In this learning objective we're going to be looking at the difference between a standard delta G in a non-standard delta G. Students are often don't see the nuances in the differences between these two so I am going to try to help you work through thinking about the difference between them. We know that this equation Delta G equal delta H minus T delta S would apply whether you are in standards state conditions or not and standard state conditions. What is the difference between standard state non-standard state? Well, let's review what it means to be in standard state conditions. That means that the pressures is at one atmosphere and if it is a solution then it will be 1 molar and when you look in tables for standard value at thermodynamics there are usually at 25 degrees Celsius but they don't have to be to be considered standard state. Now, if we started out every reactant and product in a reaction in the standard state conditions. Lets say they are all gases and every gases was at one atmosphere in there. Not just to reactants, but your reactants and your products. This system is very unlikely to be at equilibrium under those conditions. So if it's not an equilibrium that is going to proceed either to the right or is going to proceed to the left. Now remember if delta G standards is negative we know it's gonna go to the right to get to equilibrium, it is spontaneous in the forward direction. If standard GG is positive it's going to proceed to the left in order to get to equilibrium. The reverse reaction is spontaneous. Now if you wanted to know what standard delta G at a temperature other than 25 degrees Celsius then this is the equation that we would use. You would look up in tables the delta H standard and the Delta S standard. From that information these values don't change very much with temperature. Delta G is very affected by temperature. So if you put those standard delta H and standard delta S, even thought they are at 25 degrees Celsius. You put in your temperature and don't forget to use Kelvins when you do so then this is going to give you the standard delta G at a temperature other than 25 degrees Celsius. If you needed delta G at 25 degrees Celsius and its standard then you might as well just go to the tables and determine the standard delta G for the reaction and not use this equation. So we have a series of questions that I want you think through in order to figure out the connections between what is delta G standard telling you and how does it change when you leave standard state conditions. So I have a reaction and in this reaction chamber I'm starting with every gas at one atmosphere. So that's what's being represented in that picture. That is a standard a condition. We see here that the standard delta G is a negative value a -190.5. So if I started everything out at one atmosphere which statement would be true? Would the reaction shift to the right to get equilibrium, or shift to the left to get equilibrium, or neither, it is already an equilibrium? If you said A you would be correct. A negative value for a standard delta G says, and this is all it says, it is snapshots that says, if we started everything at at 1 atmosphere this reaction would proceed to the right because it's negative. Let's again start this system out at equilibrium. I mean start the system out with everything in its standard state conditions with 1 atmosphere. So we're here. We know according to this standard delta G that if we start everything and a standard state conditions the reaction will proceed to the right. Now if it gonna proceed to the right. What is going to happen to the pressure HCl? Well if you said it will increase you would be correct. The HC; pressure is going to go up. These pressures are going to go down as it proceeds to the right. So it's always going in the forward direction and reverse reaction are always happening but more of the forward is happening then the reverse when you have a -190.5 kilojoule, we have a negative value for that standard delta G. Now as soon as it starts proceeding to the right my question for you is, is it still under standard state conditions? Well no, because if the pressure is going up it is no longer one atmosphere it is greater than one atmosphere. And if its greater than one atmosphere then we are not under standard state conditions and this number of - 190.5 no longer applies. It only applies when when we are in standard state conditions. Just to think about this, does it really give us a good connection between standard and non-standard delay G? What's going to happen to the total pressure on the system as you establish equilibrium from the point at the beginning where everything is one. Starts proceeding to the right and it finally get equilibrium and the foreign reverse and reaction are happening at the same rate, what is going to happen to the total pressure? If you said stay the same, then you're correct? The pressure of HCl is going up as it proceeds to the right. The pressure of H_2 is going down the pressure Cl_2 is going down because the number of moles on the left which we have two moles and gas not just is not just moles its moles and gas there's two moles gas in the left and there's two moles a guess on the right then as this reaction shifts the total pressures not changing. Okay now here is a question I really want you to think about. When the equilibrium is finally established, ok so what's happening here? We started everything at one atmosphere. We saw that at one atmosphere we know that the standard delta G is negative that tells me this reaction is going to proceed to the right to establish equilibrium and it finally gets there. Once it establishes equilibrium what is going to be the value of the standard delta G. Will it increase to get equilibrium? Will it have decreased to get to equilibrium? Or will it stay the same? Students often miss this. So don't feel bad if you didn't you've got to get to the understanding of why would stay the same. A standard delta G is a snapshot of where where it would be if everything were at one atmosphere. So standard delta G never changes. Its value is the - 190.5 kilojoules per mole for this reaction. So as a reaction release these one atmosphere conditions we no longer have a standard state condition but the standard delta G is the snapshot of what it is when everything is that. So don't think that your standard delta G is going to change. Which Delta G without the circle it certainly will change. So our next question is to think about what is happening to Delta G. Once you establish equilibria. Will it be positive will be negative? Will be equal to 0. Well it you said it would be equal to 0 you are correct and that's exactly what happens. Delta G equals zero an equilibrium. Not standard delta G the only way standard delta G could equal 0 is if when everything was that was at one atmosphere this thing happened to be at equilibrium. Very, very unlikely. Very unlikely. Well this reaction certainly isn't it had a negative value we know it wi'll proceed to the right. We will build up more and more of these two. We will use up some of this and some this and finally get to equilibrium. When he gets the equilibrium, at that point the Delta G for the reaction is no longer standard state conditions is 0, That is true for every reaction it's Delta G is equal to 0 at equilibrium. So what happened to the Delta G? The Delta G started out when we were all at one atmosphere with this number. It gets to equilibrium Delta G non-standard is equal to 0 so the negative Delta G is increasing as the reaction keeps working its way closer and closer to equilibrium. Now we will look at the mathematical relationship between the standard in a non-standard delta G, there is a connection. So anytime you need in non-standard delta G is when you're not in standard state conditions. We have this equation help us get to a non-standard. So if we want to know what the reaction was at going to proceed to the right or proceed to the left and we had conditions other than standard state conditions then this is the equation that we would use. The R that we use in this equation is the 8.314 because it has the energy unit a joule, and you actually probably have to convert it to kilojoules before you could add those terms together. We see that a Kelvin is here in this K and that's what we have to use for the temperature and our Q is a reaction quotient. The reaction quotient if we remember for reaction is just like a K equilibrium constant. Its products and concentrations if there aqueous pressures if it is gases, but its products over reactants raised to the power of their coefficients. So there's some coefficients of the balanced equation is raised to those powers. So that's what the reaction quotient is so we can look at this reaction and say OK at this point in time there is this much product, there is this much reactant. Let's put those numbers in there and let obtain a value for a non-standard delta G. Now if everything were 1 then Q would be 1. And if Q is one the natural log of Q is 0 and this term goes away and that make sense. So we're just gonna think through this equation. In terms of what how does change in Q affect the Delta G? So let's start with a standard Delta G that's negative. This means that the reaction is spontaneous in the forward direction. if you we in standard state conditions. We put everything in a reaction chamber and this is kinda like our little questions that we went through are standard delta G was negative the reaction spontaneous in the forward direction. So the question is what would happen, we have to do, in terms of conditions? What would we have to do in terms of Q here in order to make Delta G positive? Well if the delta G is positive we know that it's no longer spontaneous in the forward direction is now spontaneous the reverse direction. How do we make Delta G positive? Well, the only way we can make delta G positive is to make this term positive. Well this term is positive when Q is what? How do we make the natural log of Q be positive value? Well if Q is greater than one it's a value that is bigger than one then the national log of Q is positive. That is the first thing. As long as Q is greater than 1 then when we take the natural log it will be positive in this term that I have circled up there is going to be a positive term. We just gotta make positive enough. So anytime that the products are greater than the reactants because remember what Q is? It is products over reactants. Remember? Raised to the power of their coefficients. So anytime products is greater than reactants then the Q is going to be bigger than 1 and a national log of Q will be a positive value and we can get it positive enough to overcome all the negative that we have here. We can turn a reaction that was spontaneous in the forward direction to be spontaneous instead in the reverse direction. Well lets go through that same mathematical logic in the revers. What if the standard up to G was positive the reaction is not spontaneous in the forward direction it is only spontaneous in reverse reaction direction if everything were under standard state conditions. So we start everything is standard state conditions. We know the reactions going to proceed to the left. That is when a positive means its spontaneous in the reverse direction. What conditions would we have to have in order to make this negative. We want this reaction proceed in the forward direction instead of in the reverse direction. Well lets think about what has to happen to this term. In order to make this reaction becomes spontaineous the forward direction this term is going to have to be very negative. It is going to have to be negative enough to overtake the positive of the standard delta G. How can we make that term negative? Well, the only way we can affect that term R cannot change it's always the same. Temperature is Kelvin and there's no such thing as a negative Kelvin so the only way we can make this term negative is to make the natural log of Q negative. So if Q is less than one, it is a fraction less than 1. Then anything can you take the natural log of a number less than 1 its negative. So how do we make Q less than one? Once again we remember that Q is products over reactants raised to the power their coefficients. As long as you have a lot of products so this is a big number. and this is a small number. So products is an smaller than the reactants and you want to have a number less than 1 and as long as it is small enough this natural log of Q would be negative enough that we would have more negative than we have positive and that would make my non-standard delta D negative and we can for force this reaction to become spontaneous in the forward direction. Now you going to be doing mathematics numbers for this it's not a big deal. You just gotta watch for signs if you're given a standard delta G and they want to know a non-standard delta G this is the equation you will pull out. Watch you units if Delta G which is typically in kilojoules is given to you and then we're going to have to make sure that this term gets converted kilojoules before we convert it. The thing you have to watch is your T and make sure it is in Kelvin. Then you will be able to easily convert between a standard delta G and a non-standard delta G. So this is the enough are learning objective number 9 in which we are seeing the connection between the Standard delta G and a non-standard delta G. We see that they're not the same thing we have to understand their differences and we have an equation that will help us convert between them.
