In this learning objective: we're going to learn how to calculate delta G of
a reaction by a different route.
We are going to define what's called standard free energy of formation.
Now and thermochemistry you had learned about the Delta H of formation.
Here at the Delta G a formation and we're going to use those values to calculate
free energy of a reaction. This equation
looks very similar to one that you've seen before, but with G's instead of
H's. So lets go through and look at this equation
first its standards state conditions. We see that little
circle there. So that means we have one molar if its solution in one atmosphere
if they are gases.
What these are defined as here are the
free energy of formation. Now the definition is very similar to the Delta
H of formation but it's the free energy change that occurs when
one mole of the compound is formed.
So it's gonna be a product, and its form from the
elements in their standard states. So that's the definition
and if you recall back to Delta H of a formation
very very similar definition. Because this definition if you have a
stable element to make that element from itself doesn't require any energy so
the Delta G of formation of
any element is equal to 0.
So now
we have that equation and we can use that equation to calculate the Delta g of a reaction.
We know that the Delta G over a reaction
when it is, negative less than zero
then it is spontaneous in a forward direction.
So I want you to first of all calculate the Delta G
of this reaction and then determine
if it is spontaneous as written.
Well the answer is yes because the Delta G of the reaction
here is a -818.0 kilojoules.
If you did not obtain that value for the Delta G the reaction
go back and look did you incorporate
the coefficient of 2 here? Here you incorporate the coefficient of 2 here as well
but I didn't give you the Delta G a formation of O_2 because I expect you
to know that it is 0.
So it doesn't matter whether you multiply by two or not because 0 times
2 is still 0. So it should be
the carbon dioxide value
and 2 times
H_2O liquid value minus
the Delta G of formation of this.
So minus in a -50.8 to get up to that value of -18
kilojoules and a negative value tells me that this reaction
is spontaneous as written.
So we're going to learn another method for calculating the Delta G of a reaction
and that is by utilizing various chemical reactions in which we know the Delta G.
We can realize that the Delta G a formation for
overall reaction is determined whether you can do it in one step
or many steps. So for reaction is this the sum of
various chemical reactions. In other words you can add various
reactions and come up with an overall reaction as a sum of those.
Then the Delta G for the overall reaction is simply the sum up the Delta G of those
various reactions.
Now you did something very similar with delta H's in what was called
Hess's law in thermochemistry. A very very similar process but we're going to see it
played out here
with Delta G. Lets see it done with this
set of information. Were given at Delta G's
for these three reactions. We know their Delta G values one two and three.
We are trying to obtain the Delta G
for this overall reaction that you see here.
So I can manipulate these three reactions below
in such a way that they add up to give me that overall reaction
then I could add up their Delta G values. So let's see if we can find pieces that we need.
We needCH_4 gas on the left hand side of the equation.
This has CH_4 gas but it's on the right hand side.
So I'm going to revers this reaction and there would be CH_4
gas producing carbon
solid which is graphite plus 2
H_2 gas. If you reverse the reaction you would change the sign
of the the Delta G. It would be a positive
55.5 kilojoules.
So we have the first piece in place.
We come to his second substance that we're trying to get into place and that is Cl_2.
If I look down at these reactions I sea that Cl_2 is in 2 of these reactions.
So I want to skip over the chlorine for now.
I am going to go to the next substance. Tthe next substance in line
is the Cl_4. Cl_4 only appears in the second reaction here.
See it here ,and it's on the correct side the equation is the correct quantity
so I am going to bring that reaction down just as I see it.
Carbon in the form of graphite
plus 2 Cl_2 gas
produces CCl_4 gas.
Since I'm using the reaction exactly as I see it I we use the Delta G
exactly as it's given.
Now I have this in place. That gave me a couple chlorines which I need 4 but
I only have 2 so far so let's see what happens next.
I need the next, look at the substances, it is
HCl gas. I need
4 moles of it. Down here is HCl gas
but there's only 2 moles of it. So I am going to double that equation
that is gonna give me 2 H_2
plus 2 Cl_2
generating 4 HCl.
That will give me a Delta G for this reaction
2 times the size the Delta G the reaction that was up there. So it is 2 times
negative 65.3
kilojoules. All right now we want to see do these reactions add up to give me
the reaction I was looking for? Lets cancel anything that is in common.
I have 3 H_2 and 2 H_2.
That gets cancelled. I have carbon solid in the form of graphite
that gets cancelled. Nothing else is cancelled. The chlorines won't cancel because
they're on the same side the equation
and that will leave me with CH_4 gas
plus 4 Cl_2 gas
yielding CCl_4 gas
plus 4 HCl gas. That
is indeed the reaction we were trying to obtain up here.
They match, therefore I can take these equations, I mean these Delta
G values that we see on the right hand side
and we can add them together and that will give me
the Delta G for this reaction. And the value is a -142,
142.4 kilojoules.
A very similar process to Hess's law.
Last thing I want to talk about in this learning objective:
is why free energy is the energy free and available to do work.
How come all the energy is not available? Lets look at this reaction.
In this reaction the Delta H is a- 95.7 that is entropy change that
is the heat given off.
All of that not available to do work, why is that?
We look at this reaction we see that we have to moles of gas
going to one mole of gas. So the Delta
S of this reaction
is going to be negative.
Less than zero, its negative.
For a reaction to be spontaneous the Delta S in the universe has to be positive.
So I have got to have a positive Delta S
of the surroundings. It has to be greater than 0.
I have to make up for the fact that the Delta S the reaction
is negative. Now let's get some words up on the screen.
So if the delta S for the reaction is less than zero, which it is here because I'm
decreasing the moles of gas. Then some of that energy.
Some of this energy here, released to to the surroundings
has got to go to increasing the entropy of the surroundings.
We have to have a least as much
for a little bit more actually a little bit more delta S's surroundings
being positive to overcome the fact that the Delta
S old the reaction was negative. So some other this heat
has to be going out to the surroundings in order to bring
up the entropy.
So let's I'm not giving calculations here but in this example
the minimum heat loss to the surrounding his going to
have to be 33.4 kilojoules.
I have to have that much in order to
overcome the fact that the reaction was negative Delta S.
I have to increase the disorder by that amount.
So this much heat is determined. Now how do I know that much?
Because that we remember that the delta S the surroundings as negative delta
H over t. Now some of that I want to be able to use as
work and the
minimum work that I can get outta this
would be 62.3. Because I take the thirty 3.4
out at this -95.7
and I can do 62.3 kilojoules of work.
So this would be, in this case, the theoretical maximum
be the theoretical maximum out
of energy of that Delta H that could be used and that is the part that is free
to do work. We are now at the end of our learning objective number 8.
Our primary purpose at this learning objective is to learn new methods for
calculating the Delta G of a reaction.
You had previously learned that Delta G is equal to delta
H minus T delta S, but here we learned about Delta G formations
and use them and sum them up to come up with the Delta G of a reaction.
We also did a stepwise process where we added equations together
to get the Delta G other overall reaction.
