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In this learning objective we're going to be looking at the difference between

Â a standard delta G in a non-standard delta G.

Â Students are often don't see the nuances in the differences between these two

Â so I am going to try to help you work through

Â thinking about the difference between them. We know that this equation Delta

Â G equal delta

Â H minus T delta S would apply whether you are in standards state

Â conditions or not and standard state conditions.

Â What is the difference between standard state non-standard state? Well, let's

Â review what it means to be in standard state conditions.

Â That means that the pressures is at one atmosphere

Â and if it is a solution

Â then it will be 1 molar and when you look in tables for standard

Â value at thermodynamics there are usually at 25 degrees Celsius

Â but they don't have to be to be considered standard state.

Â Now, if we started out

Â every reactant and product in a reaction

Â in the standard state conditions. Lets say they are all gases and every gases was at

Â one atmosphere in there.

Â Not just to reactants, but your reactants and your products.

Â This system is very unlikely to be at

Â equilibrium under those conditions. So

Â if it's not an equilibrium that is going to proceed either to the right or is going to

Â proceed to the left. Now remember

Â if delta G standards is negative we know it's gonna go to the right to get

Â to equilibrium, it is spontaneous in the forward direction.

Â If standard GG is positive it's going to proceed to the

Â left in order to get to equilibrium.

Â The reverse reaction is spontaneous. Now if you wanted to know what standard delta

Â G at a temperature

Â other than 25 degrees Celsius then this is the equation that we would use.

Â You would look up in tables the delta H standard and the Delta S standard.

Â From that information these values don't change very much with temperature.

Â Delta G is very affected by temperature. So if you put those

Â standard delta H and standard delta S, even thought they are at 25 degrees Celsius.

Â You put in your temperature and don't forget to use Kelvins when you do so

Â then this is going to give you the standard delta G at a temperature

Â other than 25 degrees Celsius. If you needed delta G at 25 degrees Celsius

Â and its standard then you might as well just go to the tables and determine the standard

Â delta G for the reaction

Â and not use this equation. So we have a series of questions that I want you think

Â through in order to figure out

Â the connections between what is delta G standard telling you

Â and how does it change when you leave standard state conditions.

Â So I have a reaction and in this reaction chamber I'm starting with

Â every gas at one atmosphere.

Â So that's what's being represented in that picture. That is a standard a condition.

Â We see here that the standard delta G is a negative value

Â a -190.5. So if I started everything out at one atmosphere which

Â statement would be true?

Â Would the reaction shift to the right to get equilibrium, or shift to the left to get equilibrium,

Â or neither, it is already an equilibrium?

Â If you said A you would be correct.

Â A negative value for a standard delta G says,

Â and this is all it says, it is snapshots that says, if we started

Â everything at at 1 atmosphere this reaction would proceed

Â to the right because it's negative. Let's again start this system out at equilibrium.

Â I mean start the system out with everything in its standard state

Â conditions with 1 atmosphere.

Â So we're here. We know according to this standard delta G that if we start

Â everything and a standard state conditions

Â the reaction will proceed to the right. Now if it gonna proceed to the right.

Â What is going to happen to the pressure HCl?

Â Well if you said it will increase

Â you would be correct. The HC; pressure is going to go up.

Â These pressures are going to go down as it proceeds to the right.

Â So it's always going in the forward direction and reverse reaction are always happening

Â but more of the forward is happening then the reverse when you have a -190.5

Â kilojoule, we have a negative value for that standard delta G. Now as soon as it

Â starts proceeding to the right

Â my question for you is, is it still under standard state conditions?

Â Well no, because if the pressure is going up it is no longer one atmosphere

Â it is greater than one atmosphere. And if its greater than one atmosphere

Â then we are not under standard state conditions and this number

Â of - 190.5 no longer applies. It

Â only applies when when we are in standard state conditions.

Â Just to think about this,

Â does it really give us a good connection between standard and non-standard delay G?

Â What's going to happen to the total pressure on the system

Â as you establish equilibrium from the point

Â at the beginning where everything is one. Starts proceeding to the right

Â and it finally get equilibrium and the foreign reverse and reaction are happening at the

Â same rate, what is going to happen to the total pressure?

Â If you said stay the same, then you're correct? The pressure of

Â HCl is going up as it proceeds to the right. The pressure of H_2 is going down

Â the pressure Cl_2 is going down

Â because the number of moles on the left which we have two moles and gas not just

Â is not just moles its moles and gas

Â there's two moles gas in the left and there's two moles a guess on the right

Â then as this reaction shifts the total pressures not changing.

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and it finally gets there. Once it

Â establishes equilibrium what is going to be the value

Â of the standard delta G. Will it increase to get equilibrium?

Â Will it have decreased to get to equilibrium? Or will it stay the same?

Â Students often miss this.

Â So don't feel bad if you didn't you've got to get to the

Â understanding of why would stay the same. A standard delta G

Â is a snapshot of where where it would be

Â if everything were at one atmosphere. So standard delta G never changes.

Â Its value is the - 190.5 kilojoules per mole for this reaction.

Â So as a reaction release these one atmosphere conditions

Â we no longer have a standard state condition

Â but the standard delta G is the snapshot of what it is when everything is that.

Â So don't think that your standard delta G is going to change.

Â Which Delta G without the circle

Â it certainly will change. So our next question is to think about

Â what is happening to Delta G. Once

Â you establish equilibria. Will it be positive will be negative?

Â Will be equal to 0. Well it you said it would be equal to 0 you are correct and that's exactly what happens.

Â Delta G equals zero an equilibrium.

Â Not standard delta G the only way

Â standard delta G could equal 0

Â is if when everything was that was at one atmosphere

Â this thing happened to be at equilibrium. Very, very unlikely.

Â Very unlikely. Well this reaction certainly isn't it had a negative value

Â we know it wi'll proceed to the right. We will build up more and more

Â of these two. We will use up some of this and some this and finally get to equilibrium.

Â When he gets the equilibrium, at that point

Â the Delta G for the reaction is no longer standard state conditions

Â is 0, That is true for every reaction it's Delta G is equal to 0

Â at equilibrium. So what happened to the Delta G?

Â The Delta G started out when we were all at one atmosphere

Â with this number. It gets to equilibrium

Â Delta G non-standard is equal to 0

Â so the negative Delta G is increasing

Â as the reaction keeps working its way closer and closer

Â to equilibrium. Now we will look at the mathematical relationship between the standard

Â in a non-standard delta G, there is a connection. So anytime you need in

Â non-standard delta G

Â is when you're not in standard state conditions.

Â We have this equation help us get to a non-standard.

Â So if we want to know what the reaction was at

Â going to proceed to the right or proceed to the left and we had conditions other than

Â standard state conditions

Â then this is the equation that we would use. The R that we use in this equation

Â is the 8.314 because it has the

Â energy unit a joule, and you actually probably have to convert it to

Â kilojoules before you could add those terms together.

Â We see that a Kelvin is here in this K and that's what we have to use for the temperature

Â and our Q is a reaction quotient.

Â The reaction quotient if we remember for reaction is just like a K

Â equilibrium constant. Its products and concentrations if there aqueous

Â pressures if it is

Â gases, but its products over reactants

Â raised to the power of their coefficients. So there's some coefficients of the

Â balanced equation is raised to those powers.

Â So that's what the reaction quotient is so

Â we can look at this reaction and say OK at this point in time

Â there is this much product, there is this much reactant. Let's put those numbers in there

Â and let obtain a value for a non-standard

Â delta G. Now if everything were 1 then Q would be 1.

Â And if Q is one the natural log of Q is 0 and this term goes away

Â and that make sense.

Â So we're just gonna think through this equation.

Â In terms of what how does change in Q affect the Delta G?

Â So let's start with a standard Delta G that's negative.

Â This means that the reaction is spontaneous in the forward direction.

Â if you we in standard state conditions. We put everything in a

Â reaction chamber

Â and this is kinda like our little questions that we went through

Â are standard delta G was negative the reaction spontaneous in the forward direction.

Â So the question is what would happen, we have to do, in terms of conditions?

Â What would we have to do in terms of Q here in order to make Delta G positive?

Â Well if the delta G is positive we know that it's no longer spontaneous in the forward

Â direction is now spontaneous the reverse direction.

Â How do we make Delta G positive? Well, the only way we can make delta G positive

Â is to make this term positive.

Â Well this term is positive when Q is what? How do we make the natural log of Q

Â be positive value? Well

Â if Q is greater than one

Â it's a value that is bigger than one then the national log of Q is positive.

Â That is the first thing. As long as Q is greater than 1 then when we take the natural

Â log it will be positive in this term

Â that I have circled up there is going to be a positive term. We just gotta make

Â positive enough.

Â So anytime that the products are greater than the reactants because remember what Q is?

Â It is products over reactants.

Â Remember? Raised to the power of their coefficients.

Â So anytime products is greater than reactants

Â then the Q is going to be bigger than 1 and a national log of Q will be a

Â positive value

Â and we can get it positive enough to overcome all the negative that we have here.

Â We can turn a reaction that was spontaneous in the forward

Â direction to be spontaneous instead

Â in the reverse direction. Well lets go through that same mathematical logic

Â in the revers. What if the standard up to G was positive

Â the reaction is not spontaneous in the forward direction

Â it is only spontaneous in reverse reaction direction

Â if everything were under standard state conditions.

Â So we start everything is standard state conditions. We know the reactions going to proceed

Â to the left. That is when a positive

Â means its spontaneous in the reverse direction.

Â What conditions would we have to have in order to make

Â this negative. We want this reaction proceed in the forward direction

Â instead of in the reverse direction. Well lets think about what has to happen to this term.

Â In order to make this reaction becomes spontaineous the forward direction

Â this term is going to have to be very negative. It is going to have to be negative enough to

Â overtake the positive of the standard delta G.

Â How can we make that term negative?

Â Well, the only way we can affect that term R cannot change it's always the same.

Â Temperature is Kelvin and there's no such thing as a negative Kelvin so the only

Â way we can make this term negative

Â is to make the natural log of Q negative.

Â So if Q is less than one, it is a fraction

Â less than 1. Then anything can you take the natural log of a number less than 1

Â its negative. So how do we make Q less than one?

Â Once again we remember that Q is products

Â over reactants raised to the power their coefficients.

Â As long as you have a lot of products so this is a big number.

Â and this is a small number. So products

Â is an smaller than the reactants and you want to have a number

Â less than 1 and as long as it is small enough

Â this natural log of Q would be negative enough that we would

Â have more negative than we have positive and that would make my

Â non-standard delta D negative and we can for force this reaction to become spontaneous

Â in the forward direction. Now you going to

Â be doing mathematics numbers for this it's not a big deal.

Â You just gotta watch for signs if you're given a standard delta G

Â and they want to know a non-standard delta G this is the equation you will pull out.

Â Watch you units if Delta G which is typically in kilojoules is given to you

Â and then we're going to have to make sure that this term gets converted

Â kilojoules before we convert it.

Â The thing you have to watch is your T and make sure it is in Kelvin.

Â Then you will be able to easily convert between a standard delta G

Â and a non-standard delta G.

Â So this is the enough are learning objective number 9 in which we are

Â seeing the connection between the Standard delta G and a non-standard delta G.

Â We see that they're not the same thing we have to understand their differences

Â and we have an equation that will help us

Â convert between them.

Â