Lecture 17. Final Exam Review
Blake, Donald R.
UCI Chem 1B is the second quarter of General Chemistry and covers the following topics: properties of gases, liquids, solids; changes of state; properties of solutions; stoichiometry; thermochemistry; and thermodynamics. 
 Okay, so this is it You know last night I was thinking That as you get older Time just goes fast. I mean these lectures seem to just drag on and on. I don't know how you feel but But the time I mean I can't believe it's been ten weeks now anyway, so So this is it so We will after this we'll see each other on Thursday morning at 8 o'clock right here Of course the people that are in this room right now are the ones that usually show up the one the empty seats are the ones that Don't always show up My suggestion folks is that you show up a little early Okay The final will be about twice as long Okay, or what should take you about twice as long to to do and in the past We've had you know about 70 minutes for the exam, so maybe we'll be quite twice long But I don't want to have you people stringing in here until 8 10 8 15 in the morning I want to pass the exam out at 8 so that doesn't mean you show up at 8 that means you should be here be in your seats And ready to take the exam and have us passing it out at 8. Otherwise It's a problem. Okay All right, well folks we're on the the home stretch here Just a little bit about the chat. Lea I've had All kinds of questions. I even had a question last night This one person tends to be a little bit on the aggressive side, but he came up to me and he said So do you anticipate that the Thursday final exam will be Pretty much like the Friday final exam And I said well I said, you know I learned my lesson on the first exam I need to make them, you know different, but I'm not that creative. I can't make it so that there's I mean There's gonna be a Clausius-Clapeyron equation or question on there and whether you got to Take something of the e or you just plug in chug or you solve for t There's gonna be a Clausius-Clapeyron on the night exam and one on the morning exam So he said what would it be possible then for you to get me to email me the exam? The Thursday exam so I can practice I Mean I'm just shocked that I mean He's not even gonna try and cheat, you know, he's not gonna even ask you anyway, so Anyway We will be grading your exams on Thursday We will hopefully post I'll do what I did before send out an email that has your scores on the That that evening sometime on Thursday night, we will not though post the key Because of the next exam okay, so the key will be posted sometime Probably Saturday Because once again, we won't start grading the the the other exam goes from seven o'clock at night until Nine o'clock a night on Friday night. I can't believe that. Okay, okay So I Have tried in this ten weeks Sometimes I point out this is a good question. Okay, this would be a good exam question I'm hoping folks that not only have you learned something about maybe how to predict a Possible question from my class, but that it will work for you later. All right Here's an example. Okay, I want you to think of how could Don ask a multiple choice question From this slide. Okay, so the chalet. I love the chalet. Okay So if you change the concentration Can you shift equilibrium and can you change the equilibrium constant? Okay, so For this one if you change the concentration of something I don't care products or reactants does that shift equilibrium? Does it change the rate constant Does it change the rate constant no the rate constant we are trying to maintain that at all cost Okay So we we expect it to stay the same if I say the rate constant is point one Then it's point one no matter what you dump into it. Okay, and so if you pull something out you dump more in it Reequilibrates so that you're at point one. All right, so that's it's called a rate constant All right, so that's it's called a rate constant for a reason. Okay, and so you do not change the rate constant All right said the the equilibrium constant By changing the the number of bowls or the concentration. How about pressure? I pushed the button too soon Okay, does it change the equilibrium constant? Okay volume and Volume and pressure I'm a little bit confused at this I mean because I I see it as if you double the volume then you've cut the pressure in half Okay, or if you cut the volume in half you've increased the pressure so those to me Should be the same. Okay, so I didn't even ask you okay about temperature in a shift equilibrium sure We have a reaction where we have a plus b goes to c plus d and if there's some some heat given off If we add heat to one side it makes things happen. Okay, or it makes things slow down So absolutely changing the temperature can shift the equilibrium. Does it change the equilibrium constant? Yes, it does that's the one thing folks That is the one thing that changes the equilibrium constant is the temperature So I have to specifically say to you you add this temperature And I think what you'll find is is that in all the examples that we've done so far It'll say it I mean some wacky temperature maybe 1280 degrees, but but you it's at a certain temperature The the k changes now we know that We've seen plots Okay, we've seen figures where it was like on the y-axis. It was k and on this axis was temperature Okay, and if you have something something that's exothermic gives off energy as you warm up things The k at low temperature once again This is low temperature here and high temperature there low k high k starts here and it kind does this Okay, because as you warm things up more heat as a product You've already got heat so it slows the case so temperature absolutely can affect k a Catalyst we're gonna talk about catalyst at the end of the lecture It does not shift the equilibrium Okay And it does not change the equilibrium constant. Okay, so this is something that that we will get into in in an hour or so So folks This is a little bit of a hint there are At least 80 people missing Okay, I can see we didn't even want to say to turn off your cell phones because there are so many empty seats Okay, so I want you to look at this Look at this right here How many when I ask a multiple choice question, how many options do I usually give you? I like five Okay, let's say five So if I were to ask a question on Multiple choice for this What could I ask could I say When I change x equilibrium changes and I would say concentration Boom boom boom boom and if I and all of a sudden which one does not change. Can you see this is a nice question? It was one answer this different from the other four How about over here? So I'm telling you right now that the chances of these two being on the exam at least one of them is about 90% Okay It will just say I Don't know how I'd say it, but it'll be on there All right, I want you to do this for me right now calculate K for the reaction of Ethanol liquid I didn't know how to make an equilibrium sign I couldn't do that so this is supposed to I Can even copy and paste sorry this is supposed to be two arrows going different different ways goes to ethanol gas and The concentration or the pressure of this gas is point three three Three three three atmospheres. I want to know what is K? You got a liquid on one side and a gas on the other Where's that guy that always knows the answers What's your answer chief right there he says that's the K right K is equal to You know products the the pressure of the products over the pressure reactants, but this Okay, this has a value of Not of one Okay, so it's the pressure here over the reactants is equal to K So in this case K is equal to point three three three no no units are necessary. Okay? Everybody understand that Okay now I Want you to calculate the K for this reaction ethanol liquid ethanol gas to 253 Torr Who's got an answer? What is it? He says it's point three three three what is with that? Why isn't it 250? He says it's point three three three what is with that? Why isn't it 253? So do me this for those of you who are a little bit confused there is 760 Torr in an atmosphere So divide seven divide that number 253 by 760 Okay Divide that number that I gave you by 760 If I've done the math correctly you get point three three three So the problem folks is that while I told you at one point that Units didn't matter and they would cancel if you only have if you have an odd number of units It doesn't work that way. Okay, if I've got You know h2 on top certain pressure and eight and and two on the bottom Then the units cancel if I've got I don't care five Torr five Torr hydrogen is the product and Two Torr of nitrogen was the reactants. It's five over two and guess what? The k is is two and a half Okay, and if I converted all that to atmospheres. I'd still get two and a half problem is is that when you just have just one thing Okay, there's nothing to divide it by those units don't cancel and then the units do matter Okay, so the units do matter so you can't have this is the same reaction You can't have two different k's with the same thing. This is point three three three Torr and there's Pascal's and there's all kinds of Things that you can use. Okay, so we have to have some normalization And that is this Okay, I So I suggest that you either commit this to memory or try and understand the concept for aqueous It's one molar Okay, one molar are the parameters to use here. This is the reference state the activity here is the number of moles Over one molar or sorry the the molality the molarity of whatever it is divided by one molar Okay For eight for gaseous. It's the atmospheric partial pressure the partial pressure is in atmospheres now folks an atmosphere is the same as a bar Okay, but I don't use bars and I don't use Pascal's Okay, I use atmospheres and I use Torr So it's very possible. You'll get a question. It says something in Torr And it's something that's the door you're going to get in atmospheres. Okay, but give it to you in atmospheres and you're good to go So once again when you set up an equation for K you must use atmospheres Unless you have the same number of moles on top and bottom And what I mean folks is that if you've got a plus b goes to c plus d and a And b is equal to c and d in terms of the coefficient then everything cancels out But gases are this for liquids and solids folks it's a one Okay, so on that last thing we could say products would be in the partial pressure we had a point three three three Okay, the liquid so it's gonna be k is equal to point three three three Over and we had a liquid over one If you want to keep track of all your your your p's and q's there, that's what you would do Okay, so it's very important for you to understand that there are times you can use Torr and there are times you cannot use Torr There are times you can use milli Millimolar and there's times you can't use millimolar All right So we're gonna talk more about Delta G today so that other stuff was was more k equilibrium and all Delta G is Equal to the overall Delta G is equal to the Delta G of the products minus the Delta G of reactants Okay for everything except for one thing it's been products minus reactants for ten weeks Okay, and that's pretty much how it's going to stay products over reactants So the reactants this is an energy diagram folks if this is energy this way I'm not sure what's this way maybe time The reactants have an energy here if they actually do this and the products are lower in energy What that means folks is that the products are lower in energy, but right here is a big plus heat Okay, if this is actually lower than this Then it's this the energy had to go somewhere Okay, the energy is going to be products plus heat. All right, so an exothermic reaction. It looks like this Endothermic would be from here to here Okay, you have to give these guys some heat to get it to be up there Certainly not spontaneous and we know that if Delta G is less than zero Okay, if you do a calculation Delta's less than zero that Q is less than K and It's going to Head that direction. All right. Yeah Delta G is equal to zero word equilibrium Okay, word equilibrium and Q equals K and if Delta G is positive It is not spontaneous and Q is actually bigger than K So you got to go from the reactants of the product side to the reactants side? So We're going to go into this a lot But I would suggest you just write it down the Delta G is equal to Delta G not plus RT log Q So that Q has come back Okay, the Q that I made it sound like early on was not all that important turns out to have reared its ugly head again So it's a good thing that I didn't trivialize it too much earlier So Delta G is equal to Delta G not plus RT log Q. We're going to be going over those several of those So don't don't panic with this And then at equilibrium When Delta G is equal to zero then Delta G not plus RT log K is equal to zero Okay do this think about it What are Delta G and K at equilibrium at 25 degrees C for this Well last night I said well, it's obviously it's this This but it's wrong Okay, I went through this big explanation of what if I had just water sitting here is it boiling look at this Does that look like it's boiling to you? That looks like boiling water to me So I said oh well It's obviously 25 degrees isn't boiling Okay, and therefore water would rather be liquid than gas at room temperature So that obviously Delta G must be equal to zero or greater than zero turns out the key word right here is equilibrium folks Equilibrium Delta G equals zero do not forget that so don't read these problems on the exam so fast like I did last night The ice that you screw up Okay, that's the key if something has said it equilibrium folks it could cut out a whole lot of work that you need to do Look at that Okay So we did a we've done several of these folks where it is One reaction at least another reaction at least another reaction So it's important for you to understand we did this I don't know I can't remember what chapter enthalpy was chapter six maybe chapter seven. I don't matter So I would like you to do this just humor me okay What is K sub one in this the way it's written right here a Going to be okay. What is K sub one? I? Hope you're gonna say that it's gonna be The concentration of B over a is equal to K sub one write that down please just write it down I have a reason for asking you to do this So here we have a going to be is K plus one So now I want you to write down what is case of two? So case of two is B goes to C and That is case of two and that should be Concentration is case of two is equal to the concentration of C over the concentration of B? And how about this one? From these first two folks You should be able to figure out what this is okay? so If you if you do so now just write K one is Equal to B over a and K two is equal to C over B Okay, multiply those two together. I Don't mean K one K two. I mean what they were equal to okay, so K one is equal to B over a Okay, and K two is equal to C over B So you know this you wrote that all of you probably wrote that case of three is equal to C over a correct But what I'm saying is is that you can show yourself that that's the case Because to get there from a to C you knew that B over a was equal to K one and C over B was equal to K two So look at this for this first one K one B over a B over a That's K one K two is C over B. That's K two And then look at these guys cancel They cancel so you're just left with C over a All right, so you could sort of work that way So in this case to find K sub three you just multiply K one times K two is equal to K three for the Enthelpiece of those reactions you add them okay So H sub three is equal to H one the delta H one plus delta H two So folks this encompasses quite a few things Okay, it encompasses the equilibrium part this part is to sort of refresh your memory on what we did in chapter Six or seven whatever it was So don't forget how to do this type of thing This is like a little bit of not like a Hess's law, but it's a I love Hess's law So once again, it's just sort of playing with with these things All right, so Delta G not of reaction is equal to the Delta H not of reaction minus T Delta S not of reaction Okay, this is equal to once you just like the previous pay Delta G of formation minus of products minus reactants and I wish I had been able to write something up here, but this if you remember folks that Delta G of reaction is equal to minus RT log K Okay, and all this is is a rearrangement of that Okay In fact, let's do this right now Write this exact thing Okay, and now we're gonna do something in reverse We're gonna do something in reverse of what we've been doing So you write that And Then I want you to take the log the natural log of both sides the natural log of both sides That's Ln I Suggest you write this down. I mean maybe you got these kind of brains that allow you to visualize these things That's wonderful. Yes. You can't visualize on the exam. I Need to see it so make sure that you're good going from visualization to putting it on the paper So what is if we take the log of both sides, what do we get? What do we get Log K is equal to That's right Okay, so what you're saying is is that you just do this you just take the log of this makes this a log K take the log of this You just end up with this Okay So you don't you only have to memorize one thing folks, okay? I tend to memorize that Delta G is equal to minus RT log K But you can play with it any way you want so folks if I give you a K And I say calculate Delta G not a reaction. That's how you do it Okay, so look at this right here R is a constant T is not a constant Delta G not is not a constant and this isn't a constant So I can if I give you two or three things here you can figure out different things Okay, so once again, this is a nice way I you know whether you want to memorize that or not as long as you memorize that Delta G not a reaction is equal to minus RT log K they both work Okay, we know this okay that if Delta G is less than zero then The reaction is going to proceed from reactants to products if Delta G is greater than zero. It's going to come back this direction Okay, what happens if we increase the temperature In a case where Delta G is less than zero Think about it. Okay, folks. I want you to I I've had students come to me and they say, you know I can't I can't visualize this or I can visualize this but I can't mathematically solve this There are sometimes when you can do it both ways Okay, when you can visualize it and you say okay, I can understand this and then the other times that you can actually prove it Yourself mathematically. All right, some of you have stronger math skills than others I just want you to think about this. You got something like this Delta G is negative What usually does that mean Not always but it usually means that there's some heat generated Okay, remember heat given off negative Delta H Delta G is equal to Delta H minus T Delta S So if you've got something that is favorable Okay, and it's spontaneous Then there oftentimes you're you're you have heat that's given off. Okay, so as you increase the temperature for something like this Is that going to make the reaction go faster or slower? I heard faster, but it's slower. Okay k decreases okay if you remember we had a plot of Delta G versus temperature and On something that was exothermic it looked like this Okay, the higher the temperature the Delta G decreased All right, or it was there's another with K like this. Okay was here and temperature was here And it went like this other ones that went like this So if you have folks if you have heat that is generated you need to think of this as adding heat right there Okay, yeah heat right there, and then you add heat is That going to force things in this direction or this direction it goes back this way Okay, if I add all of you I think understand if I add if I double the amount of D By double the amount of D. Does the reaction tend if it's an equilibrium at equilibrium? I add more D. Which way does the reaction go? Okay, so if folks What you don't see because we don't use to put in there is heat Heat is over here, and it really should be figured in on this equilibrium constant. Okay, so if we have heat And now we're to equilibrium and I add more heat what happens? Okay, so folks if you increase the temperature K tends to decrease When you have a Delta G that is smaller than zero and it'll increase here Okay, you decrease the temperature then these things just flip boom boom So once again, I want you to I've showed you figures. I've showed you plots I want you to try and understand this. This is just a conceptual thing. There's no math involved here aside from the fact That you know that the rate the K equilibrium for this is you know cease Whatever the coefficient is Concentration of C times the concentration D divided by the constant of D rotation of B and the concentration of a So if you have something else here if you add something here it tends to come back this way if you add something on this side It goes that way So Delta G is not as simple at least it's not simple for my brain, and I don't understand why I struggle with this but I Thought I understood Delta G, and now I'm not sure I do so So the Delta G of reaction Okay, and once again, this is the the molar is products minus reactants, okay? So I've Written this okay that the value of Delta G at a particular stage of the reaction. Okay once again We've had plots where it was Delta G and time and you remember how I had one where I said Okay, this is all the the reactants side starting here and the products pure products there pure reactants here We add stuff in and this is like a the reaction the progress of the reaction or whatever on this and This was Delta G and we had a Delta G that kind of came down like this and right here it hit Zero this is where equilibrium was okay, so folks on this screen right now if this is where equilibrium is on this scale Okay, there's equilibrium Do you think we have are we favoring the reactant side or the product side? Reactive doesn't mean we don't we have product we have product right because this is about I don't know a third or fourth of the Way there, so we have some products, but we have more reactants about now I Think we're not quite half about now We're starting to favor the products more more more more more more this is one where K folks is like ten thousand Okay, K is huge and that Delta G Came down like this and only right there Had a minimum of zero okay, so equilibrium was right there Okay, so the bigger the K the further this way we are this model the K I mean if K is zero point zero zero zero zero one We're right here that Delta G went And had a minimum somewhere right here or up there doesn't matter where the minimum was It's just that the minimum happened on this side So Delta G is changing during the reaction and that's what we need to be able to calculate So the value of Delta G at a particular stage of the reaction Somewhere in here, okay is the difference in the molar Gibbs free energy molar means the molar and it's at standard pressure and concentration of the Productive reactions at the partial pressure or concentrations that they have at that stage Not what you find in a book Okay, what you find in a book is you have this and you have this and you look it up You say Delta G is equal to minus 147 kilojoules per mole, okay But we're not at equilibrium. We're not at standard. Okay, not everything works at one molar or one atmosphere So what this Delta G tells us is where we are at that stage and once again is weighted by these coefficients So Delta G of reaction is equal to this is what you find in a book Okay, the reaction is this plus RT log Q Q folks when Q is equal to K. We're at equilibrium, right? If Q is smaller than K, we go this direction Q is bigger than K. We go in this direction Okay, so Q tells us something and so that the Delta G this is the this is like what you really get Okay, you remember how we talked about Delta H Okay, Delta H is the heat of reaction Problem is is that does not really tell you What whether it's even spontaneous or not? Okay, Delta H of a negative value means it's exothermic So you got a Delta H of a hundred kilojoules per mole That but that didn't tell us how much we got out we had to invoke entropy an Entropy is very dependent on temperature. So then we came up with Delta G, which is the Gibbs free energy, which is Delta H minus T Delta S So now in order to tell whether something is spontaneous you have both the enthalpy and the entropy Okay So that was Delta G now what we're finding is that this Delta this new Delta G is the one that tells us sort of where We are in this reaction and this The the Q part, okay I think this is this this is like a little bit the Delta G are is a little bit like a correction factor the further You are from standard conditions the bigger or smaller this is okay, so look at this up here if we're at standard conditions This is zero Okay As you move further away from standard conditions this starts to become a little bit more and more important So if you're way off of standard conditions then this becomes significant Okay We're gonna do a few problems. Hopefully when you walk out of here, you won't be any more confused when you walked in and Charlie put this together for me. I'm gonna put this on the the class website I'm not gonna go over it right now, but it I think it's a nice little review Because I want to get right to this question. Let's do this question This is 2.2 from your book calculating the Gibbs free energy of reaction from the reaction quotient Okay, the standard Gibbs free energy of reaction for two SO2 gas plus an O2 gas going to two SO3 gases is Delta G of reaction knot is equal to a hundred and forty one point seven kilo Joules per mole at twenty five point zero zero degrees C What is the Gibbs free energy of reaction when the partial pressure of each gas this this and this is 100 bar which is 100 atmospheres folks, okay? 100 atmospheres okay Let's answer that question first. So what is the Delta G of? Reaction, okay. Well, what is the Gibbs free energy of reaction when we have this now you're given the standard? free energy so Remember it was Delta G of reaction is equal to this Plus RT log Q It's a new reaction, so I'll give it to you. I should if I had just copied and pasted this I'd have that reaction here So don't forget Delta G R is equal to Delta G R not which is right there plus RT log Q Oh By the way, there's a mistake here in your book This should say PSO 2 SO 2 is right there that should be the PSO because that's the one that's squared That's on the the reactant side, so you've got oxygen which has a coefficient of 1. This is PO 2. This should be PSO 2 It's your book not me at this time, okay, and I can't change it Okay Okay, so Q folks that Part that is sort of the correction factor Is 0.01 Alright 0.01 Don't forget It was Delta G R is equal to Delta G are not means it standard Plus RT log Q Okay Q is that Okay log of that anything what is the log of one? Everybody take your calculators hit enter one or whatever and then hit log of it and L Ln of one is equal to zero So anything bigger than one one point one one point two six hundred and twenty five Gives you a positive value coming out of that log and in less than one point nine nine point nine zero zero point Or zero point zero one gives you a negative number Okay, so you've got a negative number Maybe not a very big number, but it's a negative number than times RT Okay, so what is the answer? Okay, well here's the first part this is what you get under standard conditions right there Delta G are not is right there That was given to you Okay, and then what they did was they multiplied R times T and they get this log of this. That's a small number Okay, I wish there was one more step in here I think this value is something like this times this is I think minus 12 kilojoules Add them together and you get this How many got that How many cheated by looking at the notes Okay So folks this is a if you cheated. Oh, yes Thought you just a slow answer. Yes Which are okay, so You look at the units Delta G is always in Jules Okay, it's always in Jules enthalpy is always in Jules and so The R that you would use for this is the R that has Jules in it, which is the 8.31 Okay 8.31 Jules per degree mole So you have that in there you multiply by T since you had per degree The T's can get rid of each other and you're just left with killed Jules per mole So your answer and it has to be look at this you got killed Jules per mole here You have to have the same units for this To give you this now don't forget that is in Jules the R is in Jules So you've got to divide by a thousand if you're going to add to this what you do not want to do is to say okay This is that this part is 2479 Jules is what it is you can't add 279 Jules to this and come up with 2000 and something so you've got to convert this to to kill Jules and then this is the answer So folks what this says yes, so I said everybody email this morning This morning, and I said no graphing calculators. We're gonna check at the door And I know I'm not happy about this, but I got some emails from people telling me that they overheard people who said they're gonna cheat They're taking a put calculation. They're gonna put equations and stuff in their calculators. That's what I heard I Told the class last night. They could use graphing calculators, and now I'm saying no Go to Target go to the student store by a ten dollar scientific calculator which still can do a whole lot more than Now I know how to use them. I'm sorry folks I wish it weren't this way But think of it this way if people are actually going to program Equations and things into their calculators They're gonna get a better score than they deserve and since this class is graded on a curve That means that their cheating is going to maybe put them above you and I do not want you to suffer So there's the cost-benefit analysis here. So you got you got to spend money ten dollars Okay, which I prefer you not have to spend and you have to get used to new calculator I'm sorry. I don't know what what what will you like to do? No calculator There's a concept So no graphing calculators, sorry none Okay, so look at the difference at standard conditions This would be a hundred and forty one negative hundred and forty one kilojoules But under the conditions that we were in invoking on this it went up even greater Okay, it decreased even more. So this is what we've got. Okay, so to try and convince you of this with before we had 100 atmospheres for everything, okay Do it with one atmosphere for everything One atmosphere do it I'll go back to the The previous page all products and reactions are at one atmosphere Who's got the answer yes you Negative one forty one Why because folks if Q is one times one over one times one or whatever It's one Log of one is zero so this whole thing falls out and so the delta g of reaction is equal to del G not of reaction Okay, try one and a half atmospheres one and a half atmospheres for everything try it So one and a half atmospheres oops where they go So it's going to be one and a half squared times one and a half squared times one and a half So I think your answer is going to be one over one and a half The Q will be one over one and a half Because I got a square there and a square there so those just cancel so she's gonna be one over 1.5 What is the log of one over 1.5? What is the log of one over 1.5? How many don't understand why I'm asking that question yes Yes No, no, no, no I already know that I'm going to be doing a reaction No, no, no, no I already said Okay So once you put things into this standard state, okay, there are no units It's all divided by one atm That yes, you're right. That's we have to kind of do that. Otherwise you're right you end up with you know Unless you have the exact same number of moles of gas and then you end up with a unit in the log So it's everything it's going to be a hundred atmospheres divided by one atmosphere gives you a hundred That's why there's just hundreds here and no units. Yes I don't care what you give me but don't forget they got to be the same Okay, if you want to give it to me in joules, then you got to multiply by thousand that hundred and forty one So that's a big number Okay, so what is q? What is the log of one point five one over one point five? Negative four something huh, so what is that then what if we then plug that in what do we get for the RT log q part? So Right there, what do we get for this? Negative one So negative one is what we get if we have a half of an outlet one and a half atmospheres Okay, one and a half atmospheres is pretty darn close to standard And therefore the correction this whole thing is negative one Whereas at a hundred atmospheres for each thing it was a negative 13 or something like that 12 Okay, so once again, this is the correction factor. This is the correction that says the further we are away from from standard state The bigger this thing or the the bigger or smaller depending this is going to be okay So hopefully understand when and how to use the Delta G thing do this one I Predicting the value of K from the standard Gibbs free energy of reaction at 25 degrees C The standard Gibbs free energy of reaction for half hydrogen gas plus a half iodine gas goes to a whole iodine Hi is one plus 1.70 kilojoules per mole calculate the equilibrium constant for this reaction So You've got temperature and you've got G So think back to an equation that I gave you Where you could calculate K So log K is equal to minus Delta G R over or Delta G of reaction over RT. So you've got this You know what this is and you know what that is? So you just plug this in 1.7 times 10 to the 3 now we're in jewel we've converted out of jewels instead of kilojoules Okay, because then this is in jewels So then we just have this in jewels this in jewels this in K and you come up with the log K is equal to minus 0.685 Yes We're gonna get there it's a good question Okay, so What they did was they just went ahead and solved they remembered the equation okay, that They sort of rearranged it because I was remember it as Delta G is equal to minus RT log K But doesn't matter remember however you want and just plug stuff in and then you just solve for X All right, so in this case they did it where log K is equal to this Okay, now Well, I want just what K is not what the log of K is So I just take this side of the e-power and this side of the e-power. There's a button on your calculator for that And you get this e to the log X is equal to X so K is equal to point five How many got that yes? This no these are just to try and show you You know the relationship between K and whatever no you will not have to draw that I will not ask you a question about that So how many got that point five? Don't forget this Okay, do this one suppose that in an equilibrium mixture of HCl Cl2 and H2 the partial pressure of H2 is 4.2 Millipascals once again, I'm not going to give you a milli-pascals. I'm going to give you atmospheres or Tor For cl2 it is 8.3. What is the partial pressure of HCl at 500 degrees? K given that the equilibrium constant so you're giving equilibrium constant is 4 times 10 of the 18 and Here's the reaction H2 plus HCl goes to 28 plus Cl goes to 2 HCl right there. That's your equation right there So you want to calculate the equilibrium composition so what we're asking for is what is the partial pressure of this? Okay Bless you, okay, so You got a K You got these little wacky things. I'm just going to tell you right now that 4.2 Millipascals is equal to 4.2 times 10 of the minus 8 atmospheres So this is 4.2 times 10 of the minus 8 atmospheres and this is 8.3 times 10 of the minus 8 atmospheres It's a conversion you're not going to have to do on the exam You Okay, so what we know is that K is equal to the pressure of the pressure of hydrogen times the pressure of Cl. So we can rearrange this and say we want to know what is the pressure of HCl. So we just take the HCl, we move these guys to the other side and we just say that K is now equal to the pressure of hydrogen times the pressure of Cl2, okay, and they do too many steps for me, but because HCl is squared here then, then they take the square root of that whole mess. So here you've got the pressure of HCl is equal to, this is the value you are given for the K, this is the pressure of the hydrogen, this is the pressure of the chlorine, and because we are looking for this, this is the square, we take the square root of that and it is equal to about 120. So K is equal to about, I think I got 118. Yes? Pressure has to be 8m. Pressure has to be 8m, yeah. 8m is a bar, okay. However, if, yeah, let's leave it like that. Yes? Loud. At a 2, why would you do that? What do you mean a 2? A coefficient of 2, because it's 1 and 1 and 1. Okay, okay, he's asking if instead you had done, instead of a half and a half of the coefficient, so you could have just gone 1, 1 and 2. I'm asking you, do you think it matters? The same answer. You get the same answer? Yes. Okay, then you can do it. So can we do this problem, folks? Okay, let's do this. This is an easy one. All we're doing is we're trying to figure out Q. Unfortunately, we're doing Q. Don't be fooled, folks. Look at this. The reaction, wherever the reaction is, oh never mind, there's no trick. I thought there was a liquid or a solid in there, but I guess it's not. Oh, no, iodine gas, okay. So here, folks, they give you a K. You can't answer this problem if you don't know what the K is, okay? Because the question is which way is the reaction going to go? So you figure out Q, if Q is bigger than K, then it goes, if Q is bigger than K, it goes this direction. If Q is smaller than K, it goes that direction. So just plug everything in. K is 46. What does that tell you, folks? What does that tell you? If I gave you this and I said draw, draw the delta G versus temperature, delta G is here, temperature is there. It would go way over here. 46 is a pretty good size number. So it would minimize, it would reach the equilibrium somewhere over in this area. If I said 4,600, it would be even further over. So if I said 4,600, it would be even further over. So, 4,600, it would be even further over. So in this case, folks, we had up with a Q of 1. Q of 1, K of 46. Which way is the reaction going to go? Toward the products. That's right. So folks, I'm not trying to confuse you, but on a question like this, where you have the same number of moles, you've got two moles of gas on top and two moles of gas on the bottom. If I had said that all gases are at a pressure of 55 Torr, 55 Torr, okay, instead of 0.55, guess what? You got Torr squared, you got Torr times Torr, you got Torr squared. Everything cancels. It's still going to be, it's going to be 55 squared over 55 and 55, you still get a Q of 1, okay, so you don't, in some cases, you don't have to do the conversion. You just have to do the conversion when there are different number of moles on top and bottom. Okay, so the reaction is going to proceed toward the products. Something quick about catalysts. For catalyst folks, if we have AB here and we're going to CD, the distance between here and here, okay, that is how much energy you get off, all right? So you've got to climb this hill, this activation energy, and then you go down there, okay? So about this much energy is given off going from AB to C plus D. What a catalyst does, it just makes the path easier, okay? The difference between here and here is the exact same, okay? It's a little bit like if I said, let me see, that I want you to go from here to Babua Island, okay? Here to Babua Island, but the path of A plus B goes to C plus D that I'm asking half of you to take is you go to Babua Island from this room, but you have to climb over Saddleback Mountain. So your path is from here down to there, over the mountain, and then to Babua Island, that's this, okay? Or there's the one where you leave this building and you walk to the beach, that's this one, okay? Now the amount of energy, the difference in potential energy between you being here and a new port beach doesn't matter whether you went to Big Bear Mountain or Kilimanjaro, okay? The catalyst provides a different pathway, okay? And this, believe it or not, this little hill right here is really what determines how fast something goes. But this is really a tall one, okay? Then we're going to stay more where A and B is, okay? Not get to C and D. Even though you can see going from here to here makes thermodynamic sense, but if this hill is too big, you just don't do it. A catalyst allows a different path. So in engineering folks, using a catalyst for this can sometimes save you millions or billions of dollars or make you millions and billions because you've actually been able to come up with a way to turn sand into gold, okay? Or coal into diamonds, okay? Coal to diamonds folks, this will go way up there, okay? You can come up with a different way to make diamonds, you know, good for you. So it doesn't change the K, nor does it shift the equilibrium in the system. It just allows a different path. I said all this. Man, I'm good. Okay, so this is it. I think Josette was on a trip this, yeah. So she's not in this picture. Anyway, so folks, I hope you've learned something. I will see you all on Thursday, I guess a week. A week from today, but a little earlier. Study hard. Thank you.
 Okay, so this is it  You know last night I was thinking  That as you get older  Time just goes fast. I mean these lectures seem to just drag on and on. I don't know how you feel but  But the time I mean I can't believe it's been ten weeks now anyway, so  So this is it so  We will after this we'll see each other on Thursday morning at 8 o'clock right here  Of course the people that are in this room right now are the ones that usually show up the one the empty seats are the ones that  Don't always show up  My suggestion folks is that you show up a little early  Okay  The final will be about twice as long  Okay, or what should take you about twice as long to to do and in the past  We've had you know about 70 minutes for the exam, so maybe we'll be quite twice long  But I don't want to have you people stringing in here until 8 10 8 15 in the morning  I want to pass the exam out at 8 so that doesn't mean you show up at 8 that means you should be here be in your seats  And ready to take the exam and have us passing it out at 8. Otherwise  It's a problem. Okay  All right, well folks we're on the the home stretch here  Just a little bit about the chat. Lea  I've had  All kinds of questions. I even had a question last night  This one person tends to be a little bit on the aggressive side, but he came up to me and he said  So do you anticipate that the Thursday final exam will be  Pretty much like the Friday final exam  And I said well I said, you know I learned my lesson on the first exam  I need to make them, you know different, but I'm not that creative. I can't make it so that there's I mean  There's gonna be a Clausius-Clapeyron equation or question on there and whether you got to  Take something of the e or you just plug in chug or you solve for t  There's gonna be a Clausius-Clapeyron on the night exam and one on the morning exam  So he said what would it be possible then for you to get me to email me the exam?  The Thursday exam so I can practice  I  Mean I'm just shocked that I  mean  He's not even gonna try and cheat, you know, he's not gonna even ask you anyway, so  Anyway  We will be grading your exams on Thursday  We will hopefully post I'll do what I did before send out an email that has your  scores on the  That that evening sometime on Thursday night, we will not though post the key  Because of the next exam okay, so the key will be posted sometime  Probably Saturday  Because once again, we won't start grading the the the other exam goes from seven o'clock at night until  Nine o'clock a night on Friday night. I can't believe that. Okay, okay  So I  Have tried in this ten weeks  Sometimes I point out this is a good question. Okay, this would be a good exam question  I'm hoping folks that not only have you learned something about maybe how to predict a  Possible question from my class, but that it will work for you later. All right  Here's an example. Okay, I want you to think of how could Don ask a multiple choice question  From this slide. Okay, so the chalet. I love the chalet. Okay  So if you change the concentration  Can you shift equilibrium and can you change the equilibrium constant? Okay, so  For this one if you change the concentration of something I don't care products or reactants does that shift equilibrium?  Does it change the rate constant  Does it change the rate constant no the rate constant we are trying to maintain that at all cost  Okay  So we we expect it to stay the same if I say the rate constant is point one  Then it's point one no matter what you dump into it. Okay, and so if you pull something out you dump more in it  Reequilibrates so that you're at point one. All right, so that's it's called a rate constant  All right, so that's it's called a rate constant for a reason. Okay, and so you do not change the rate constant  All right said the the equilibrium constant  By changing the the number of bowls or the concentration. How about pressure?  I pushed the button too soon  Okay, does it change the equilibrium constant?  Okay  volume and  Volume and pressure I'm a little bit confused at this  I mean because I I see it as if you double the volume then you've cut the pressure in half  Okay, or if you cut the volume in half you've increased the pressure so those to me  Should be the same. Okay, so I didn't even ask you okay about temperature  in a shift equilibrium  sure  We have a reaction where we have a plus b goes to c plus d and if there's some some heat given off  If we add heat to one side it makes things happen. Okay, or it makes things slow down  So absolutely changing the temperature can shift the equilibrium. Does it change the equilibrium constant?  Yes, it does that's the one thing folks  That is the one thing that changes the equilibrium constant is the temperature  So I have to specifically say to you you add this temperature  And I think what you'll find is is that in all the examples that we've done so far  It'll say it I mean some wacky temperature maybe  1280 degrees, but but you it's at a certain temperature  The the k changes now we know that  We've seen plots  Okay, we've seen figures where it was like on the y-axis. It was k and on this axis was temperature  Okay, and if you have something something that's exothermic gives off energy as you warm up things  The k at low temperature once again  This is low temperature here and high temperature there low k high k starts here and it kind does this  Okay, because as you warm things up more heat as a product  You've already got heat so it slows the case so temperature absolutely can affect k a  Catalyst we're gonna talk about catalyst at the end of the lecture  It does not shift the equilibrium  Okay  And it does not change the equilibrium constant. Okay, so this is something that that we will get into in in an hour or so  So folks  This is a little bit of a hint there are  At least 80 people missing  Okay, I can see we didn't even want to say to turn off your cell phones because there are so many empty seats  Okay, so I want you to look at this  Look at this right here  How many when I ask a multiple choice question, how many options do I usually give you?  I like five  Okay, let's say five  So if I were to ask a question on  Multiple choice for this  What could I ask could I say  When I change x equilibrium changes and I would say concentration  Boom boom boom boom and if I and all of a sudden which one does not change. Can you see this is a nice question?  It was one answer this different from the other four  How about over here?  So I'm telling you right now that the chances of these two being on the exam at least one of them is about 90%  Okay  It will just say I  Don't know how I'd say it, but it'll be on there  All right, I want you to do this for me right now calculate K  for the reaction of  Ethanol liquid I didn't know how to make an equilibrium sign I couldn't do that so this is supposed to I  Can even copy and paste sorry this is supposed to be two arrows going different different ways  goes to ethanol gas and  The concentration or the pressure of this gas is point three three  Three three three atmospheres. I want to know what is K?  You got a liquid on one side and a gas on the other  Where's that guy that always knows the answers  What's your answer chief right there he says that's the K  right  K is equal to  You know products the the pressure of the products over the pressure reactants, but this  Okay, this has a value of  Not of one  Okay, so it's the pressure here over the reactants is equal to K  So in this case K is equal to point three three three no no units are necessary. Okay?  Everybody understand that  Okay now I  Want you to calculate the K for this reaction ethanol liquid ethanol gas to 253 Torr  Who's got an answer?  What is it?  He says it's point three three three what is with that?  Why isn't it  250?  He says it's point three three three what is with that? Why isn't it 253?  So do me this for those of you who are a little bit confused there is  760 Torr in an atmosphere  So divide seven divide that number 253 by 760  Okay  Divide that number that I gave you by 760  If I've done the math correctly you get point three three three  So the problem folks is that while I told you at one point that  Units didn't matter and they would cancel if you only have if you have an odd number of units  It doesn't work that way. Okay, if I've got  You know h2 on top certain pressure and eight and and two on the bottom  Then the units cancel if I've got I don't care five Torr five Torr hydrogen is the product and  Two Torr of nitrogen was the reactants. It's five over two and guess what?  The k is is two and a half  Okay, and if I converted all that to atmospheres. I'd still get two and a half problem is is that when you just have just one thing  Okay, there's nothing to divide it by those units don't cancel and then the units do matter  Okay, so the units do matter so you can't have this is the same reaction  You can't have two different k's with the same thing. This is point three three three Torr and there's Pascal's and there's all kinds of  Things that you can use. Okay, so we have to have some  normalization  And that is this  Okay, I  So I suggest that you  either commit this to memory or try and understand the concept for aqueous  It's one molar  Okay, one molar are the parameters to use here. This is the reference state the activity here is the number of moles  Over one molar or sorry the the molality the molarity of whatever it is divided by one molar  Okay  For eight for gaseous. It's the atmospheric partial pressure the partial pressure is in atmospheres now folks an atmosphere is the same as a bar  Okay, but I don't use bars and I don't use Pascal's  Okay, I use atmospheres and I use Torr  So it's very possible. You'll get a question. It says something in Torr  And it's something that's the door you're going to get in atmospheres. Okay, but give it to you in atmospheres and you're good to go  So once again when you set up an equation for K you must use atmospheres  Unless you have the same number of moles on top and bottom  And what I mean folks is that if you've got a plus b goes to c plus d and  a  And b is equal to c and d in terms of the coefficient then everything cancels out  But gases are this for liquids and solids folks it's a one  Okay, so on that last thing we could say products would be in the partial pressure we had a point three three three  Okay, the liquid so it's gonna be k is equal to point three three three  Over and we had a liquid over one  If you want to keep track of all your your your p's and q's there, that's what you would do  Okay, so it's very important for you to understand that there are times you can use Torr and there are times you cannot use Torr  There are times you can use milli  Millimolar and there's times you can't use millimolar  All right  So we're gonna talk more about Delta G today so that other stuff was was more k equilibrium and all  Delta G is  Equal to the overall Delta G is equal to the Delta G of the products minus the Delta G of reactants  Okay for everything except for one thing it's been products minus reactants for ten weeks  Okay, and that's pretty much how it's going to stay products over reactants  So the reactants this is an energy diagram folks if this is energy this way  I'm not sure what's this way maybe time  The reactants have an energy here if they actually do this and the products are lower in energy  What that means folks is that the products are lower in energy, but right here is a big plus heat  Okay, if this is actually lower than this  Then it's this the energy had to go somewhere  Okay, the energy is going to be products plus heat. All right, so an exothermic reaction. It looks like this  Endothermic would be from here to here  Okay, you have to give these guys some heat to get it to be up there  Certainly not spontaneous and we know that if Delta G is less than zero  Okay, if you do a calculation Delta's less than zero that Q is less than K and  It's going to  Head that direction. All right. Yeah Delta G is equal to zero word equilibrium  Okay, word equilibrium and Q equals K and if Delta G is positive  It is not spontaneous and Q is actually bigger than K  So you got to go from the reactants of the product side to the reactants side?  So  We're going to go into this a lot  But I would suggest you just write it down the Delta G is equal to Delta G not plus RT log Q  So that Q has come back  Okay, the Q that I made it sound like early on was not all that important turns out to have reared its ugly head again  So it's a good thing that I didn't trivialize it too much earlier  So Delta G is equal to Delta G not plus RT log Q. We're going to be going over those several of those  So don't don't panic with this  And then at equilibrium  When Delta G is equal to zero then Delta G not plus RT log K is equal to zero  Okay do this think about it  What are Delta G and K at equilibrium at 25 degrees C for this  Well last night I said well, it's obviously it's this  This but it's wrong  Okay, I went through this big explanation of what if I had just water sitting here is it boiling look at this  Does that look like it's boiling to you?  That looks like boiling water to me  So I said oh well  It's obviously 25 degrees isn't boiling  Okay, and therefore water would rather be liquid than gas at room temperature  So that obviously Delta G must be equal to zero or greater than zero turns out the key word right here is equilibrium folks  Equilibrium Delta G equals zero do not forget that so don't read these problems on the exam so fast like I did last night  The ice that you screw up  Okay, that's the key if something has said it equilibrium folks it could cut out a whole lot of work that you need to do  Look at that  Okay  So we did a we've done several of these folks where it is  One reaction at least another reaction at least another reaction  So it's important for you to understand we did this  I don't know I can't remember what chapter enthalpy was chapter six maybe chapter seven. I don't matter  So I would like you to do this just humor me  okay  What is K sub one in this the way it's written right here a  Going to be okay. What is K sub one? I?  Hope you're gonna say that it's gonna be  The concentration of B over a is equal to K sub one write that down please just write it down  I have a reason for asking you to do this  So here we have a going to be is K plus one  So now I want you to write down what is case of two?  So case of two is B goes to C and  That is case of two and that should be  Concentration is case of two is equal to the concentration of C over the concentration of B?  And how about this one? From these first two folks  You should be able to figure out what this is okay?  so  If you if you do so now just write K one is  Equal to B over a and K two is equal to C over B  Okay, multiply those two together. I  Don't mean K one K two. I mean what they were equal to okay, so K one is equal to B over a  Okay, and K two is equal to C over B  So you know this you wrote that all of you probably wrote that case of three is equal to C over a correct  But what I'm saying is is that you can show yourself that that's the case  Because to get there from a to C you knew that B over a was equal to K one and C over B was equal to K two  So look at this for this first one K one B over a B over a  That's K one K two is C over B. That's K two  And then look at these guys cancel  They cancel so you're just left with C over a  All right, so you could sort of work that way  So in this case to find K sub three you just multiply K one times K two is equal to K three  for the  Enthelpiece of those reactions you add them okay  So H sub three is equal to H one the delta H one plus delta H two  So folks this encompasses quite a few things  Okay, it encompasses the equilibrium part this part is to sort of refresh your memory on what we did in chapter  Six or seven whatever it was  So don't forget how to do this type of thing  This is like a little bit of not like a Hess's law, but it's a I love Hess's law  So once again, it's just sort of playing with with these things  All right, so  Delta G not of reaction is equal to the Delta H not of reaction minus T Delta S not of reaction  Okay, this is equal to once you just like the previous pay Delta G  of formation minus of products minus reactants and  I wish I had been able to write something up here, but this if you remember folks that  Delta G of reaction is equal to minus RT log K  Okay, and all this is is a rearrangement of that  Okay  In fact, let's do this right now  Write this exact thing  Okay, and now we're gonna do something in reverse  We're gonna do something in reverse of what we've been doing  So you write that  And  Then I want you to take the log the natural log of both sides the natural log of both sides  That's Ln I  Suggest you write this down. I mean maybe you got these kind of brains that allow you to visualize these things  That's wonderful. Yes. You can't visualize on the exam. I  Need to see it so make sure that you're good going from visualization to putting it on the paper  So what is if we take the log of both sides, what do we get?  What do we get  Log K is equal to  That's right  Okay, so what you're saying is is that you just do this you just take the log of this makes this a log K take the log of this  You just end up with this  Okay  So you don't you only have to memorize one thing folks, okay?  I tend to memorize that Delta G is equal to minus RT log K  But you can play with it any way you want so folks if I give you a K  And I say calculate Delta G not a reaction. That's how you do it  Okay, so look at this right here  R is a constant  T is not a constant Delta G not is not a constant and this isn't a constant  So I can if I give you two or three things here you can figure out different things  Okay, so once again, this is a nice way  I you know whether you want to memorize that or not as long as you memorize that Delta G not a reaction is equal to minus  RT log K they both work  Okay, we know this okay that if Delta G is less than zero then  The reaction is going to proceed from reactants to products if Delta G is greater than zero. It's going to come back this direction  Okay, what happens if we increase the temperature  In a case where Delta G is less than zero  Think about it. Okay, folks. I want you to I I've had students come to me and they say, you know  I can't I can't visualize this or I can visualize this but I can't mathematically solve this  There are sometimes when you can do it both ways  Okay, when you can visualize it and you say okay, I can understand this and then the other times that you can actually prove it  Yourself mathematically. All right, some of you have stronger math skills than others  I just want you to think about this. You got something like this Delta G is negative  What usually does that mean  Not always but it usually means that there's some heat generated  Okay, remember heat given off negative Delta H Delta G is equal to Delta H minus T Delta S  So if you've got something that is favorable  Okay, and it's spontaneous  Then there oftentimes you're you're you have heat that's given off. Okay, so as you increase the temperature for something like this  Is that going to make the reaction go faster or slower?  I heard faster, but it's slower. Okay k decreases  okay  if you remember we had a plot of  Delta G versus temperature and  On something that was exothermic it looked like this  Okay, the higher the temperature the Delta G decreased  All right, or it was there's another with K like this. Okay was here and temperature was here  And it went like this other ones that went like this  So if you have folks if you have heat that is generated you need to think of this as adding heat right there  Okay, yeah heat right there, and then you add heat is  That going to force things in this direction or this direction it goes back this way  Okay, if I add all of you I think understand if I add if I double the amount of D  By double the amount of D. Does the reaction tend if it's an equilibrium at equilibrium?  I add more D. Which way does the reaction go?  Okay, so if folks  What you don't see because we don't use to put in there is heat  Heat is over here, and it really should be figured in on this equilibrium constant. Okay, so if we have heat  And now we're to equilibrium and I add more heat what happens?  Okay, so folks if you increase the temperature K tends to decrease  When you have a Delta G that is smaller than zero and it'll increase here  Okay, you decrease the temperature then these things just flip boom boom  So once again, I want you to I've showed you figures. I've showed you plots  I want you to try and understand this. This is just a conceptual thing. There's no math involved here aside from the fact  That you know that the rate the K equilibrium for this is you know cease  Whatever the coefficient is  Concentration of C times the concentration D divided by the constant of D rotation of B and the concentration of a  So if you have something else here if you add something here it tends to come back this way if you add something on this side  It goes that way  So  Delta G is not as simple at least it's not simple for my brain, and I don't understand why I struggle with this  but I  Thought I understood Delta G, and now I'm not sure I do so  So the Delta G of reaction  Okay, and once again, this is the the molar is products minus reactants, okay?  So I've  Written this okay that the value of Delta G at a particular stage of the reaction. Okay once again  We've had plots where it was Delta G and time and you remember how I had one where I said  Okay, this is all the the reactants side starting here and the products pure products there pure reactants here  We add stuff in and this is like a the reaction  the progress of the reaction or whatever on this and  This was Delta G and we had a Delta G that kind of came down like this and right here it hit  Zero this is where equilibrium was okay, so folks on this screen right now if this is where equilibrium is on this scale  Okay, there's equilibrium  Do you think we have are we favoring the reactant side or the product side?  Reactive doesn't mean we don't we have product we have product right because this is about I don't know a third or fourth of the  Way there, so we have some products, but we have more reactants about now I  Think we're not quite half  about now  We're starting to favor the products more more more more more more this is one where K folks is like ten thousand  Okay, K is huge and that Delta G  Came down like this and only right there  Had a minimum of zero okay, so equilibrium was right there  Okay, so the bigger the K the further this way we are this model the K  I mean if K is zero point zero zero zero zero one  We're right here that Delta G went  And had a minimum somewhere right here or up there doesn't matter where the minimum was  It's just that the minimum happened on this side  So Delta G is changing during the reaction and that's what we need to be able to calculate  So the value of Delta G at a particular stage of the reaction  Somewhere in here, okay is the difference in the molar Gibbs free energy  molar means the molar and it's at standard  pressure and concentration of the  Productive reactions at the partial pressure or concentrations that they have at that stage  Not what you find in a book  Okay, what you find in a book is you have this and you have this and you look it up  You say Delta G is equal to minus  147 kilojoules per mole, okay  But we're not at equilibrium. We're not at standard. Okay, not everything works at one molar or one atmosphere  So what this Delta G tells us is where we are at that stage and once again is weighted by these coefficients  So  Delta G of reaction is equal to this is what you find in a book  Okay, the reaction is this  plus RT log Q  Q folks when Q is equal to K. We're at equilibrium, right?  If Q is smaller than K, we go this direction  Q is bigger than K. We go in this direction  Okay, so Q tells us something and so that the Delta G this is the this is like what you really get  Okay, you remember how we talked about Delta H  Okay, Delta H is the heat of reaction  Problem is is that does not really tell you  What whether it's even spontaneous or not?  Okay, Delta H of a negative value means it's exothermic  So you got a Delta H of a hundred kilojoules per mole  That but that didn't tell us how much we got out we had to invoke  entropy an  Entropy is very dependent on temperature. So then we came up with Delta G, which is the Gibbs free energy, which is  Delta H minus T Delta S  So now in order to tell whether something is spontaneous you have both the enthalpy and the entropy  Okay  So that was Delta G now what we're finding is that this Delta this new Delta G is the one that tells us sort of where  We are in this reaction and this  The the Q part, okay  I think this is this this is like a little bit the Delta G are is a little bit like a correction factor the further  You are from standard conditions the bigger or smaller this is okay, so look at this up here if we're at standard conditions  This is zero  Okay  As you move further away from standard conditions this starts to become a little bit more and more important  So if you're way off of standard conditions then this becomes  significant  Okay  We're gonna do a few problems. Hopefully when you walk out of here, you won't be any more confused when you walked in and  Charlie put this together for me. I'm gonna put this on the the class website  I'm not gonna go over it right now, but it I think it's a nice little review  Because I want to get right to this question. Let's do this question  This is  2.2 from your book calculating the Gibbs free energy of reaction from the reaction quotient  Okay, the standard Gibbs free energy of reaction for two SO2 gas plus an O2 gas going to two SO3 gases is  Delta G of reaction knot is equal to a hundred and forty one point seven kilo Joules per mole at twenty five point zero zero degrees C  What is the Gibbs free energy of reaction when the partial pressure of each gas this this and this is  100 bar which is 100 atmospheres folks, okay?  100 atmospheres  okay  Let's answer that question first. So what is the Delta G of?  Reaction, okay. Well, what is the Gibbs free energy of reaction when we have this now you're given the standard?  free energy so  Remember it was Delta G of reaction is equal to this  Plus RT log Q  It's a new reaction, so I'll give it to you. I should if I had just copied and pasted this I'd have that reaction here  So don't forget Delta G R is equal to Delta G R not which is right there  plus RT log Q  Oh  By the way, there's a mistake here in your book  This should say PSO 2  SO 2 is right there that should be the PSO because that's the one that's squared  That's on the the reactant side, so you've got oxygen which has a coefficient of 1. This is PO 2. This should be PSO 2  It's your book not me at this time, okay, and I can't change it  Okay  Okay, so Q folks that  Part that is sort of the correction factor  Is  0.01  Alright  0.01  Don't forget  It was Delta G R is equal to Delta G are not  means it standard  Plus RT log Q  Okay  Q is that  Okay log of that anything what is the log of one?  Everybody take your calculators hit enter one or whatever and then hit log of it and L Ln of one is equal to zero  So anything bigger than one one point one one point two six hundred and twenty five  Gives you a positive value coming out of that log and in less than one point nine nine point nine zero zero point  Or zero point zero one gives you a negative number  Okay, so you've got a negative number  Maybe not a very big number, but it's a negative number than times RT  Okay, so what is the answer?  Okay, well here's the first part this is what you get under standard conditions right there Delta G are not is right there  That was given to you  Okay, and then what they did was they multiplied R times T and they get this log of this. That's a small number  Okay, I wish there was one more step in here  I think this value is something like this times this is I think minus 12 kilojoules  Add them together and you get this  How many got that  How many cheated by looking at the notes  Okay  So folks this is a if you cheated. Oh, yes  Thought you just a slow answer. Yes  Which are okay, so  You look at the units Delta G is always in Jules  Okay, it's always in Jules  enthalpy is always in Jules and so  The R that you would use for this is the R that has Jules in it, which is the 8.31  Okay 8.31 Jules per degree mole  So you have that in there you multiply by T since you had per degree  The T's can get rid of each other and you're just left with killed Jules per mole  So your answer and it has to be look at this you got killed Jules per mole here  You have to have the same units for this  To give you this now don't forget that is in Jules  the R is in Jules  So you've got to divide by a thousand if you're going to add to this what you do not want to do is to say okay  This is that this part is  2479  Jules is what it is you can't add 279 Jules to this and come up with  2000 and something so you've got to convert this to to kill Jules and then this is the answer  So folks what this says yes, so I said everybody email this morning  This morning, and I said no graphing calculators. We're gonna check at the door  And I know I'm not happy about this, but I got some emails from people telling me that they overheard people who said they're gonna cheat  They're taking a put calculation. They're gonna put equations and stuff in their calculators. That's what I heard I  Told the class last night. They could use graphing calculators, and now I'm saying no  Go to Target go to the student store by a ten dollar scientific calculator which still can do a whole lot more than  Now I know how to use them. I'm sorry folks  I wish it weren't this way  But think of it this way if people are actually going to program  Equations and things into their calculators  They're gonna get a better score than they deserve and since this class is graded on a curve  That means that their cheating is going to maybe put them above you and I do not want you to suffer  So there's the cost-benefit analysis here. So you got you got to spend money  ten dollars  Okay, which I prefer you not have to spend and you have to get used to new calculator  I'm sorry. I don't know what what what will you like to do?  No calculator  There's a concept  So no graphing calculators, sorry  none  Okay, so look at the difference at standard conditions  This would be a hundred and forty one negative hundred and forty one kilojoules  But under the conditions that we were in invoking on this it went up even greater  Okay, it decreased even more. So this is what we've got. Okay, so  to try and convince you of this with before we had  100 atmospheres for everything, okay  Do it with one atmosphere for everything  One atmosphere do it  I'll go back to the  The previous page all products and reactions are at one atmosphere  Who's got the answer yes you  Negative one forty one  Why because folks if  Q is one times one over one times one or whatever  It's one  Log of one is zero so this whole thing falls out and so the delta g of reaction is equal to del G not of reaction  Okay, try one and a half atmospheres one and a half atmospheres for everything try it  So one and a half atmospheres oops where they go  So it's going to be one and a half squared times one and a half squared times one and a half  So I think your answer is going to be one over one and a half  The Q will be one over one and a half  Because I got a square there and a square there so those just cancel so she's gonna be one over  1.5  What is the log of one over 1.5?  What is the log of one over 1.5?  How many don't understand why I'm asking that question yes  Yes  No, no, no, no I already know that  I'm going to be doing a reaction  No, no, no, no I already said  Okay  So once you put things into this standard state, okay, there are no units  It's all divided by one atm  That yes, you're right. That's we have to kind of do that. Otherwise you're right you end up with you know  Unless you have the exact same number of moles of gas and then you end up with a unit in the log  So it's everything it's going to be a hundred atmospheres divided by one atmosphere gives you a hundred  That's why there's just hundreds here and no units. Yes  I don't care what you give me but don't forget they got to be the same  Okay, if you want to give it to me in joules, then you got to multiply by thousand that hundred and forty one  So that's a big number  Okay, so what is q?  What is the log of one point five one over one point five?  Negative four something huh, so what is that then what if we then plug that in what do we get for the RT log q part?  So  Right there, what do we get for this?  Negative one  So negative one is what we get if we have a half of an outlet one and a half atmospheres  Okay, one and a half atmospheres is pretty darn close to standard  And therefore the correction this whole thing is negative one  Whereas at a hundred atmospheres for each thing it was a negative  13 or something like that 12  Okay, so once again, this is the correction factor. This is the correction that says the further we are away from from standard state  The bigger this thing or the the bigger or smaller depending this is going to be okay  So hopefully understand when and how to use the Delta G thing do this one  I  Predicting the value of K from the standard Gibbs free energy of reaction at 25 degrees C  The standard Gibbs free energy of reaction for half hydrogen gas plus a half iodine gas goes to a whole iodine  Hi is one plus  1.70 kilojoules per mole calculate the equilibrium constant for this reaction  So  You've got temperature and you've got G  So think back to an equation that I gave you  Where you could calculate K  So log K is equal to minus Delta G R over or Delta G of reaction over RT. So you've got this  You know what this is and you know what that is?  So you just plug this in 1.7 times 10 to the 3 now we're in jewel we've converted out of jewels instead of kilojoules  Okay, because then this is in jewels  So then we just have this in jewels this in jewels this in K and you come up with the log K is equal to minus  0.685  Yes  We're gonna get there it's a good question  Okay, so  What they did was they just went ahead and solved they remembered the equation  okay, that  They sort of rearranged it because I was remember it as Delta G is equal to minus RT log K  But doesn't matter remember however you want and just plug stuff in and then you just solve for X  All right, so in this case they did it where log K is equal to  this  Okay, now  Well, I want just what K is not what the log of K is  So I just take this side of the e-power and this side of the e-power. There's a button on your calculator for that  And you get this e to the log X is equal to X so K is equal to point five  How many got that yes?  This no these are just to try and show you  You know the relationship between K and whatever no you will not have to draw that I will not ask you a question about that  So how many got that point five?  Don't forget this  Okay, do this one suppose that in an equilibrium mixture of HCl Cl2 and H2 the partial pressure of H2 is  4.2  Millipascals once again, I'm not going to give you a milli-pascals. I'm going to give you atmospheres or Tor  For cl2 it is 8.3. What is the partial pressure of HCl at 500 degrees?  K given that the equilibrium constant so you're giving equilibrium constant is 4 times 10 of the 18 and  Here's the reaction H2 plus HCl goes to 28 plus Cl goes to 2 HCl right there. That's your equation right there  So you want to calculate the equilibrium composition so what we're asking for is what is the partial pressure of this?  Okay  Bless you, okay, so  You got a K  You got these little wacky things. I'm just going to tell you right now that 4.2  Millipascals is equal to 4.2 times 10 of the minus 8 atmospheres  So this is 4.2 times 10 of the minus 8 atmospheres and this is 8.3 times 10 of the minus 8 atmospheres  It's a conversion you're not going to have to do on the exam  You  Okay, so what we know is that K is equal to the pressure  of the pressure of hydrogen times the pressure of Cl.  So we can rearrange this and say we want to know what is the pressure of HCl.  So we just take the HCl, we move these guys to the other side and we just say that K is  now equal to the pressure of hydrogen times the pressure of Cl2, okay, and they do too  many steps for me, but because HCl is squared here then, then they take the square root  of that whole mess.  So here you've got the pressure of HCl is equal to, this is the value you are given  for the K, this is the pressure of the hydrogen, this is the pressure of the chlorine, and  because we are looking for this, this is the square, we take the square root of that and  it is equal to about 120.  So K is equal to about, I think I got 118.  Yes?  Pressure has to be 8m.  Pressure has to be 8m, yeah.  8m is a bar, okay.  However, if, yeah, let's leave it like that.  Yes?  Loud.  At a 2, why would you do that?  What do you mean a 2?  A coefficient of 2, because it's 1 and 1 and 1.  Okay, okay, he's asking if instead you had done, instead of a half and a half of the  coefficient, so you could have just gone 1, 1 and 2.  I'm asking you, do you think it matters?  The same answer.  You get the same answer?  Yes.  Okay, then you can do it.  So can we do this problem, folks?  Okay, let's do this.  This is an easy one.  All we're doing is we're trying to figure out Q.  Unfortunately, we're doing Q.  Don't be fooled, folks.  Look at this.  The reaction, wherever the reaction is, oh never mind, there's no trick.  I thought there was a liquid or a solid in there, but I guess it's not.  Oh, no, iodine gas, okay.  So here, folks, they give you a K. You can't answer this problem if you don't know what  the K is, okay?  Because the question is which way is the reaction going to go?  So you figure out Q, if Q is bigger than K, then it goes, if Q is bigger than K, it goes  this direction.  If Q is smaller than K, it goes that direction.  So just plug everything in.  K is 46.  What does that tell you, folks?  What does that tell you?  If I gave you this and I said draw, draw the delta G versus temperature, delta G is here,  temperature is there.  It would go way over here.  46 is a pretty good size number.  So it would minimize, it would reach the equilibrium somewhere over in this area.  If I said 4,600, it would be even further over.  So if I said 4,600, it would be even further over.  So, 4,600, it would be even further over.  So in this case, folks, we had up with a Q of 1.  Q of 1, K of 46.  Which way is the reaction going to go?  Toward the products.  That's right.  So folks, I'm not trying to confuse you, but on a question like this, where you have the  same number of moles, you've got two moles of gas on top and two moles of gas on the bottom.  If I had said that all gases are at a pressure of 55 Torr, 55 Torr, okay, instead of 0.55,  guess what?  You got Torr squared, you got Torr times Torr, you got Torr squared.  Everything cancels.  It's still going to be, it's going to be 55 squared over 55 and 55, you still get a Q  of 1, okay, so you don't, in some cases, you don't have to do the conversion.  You just have to do the conversion when there are different number of moles on top and bottom.  Okay, so the reaction is going to proceed toward the products.  Something quick about catalysts.  For catalyst folks, if we have AB here and we're going to CD, the distance between here  and here, okay, that is how much energy you get off, all right?  So you've got to climb this hill, this activation energy, and then you go down there, okay?  So about this much energy is given off going from AB to C plus D. What a catalyst does,  it just makes the path easier, okay?  The difference between here and here is the exact same, okay?  It's a little bit like if I said, let me see, that I want you to go from here to Babua  Island, okay?  Here to Babua Island, but the path of A plus B goes to C plus D that I'm asking half of  you to take is you go to Babua Island from this room, but you have to climb over Saddleback  Mountain.  So your path is from here down to there, over the mountain, and then to Babua Island, that's  this, okay?  Or there's the one where you leave this building and you walk to the beach, that's this one,  okay?  Now the amount of energy, the difference in potential energy between you being here and  a new port beach doesn't matter whether you went to Big Bear Mountain or Kilimanjaro,  okay?  The catalyst provides a different pathway, okay?  And this, believe it or not, this little hill right here is really what determines how fast  something goes.  But this is really a tall one, okay?  Then we're going to stay more where A and B is, okay?  Not get to C and D. Even though you can see going from here to here makes thermodynamic  sense, but if this hill is too big, you just don't do it.  A catalyst allows a different path.  So in engineering folks, using a catalyst for this can sometimes save you millions or  billions of dollars or make you millions and billions because you've actually been able  to come up with a way to turn sand into gold, okay?  Or coal into diamonds, okay?  Coal to diamonds folks, this will go way up there, okay?  You can come up with a different way to make diamonds, you know, good for you.  So it doesn't change the K, nor does it shift the equilibrium in the system.  It just allows a different path.  I said all this.  Man, I'm good.  Okay, so this is it.  I think Josette was on a trip this, yeah.  So she's not in this picture.  Anyway, so folks, I hope you've learned something.  I will see you all on Thursday, I guess a week.  A week from today, but a little earlier.  Study hard.  Thank you.
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Okay, so this is it You know last night I was thinking That as you get older Time just goes fast. I mean these lectures seem to just drag on and on. I don't know how you feel but But the time I mean I can't believe it's been ten weeks now anyway, so So this is it so We will after this we'll see each other on Thursday morning at 8 o'clock right here Of course the people that are in this room right now are the ones that usually show up the one the empty seats are the ones that Don't always show up My suggestion folks is that you show up a little early Okay The final will be about twice as long Okay, or what should take you about twice as long to to do and in the past We've had you know about 70 minutes for the exam, so maybe we'll be quite twice long But I don't want to have you people stringing in here until 8 10 8 15 in the morning I want to pass the exam out at 8 so that doesn't mean you show up at 8 that means you should be here be in your seats And ready to take the exam and have us passing it out at 8. Otherwise It's a problem. Okay All right, well folks we're on the the home stretch here Just a little bit about the chat. Lea I've had All kinds of questions. I even had a question last night This one person tends to be a little bit on the aggressive side, but he came up to me and he said So do you anticipate that the Thursday final exam will be Pretty much like the Friday final exam And I said well I said, you know I learned my lesson on the first exam I need to make them, you know different, but I'm not that creative. I can't make it so that there's I mean There's gonna be a Clausius-Clapeyron equation or question on there and whether you got to Take something of the e or you just plug in chug or you solve for t There's gonna be a Clausius-Clapeyron on the night exam and one on the morning exam So he said what would it be possible then for you to get me to email me the exam? The Thursday exam so I can practice I Mean I'm just shocked that I mean He's not even gonna try and cheat, you know, he's not gonna even ask you anyway, so Anyway We will be grading your exams on Thursday We will hopefully post I'll do what I did before send out an email that has your scores on the That that evening sometime on Thursday night, we will not though post the key Because of the next exam okay, so the key will be posted sometime Probably Saturday Because once again, we won't start grading the the the other exam goes from seven o'clock at night until Nine o'clock a night on Friday night. I can't believe that. Okay, okay So I Have tried in this ten weeks Sometimes I point out this is a good question. Okay, this would be a good exam question I'm hoping folks that not only have you learned something about maybe how to predict a Possible question from my class, but that it will work for you later. All right Here's an example. Okay, I want you to think of how could Don ask a multiple choice question From this slide. Okay, so the chalet. I love the chalet. Okay So if you change the concentration Can you shift equilibrium and can you change the equilibrium constant? Okay, so For this one if you change the concentration of something I don't care products or reactants does that shift equilibrium? Does it change the rate constant Does it change the rate constant no the rate constant we are trying to maintain that at all cost Okay So we we expect it to stay the same if I say the rate constant is point one Then it's point one no matter what you dump into it. Okay, and so if you pull something out you dump more in it Reequilibrates so that you're at point one. All right, so that's it's called a rate constant All right, so that's it's called a rate constant for a reason. Okay, and so you do not change the rate constant All right said the the equilibrium constant By changing the the number of bowls or the concentration. How about pressure? I pushed the button too soon Okay, does it change the equilibrium constant? Okay volume and Volume and pressure I'm a little bit confused at this I mean because I I see it as if you double the volume then you've cut the pressure in half Okay, or if you cut the volume in half you've increased the pressure so those to me Should be the same. Okay, so I didn't even ask you okay about temperature in a shift equilibrium sure We have a reaction where we have a plus b goes to c plus d and if there's some some heat given off If we add heat to one side it makes things happen. Okay, or it makes things slow down So absolutely changing the temperature can shift the equilibrium. Does it change the equilibrium constant? Yes, it does that's the one thing folks That is the one thing that changes the equilibrium constant is the temperature So I have to specifically say to you you add this temperature And I think what you'll find is is that in all the examples that we've done so far It'll say it I mean some wacky temperature maybe 1280 degrees, but but you it's at a certain temperature The the k changes now we know that We've seen plots Okay, we've seen figures where it was like on the y-axis. It was k and on this axis was temperature Okay, and if you have something something that's exothermic gives off energy as you warm up things The k at low temperature once again This is low temperature here and high temperature there low k high k starts here and it kind does this Okay, because as you warm things up more heat as a product You've already got heat so it slows the case so temperature absolutely can affect k a Catalyst we're gonna talk about catalyst at the end of the lecture It does not shift the equilibrium Okay And it does not change the equilibrium constant. Okay, so this is something that that we will get into in in an hour or so So folks This is a little bit of a hint there are At least 80 people missing Okay, I can see we didn't even want to say to turn off your cell phones because there are so many empty seats Okay, so I want you to look at this Look at this right here How many when I ask a multiple choice question, how many options do I usually give you? I like five Okay, let's say five So if I were to ask a question on Multiple choice for this What could I ask could I say When I change x equilibrium changes and I would say concentration Boom boom boom boom and if I and all of a sudden which one does not change. Can you see this is a nice question? It was one answer this different from the other four How about over here? So I'm telling you right now that the chances of these two being on the exam at least one of them is about 90% Okay It will just say I Don't know how I'd say it, but it'll be on there All right, I want you to do this for me right now calculate K for the reaction of Ethanol liquid I didn't know how to make an equilibrium sign I couldn't do that so this is supposed to I Can even copy and paste sorry this is supposed to be two arrows going different different ways goes to ethanol gas and The concentration or the pressure of this gas is point three three Three three three atmospheres. I want to know what is K? You got a liquid on one side and a gas on the other Where's that guy that always knows the answers What's your answer chief right there he says that's the K right K is equal to You know products the the pressure of the products over the pressure reactants, but this Okay, this has a value of Not of one Okay, so it's the pressure here over the reactants is equal to K So in this case K is equal to point three three three no no units are necessary. Okay? Everybody understand that Okay now I Want you to calculate the K for this reaction ethanol liquid ethanol gas to 253 Torr Who's got an answer? What is it? He says it's point three three three what is with that? Why isn't it 250? He says it's point three three three what is with that? Why isn't it 253? So do me this for those of you who are a little bit confused there is 760 Torr in an atmosphere So divide seven divide that number 253 by 760 Okay Divide that number that I gave you by 760 If I've done the math correctly you get point three three three So the problem folks is that while I told you at one point that Units didn't matter and they would cancel if you only have if you have an odd number of units It doesn't work that way. Okay, if I've got You know h2 on top certain pressure and eight and and two on the bottom Then the units cancel if I've got I don't care five Torr five Torr hydrogen is the product and Two Torr of nitrogen was the reactants. It's five over two and guess what? The k is is two and a half Okay, and if I converted all that to atmospheres. I'd still get two and a half problem is is that when you just have just one thing Okay, there's nothing to divide it by those units don't cancel and then the units do matter Okay, so the units do matter so you can't have this is the same reaction You can't have two different k's with the same thing. This is point three three three Torr and there's Pascal's and there's all kinds of Things that you can use. Okay, so we have to have some normalization And that is this Okay, I So I suggest that you either commit this to memory or try and understand the concept for aqueous It's one molar Okay, one molar are the parameters to use here. This is the reference state the activity here is the number of moles Over one molar or sorry the the molality the molarity of whatever it is divided by one molar Okay For eight for gaseous. It's the atmospheric partial pressure the partial pressure is in atmospheres now folks an atmosphere is the same as a bar Okay, but I don't use bars and I don't use Pascal's Okay, I use atmospheres and I use Torr So it's very possible. You'll get a question. It says something in Torr And it's something that's the door you're going to get in atmospheres. Okay, but give it to you in atmospheres and you're good to go So once again when you set up an equation for K you must use atmospheres Unless you have the same number of moles on top and bottom And what I mean folks is that if you've got a plus b goes to c plus d and a And b is equal to c and d in terms of the coefficient then everything cancels out But gases are this for liquids and solids folks it's a one Okay, so on that last thing we could say products would be in the partial pressure we had a point three three three Okay, the liquid so it's gonna be k is equal to point three three three Over and we had a liquid over one If you want to keep track of all your your your p's and q's there, that's what you would do Okay, so it's very important for you to understand that there are times you can use Torr and there are times you cannot use Torr There are times you can use milli Millimolar and there's times you can't use millimolar All right So we're gonna talk more about Delta G today so that other stuff was was more k equilibrium and all Delta G is Equal to the overall Delta G is equal to the Delta G of the products minus the Delta G of reactants Okay for everything except for one thing it's been products minus reactants for ten weeks Okay, and that's pretty much how it's going to stay products over reactants So the reactants this is an energy diagram folks if this is energy this way I'm not sure what's this way maybe time The reactants have an energy here if they actually do this and the products are lower in energy What that means folks is that the products are lower in energy, but right here is a big plus heat Okay, if this is actually lower than this Then it's this the energy had to go somewhere Okay, the energy is going to be products plus heat. All right, so an exothermic reaction. It looks like this Endothermic would be from here to here Okay, you have to give these guys some heat to get it to be up there Certainly not spontaneous and we know that if Delta G is less than zero Okay, if you do a calculation Delta's less than zero that Q is less than K and It's going to Head that direction. All right. Yeah Delta G is equal to zero word equilibrium Okay, word equilibrium and Q equals K and if Delta G is positive It is not spontaneous and Q is actually bigger than K So you got to go from the reactants of the product side to the reactants side? So We're going to go into this a lot But I would suggest you just write it down the Delta G is equal to Delta G not plus RT log Q So that Q has come back Okay, the Q that I made it sound like early on was not all that important turns out to have reared its ugly head again So it's a good thing that I didn't trivialize it too much earlier So Delta G is equal to Delta G not plus RT log Q. We're going to be going over those several of those So don't don't panic with this And then at equilibrium When Delta G is equal to zero then Delta G not plus RT log K is equal to zero Okay do this think about it What are Delta G and K at equilibrium at 25 degrees C for this Well last night I said well, it's obviously it's this This but it's wrong Okay, I went through this big explanation of what if I had just water sitting here is it boiling look at this Does that look like it's boiling to you? That looks like boiling water to me So I said oh well It's obviously 25 degrees isn't boiling Okay, and therefore water would rather be liquid than gas at room temperature So that obviously Delta G must be equal to zero or greater than zero turns out the key word right here is equilibrium folks Equilibrium Delta G equals zero do not forget that so don't read these problems on the exam so fast like I did last night The ice that you screw up Okay, that's the key if something has said it equilibrium folks it could cut out a whole lot of work that you need to do Look at that Okay So we did a we've done several of these folks where it is One reaction at least another reaction at least another reaction So it's important for you to understand we did this I don't know I can't remember what chapter enthalpy was chapter six maybe chapter seven. I don't matter So I would like you to do this just humor me okay What is K sub one in this the way it's written right here a Going to be okay. What is K sub one? I? Hope you're gonna say that it's gonna be The concentration of B over a is equal to K sub one write that down please just write it down I have a reason for asking you to do this So here we have a going to be is K plus one So now I want you to write down what is case of two? So case of two is B goes to C and That is case of two and that should be Concentration is case of two is equal to the concentration of C over the concentration of B? And how about this one? From these first two folks You should be able to figure out what this is okay? so If you if you do so now just write K one is Equal to B over a and K two is equal to C over B Okay, multiply those two together. I Don't mean K one K two. I mean what they were equal to okay, so K one is equal to B over a Okay, and K two is equal to C over B So you know this you wrote that all of you probably wrote that case of three is equal to C over a correct But what I'm saying is is that you can show yourself that that's the case Because to get there from a to C you knew that B over a was equal to K one and C over B was equal to K two So look at this for this first one K one B over a B over a That's K one K two is C over B. That's K two And then look at these guys cancel They cancel so you're just left with C over a All right, so you could sort of work that way So in this case to find K sub three you just multiply K one times K two is equal to K three for the Enthelpiece of those reactions you add them okay So H sub three is equal to H one the delta H one plus delta H two So folks this encompasses quite a few things Okay, it encompasses the equilibrium part this part is to sort of refresh your memory on what we did in chapter Six or seven whatever it was So don't forget how to do this type of thing This is like a little bit of not like a Hess's law, but it's a I love Hess's law So once again, it's just sort of playing with with these things All right, so Delta G not of reaction is equal to the Delta H not of reaction minus T Delta S not of reaction Okay, this is equal to once you just like the previous pay Delta G of formation minus of products minus reactants and I wish I had been able to write something up here, but this if you remember folks that Delta G of reaction is equal to minus RT log K Okay, and all this is is a rearrangement of that Okay In fact, let's do this right now Write this exact thing Okay, and now we're gonna do something in reverse We're gonna do something in reverse of what we've been doing So you write that And Then I want you to take the log the natural log of both sides the natural log of both sides That's Ln I Suggest you write this down. I mean maybe you got these kind of brains that allow you to visualize these things That's wonderful. Yes. You can't visualize on the exam. I Need to see it so make sure that you're good going from visualization to putting it on the paper So what is if we take the log of both sides, what do we get? What do we get Log K is equal to That's right Okay, so what you're saying is is that you just do this you just take the log of this makes this a log K take the log of this You just end up with this Okay So you don't you only have to memorize one thing folks, okay? I tend to memorize that Delta G is equal to minus RT log K But you can play with it any way you want so folks if I give you a K And I say calculate Delta G not a reaction. That's how you do it Okay, so look at this right here R is a constant T is not a constant Delta G not is not a constant and this isn't a constant So I can if I give you two or three things here you can figure out different things Okay, so once again, this is a nice way I you know whether you want to memorize that or not as long as you memorize that Delta G not a reaction is equal to minus RT log K they both work Okay, we know this okay that if Delta G is less than zero then The reaction is going to proceed from reactants to products if Delta G is greater than zero. It's going to come back this direction Okay, what happens if we increase the temperature In a case where Delta G is less than zero Think about it. Okay, folks. I want you to I I've had students come to me and they say, you know I can't I can't visualize this or I can visualize this but I can't mathematically solve this There are sometimes when you can do it both ways Okay, when you can visualize it and you say okay, I can understand this and then the other times that you can actually prove it Yourself mathematically. All right, some of you have stronger math skills than others I just want you to think about this. You got something like this Delta G is negative What usually does that mean Not always but it usually means that there's some heat generated Okay, remember heat given off negative Delta H Delta G is equal to Delta H minus T Delta S So if you've got something that is favorable Okay, and it's spontaneous Then there oftentimes you're you're you have heat that's given off. Okay, so as you increase the temperature for something like this Is that going to make the reaction go faster or slower? I heard faster, but it's slower. Okay k decreases okay if you remember we had a plot of Delta G versus temperature and On something that was exothermic it looked like this Okay, the higher the temperature the Delta G decreased All right, or it was there's another with K like this. Okay was here and temperature was here And it went like this other ones that went like this So if you have folks if you have heat that is generated you need to think of this as adding heat right there Okay, yeah heat right there, and then you add heat is That going to force things in this direction or this direction it goes back this way Okay, if I add all of you I think understand if I add if I double the amount of D By double the amount of D. Does the reaction tend if it's an equilibrium at equilibrium? I add more D. Which way does the reaction go? Okay, so if folks What you don't see because we don't use to put in there is heat Heat is over here, and it really should be figured in on this equilibrium constant. Okay, so if we have heat And now we're to equilibrium and I add more heat what happens? Okay, so folks if you increase the temperature K tends to decrease When you have a Delta G that is smaller than zero and it'll increase here Okay, you decrease the temperature then these things just flip boom boom So once again, I want you to I've showed you figures. I've showed you plots I want you to try and understand this. This is just a conceptual thing. There's no math involved here aside from the fact That you know that the rate the K equilibrium for this is you know cease Whatever the coefficient is Concentration of C times the concentration D divided by the constant of D rotation of B and the concentration of a So if you have something else here if you add something here it tends to come back this way if you add something on this side It goes that way So Delta G is not as simple at least it's not simple for my brain, and I don't understand why I struggle with this but I Thought I understood Delta G, and now I'm not sure I do so So the Delta G of reaction Okay, and once again, this is the the molar is products minus reactants, okay? So I've Written this okay that the value of Delta G at a particular stage of the reaction. Okay once again We've had plots where it was Delta G and time and you remember how I had one where I said Okay, this is all the the reactants side starting here and the products pure products there pure reactants here We add stuff in and this is like a the reaction the progress of the reaction or whatever on this and This was Delta G and we had a Delta G that kind of came down like this and right here it hit Zero this is where equilibrium was okay, so folks on this screen right now if this is where equilibrium is on this scale Okay, there's equilibrium Do you think we have are we favoring the reactant side or the product side? Reactive doesn't mean we don't we have product we have product right because this is about I don't know a third or fourth of the Way there, so we have some products, but we have more reactants about now I Think we're not quite half about now We're starting to favor the products more more more more more more this is one where K folks is like ten thousand Okay, K is huge and that Delta G Came down like this and only right there Had a minimum of zero okay, so equilibrium was right there Okay, so the bigger the K the further this way we are this model the K I mean if K is zero point zero zero zero zero one We're right here that Delta G went And had a minimum somewhere right here or up there doesn't matter where the minimum was It's just that the minimum happened on this side So Delta G is changing during the reaction and that's what we need to be able to calculate So the value of Delta G at a particular stage of the reaction Somewhere in here, okay is the difference in the molar Gibbs free energy molar means the molar and it's at standard pressure and concentration of the Productive reactions at the partial pressure or concentrations that they have at that stage Not what you find in a book Okay, what you find in a book is you have this and you have this and you look it up You say Delta G is equal to minus 147 kilojoules per mole, okay But we're not at equilibrium. We're not at standard. Okay, not everything works at one molar or one atmosphere So what this Delta G tells us is where we are at that stage and once again is weighted by these coefficients So Delta G of reaction is equal to this is what you find in a book Okay, the reaction is this plus RT log Q Q folks when Q is equal to K. We're at equilibrium, right? If Q is smaller than K, we go this direction Q is bigger than K. We go in this direction Okay, so Q tells us something and so that the Delta G this is the this is like what you really get Okay, you remember how we talked about Delta H Okay, Delta H is the heat of reaction Problem is is that does not really tell you What whether it's even spontaneous or not? Okay, Delta H of a negative value means it's exothermic So you got a Delta H of a hundred kilojoules per mole That but that didn't tell us how much we got out we had to invoke entropy an Entropy is very dependent on temperature. So then we came up with Delta G, which is the Gibbs free energy, which is Delta H minus T Delta S So now in order to tell whether something is spontaneous you have both the enthalpy and the entropy Okay So that was Delta G now what we're finding is that this Delta this new Delta G is the one that tells us sort of where We are in this reaction and this The the Q part, okay I think this is this this is like a little bit the Delta G are is a little bit like a correction factor the further You are from standard conditions the bigger or smaller this is okay, so look at this up here if we're at standard conditions This is zero Okay As you move further away from standard conditions this starts to become a little bit more and more important So if you're way off of standard conditions then this becomes significant Okay We're gonna do a few problems. Hopefully when you walk out of here, you won't be any more confused when you walked in and Charlie put this together for me. I'm gonna put this on the the class website I'm not gonna go over it right now, but it I think it's a nice little review Because I want to get right to this question. Let's do this question This is 2.2 from your book calculating the Gibbs free energy of reaction from the reaction quotient Okay, the standard Gibbs free energy of reaction for two SO2 gas plus an O2 gas going to two SO3 gases is Delta G of reaction knot is equal to a hundred and forty one point seven kilo Joules per mole at twenty five point zero zero degrees C What is the Gibbs free energy of reaction when the partial pressure of each gas this this and this is 100 bar which is 100 atmospheres folks, okay? 100 atmospheres okay Let's answer that question first. So what is the Delta G of? Reaction, okay. Well, what is the Gibbs free energy of reaction when we have this now you're given the standard? free energy so Remember it was Delta G of reaction is equal to this Plus RT log Q It's a new reaction, so I'll give it to you. I should if I had just copied and pasted this I'd have that reaction here So don't forget Delta G R is equal to Delta G R not which is right there plus RT log Q Oh By the way, there's a mistake here in your book This should say PSO 2 SO 2 is right there that should be the PSO because that's the one that's squared That's on the the reactant side, so you've got oxygen which has a coefficient of 1. This is PO 2. This should be PSO 2 It's your book not me at this time, okay, and I can't change it Okay Okay, so Q folks that Part that is sort of the correction factor Is 0.01 Alright 0.01 Don't forget It was Delta G R is equal to Delta G are not means it standard Plus RT log Q Okay Q is that Okay log of that anything what is the log of one? Everybody take your calculators hit enter one or whatever and then hit log of it and L Ln of one is equal to zero So anything bigger than one one point one one point two six hundred and twenty five Gives you a positive value coming out of that log and in less than one point nine nine point nine zero zero point Or zero point zero one gives you a negative number Okay, so you've got a negative number Maybe not a very big number, but it's a negative number than times RT Okay, so what is the answer? Okay, well here's the first part this is what you get under standard conditions right there Delta G are not is right there That was given to you Okay, and then what they did was they multiplied R times T and they get this log of this. That's a small number Okay, I wish there was one more step in here I think this value is something like this times this is I think minus 12 kilojoules Add them together and you get this How many got that How many cheated by looking at the notes Okay So folks this is a if you cheated. Oh, yes Thought you just a slow answer. Yes Which are okay, so You look at the units Delta G is always in Jules Okay, it's always in Jules enthalpy is always in Jules and so The R that you would use for this is the R that has Jules in it, which is the 8.31 Okay 8.31 Jules per degree mole So you have that in there you multiply by T since you had per degree The T's can get rid of each other and you're just left with killed Jules per mole So your answer and it has to be look at this you got killed Jules per mole here You have to have the same units for this To give you this now don't forget that is in Jules the R is in Jules So you've got to divide by a thousand if you're going to add to this what you do not want to do is to say okay This is that this part is 2479 Jules is what it is you can't add 279 Jules to this and come up with 2000 and something so you've got to convert this to to kill Jules and then this is the answer So folks what this says yes, so I said everybody email this morning This morning, and I said no graphing calculators. We're gonna check at the door And I know I'm not happy about this, but I got some emails from people telling me that they overheard people who said they're gonna cheat They're taking a put calculation. They're gonna put equations and stuff in their calculators. That's what I heard I Told the class last night. They could use graphing calculators, and now I'm saying no Go to Target go to the student store by a ten dollar scientific calculator which still can do a whole lot more than Now I know how to use them. I'm sorry folks I wish it weren't this way But think of it this way if people are actually going to program Equations and things into their calculators They're gonna get a better score than they deserve and since this class is graded on a curve That means that their cheating is going to maybe put them above you and I do not want you to suffer So there's the cost-benefit analysis here. So you got you got to spend money ten dollars Okay, which I prefer you not have to spend and you have to get used to new calculator I'm sorry. I don't know what what what will you like to do? No calculator There's a concept So no graphing calculators, sorry none Okay, so look at the difference at standard conditions This would be a hundred and forty one negative hundred and forty one kilojoules But under the conditions that we were in invoking on this it went up even greater Okay, it decreased even more. So this is what we've got. Okay, so to try and convince you of this with before we had 100 atmospheres for everything, okay Do it with one atmosphere for everything One atmosphere do it I'll go back to the The previous page all products and reactions are at one atmosphere Who's got the answer yes you Negative one forty one Why because folks if Q is one times one over one times one or whatever It's one Log of one is zero so this whole thing falls out and so the delta g of reaction is equal to del G not of reaction Okay, try one and a half atmospheres one and a half atmospheres for everything try it So one and a half atmospheres oops where they go So it's going to be one and a half squared times one and a half squared times one and a half So I think your answer is going to be one over one and a half The Q will be one over one and a half Because I got a square there and a square there so those just cancel so she's gonna be one over 1.5 What is the log of one over 1.5? What is the log of one over 1.5? How many don't understand why I'm asking that question yes Yes No, no, no, no I already know that I'm going to be doing a reaction No, no, no, no I already said Okay So once you put things into this standard state, okay, there are no units It's all divided by one atm That yes, you're right. That's we have to kind of do that. Otherwise you're right you end up with you know Unless you have the exact same number of moles of gas and then you end up with a unit in the log So it's everything it's going to be a hundred atmospheres divided by one atmosphere gives you a hundred That's why there's just hundreds here and no units. Yes I don't care what you give me but don't forget they got to be the same Okay, if you want to give it to me in joules, then you got to multiply by thousand that hundred and forty one So that's a big number Okay, so what is q? What is the log of one point five one over one point five? Negative four something huh, so what is that then what if we then plug that in what do we get for the RT log q part? So Right there, what do we get for this? Negative one So negative one is what we get if we have a half of an outlet one and a half atmospheres Okay, one and a half atmospheres is pretty darn close to standard And therefore the correction this whole thing is negative one Whereas at a hundred atmospheres for each thing it was a negative 13 or something like that 12 Okay, so once again, this is the correction factor. This is the correction that says the further we are away from from standard state The bigger this thing or the the bigger or smaller depending this is going to be okay So hopefully understand when and how to use the Delta G thing do this one I Predicting the value of K from the standard Gibbs free energy of reaction at 25 degrees C The standard Gibbs free energy of reaction for half hydrogen gas plus a half iodine gas goes to a whole iodine Hi is one plus 1.70 kilojoules per mole calculate the equilibrium constant for this reaction So You've got temperature and you've got G So think back to an equation that I gave you Where you could calculate K So log K is equal to minus Delta G R over or Delta G of reaction over RT. So you've got this You know what this is and you know what that is? So you just plug this in 1.7 times 10 to the 3 now we're in jewel we've converted out of jewels instead of kilojoules Okay, because then this is in jewels So then we just have this in jewels this in jewels this in K and you come up with the log K is equal to minus 0.685 Yes We're gonna get there it's a good question Okay, so What they did was they just went ahead and solved they remembered the equation okay, that They sort of rearranged it because I was remember it as Delta G is equal to minus RT log K But doesn't matter remember however you want and just plug stuff in and then you just solve for X All right, so in this case they did it where log K is equal to this Okay, now Well, I want just what K is not what the log of K is So I just take this side of the e-power and this side of the e-power. There's a button on your calculator for that And you get this e to the log X is equal to X so K is equal to point five How many got that yes? This no these are just to try and show you You know the relationship between K and whatever no you will not have to draw that I will not ask you a question about that So how many got that point five? Don't forget this Okay, do this one suppose that in an equilibrium mixture of HCl Cl2 and H2 the partial pressure of H2 is 4.2 Millipascals once again, I'm not going to give you a milli-pascals. I'm going to give you atmospheres or Tor For cl2 it is 8.3. What is the partial pressure of HCl at 500 degrees? K given that the equilibrium constant so you're giving equilibrium constant is 4 times 10 of the 18 and Here's the reaction H2 plus HCl goes to 28 plus Cl goes to 2 HCl right there. That's your equation right there So you want to calculate the equilibrium composition so what we're asking for is what is the partial pressure of this? Okay Bless you, okay, so You got a K You got these little wacky things. I'm just going to tell you right now that 4.2 Millipascals is equal to 4.2 times 10 of the minus 8 atmospheres So this is 4.2 times 10 of the minus 8 atmospheres and this is 8.3 times 10 of the minus 8 atmospheres It's a conversion you're not going to have to do on the exam You Okay, so what we know is that K is equal to the pressure of the pressure of hydrogen times the pressure of Cl. So we can rearrange this and say we want to know what is the pressure of HCl. So we just take the HCl, we move these guys to the other side and we just say that K is now equal to the pressure of hydrogen times the pressure of Cl2, okay, and they do too many steps for me, but because HCl is squared here then, then they take the square root of that whole mess. So here you've got the pressure of HCl is equal to, this is the value you are given for the K, this is the pressure of the hydrogen, this is the pressure of the chlorine, and because we are looking for this, this is the square, we take the square root of that and it is equal to about 120. So K is equal to about, I think I got 118. Yes? Pressure has to be 8m. Pressure has to be 8m, yeah. 8m is a bar, okay. However, if, yeah, let's leave it like that. Yes? Loud. At a 2, why would you do that? What do you mean a 2? A coefficient of 2, because it's 1 and 1 and 1. Okay, okay, he's asking if instead you had done, instead of a half and a half of the coefficient, so you could have just gone 1, 1 and 2. I'm asking you, do you think it matters? The same answer. You get the same answer? Yes. Okay, then you can do it. So can we do this problem, folks? Okay, let's do this. This is an easy one. All we're doing is we're trying to figure out Q. Unfortunately, we're doing Q. Don't be fooled, folks. Look at this. The reaction, wherever the reaction is, oh never mind, there's no trick. I thought there was a liquid or a solid in there, but I guess it's not. Oh, no, iodine gas, okay. So here, folks, they give you a K. You can't answer this problem if you don't know what the K is, okay? Because the question is which way is the reaction going to go? So you figure out Q, if Q is bigger than K, then it goes, if Q is bigger than K, it goes this direction. If Q is smaller than K, it goes that direction. So just plug everything in. K is 46. What does that tell you, folks? What does that tell you? If I gave you this and I said draw, draw the delta G versus temperature, delta G is here, temperature is there. It would go way over here. 46 is a pretty good size number. So it would minimize, it would reach the equilibrium somewhere over in this area. If I said 4,600, it would be even further over. So if I said 4,600, it would be even further over. So, 4,600, it would be even further over. So in this case, folks, we had up with a Q of 1. Q of 1, K of 46. Which way is the reaction going to go? Toward the products. That's right. So folks, I'm not trying to confuse you, but on a question like this, where you have the same number of moles, you've got two moles of gas on top and two moles of gas on the bottom. If I had said that all gases are at a pressure of 55 Torr, 55 Torr, okay, instead of 0.55, guess what? You got Torr squared, you got Torr times Torr, you got Torr squared. Everything cancels. It's still going to be, it's going to be 55 squared over 55 and 55, you still get a Q of 1, okay, so you don't, in some cases, you don't have to do the conversion. You just have to do the conversion when there are different number of moles on top and bottom. Okay, so the reaction is going to proceed toward the products. Something quick about catalysts. For catalyst folks, if we have AB here and we're going to CD, the distance between here and here, okay, that is how much energy you get off, all right? So you've got to climb this hill, this activation energy, and then you go down there, okay? So about this much energy is given off going from AB to C plus D. What a catalyst does, it just makes the path easier, okay? The difference between here and here is the exact same, okay? It's a little bit like if I said, let me see, that I want you to go from here to Babua Island, okay? Here to Babua Island, but the path of A plus B goes to C plus D that I'm asking half of you to take is you go to Babua Island from this room, but you have to climb over Saddleback Mountain. So your path is from here down to there, over the mountain, and then to Babua Island, that's this, okay? Or there's the one where you leave this building and you walk to the beach, that's this one, okay? Now the amount of energy, the difference in potential energy between you being here and a new port beach doesn't matter whether you went to Big Bear Mountain or Kilimanjaro, okay? The catalyst provides a different pathway, okay? And this, believe it or not, this little hill right here is really what determines how fast something goes. But this is really a tall one, okay? Then we're going to stay more where A and B is, okay? Not get to C and D. Even though you can see going from here to here makes thermodynamic sense, but if this hill is too big, you just don't do it. A catalyst allows a different path. So in engineering folks, using a catalyst for this can sometimes save you millions or billions of dollars or make you millions and billions because you've actually been able to come up with a way to turn sand into gold, okay? Or coal into diamonds, okay? Coal to diamonds folks, this will go way up there, okay? You can come up with a different way to make diamonds, you know, good for you. So it doesn't change the K, nor does it shift the equilibrium in the system. It just allows a different path. I said all this. Man, I'm good. Okay, so this is it. I think Josette was on a trip this, yeah. So she's not in this picture. Anyway, so folks, I hope you've learned something. I will see you all on Thursday, I guess a week. A week from today, but a little earlier. Study hard. Thank you.
That is the one thing that changes the reference state the activity here is the number are this for liquids and solids folks it's a one two SO3 gases is Delta G of reaction knot is