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In this learning objective: we're going to learn how to calculate delta G of

Â a reaction by a different route.

Â We are going to define what's called standard free energy of formation.

Â Now and thermochemistry you had learned about the Delta H of formation.

Â Here at the Delta G a formation and we're going to use those values to calculate

Â free energy of a reaction. This equation

Â looks very similar to one that you've seen before, but with G's instead of

Â H's. So lets go through and look at this equation

Â first its standards state conditions. We see that little

Â circle there. So that means we have one molar if its solution in one atmosphere

Â if they are gases.

Â What these are defined as here are the

Â free energy of formation. Now the definition is very similar to the Delta

Â H of formation but it's the free energy change that occurs when

Â one mole of the compound is formed.

Â So it's gonna be a product, and its form from the

Â elements in their standard states. So that's the definition

Â and if you recall back to Delta H of a formation

Â very very similar definition. Because this definition if you have a

Â stable element to make that element from itself doesn't require any energy so

Â the Delta G of formation of

Â any element is equal to 0.

Â So now

Â we have that equation and we can use that equation to calculate the Delta g of a reaction.

Â We know that the Delta G over a reaction

Â when it is, negative less than zero

Â then it is spontaneous in a forward direction.

Â So I want you to first of all calculate the Delta G

Â of this reaction and then determine

Â if it is spontaneous as written.

Â Well the answer is yes because the Delta G of the reaction

Â here is a -818.0 kilojoules.

Â If you did not obtain that value for the Delta G the reaction

Â go back and look did you incorporate

Â the coefficient of 2 here? Here you incorporate the coefficient of 2 here as well

Â but I didn't give you the Delta G a formation of O_2 because I expect you

Â to know that it is 0.

Â So it doesn't matter whether you multiply by two or not because 0 times

Â 2 is still 0. So it should be

Â the carbon dioxide value

Â and 2 times

Â H_2O liquid value minus

Â the Delta G of formation of this.

Â So minus in a -50.8 to get up to that value of -18

Â kilojoules and a negative value tells me that this reaction

Â is spontaneous as written.

Â So we're going to learn another method for calculating the Delta G of a reaction

Â and that is by utilizing various chemical reactions in which we know the Delta G.

Â We can realize that the Delta G a formation for

Â overall reaction is determined whether you can do it in one step

Â or many steps. So for reaction is this the sum of

Â various chemical reactions. In other words you can add various

Â reactions and come up with an overall reaction as a sum of those.

Â Then the Delta G for the overall reaction is simply the sum up the Delta G of those

Â various reactions.

Â Now you did something very similar with delta H's in what was called

Â Hess's law in thermochemistry. A very very similar process but we're going to see it

Â played out here

Â with Delta G. Lets see it done with this

Â set of information. Were given at Delta G's

Â for these three reactions. We know their Delta G values one two and three.

Â We are trying to obtain the Delta G

Â for this overall reaction that you see here.

Â So I can manipulate these three reactions below

Â in such a way that they add up to give me that overall reaction

Â then I could add up their Delta G values. So let's see if we can find pieces that we need.

Â We needCH_4 gas on the left hand side of the equation.

Â This has CH_4 gas but it's on the right hand side.

Â So I'm going to revers this reaction and there would be CH_4

Â gas producing carbon

Â solid which is graphite plus 2

Â H_2 gas. If you reverse the reaction you would change the sign

Â of the the Delta G. It would be a positive

Â 55.5 kilojoules.

Â So we have the first piece in place.

Â We come to his second substance that we're trying to get into place and that is Cl_2.

Â If I look down at these reactions I sea that Cl_2 is in 2 of these reactions.

Â So I want to skip over the chlorine for now.

Â I am going to go to the next substance. Tthe next substance in line

Â is the Cl_4. Cl_4 only appears in the second reaction here.

Â See it here ,and it's on the correct side the equation is the correct quantity

Â so I am going to bring that reaction down just as I see it.

Â Carbon in the form of graphite

Â plus 2 Cl_2 gas

Â produces CCl_4 gas.

Â Since I'm using the reaction exactly as I see it I we use the Delta G

Â exactly as it's given.

Â Now I have this in place. That gave me a couple chlorines which I need 4 but

Â I only have 2 so far so let's see what happens next.

Â I need the next, look at the substances, it is

Â HCl gas. I need

Â 4 moles of it. Down here is HCl gas

Â but there's only 2 moles of it. So I am going to double that equation

Â that is gonna give me 2 H_2

Â plus 2 Cl_2

Â generating 4 HCl.

Â That will give me a Delta G for this reaction

Â 2 times the size the Delta G the reaction that was up there. So it is 2 times

Â negative 65.3

Â kilojoules. All right now we want to see do these reactions add up to give me

Â the reaction I was looking for? Lets cancel anything that is in common.

Â I have 3 H_2 and 2 H_2.

Â That gets cancelled. I have carbon solid in the form of graphite

Â that gets cancelled. Nothing else is cancelled. The chlorines won't cancel because

Â they're on the same side the equation

Â and that will leave me with CH_4 gas

Â plus 4 Cl_2 gas

Â yielding CCl_4 gas

Â plus 4 HCl gas. That

Â is indeed the reaction we were trying to obtain up here.

Â They match, therefore I can take these equations, I mean these Delta

Â G values that we see on the right hand side

Â and we can add them together and that will give me

Â the Delta G for this reaction. And the value is a -142,

Â 142.4 kilojoules.

Â A very similar process to Hess's law.

Â Last thing I want to talk about in this learning objective:

Â is why free energy is the energy free and available to do work.

Â How come all the energy is not available? Lets look at this reaction.

Â In this reaction the Delta H is a- 95.7 that is entropy change that

Â is the heat given off.

Â All of that not available to do work, why is that?

Â We look at this reaction we see that we have to moles of gas

Â going to one mole of gas. So the Delta

Â S of this reaction

Â is going to be negative.

Â Less than zero, its negative.

Â For a reaction to be spontaneous the Delta S in the universe has to be positive.

Â So I have got to have a positive Delta S

Â of the surroundings. It has to be greater than 0.

Â I have to make up for the fact that the Delta S the reaction

Â is negative. Now let's get some words up on the screen.

Â So if the delta S for the reaction is less than zero, which it is here because I'm

Â decreasing the moles of gas. Then some of that energy.

Â Some of this energy here, released to to the surroundings

Â has got to go to increasing the entropy of the surroundings.

Â We have to have a least as much

Â for a little bit more actually a little bit more delta S's surroundings

Â being positive to overcome the fact that the Delta

Â S old the reaction was negative. So some other this heat

Â has to be going out to the surroundings in order to bring

Â up the entropy.

Â So let's I'm not giving calculations here but in this example

Â the minimum heat loss to the surrounding his going to

Â have to be 33.4 kilojoules.

Â I have to have that much in order to

Â overcome the fact that the reaction was negative Delta S.

Â I have to increase the disorder by that amount.

Â So this much heat is determined. Now how do I know that much?

Â Because that we remember that the delta S the surroundings as negative delta

Â H over t. Now some of that I want to be able to use as

Â work and the

Â minimum work that I can get outta this

Â would be 62.3. Because I take the thirty 3.4

Â out at this -95.7

Â and I can do 62.3 kilojoules of work.

Â So this would be, in this case, the theoretical maximum

Â be the theoretical maximum out

Â of energy of that Delta H that could be used and that is the part that is free

Â to do work. We are now at the end of our learning objective number 8.

Â Our primary purpose at this learning objective is to learn new methods for

Â calculating the Delta G of a reaction.

Â You had previously learned that Delta G is equal to delta

Â H minus T delta S, but here we learned about Delta G formations

Â and use them and sum them up to come up with the Delta G of a reaction.

Â We also did a stepwise process where we added equations together

Â to get the Delta G other overall reaction.

Â