5.09 Calculating Standard Free Energy Change

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From the course by University of Kentucky

Advanced Chemistry

333 ratings

University of Kentucky

Advanced Chemistry

333 ratings

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.

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 Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

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.

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