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We are ready for 10th and final learning objective:

of the thermodynamics unit. In this unit we're going to be seeing the

connection between the standard delta G

and the K value the equilibrium constant value for our reaction.

We going to derive the equation that we will use first

and then practice using the equation in some example problems.

Begin by answering

this question we have

this black equation sitting here. In this equation

which we learn in last learning object we see connection between

non-standard and standard K. We are going let this reaction run until he gets to

equilibrium and once equilibrium is established which one of those

statements is correct.

Did you say that Delta G would be zero and Q would equal K?

Well, that is correct. When we get to equilibrium

We know that to the Delta G is equal to 0.

The relationship between Q and K is they are both

the concentration of products, or pressure if it is gases,

over reactants raised to power of their coefficients. Whatever those coefficients

might be.

If you are not at equilibrium they were putting in

these values wherever we are but it's going to keep on running in till finally

gets to equilibrium

and we will remove these little initials

statement there and we are now at K.

So this is that correct statement now we're going to take this equation in

we and going to do some

substitution. Where we see is non-standard delta G we will put

zero in, and where we see a Q we will put a K in.

This is going to give us this equation. If take this equation rearrange it a bit

we're going to obtain this equation. Standard delta G equals minus

RT natural log of K. This equation

is one that you will want to learn. You're going to make sure that

if you're dealing with gases that you are working with the K_p.

If you are dealing with solutions you will use K_c.

That is a must.

A review of K_p is that you have to put

things in atmospheres when you are obtaining a K_p.

For K_c you put things in molarity.

With this slide we are just going to be looking at the connections between

the standard delta G and the K value and the way this reaction proceeds.

If we have a big situation in which we're starting at standard state conditions

P_a would be equal to P_b and they would both be equal to 1.

One atmosphere because that standard state conditions.

So that is where the dashed line is located. If we look down here at the bottom

if this reaction is starting here

is going to want to proceed towards the minimum energy.

It's going to be decreasing and in the direction going towards the right.

In which we're going to be producing more B

and less A, and that corresponds with a K greater than one.

K is greater than one we know it's going to proceed to products

in order to get to the equilibrium.

So when Delta G is less than zero.

The equilibrium is going to favor those products. So we have a connection between

a negative standard delta F and a positive K.

Up here in the top graph we have a situation in

which K is less than one. What we learned in equilibria

sections that if K is less than one

the reaction has to proceed towards reactants, its gonna go this way.

So it's heading in the direction making more A and less B.

The energy is going to decrease as it goes in that direction until we

get to that minimum.

So when we a start in standard state conditions the standard delta G

would be greater than 0.

The equilibrium is going to favor the reactants, its gonna proceed

to the left until it gets to the minimum value

there. So the reverse reaction is spontaneous we have these connections

between the K

and standard delta G. Look at it mathematically

with a statement up hear. Let's begin with a K

that is greater than one. If K is greater than one

the natural log of K is going to be positive.

If the natural log of K is positively know the temperature is always positive.

R is a point 314 that's positive.

Then we change the sign that's going to give me a negative

Delta G both a K greater than one

and negative Delta G means the products are favored.

It will proceed to the right in order to reach equilibrium.

Once it reaches equilibrium you will have more products.

If K were equal to 1, natural log of K is 0 and delta G standard is 0

that means is reaction is already an equilibrium

and the products the reactants are equally favored.

If K is less than one,

if K is less than one let's consider this

if this is less that 1 then this is going to be negative.

If this is negative in we change the sign

this standard delta G is going to be positive. This

reaction is going to proceed towards the reactants.

If it proceed towards the reactants and then eventual get to equilibrium

once it gets to equilibrium the reactants are gonna be favored

over products.

So here we have this equation once again and really

this equation has one big job. If you know

delta G you can obtain K. If you know K you can obtain

standard delta G is its job. You have to this is standard state conditions

to go between the Delta G and the K.

So we are going practice this couple times. First one

we going to calculate standard delta G. If we going to calculate standard delta G then we are going to

need to know K.

We have to obtain K.

The K for this reactions is known as K_sp because it is the solubility of this.

So we are going to obtain

K_sp for this reaction. Now how we going to obtain K_sp from the information given?

Well, first thing you're gonna need to know is the molar solubility.

They gave me in the solubility in grams per liter but I cannot obtain a K

value unless I know it in

moles per liter. So the first thing I want to do the 6.7 x 10^-3 grams per liter

can be converted to moles per liter by knowing the molar mass

of the silver phosphate. The molar mass is 418.6 grams per mole

and this will give me a molar solubility up 1.6 x 10 ^ -5 molars.

Lets write this reaction for this

silver phosphate dissolving in water.

AgPO_4 is a solid we place it into water

and we obtain this equilibrium there will be 3 Ag+

aqueous and 1 phosphate

aqueous. No we are going to place

some of the solid into solution

or into the water and there will be none dissolved.

When they have given me the molar solubility they are telling me the amount

this silver phosphate

that dissolves 1.6 x 10 ^ -5.

This is what molar solubility is. This is an an ICE table that I'm creating here

and so for every one of these dissolved we're going to produce

3 of the silver.

We are going to produce 1 of the phosphate.

So that's a change line. Now this a tiny amount so there will be some of the

solid sitting on the bottom.

I would have 4.8 x 10 ^ -5 for the silver ion

and 1.6 x 10 ^ -5 for the phosphate ion.

Now the K_sp we know is equal to the silver concentration cubed

times the phosphate concentrations we can just plug those E numbers into this equation.

4.8 x 10 ^ -5 will need to be cubed

and 1.6 x 10 ^ -5.

When I multiply those values out I will obtain a value of 1.8 x 10 ^ -18. So that is a K_sp.

Well, once we K_sp we can get delta G. We know that standard delta G

is equal to minus RT

natural log K and in this case the it is a K_sp.

R is 8.314 joules per moles Kelvin. Temperature at 25 degrees Celsius is 298 Kelvin.

Natural log of K_sp which was 1.8 X 10 ^ -18

this will give me a Delta G

of 1.01 x 10 ^ 5

joules per mole notice the Kelvins cancel

or this would be a 101 kilojoules per mole.

In our next problem we're going to try to obtain

a value for K_p. So if we are going to obtain a value for K_p

seeing what kind of information they have below

we could first determine the Delta G.

I don't just mean any Delta G

we need the standard Delta G. So lets obtain the standard delta G from the

standard delta G of formations.

The delta G standard for the reaction that is listed up there

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Okay so now lets plug our numbers in.

Because R is in kilojoules per mole Kelvin, I'm going to put Delta G

in joules so that's going to be - 39,000 joules per mole. 8.314

jewels per mole Kelvin and

temperature 25 is 298 Kelvin

is equal to the natural law of K. This will give me a value

-15.7 now to obtain

K. If this is true that we see here

how do we get K? [I'm gonna move up to the small space up here.

and do it in red so we can keep track of it] The natural log of K

is equal to -15.7. We need to take

E to both sides because that will cancel

the natural log and leave me with the K.

This is equal to and 1.5 x 10 ^-7.

Now since these are gases, I know that this is going to be a K_p value that is obtained.

Now we have our answer. So these two example problems get me between

the standard delta G and the K. The last thing we want to look at

is a relationship between temperature and equilibrium constant.

If you recall way back when we first started talking about

equilibria we said that K changes with temperature.

Up here in this space at the top we will about that connection.

If we had a reversible reaction like A

going to B. We learned that

whether or not it's endothermic or exothermic will affect the value of K.

If the reaction is endothermic

and we were to raise the temperature

we know that this would shift the reaction to the right and K

would increase. It would get larger.

On the other hand if we took the heat

and we put the heat on the right and said this reaction is

exothermic instead. As we raised the temperature

what would happen is the equilibrium which shift to the left

and the K we get smaller.

So we saw that connection. We are going to derive an equation and see the connection

mathematically here.

First of all look at this what I've written. How in the world?

Where did this come from? That is the question. We see the equation

well this half is equal to standard delta G. And this half

is equal to standard delta G by another equation. So we have just set them equal to

each other.

Now let sod some manipulations. What have I done here? I have

simply divided by RT. Lets also

multiply through by -1. So that is going to change that sign, change that sign, and change that sign.

Now let's do some regrouping.

Take out the T's and we have this. I've taken out

T's they're entirely on the right-hand terms. So the T's have gone away.

I have factored out the T in the first term

we have 1 over t there and when I do that

I have an equation that in the slope intercept form a line.

Y = MX + B.

So that tells me that if I plot along

the y-axis the natural log of K.

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So we take this equation in this form we see it in the slope intercept form

of a line. That means that if we were to plot

along the y-axis

the natural law of K

and along the x-axis one over temperature

we would get a straight line. That straight line would have a slope

equal to this. Now it's not always a positive slope

or always a negative slope. If Delta H were positive

and it was an endothermic reaction then the slope would be negative.

If on the other hand it was an exothermic reaction

and delta H was negative then we would have a positive slope.

So sometimes the slope will be negative sometimes positive but we have

relationship between

the K value and the temperature.

Anytime you have a straight line it only takes two points to find that line

so the two-point equation would look like this but again

how T and K relates is dependent upon whether this

term or this energy is positive or negative.

Now we won't do any work with this equation I'm not gonna have you do any

mathematical example to this

problems with it. I just wanted to revisit the notion

of the K value and the temperature value that we learn way back in

equilibrium unit.

See that because of delta G

we can make that relationship between how temperature affects K

not just that temperature affects K.

So here's a summary of all the equations basically that we used in this unit.

We learned first of all about Delta S

and we learn how to calculate delta S the we learned about Delta G

and to find it and we have so many equations that have a Delta G in it.

Where students really struggle is to understand which one

they should use. So as your working through the example problems be thinking about

why is that the right equation to use. So that when you come to the

assessment at the end you'll be able to draw from the right equation and be able

to work it.

This is the into the learning objective 10 in which we have seen the connection

between standard delta G

and K. It is also the end are thermodynamics unit

in its entirety. So work through the problems the practice problems are

presented to you

and then work on your assessment.