A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.
5.11 Comparing ∆G° and K
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From the course by University of Kentucky
Advanced Chemistry
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From the lesson
Thermodynamics
The overarching theme of thermodynamics is the prediction of whether a reaction will occur spontaneously under a certain set of conditions. Entropy and Free Energy are defined and utilized for this purpose.
Meet the Instructors
Dr. Allison SoultLecturer
Chemistry
Dr. Kim WoodrumSr. Lecturer
Chemistry
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
would be 1 because there's a 1 in front the PCl_3
1 times the delta G of formation
of PCl_3 +
1 times the delta G of formation
of Cl_2- so it is product minus reactants
one times the delta G of formation
of PCl_5. So lets plug those numbers in.
PCl_3 was a -286 kilojoules per mole.
They didn't give it to me for coined
Delta G formation for any element is 0.
Element in its standard state. And then we will subtract the delta G formation of PCl_5
we are going to subtract a -325
kilojoules per mole. This is going to give us a value
of 39.0
a positive value 39.0 kilojoules
per mole. Now we are ready to obtain
K. We know that standard delta G equals
-RT natural log of K. Let's solve for
natural log of k first. Lets divide both sides by -RT.
So a negative delta G
over RT is equal to the natural log of K.
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.
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.
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