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
