okay welcome to thermodynamics continue the second part so now we're gonna actually calculate some stuff we're gonna find Delta G in a various ways so first let's look at G itself G is a state function much like H enthalpy was a state function so a state function if you recall only depends on its current state it doesn't matter its path independent it doesn't matter how it got there the quantity only depends on its current state state function now that's important because that makes the Delta really easy because it's just final minus initial because if the final condition only contains can only that current state matters and same with the initial condition and the paths in between doesn't matter then final minus initial that's the change right so certain things are state functions and oftentimes those are given capital letters as their symbols so that's a good clue as to what might be a state function of what might not but you can also think about it as well so for example if I was going to travel to Los Cabos and I wanted to know my change in elevation well it would just be final minus initial right where my final elevation - my initial elevation that's my change in elevation and that would be a state function now the energy it takes to get there would not be a state function because that would depend on the path I took did i drive my car over highway 17 did i ride my bike over mountain charlie how did I do it what path did I take what mode did I take and that would matter so that would not be a state function so but Delta G is a state function Delta H was a state function we spent an entire chapter chapter nine calculating Delta H we can use many of those same methods to find Delta G as well and so we're gonna do that again in this chapter but adjust for Delta G instead of Delta H for your reference the chapter that we did all the Delta H is in that had those methods was thermochemistry whose chapter 9 and opens that was in chem 1a I mean also we're gonna need some thermodynamic data so that's an appendix G and in the back of the book okay let's take our first method okay the first method we're gonna do is Delta G from standard free energies of formation so we did this for Delta H from standard enthalpies of formation but now rich in do it for free energies instead so that's really the only thing that's different from this method that we did before so one of the things we have to realize when we're using these is what reaction this is for because this isn't a table and we can look up the value but we need to know what the reaction is that it corresponds to so that's important so this reaction without little f means formation and formation is the reaction that produces one mole of the species so that's gonna be your product one mole of that whatever species and then from the constituent elements in their natural state so from these as elements in their state that they occur under normal conditions so this may require fractional coefficients on the I forgot to write down fractional coefficients on the reactant side all right so let's let's do that so now we also when we're looking in the table we have to be careful of the phase the phase is written because the formation the amount of energy for a gas is gonna be different than it is for a liquid or a solid so we have to look at those phases as well and make sure that those matched when I'm plugging the wrong number for the wrong phase because there would be some energy associated with that phase change right and that would be included in that number okay so if this is the Delta G of standard free energy of formation for ammonia gas then I know on the product side of this I have ammonia gas and the coefficient on that has to be one okay now on this side on the reactant side I'm just gonna put the elements so this element is nitrogen and that comes as a diatomic gas so n2 plus and I'm gonna live a little space here for coefficients hydrogen also comes as a diatomic gas so we'll put that in there and now this is the this is this part right here this has to be a 1 so these may be fractional these coefficients normally we don't do that normally we multiply through so that they're all whole number coefficients but in this case we're gonna have to because that's defined as being one that limits us to what these have to be so and you only need one end here I have two ends here so that's gonna have to be 1/2 n 2 and then here I need three hydrogen's I only have two so this is gonna have to be three halves H 2 so this is the reaction that if I lift up Delta G formation for ammonia gas this is the reaction that corresponds to and you have to be able to put that in yourself okay now let's try another one so here we have sodium hydrogen carbonate solid so this is baking soda right so let's do that so I know on the product side I'm gonna have n hco3 solid okay so now on this side the elements so I have sodium now this is the this is the tricky part right when we think of sodium in its natural state we think of sodium ions because that's how they occur but that's not an element that's an ion it's been ionized so the element is actually sodium metal gate don't want to get too near that all right now hydrogen again is h2 gas leave myself enough room carbon now carbon is gonna come as a solid but here's the tricky part there are two different solids that are common graphite and diamond hmm so what's the natural state at one atmosphere I'll give you a hint diamonds aren't forever at one atmosphere they're turning into graphite so this is actually solid graphite plus some oxygens which are gonna be gas you never leave myself enough room all right so that's a one that's easy that's a well it's just like up here except it's gonna be a half because that'll give me one hydrogen this is one and then this oxygen like up here is gonna be three halves so I get sodium solid half a hydrogen gas the carbon as graphite solid and the rehab is of an oxygen gas okay so we should be able to go back and forth just remember the key things that's always got to be a1 and these have to be elements all right so let's go on the next thing is Delta G of formation is full then the tables are from one mole of the product we already talked about that for one mole but it's also for the formation reaction as written so for example if you look this up in a table what you're gonna see is that the Delta G is the standard free energy of formation for this species it's listed as negative 850 1.0 kilojoules per mole it doesn't say per mole of what but it's implied in the table that that's per one mole of that product so I could write that as a conversion factor this many kilojoules for one mole that substance and I could use it right side up or I could use it upside down if I wanted to use it in stoichiometry as a conversion factor and that's for the species but if I look at the overall equation I can actually do this in terms of moles of any of the species because it's for the reaction as written and as written it's for one mole of that but it's also using one mole of this 1/2 a mole of this 1 mole of this and 3 house moles of that so I can also use those species in there I just have to include their coefficients in there to correct the number for moles of that species so it's really a molar ratio for any species in the reaction based on its coefficients so let's put those other ones in there okay so per mole of this or the same number for one mole of sodium because coefficients also long or the same number for half a mole of h2 now when you actually divide that that 851 by the half it's gonna change the number right so per mole of hydrogen this would be a different number because the coefficient isn't 1 in the reaction as written here is also 1/2 1 so that's the same number and then here that's the same number prefer three halves moles so when you divide that out it's going to change the number per mole of oxygen but it's still based on this number which is four again the reaction is written and you're gonna hear me say that over and over again for the reaction is written for the reaction is written and what that means is the per mole of what depends on what species you're talking about ok next what if it's an element in its natural state so here I have ammonia and I have these compounds what if I had Delta G formation for an element oftentimes those don't keep up here on the table and you're thinking well why why methyls in the table well what's the difference between an energy between an element one mole of an element in its natural state and the element in its natural state nothing so that's why they're not in there because they are zero so elements in their natural state have a free energy Santa Fe and appear in formation of zero kilojoules per mole okay so I've written that in elements in their natural state have a Delta G of formation and earth free energy of formation of zero kilojoules per mole because there is no difference between the element in its natural state and the element in his natural state zero but don't forget it cuz oftentimes you'll seem like you're missing information but that's be gay oh wait that's an element you don't have to tell me cuz I already know what it is okay now the equation that we're gonna use for these to solve from the free energies of formation is that the Delta G standard Delta G for the reaction as written that we're doing is going to be equal to the sum of the number of moles times the Delta G formation for each of the products I'm going to do that multiplication for each of the products its coefficient times its Delta G from the table for each of the products and Adam all app I'm gonna do the same thing for the reactants and I'm gonna subtract this sum from this sum so we've done this before but let's do a problem that uses it all right so our first example problem so we're gonna use this table of the standard free energies of formation for these species to figure out the Delta G for a specific reaction and they want us to find the Delta G for the reaction that is the combustion of liquid benzene and they want us they want the answer in kilojoules per mole of benzene so if I look at a Delta G in a table for this reaction it's gonna be per mole of benzene so given the table here now he also gives a hint write a balanced equation first for the combustion we have to know what the reaction is that we're solving for and then determine the Delta G for that reaction as written again here's your as written and then per mole of benzene because the coefficient on benzene might not be one in that but then we can factor that in to get our final answer okay so first we have to write the reaction so on the reactant side this is a combustion reaction so we kinda have to remember some of these things from chem 1a we're gonna have c6h6 liquid I'm gonna leave some space right there look at that and this is gonna go to what are our products well its combustion of a hydrocarbon so first of all I'm gonna be combining it with an oxidizer which is gonna be oxygen gas and then our products are gonna be co2 and h2o so co2 and that's gonna come off as a gas it's gonna be hot and h2o also coming off as a gas alright now what I need to do then is balance these so I'm gonna do that you can balance by trial and error but I find it's much easier to do it systematically so I'm gonna look for species that aren't in a whole lot of different elements that aren't in a whole lot of different species because like for example right now I don't know what to make this coefficient so there's oxygen in all of these other species so I don't want to start there I'm gonna start with something else like carbon or hydrogen because carbon is only here in here hydrogen is only here in here so I can figure out what those coefficients with lengthy I'm gonna do this in pencil because I might have to erase all right so six so that would have to be a six H is six so that would have to be a three all right now at this point I'm running into a problem and this is very common when we're balancing equations if I go an out to the oxygen no matter what coefficient I put here this is gonna be an even number of oxygens but on this side that's an even number but this is an odd number so I have an even odd issue right I have an even number on one side and an odd number on the other when that happens the easiest thing to do I don't want to double everything automatically I just want to double the species that's making me on the one side that's odd which is this one right so if I double that one now I'll have an even number for that species and I can do that okay and I have to go back and fix my hydrogen's though so I'm gonna have to double this and that's gonna affect my carbons so I'm gonna have to double this one but notice I didn't double everything I just doubled the things I had to starting with the species that was making me odd and now I can have an even number of oxygens even number of oxygens I can make this work so I see I have 24 and another 6 is 30 so this is gonna have to be 15 oh twos alright whenever I do this I want to check mostly because I don't want to make a mistake I'm gonna do all of this work after this based on the wrong coefficients I don't also I want to get the point for the check so check so my elements carbon hydrogen on oxygen see H and O and I also want to check my overall charge on both sides to make sure that that's not messed up in any way so I have to conserve mass on both sides I also have to conserve charge on both sides so and that's going to be important when we get into the electrochemistry chapter next so let's see I have 12 carbons and on the other side I have 12 carbons Jack I have 12 hydrogen's and on the other side I have 12 hydrogen's check I've got 30 oxygens 24 plus another 6 is 30 as well so that checks out my charge overall on this side is 0 my charge overall on this side is 0 as well boom checks out ok now we're ready to go on and use our equations so let's write that in ok so I wrote my variable equation in here just like I had it before the standard free energy change for the reaction and then these two terms for the products minus the reactants and as - reactants right because change is final minus initial this is the final this is the initial right all right so and what I'm going to do is I'm just going to take these and put them in a square bracket and then I'm going to subtract all of these in a square bag those up in another square bracket and then subtract the two you so let's put our units in as well so we're doing the products first so n is the number of moles so the coefficient and then this is the value from the table so you want to do with units as well so this is we'll do carbon dioxide first so this is 12 moles of co2 and I'm gonna multiply that by its free energy of formation so well put them right here negative 3 94.4 and the units on that are kilojoules per mole alright these molds are gonna cancel this is actually kilojoule per mole of carbon dioxide gas it's okay if you don't write it there as long as you at least write it there okay now Plus this is a sum the same thing for the next product so 6 moles of h2o and I'm also gonna let you leave the phases off of here because you could presumably have it up here in this reaction and it's standard free energy of formation negative two to eight point six kilojoules per mole alright so that's all my products now I'm gonna add up the same thing for my reactants and then subtract the two terms so for my reactants I have two moles of c6h6 and it's 120 4.5 kilojoules per mole plus because it's the sum the other reactant 15 moles of o2 times oh they didn't give me that one mmm well there's something going on there's a reason they didn't give it to me because I should know what it is because this is the element in its natural state so that is zero kilojoules per mole am I gonna make you write that in yes but just there and then it's gone okay so now bring this down again Delta G nought reaction equals so this right here works out to be negative 6100 4.4 kilojoules thousand one hundred and four point four kilojoules and this works out to be positive 249 kilojoules point zero kilojoules which equals negative six thousand three hundred and fifty three point four kilojoules okay now this is for the reaction as written and I'm going to write that over and over again right so for the reaction as it was written here with those coefficients okay so that is meaning that that is the amount for two moles of this 15 moles of this 12 miles now write that out in a second before we do that I want to point out a few interesting things here first of all these are whole numbers so when I'm multiplying by whole numbers it's just addition I'm just adding this up twice I'm adding this at 12 times so my sig figs in here are gonna be by the column it's just like we would for addition rules right and that's exactly zero so I'm gonna keep to my tenth just like I had in all the other answers so that's the sig figs the units notice moles cancel moles cancel so I'm not left in kilojoules per mole I am left in just kilojoules and the reason for that is because that is the kilojoules for the reaction as written right I can't specify per mole until I say per mole of what per mole of this is going to be different than per mole of this which is gonna be different than from all of this more per mole about so this is for the reaction as it's written with these coefficients so this is per two moles of benzene or 15 moles of o2 or 12 moles of co2 or 6 moles of h2o and I could use this as a conversion factor with any of these depending on what species I had and what species I wanted to calculate something for so if I was trying it for example figuring out you know for the co2 produced I could use this if I was gonna do something for the oxygen consumed I could use this or for the fuel consumed I could use that alright so we're almost there now the only thing we have to do is it asks for it per mole of benzene and this isn't per mole of benzene this is per every two moles of benzene so I gotta fix that so negative 6 3 5 3.4 kilojoules is for every two moles of benzene so if I want this for every mole of benzene actually have to divide this out so then my answer becomes negative 3176 point 7 kilojoules per mole of c6h6 boom there's my answer all right now look at that sign that is negative right so the combustion of benzene liquid benzene is exergonic exergonic the free energy camembert's is the free energy the free energy goes down which makes sense right because we know combustion is spontaneous now let based on that let's look at this for most species in the table you're gonna see negative free energies of formation right and that's because energy was released when those compounds formed and that's why those compounds formed because the compound is lower in energy in free energy than the elements were individually so they form these compounds now some species though do have positive Delta G's of formation and those ones are great because we can get a lot of energy out of those when we convert them into other species that have negative Delta HS i-search Delta G's of formation okay well let's go on to our next way to find Delta G okay so finding Delta G for the reaction that were interested in from the Delta G's of other reactions that we may know or are given so Delta G for the reactions that for reactions that add up to the overall reaction of interest what we're looking for can be summed up to find the Delta G reaction for the overall reaction of interest so if I know other reactions and their Delta G there those reactions add up to my reaction then their Delta G's add up to my Delta G the thing that's cool about this is it does not even have to be the actual mechanism it doesn't matter what reactions you use as long as they add up to the overall reaction it doesn't actually have to be a mechanism because Delta G is Jesus state function so all that matters is final and initial the pathway doesn't matter so any path that gets me there is one I can use so we can use that and then this is exactly like s's law for Delta H we've already done this in chem 1a we did a bunch of Hess law Hess's law problems this is exactly the same thing the only difference is there's a G after the Delta instead of an H after the Delta but otherwise it's exactly the same so when I teach that and 1a I teach the students a problem solving strategy much like balancing an equation we can do it by trial and error and eventually we will get it but it's much better and faster it's much more efficient if we use a systematic method to solve this so I'm gonna give you that method ok our problem solving strategy just like we did when we're balancing the equation we want to pick things that aren't in every species or else we don't or every reaction cuz then we don't know what to change so we're gonna start by looking for unique species unique species I'm writing this in green and I'll do that color coding later again to I know doesn't show quite as well in the video as it does to my eyes but I see this green use unique species and unique species or species that only appear in one of the given reactions and the overall reaction of interest so you're gonna be given a bunch of different reactions and they're Delta G's and then you're gonna have to find the Delta G of your reaction of interest right and so the little romp is to only appear in one of those reactions and the overall reaction and you're gonna use those unique species to identify the direction and the multiplier for as many of the given reactions as possible so is it going in the four direction or unnie do I need to flip it around and make it go in the reverse direction so which direction is the reaction going and then also the multiplier do I need to multiply that reaction through by two or three halves twenty-seven sixty fourths I don't know but whatever I needed that multiplier now once I fix as many of the reactions Direction forward or backward and how much I have to multiply through as I can then it's much easier because then I've just tweaked the last little things so then I'm gonna use the other given reactions to adjust the coefficients so if I have like you know - and I need five I can adjust it using the other reactions or maybe I just wanna cancel out some unwanted species that are in the given reactions but not the overall reaction and do whatever I need to do to make the sum be the overall reaction of interest and then the Delta G's have to wrap corrected them will add up to the Delta G for the reaction overall so this will make a lot more sense once we do an example but it really is the fastest way to do it because then we have most of the equations locked down the reactions lock down and then we just have to use just a couple now when you have these problems you're gonna be given you're gonna be given a bunch of reaction and an overall reaction to solve for some of the problems are tricky because they have extra reactions you don't need so they may give you more reactions than you need so don't assume that you're gonna need every piece of information you're given also for some of them I've seen they there's more than one given the reactions that give you there's more than one way to do it and so that's interesting as well so but following this you should be okay so let's do an actual problem okay lecture example problem number two so we're given these reactions and they're Delta G's and these are again for the reaction is written so this this is kilojoules for this reaction so that's kilojoules for every two moles of this or kilojoules for every three moles of this right okay and then we want to calculate this is our reaction of interest is we're gonna calculate the Delta G for this reaction now you notice here we have atomic oxygen in this reaction that might be hard to calculate by itself because you know down here we don't have a whole lot of atomic oxygen it's usually molecular Oh too right so this might be hard to actually measure the lab directly but if we can measure these other things well then we can add those up to find this which is why this is really useful now this would be a reaction that might happen in the upper atmosphere where high-energy sunlight is blowing out two atoms into molecules into separate atoms and we would have atomic oxygen up there so all right so let's go ahead and do this and now I'm gonna split this up and I'm gonna rewrite these reactions do you have to rewrite them yes and then we're gonna put everything else in so I'm gonna make a little table here so here I'm gonna put which reaction one two or three what number which direction forward or reverse and then what my multiplier is that I need to make this work and then then I'm going to apply those and rewrite the reaction with that direct in that direction with that multiplier and then I'm going to apply those to the Delta G's that were given to me here and then I can add everything up and again I'm gonna stop the unique species now this last part I didn't talk about but if I switch the direction of the reaction I've got to switch the sign on Delta G because if it's endergonic in one direction it's exergonic and the other and vice versa it's the same magnitude the same number just opposite sign going the opposite direction is energy released or is energy absorbed okay and then if I multiply this this is actually per mole right this is per mole of this promoter that this is actually an extensive property the amount matters if I react twice as much the energy change is gonna be twice as great so if I multiply this through I change the coefficients I also have to change this so we'll also multiply that by the multiplier as well so let's do that so now let's find unique species so I'm just going to look down here okay I know is that a unique species it's there nope nope okay n o is a unique species so I'm gonna start with that so I'm gonna start with reaction number three and then using my unique species here I have one n O in overall I also have one n L so I'm just gonna use this reaction as written right so this is gonna go in the forward direction and it's going to be x one so that's pretty easy so I'm just gonna rewrite it and I'm gonna put that in color just so that you can see the dots of unique species let's see that is this is not a unique species right that's in several reactions this look there's another unique species no.2 is only in this reaction also and not in any of the other ones so that's also unique species so let's see + o3 gas which is ozone goes to no.2 which is also a unique species face is included in that because you might have like liquid water and gaseous water those don't cancel out Rex a not the same phase so the phase is part of that and then oh - let's see Oh twos not in the overall reaction and it's not meeting species so I'm just gonna do that plus o2 gas now this one I wrote the reaction exactly as written so this Delta G is still exactly what they gave me all right no changes that's that's good so now I know this reaction goes forward and the multiplier is one so just like they gave it to me now let's see if there's any other unique species in here we got this one in this one what about this one yep nope yeah so that's a unique species as well oxygen atomic oxygen is also unique species now notice that here it's on the reactant side and here it's on the product side so I'm gonna have to take this and flip it so next I'm going to go to reaction 2 and I'm gonna do it in the reverse direction we're gonna put this on this side and this on this side and notice that that coefficients don't match here I have one here I have two so I'm also gonna have to multiply this through by 1/2 because that'll make this coefficient 1 and then make this coefficient 1/2 so I'm gonna write rewrite reaction 2 in the reverse direction multiplied through by 1/2 so I get my one atomic oxygen now because I multiply through by 1/2 and that goes to 1 half o2 gas right and then I'm gonna use that Delta G but since I change the direction and I multiply through I also got to do that - Delta G so I'm gonna put minus 1/2 times what they gave me here 463 kilojoules right the - for the changing the direction and the half we're multiplying through you could multiply that in right away but I like to keep it on the outside so I make sure that I already did it and I can see if the numbers match and what I'm doing to them but you know as long as you get the right numbers it's fine all right now I've used up it turns out all three of these were unique species and I've used them up there's nothing else so now the only thing I have left to do this doesn't quite add up to this yet right I have these extra oxygen molecular oxygen so those aren't in the overall reaction I have ozone that's not in the overall reaction so I'm gonna have to use another reaction to cancel those out and to adjust it so that they do add up to the overall reaction so let's see I'm trying to get rid of ozone on the reactant side and I'm trying to get rid of molecular oxygen on the product side so here all little s I got ozone and I got molecular oxygen the only problem is if I want this to cancel this has to be on the other side if I want this to cancel this has to be on the other side so for reaction number one I'm also gonna do it in the reverse direction so that my species are on the opposite sides and they cancel and then let's see how many ozone is don't need to cancel I only need to cancel one and I have two here what happens if I multiply this through by 1/2 then I have one there sounds good but then I have three halves oxygen well how many oxygens am I trying to cancel a half and a whole which is two halves so a total of three house oh it's gonna work out great hmm it's almost like they plan that when they wrote this so I'm gonna do it in reverse and I have to multiply through by half again just like the other one so now in reverse now I have three halves Oh - gasps and I just have one oh three yes one ozone and again I'm gonna take the number they gave me this negative 326 kilojoules but I got to multiply by a negative because I changed the direction and a half because I multiply through by 1/2 all right let's see if this adds up so we'll cancel some of these out so this ozone Counsel's this ozone this three-halves oxygen cancels this half in this other hole for a total of 3 abs so now I just bring back down what I have left and I have a no gas plus atomic oxygen gas goes to no.2 gas is that what I wanted it is so now these add up to that and what I get is negative 267 0.5 kilojoules okay now in terms of sig figs that's not going to fly because these were to the units to the units to the units I'm doing addition so I can only keep to the units this is gonna round up to negative 268 kilojoules all right now you might be thinking oh wait what about that 5 there does that round does that even odd rounding rule come in it turns out when you have a fight with nothing after it you do use the even-odd rule in this case it wouldn't apply anyway so that's odd but in this case that is exactly 5 because those are exactly 1/2 so there's no uncertainty so even if this was even it would still round up in this case in case you like to think about those things ok now again again what am I gonna say here this is for the reaction as written that means it's for 4 per one mole of n o 1 mole of O or one mole of no.2 and all these coefficients are 1 so it ends up being the same but for all of even any of those species and we could use this as a conversion factor in stoichiometry so for example I could say this is negative 268 kilojoules per se one mole of an O or I could also use it upside down one mole of an O is negative 268 kilojoules I can use this as conversion factors to get from moles which I could get from grams or more liters or anything like that into an energy or from an energy into moles and then into masses or volumes or whatever I wanted to do all right okay let's go on to things left this one the relationship between Delta G and other thermo dynamic values you might have figured this out but all of these values have to be related to each other because the Delta G is going towards the point of lowest free energy right and where is that well that's equilibrium and what describes equilibrium well the equilibrium constant so does it have to be a relationship between Delta G and and K yes there totally does so that's the first one we're gonna do and so the equilibrium constant which is capital K remember capital K for equilibrium constant lowercase K for kinetics rate constant all right so that relationship is this Delta G equals now okay there's gonna be a K on this side what else do you think is gonna be on that side mmm based on all of the reactions we've done so far I'm gonna guess a natural log so yep there's a natural log in there and then what else well temperature appears to be in everything and again that would be the Kelvin temperature what else the universal constant yeah that's got to be in there and then it turns out there's a minus sign in as well and that's it Delta G the standard change in free energy is equal to minus RT times the natural log of the equilibrium constant all right now let's put in a few things here on the side here's some notes R equals 8 point 3 1 4 4 6 2 joules per Kelvin mole be careful this is Kelvin this is the equilibrium constant they're both capital case temperature in Kelvin right just like any gas problem or anything like that temperature has to be in Kelvin and then one more thing don't forget we got to be sure this is in joules this is typically given in kilojoules we're gonna have to convert one to the other so that we have the unit's match right so be careful with the Joules to kilojoules conversion if you forget to do that you're gonna be off alright and I'm gonna put a little note here to see lecture example problem number 3 which we're gonna do in a minute all right let's do another relationship while we're on this topic you haven't heard of this one yet maybe but we're gonna we're leading right into it in the next chapter the standard cell potential and that symbol is e naught cell the standard cell potential all right this is a voltage this is a voltage and we're gonna do the next chapters electrochemistry we're spending a lot of time on that so potential is measured in volts which are joules per Coulomb and we'll write that down in the stack so again Delta G is gonna be related to that potential and we're gonna have this potential in there on this side what else do you think is in there well this is electrical so I'm thinking Faraday's constant and the number of moles of electrons transferred this is going to be for redox right exchanging electrons so I got to consider somehow how many electrons are exchanged and how much charge those electrons have and figure that all out so for this one a few things little notes here on the side and just like I said is equal to the number of moles of electrons exchanged so our two moles lost in electrons lost in two moles electrons gained or three moles or one mole what's that F is Faraday's constant and this is ninety six thousand four hundred eighty five point three four and the units are coulombs per mole of electron Coulomb is the SI unit of charge it's got a symbol of a capital C make sure it's a capital C and not a lowercase C because it's a lowercase C speed of light so we want to do that now a couple other things I'm gonna say that E is in volts and a vault is a Joule per Coulomb so how much energy per unit of charge and then one more thing here just like in this one we got to be careful about joules and kilojoules because this is going to be in joules right joules per Coulomb and this is in kilojoules per mole so we got to be careful with that so watch for your joule to kilojoule conversion and then for this one I'm gonna say see lecture example problem for which we'll do in a minute - all right there's one more thing I want to do before we go on and do these these example problems and that's the last topic so far everything has been standard standard it's always been for standard standard change in free energy for the reaction the chant standard change in free energy for the formation reaction so what if it's not at standard conditions it's not a big deal it just takes a real small adjustment so let's look at that and then we'll do these two examples ok so Delta G at non standard conditions okay for this Delta G at whatever conditions you have now is gonna equal the Delta G under standard conditions with a correction factor and that correction factor looks remarkably like this except that it's positive so positive RT and again the natural law but this time it's not the natural log of K it's the natural log of Q right in standard conditions everything's got to be one molar but you might not be one molar so you gotta throw those in there and you might not be the equilibrium so that might not be what you where you are either so that takes care of any conditions concentrations you might have and then any temperature that you might have and it will correct the standard value to the actual value okay so a couple of things to think about this there is a version of this where we have log base ten of Q with the correction factor coefficient of 2.303 but we're all good with natural log so I don't think we need to do that but you may see a different version as a coefficient here that's a number that's because it's log base ten instead of log nice II Thompson would love that but not us all right so a couple of notes for this at standard conditions what do you think is true well at standard conditions what is Q at standard conditions Q equals one right because if everything is one molar or or if it's a pressure it's at one bar one atmosphere then Q equals one no matter what the exponents are and how many terms are are everything's one it's gonna be one so Q equals one that means this whole term drops out because I match the log of one is zero so then this times 0 this whole term is gone so if that's true then Delta G equals Delta G standard does that make sense yes because the actual Delta G would be the standard Delta G if you were at two hundred conditions right so the only just turn councils out that's not there all right what about at equilibrium well then Q equals K therefore Delta G actual equals zero kilojoules per mole all right where do I give that ok so if Q equals K and this is equal to this term but negative and I plug K in they're the same term but positive they're exactly the same magnitude but opposite in sign they're going to cancel each other out and Delta G equals zero which makes sense because you're at equilibrium all right so it isn't a big deal if it's non-standard conditions we just have to make this little correction plug that in and we're good to go okay now let's go back and do lecture example problem three and four and we'll be done with this chapter okay let's for example problem number three so they want me to find an equilibrium constant so I'm gonna have to use that relationship we just saw in the previous page for this reaction oops I didn't write the reaction down for this reaction okay so my strategy here and they give me a table of Delta G's formation so I shouldn't have to combine two different things to solve this right first I'm gonna use the Delta G's of formation to find the Delta G for this reaction as written and then once I have that I can plug it into my equation that relates Delta G and K to get K okay cuz this is just for one species I need it for the whole reaction as written so we'll do that we'll combine two of the things we've done so I'm gonna start by writing out my variable equation alright and I'm gonna start plugging into that so I have two moles of ammonia so the Delta G for my reaction is written is gonna be two moles of NH three times they gave me that negative sixteen point five kilojoules per mole yeah that's my only product so that sums done it was easy - same thing for the reactants so one mole and two they didn't give me that but it's an element in its natural state zero kilojoules per mole plus one mole of h2 didn't give me that either that is also zero kilojoules per mole okay so that goes up there sorry I ran out of room and then well that's pretty easy alright so Delta G for the reaction close that circle there is just equal to negative 33.0 kilojoules all right and again a little note for the reaction as written okay so this reaction right here so that is per one mole of n2 three moles of h2 or two moles of NH three alright but it's for that reaction is written if you want to find K for this reaction has written then we're gonna use this whole value right here we'll just say per mole but we won't specify per mole of what per mole per unit mole okay all right so now we have to go on and figure out now that we know Delta G we have to relate that to K so I can say that this Delta G for the reaction is equal to minus RT ln okay my variable equation and well I'm solving for this so I'm going to bring this down to the other side and then I'm gonna take the anti natural log so e to the both sides alright so what I'm going to end up with is that K equals e to the Delta G naught of the reaction / - Artie ants I'm gonna see how I got that I just divided both sides by negative RT and then took the antilog each other so then all what was on the side is now in the exponent and this canceled out the natural log on the other side so I just had K and now I'm gonna plug in my numbers so still eat to that this is negative 33.0 kilojoules per mole we're just not specifying from a lot but per unit mole and then well that's gonna be a problem because I'm gonna be plugging in are down here and that's in joules so let's just change this to two joules so one kilojoule is 10 to the third jewels kilo literally means 10 to the third that's on the top and then the bottom I have minus R so minus 8 point 3 1 4 4 6 2 joules per K mole and I have T which was 25 Celsius plus 273 point one five is two ninety eight point one five Kelvin and that's all in the exponent Y all right notice to all my units cancel out kilojoules joules moles Kelvin everything cancel out so there's this is unitless inside here and I actually get that this as e to the thirteen point three one two zero five 102 let's see I have five there's a lot of sig figs three three so three sig figs boom okay the reason I'm doing the sig figs here is because that's gonna affect my sig figs and my answer so when I take and a log of something the number of sig figs becomes the number of decimal places and only those are significant go on the other way as an exponent only the significant decimal places become the number of sig figs so I'm gonna add one so I get here that K equals x times 10 to the fifth okay all right now notice how I wrote K equals cuz k's are unitless right and then also this is kind of interesting you might be wondering these are all gases is that a KP or a KC hmm they didn't give me any information to know if it's a KP or KC like if they give me if I was just solving for K and then giving me pressures it'd be a KP and if they'd give me a molarity is it be a KC but in this case they didn't give me information but in general if it's a gas it's gonna be a KP so this is a KP because it doesn't say it isn't so it's a KP now if you recall if you could if you wanted the KC wouldn't be a problem to get that from the KP if you remember from the equilibrium chapter which was openstax chapter 13 we had the relationship between those two so K P is equal to K C times RT to the Delta n power and this Delta n was the change in number of moles of gas in the reaction so the number of moles of gas in the products minus the number of moles of gas in the reactant so in this case it'd be 2 moles -4 moles so this would actually be to the negative 2 power right in this case and so we could plug in here plug in there and then solve for that alright let's do a Part B to this one more thing I want to point out before we go on to Part B though is look at this number negative 33.0 kilojoules that's just twice this number does that make sense it actually does because look at this reaction it's almost the formation reaction you have the species and you have the elements in their natural state the only difference is that this is 4 moles of this whereas the formation reaction is only for one mole of that so this literally is just double the formation reaction so it makes sense that this number is twice this number right here all right now for reals let's go on to Part B okay now Part B isn't actually related to any of this stuff that we just been doing now but I just want to tie it in to what we've already learned about this so if the Delta H is negative if it's exothermic for the reaction is written then is K gonna increase or decrease with increasing temperature right so we still want to tie in all the old concepts so let's rewrite that equation and 2 plus 3 H 2 going to 2 and H 3 and I'm gonna write this in to the equation so Delta H is negative so that means it's exothermic so I'm gonna write heat in as a product and I could write the word heat or I could put in the actual value if I wanted to if I knew it I could instead of writing the generic heat I could just put in the value and this again is for the reaction as written so this is for every one mole of this for every 3 moles of this or for every 2 moles of that there molar ratios right ok so now we're gonna look at this in a little shot li a kind of way right so stress shift we don't really need the EQ that's enough but we can just put that in there to make it feel comfortable and familiar all right so the question is asking me how is K gonna change if the temperature goes up so again a heat and temperature aren't the same thing but at higher temperatures there's more heat available so we'll put an upward arrow as the stress over that and it's on the product side so I increase something on the product side is gonna shift away to use it up so it's gonna shift it back towards the reactants what does that tell me about K well remember that K equals products over reactants and the other concentrations and there to some power right and we could plug in the actual numbers in there but this is all we need to know if when the temperature goes up I shift from products back to reactants that means that K is gonna go down as the temperature goes up so I can put that K decreases as T increases and vice versa if the temperature goes down K goes oh my god okay so just tying this back into some stuff we've already done okay next example problem all right let's your example problem number four the last one Part A so it's gonna be a part beat but Part A find the standard cell potential for the unbalanced redox reaction below from the provided standard free energies of formation okay so we have two reactions they're both written as reductions because we're gaining electrons gaining electrons so those are both written as reductions they're not gonna both be reductions one's gonna have to be flipped and be in oxidation but these are typically given values for their reductions and then the oxidation is we just switch the sign so those are given to us as well we don't need those for this part but we'll need that for Part B which is why I gave it to you when we do get to the electrochemistry chapter you will be able to figure out which one goes as a reduction and which one ends up being an oxidation but at this point we don't know that yet so in the hint I'm telling you that reaction one goes forward it's getting reduced which means reaction two is gonna have to be going the other way getting oxidized alright so now I can also give you a hint determine the Delta G's for each of those reactions one and two and then use a process similar to the houses law so flip them around multiply through whatever you got to do to make the electrons match and then out of the Delta G's for reaction 1 and reaction 2 to get the reaction the Delta G for the overall reaction okay so that's not too bad so again it's going to be like two of the problems we've done before we're gonna use the free energies of formation and then the Hess's law all right so let's write the let's write that first reaction variable equation down okay so here's my variable equation I'm actually gonna have to do this twice that accidentally started right here but I wasn't ready yet this is just the variable equation I'm actually have to do is twice once for this reaction and once for this reaction I'm not gonna have to rewrite the variable equation but I'm gonna want to specify which reaction I'm talking about so let's do the first one first so Delta G for reaction one as written right now you put my one in there all right so that's going to be the sum of the products and the reactants so here on the product side I've got one mole of a now times 87.6 kilojoules per mole and moles cancel plus next product two moles of h2o times liquid-liquid good negative 2/3 7.2 kilojoules per mole was just checking to make sure that my face is actually matched so I wasn't pulling the wrong thing out of the table those are both of my products now the same thing for my reactants so one mole and o3 - nitrate aqueous nitrate aqueous negative 1 1 1.3 kilojoules per mole and to that I'm going to add 4 moles of hydrogen ions aqueous at zero zero point zero kilojoules per mole okay and I get my Delta G for reaction one is equal to negative two seventy five point five kilojoules again just a reminder the moles cancel so this is kilojoules before the reaction is written so four one four three one or two moles of any of those species that's reaction one same thing for reaction to that this is gonna be a little easier my products have got one mole of copper solid hmm no copper solid in here oh wait element natural state mmm zero kilojoules per mole that's it on the product side on the reactant side one mole of copper two ions and hoop there we go sixty five point five kilojoules per mole and so my Delta G for reaction two standard there is going to be well just negative sixty five point five kilojoules again for this reaction has written with coefficients of one two or one okay so now we know those now let's go on to the Hess's law part okay so I've set up my table here it's like I'm gonna died in the previous one this is a little different because I don't know what the overall reaction is they didn't tell me they gave me the they reactions that up to it but not the total so hmm that makes it tough well so I can't really look for unique species between the overall and and these these other reactions because I don't know what the overall is but they did give me one piece of information that's gonna be really useful and that is that the nitrate is being reduced so I know that reaction one is going in the our direction now I don't know the multiplier well the only thing I have to go off of here is that in this reaction I have two moles of electrons and in this reaction I have three moles of electrons this is gonna have to be flipped and go the other way we'll get to that next but the other thing that's gonna have to match is the number of electrons are gonna have to cancel out the number of electrons lost by the copper is gonna have to equal the number of electrons gained by the nitrate so um let's that's gonna have to be multiplied through by two and that's gonna have to be more flipped and multiplied through by three so let's do that so this is going to be multiplied through by two okay so I'm gonna have to a no three minuses those are aqueous plus 8 h plus aqueous plus six electrons goes to two n Oh gas plus four h2o liquid all right now I did it forward so I'm gonna use this value right here my negative two seventy five point five kilojoules and I did it forwards and I have to change a sign but I did multiply through by two so it's gonna be twice that all right now for reaction two and this this isn't my unique species but this is how I was able to determine what I have to do right now reaction 2 is gonna have to go in reverse and we said we had to multiply through by three to get the number of moles of electrons to match so I'm gonna have three coppers on this side solids and on this side I'm gonna have three copper two ions aqueous and six electrons and then I did reverse it so it's gonna be - and then multiply by three times this one negative sixty five point five kilojoules all right let's see if this adds up well so this is gonna cancel with this and which it should or else I've done something wrong and I can bring it down so actually I'm for the first time seeing the overall redox reaction here so - no.3 minus aqueous does anything cancel no no okay so plus a h plus aqueous those cancelled plus three copper solids goes to - I know gas plus 4 h2o liquid plus three copper two ions aqueous okay I think I got everything right I might want to actually do a little check on there to see if I did if it's still balanced so we could do that okay so I've nitrogen oxygen hydrogen and copper so nitrogen I have two that's it and two oxygen I got six here that's it so two and four more six on that side as well hydrogens at eight and two times four is eight yeah he and coppers I have three and three and this time I charge my overall charge on this side I have eight pluses and two minuses right so it's gonna be six plus and on this side I have three times two plus I also have six plus the charge doesn't have to be zero on both sides it just has to be equal on both sides and in this case it is the same on both sides so that checks out I didn't screwed up anything in the balancing that's all good now I'm gonna bring these together and I'm gonna get negative three hundred and fifty four point five and it's still kilojoules okay all right so that's that's part of it now I have to figure out what the standard cell potential is so this is my Delta G for the overall reaction now I got to figure out the standard cell potential so I'm gonna write that equation down Delta G is equal to minus n f e naught so okay so we're gonna use this relationship that we put down earlier to get from the standard free energy for the reaction to the standard cell potential now I could have done this in the one I wrote before I didn't have reaction here I didn't have cell here because I could do it if I'd use the Delta G for just like say the reduction half-reaction I could find the potential for just the reduction half if I did it for the oxidation half I could do it for the oxidation half in this case I'm doing it for both halves the reduction and the oxidation so this is the full cell both half cells together and this is the full reaction as we've written in here okay now I can go ahead and plug in my numbers so this here was negative 350 4.5 kilojoules I could put kilojoules per mole I don't have to but I could it would be firm for the reaction as written per unit mole none of these are actually have coefficients of one though so it's a little bit weird but I could do that if I wanted to just realizing later on if I wanted to be a specific species and I have to account for the coefficient or in this case I can just leave it off two equals minus n the number of moles of electrons transferred by six six so that's six moles of electrons times Faraday's constant the ninety six thousand four hundred eighty five point three four coulombs capital C per every one mole of electrons and then I still have my variable that I'm solving for right here I'm gonna bring this the other side to solve for e cell and I get that e naught cell is equal to zero point six one two four joules per Coulomb right so what's I forgot something on this side I can't leave that in kilojoules I want to change it to joules so I'm gonna do ten one kilojoule it's ten to the third joules right so now I have joules per Coulomb right but remember a volt is equal to a Joule per Coulomb so this is equal to zero point six one two four volts so six tenths of a volt is the potential for this overall reaction by the way if if you had me or as a rubout for chem 1a and the first day we dissolve a penny and nitric acid this is that reaction the coppers the penny the nitrates the oxidizing agent from the nitric acid and then here's the protons from the acid right and we ended up with smog and pretty blue solution all right so that's that now if I could put in the mole here if I put in that mall I would have been left with an extra mole on the bottom that actually wouldn't matter because potential is not an extensive property it is an intensive that means the amount doesn't matter so if this is joules per Coulomb or joules per Coulomb per mole it doesn't matter that's very different than Delta G which is an extensive property right this per mole matters per mole of what if I double the amount the amount of energy change is going to double right but in terms of potential potentials different if I double the amount of the reagents the potential is still the same which is interesting it's kind of like density if I have twice as big an object the density is still the same because you're dividing mass by volume you have a bigger mass in a bigger volume it doesn't change it voltage potential is also an extensive prophecy so per mole if you ever get that you don't have to worry about it it's still the same value for one mole or for a billion moles all right let's do Part B on this one and Part B is something you're gonna learn soon you have learned yet but I just want to compare here okay so in Part B is asking me to compare the value from Part A that we just got to that standard cell potential that I would get if I use the table of standard reduction potentials so I'm gonna go back up to the original problem because they gave me the standard reduction potentials for these reactions okay and the one thing that the piece of information that's missing is how to relate those so I'm gonna say that e cell is equal to the standard reduction potential for the reduction half-reaction plus the standard potential for the oxidation half-reaction so if you add up the potential for one half and the potential for the other half that gives you the total right the half cells add up to the full cell okay so so far so good each cell is equal to by the way you should be able to go back and forth standard reduction potentials is this so from the symbol to the written name and back and forth so you don't get confused if they ask you in the problem for the symbol or they ask you for the event with words you can you know which each means alright so the standard cell potential is equal to the standard reduction potential for the reduction half-reaction which was the nitrate and that's going to go forward so that's zero point nine six volts plus now the interesting thing here is that notice I didn't multiply this I had to multiply this through by - I didn't multiply this through by - because again it's an intensive property it doesn't matter the amount if I do one or two or three moles it's going to be the same however the sign still matters so if I switch the direction I still have to switch the sign so for this one I multiply it through by three but I don't care about that but I also switched the direction I do care about that because this was a reduction potential so the oxidation potential going the other way is gonna be the same number but opposite in sign I guess that's the oxidation and that was the reduction but I don't want apply through by two and three and so I get that e naught cell is equal to 0.62 volts all right so how do those two compare they actually compare pretty well now notice that they don't compare in this digit they're different this were actually rounds down to six one and this is six two but that's still okay because in this table this is the the hundreds column is the uncertain digit so this is 0.6 two plus or minus one in the uncertain digit right so those two do match within the sig figs of this one all right well that is it hopefully that gets you through this chapter and the practice problems and let me know if you have any questions