all right so last thing to talk about is just you know applicability that's a word almost sure 95% sure of Delta G so how much can Delta G tell us all right and sort of real-world sitting areas or what do we need to think about if we want to apply it to a different scenario so this water and local phase going to water in the gas phase that is the physical process for water evaporating all right and if you calculate Delta G for this process we would get plus eight point five nine kilojoules per mole what does that mean non-spontaneous so that means it's not gonna happen all right what do we know does water evaporate yes water does evaporate so what's the deal well the problem is most of the time we're calculating this in our you know the general chemistry we're doing under standard state conditions which means 25 degrees Celsius okay yeah that's that's everyday real life concentration of solutions 1 molar we don't have a solution so we don't worry about that partial pressure is equal one atmosphere oh that's different okay the partial pressure of water in the gas phase is not one atmosphere okay and so what we have to do is we're gonna have to calculate Delta G at non standard state conditions all right so for to calculate Delta G at non standard state conditions all right you got Delta G at standard state conditions all right that's not too bad our that's the ideal gas constant and we're gonna need it in terms of energy so we're gonna use 8.314 joules per mole Kelvin T what do we think that means temperature Kelvin natural log times Q now you've seen uppercase Q before but we kind of just didn't spend a lot of time on it that was called our reaction quotient and we talked about that in equilibrium when we're discussing equilibrium reaction quotient was our equilibrium constant more not at equilibrium so products over reactants right so the K uppercase K just not at equilibrium because we're not at equilibrium if in my head it's a constant idea so the question was where the 8.314 so that's the ideal gas constant in units of joules and you will find it on your equation sheet because someone has provided it for you you're welcome alright so when we're looking at this okay so obviously this was K if it's an equilibrium products over reactants but this was waters in liquid phase here so all that would be would be concentration of water of course we could also put it in terms of partial pressure and that's what they're doing here so that's a reaction quotient so it's not an equilibrium partial pressure of water liquids and salt liquid water and solids just liquid water alright so it turns out that if for standard state conditions the partial pressure of water would be one atmosphere and it would be not spontaneous water wouldn't evaporate if it was that much water in the gas phase just wouldn't go up anymore okay you know in every day normal conditions the partial pressure of water in the atmosphere is you know we read a it's highly dependent on you know time add a geographic location but on average you can usually assume that it's partial pressure of waters like something like five times 10 to the negative third atmospheres so like a Milla five Milla atmospheres and so at that partial pressure guess what this equation causes this to be negative hey water evaporates if you didn't know now you know and knowings half the battle I don't know what the other half is they never told me let's try to use that equation on this example so we're gonna calculate Delta G under non-standard a state conditions for this reaction next Roman oxide plus oxygen go into no.2 we've got the standard say Delta G negative 70 1.2 kilojoules now we're just gonna need to calculate it under non c'è interstate condition all right so our new handy dandy equation is Delta G equals Delta G standard state conditions plus R T natural log of Q R is a constant which for this one since we're using it in terms of energy we're going to use the 8.314 joules per mole Kelvin that's what you're given that's what you usually find if you look up the ideal gas constant with units of energy you only problem is Delta G's in units of kilojoules so when we add them we're gonna have to make them match so usually I'm gonna have to convert this to kilojoules so in one kilojoule there are 1000 joules so that's gonna be 8.314 times 10 to the negative third kilojoules per mole Kelvin nothing changes there the other thing we're going to calculate is Q we're gonna put that all right here hmm yeah sure I'm going to yes you're right I'm living life on the edge will I make it on this page yes all right so Q is the same thing as K right just not under equilibrium conditions so it will be the concentration or in this case they're giving us partial pressure so we can use partial pressures partial pressure of the products no.2 squared because the coefficient right divided by the partial pressure of N o squared as well times the partial pressure of oxygen not squirt yes you could absolutely so you could have converted this to joules and then added them you would have got your answer in joules that's fine quite legal all right so let's plug in our numbers so we're going to get a no twos two atmospheres squared all over zero point one zero squared times 0.1 as well anyways cue will equal so 2 squared 4 divided by 0.01 times point one some point oh one divided by 0.001 that makes it bigger so four thousand four thousand what are my units they've kind of looks like it should be one over ATM but remember this is equilibrium constant and it comes out to be unitless so we'd all been eaten it's on cue okay all right so now I can finally go back in and calculate my Delta G under non standard state conditions all right so I got Delta G standard a negative 70 1.2 kilojoules plus the ideal gas constant which I just converted to kilojoules per mole Kelvin times temperature what's my temperature 298 good times the natural log of 4,000 and we get- 50.7 my units are gonna be kilojoules everything else cancelled out old can't loan know that's it moles well technically I mean you can think of it as kilojoules per mole but you we are not going to write it as just because we normally don't because you have to go to the stoichiometry the coefficients to figure out the per mole basis so seventy one point two kilojoules per mole box again seventy one point two kilo joules per two moles of no.2 I we try not to write it unless we're gonna need it alright so the question that's asked here is actually not just that number I mean that number is fine I like it we worked hard for it but the question that is posed here is this reaction more or less spontaneous what do you think that means so it's a question of the energy alright so is it more or less spontaneous under these conditions than understand or say conditions what do we know about the standard state conditions all we know is that a had previously had a Delta G of negative 70 1.2 now with these different concentrations it has a concert or Delta G of negative fifty point seven okay so obviously it's a different number it's still negative so it's still spontaneous but is there more or less spontaneous okay what's good for spot what what for a spontaneous reaction what's Delta G negative so do you think a more negative number is more spontaneous or a less negative number more sponte so the more negative the value is the quote-unquote more spontaneous it is so if it's last negative closer to zero it would be less spontaneous for some reason I wanted to do it right inclusive I wanted to feel fancy a better question would have been can you get more or less work out of this system because remember that's what Delta G was originally meant to that's why I call its free energy how much energy can you get out of this to do work and so pretty at standard state conditions you can get 70 1.2 kilojoules to do work now you can only get 50 point seven less work it can do less work