:03

process to process once isothermal compression process to Zeisel volumetric heating OK we work results on the case so just looking at the signs in this particular

34:14

case it was not possible to predict definitely whether the answer would be whether Delta as would be positive or negative for the process because 1 term is negative

34:23

and the other terms positive rights on this case she can't really predict that the answers will be positive or negative before you do the

34:31

calculations right if they were

34:33

both positive you could say yes that's deftly parts so process 1 a compression a constant temperature that's this guy only using this expression here to calculate remember we pointed out in this particular case these 2 hours do not cancel here were calculating using odds to calculate and so you don't want to cancel the

34:57

odds the unit's plug in the

35:03

numbers this is the dealt asked that we get for the 1st process right negative we predicted there would be a good idea indicated we could have done

35:16

it right after this compression were 25 C in 5 cubic centimeters process to seating a constant volume to 45 degrees C. White that's this equation right here where do you see the equals 3 or over to right Morgan assume this is an ideal this argon gas is behaving ideally so we need to multiplied by the number of malls and that's what we're doing right here take in all the numbers this is the Delta as that we get we predicted would be positive it's a pretty small number but it is positive small and positive

35:51

I so that we just add these

35:53

2 Delta as this together and we should get the composite process that's the total bill that we can say that the process is not spontaneous based on the

36:04

side of adult calculate OK

36:11

so that's the whirlwind

36:12

tour through the problem was Freddie questions on problems 1 calculating Dell s saying something about a Oconnell cycle he work internal energy user concepts that were many of them also won major 1 but really no escaping that OK

36:46

so problem Tuesday be about begins energy reactions heart of individual Reagan's product is contributing G the Gibbs energy but we looked at this curve endlessly what we deduce about the equilibrium state win look at the mixing of 2 isomers of hydrogen we did some algebra we derive something called the chemical potential ways to the chemical potential Is the Gibbs

37:16

energy that we contribute to a single chemical species right it's the Gibbs energy that we can assign to that it's the contribution of a single chemical species to the Gibbs energy that's what that

37:29

equation means even Bell there's a partial derivative here with respect a number of malls you look at that it's just puzzling right

37:37

the way I think about this partial derivative as you already know as I think about it as the scale of this told g the system ended the partial derivative of the Gibbs energy with respect to the malls of the blues stuff you get by just thinking about removing a little

37:54

bit of the blues stuff making a

37:56

measurement of the total Gibbs energy and the difference before and after you remove that little bit of stuff that's the partial derivative of the gives energy with respect to the Blues W. all the green stuff in the on stuff constant this may seem

38:12

childish but without this picture it really doesn't make intuitive sense to me but with this picture I understand perfectly the

38:24

OK so about this chemical potential basically Everything that's true for the Gibbs energies also true for the chemical potential I don't think there's any

38:36

exceptions to that except that the chemical potentials always referring To as saying it's that gives energy that we assigned to a single chemical species so

38:46

the units are energy per mall pressure dependence is the same as G temperature dependence is the same as Jr and so the concentration is something that we hadn't talked about it but what is the dependence of the chemical potential on the concentration of that species we mentioned that there's this thing called the activity was said the activity has 2 parts to it there's no activity coefficient of concentration in this equation right here

39:10

you can see very clearly how the chemical

39:13

potential relates to the concentration right through it directly related to the log of the concentration and in the

39:23

dilute limits right if any is

39:25

very dilute this scam is equal to 1 and we can see very clearly what the relationship is between concentration and the chemical potential this activity coefficient In the dilute limit is equal to 1 otherwise it's going to be less than 1 there's very few exceptions to that but we haven't

39:45

said hardly anything about the activity coefficient you're not going to have to answer questions about activity effects on mid-term to fight but you should

39:57

understand that this is the relationship between chemical potential in the activity that's that's where we get the concentration dependence is from that OK so we already understand that Gibbs energy is minimized upon you approach the equilibrium of this issue of the Boeing other the Gibbs energy is an important issue to us it's

40:24

important to understand qualitatively where that comes from because if we didn't have that Boeing effect right there was just a straight line that

40:33

connected react in some products there would be no possibility for equilibrium to exist

40:39

or if you this this system's

40:41

who go downhill intermediate products that for equilibrium exists where will we know that's not

40:47

true right equilibrium exists

40:49

some intermediate mixture of reactants and products that depends on what the Gibbs energy air and so we did this little thought experiment where we mixed oxygen and nitrogen and we did that we concluded that we're going done mixing these 2 things are going to be halfway between them in terms of the composition of the system and we can then calculate what the change in the Gibbs energy for that 1 mixture of hydrogen and oxygen and we got minors 1717 jewels that's nowhere near the line between these 2 guys it's way they down here that means that there's a lot of Boeing in the system and where it comes from it is not dealt a Shh because for art

41:37

for an ideal gas Delta agent of mixing is 0 but there are no forces acting between gas molecules right so there's no heat released or formed when you expand or contract an ideal gas there forces acting between those molecules so Delta major

41:55

0 but the entropy of mixing is not right this is 2 times don't that's the entropy of mixing acting on the system right and delta G is

42:09

Delta H minus the Delta S isn't it

42:15

Fiji's Delta each minus the Delta S and so consequently that's where this comes from it's the entropy that's driving this and bowling but the Gibbs energy right it's the entropy that allows there to be an equilibrium

42:30

state that is not all products for all reactants right it's somewhere in between so we

42:40

then we can the Gibbs energy in terms of this extended the reaction and we concluded that we can derive an expression for the

42:49

reaction Gibbs energy

42:53

but which is just the slope of this curve at any point right and we've redenominated the horizontal axis here in terms of the extent of the reaction this symbol that I can't even figure out what it is Martin got equilibrium down here we've got a positive reaction Delta geez here negative reaction Delta you so the 3 possibilities if there reaction Gibbs energies less than 0 the reaction gives

43:24

energy is less than 0 it means the slope is negative the forward

43:27

reaction spontaneous gas four-year were spontaneously and on this direction if we're here were spontaneously going go in this direction of here we're not going anywhere and that's always says OK so we have not learning more about where equilibrium is located in terms of the progress of the reaction and so we derive an expression that allows us to say you about that we derived from this very important expression right the standard reaction Gibbs

44:10

energy equal minus art you belong to a that's a big that's an equilibrium constant but the equilibrium constant of the system is related to the standard reaction Gibbs energy so yes the reaction the standard reaction Gibbs energies 0 we expect equilibrium constant to be right in between react some products in other words halfway between react in some products the words

44:42

the numerator of the equilibrium constant equals the denominator in equilibrium constant the queue for the

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reaction is equal to 1 there's many different ways that I can say the same thing OK equilibrium constant would be 1 in this case all right and if we this b is more stable a and we get the skewing we still have this bowling and brought about by by the entropy of mixing by then we can conclude

45:13

if the Standard reaction gives energy is less than 0 right here the standard energy is less than 0 the equilibrium constants can be greater than 1 and if it's less than 0 it's going to be less than 1 Pressel

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qualitatively we want us to make sense but we want to make

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sense that this curve tells us something about what the equilibrium constants gonna be are qualitatively we wanna be able to connect this picture with the equilibrium constant the OK so we can also do that quantitatively so Tamil methanol reformer can make predictions about which

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way the reaction is going to go off we know what the composition of the system as we know the equilibrium constant is we can't say the reactions shift this way it's going to shift this way we can even say what the final concentrations of each reacted and product will be at equilibrium we have to go all the way back to kill 1 to remember how to do that alright but this might be the 1st time that you've been asked to do that

46:28

since Kim once I don't know seal plus 2 H 2 gives methanol right expression procure calculated Q is just the equilibrium constant expression where these partial pressures are not necessarily equilibrium partial pressures that's all Q choose the equilibrium constant not equilibrium right so you have to take the partial pressures that you're given calculate you from them right you're not calculating equilibrium constant unless these are equilibrium pressures in which case you work

47:04

the equilibrium pressures and this would actually be the equilibrium

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constant but in this case report all the numbers immediate . 0 1 equilibrium constants 6 times 10 minus-3 so we compared decay we skew was bigger 1 is that mean it means that where

47:22

we are now is to product rich excuse higher than K because Protestant numerator of Q but if Q is higher than K and means for product Rich wrote relative to where we want to be at equilibrium In

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other words were here equilibrium constants here product wretch where to the right of equilibrium OK so obviously what's going to happen is wouldn't wanna go downhill in this direction we're not going to make methanol organ maker last nite had already make more these guys but

47:57

the answer to the question and if we

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wanted to we could calculate the reaction gives energy at that point right at every point along this curve we can if we wanted we know the composition of the system we can

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calculate the reaction gives energy it's just equal the various

48:16

right we just plug in for Q if we know the standard reaction gives energy wide and we can talk that gives energy at any point that we we did here endit we concluded that its positive while would have to be but the slope of this tangent here is obviously positive at that point for putting

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a positive number we knew we would know we made a mistake OK

48:42

and Silvester says we can change the conditions we can move the position of equilibrium and instead of here we can be over here now the reaction is going to proceed toward making a little bit more methanol where to go from here to here not a

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huge amount but a little bit this is what problem Tuesday look like write some version of these types of problems testing these concepts right here right any extra credit problem is right up the mid-term 1 looking a lot of