1Introduction

1.1Themodynamics II Overview

Thermodynamics II

Presented by Dr. Jonathan Agger

Welcome everyone to the second module of our course in Basic Physical Chemistry. In Thermodynamics I we looked at the Zeroth and First Laws of Thermodynamics. In this module we shall look at the remaining two laws. The Second Law of Thermodynamics is vital for understanding what makes chemical reactions go, it even explains why coffee goes cold and bedrooms always get messier. As in Thermodynamics I, there is a formative 10 question multiple choice test to gain feedback on your progress. Your assessment for this section of the course will comprise a final 10 question multiple choice test and the virtual practical. Good luck with this module and I hope you enjoy it.

Additional material

链接：How a sandcastle reveals the end of all things

Professor Brian Cox builds sandcastles in the Namib Desert to explain why time travels in one direction. It is a result of a phenomenon called entropy; a law of physics that tells us any system tends towards disorder. From Wonders of the Universe, BBC2.
Once you have completed all of the video lectures, you should be in a position to complete the assignments for Thermodynamics I & II. The first is a quiz on the material presented in Thermodynamics II. You will complete an assessed quiz similar covering the quizzes in the first two modules and you will also complete a virtual laboratory experiment

1.2Thermodynamics II Introduction

0:05
Welcome to week two, of our course in basic Physical Chemistry. In week one, we looked at the zeroth and first laws of thermodynamics. This week, we shall look at the remaining two laws. The second law of thermodynamics is vital for understanding what makes chemical reactions go. It is governed by the concept of entropy, which is a measure of disorder. It explains why coffee goes cold, why buildings crumble, and why bedrooms always spontaneously get messier. The second law is so profound, that we are actually going to cover it in three parts, subtitled Entropy, Spontaneity, and Equilibrium.
0:45
Once we've got to grips with entropy we are then going to study the third law of thermodynamics. Which defines what we mean by zero entropy.
0:54
Finally, we're going to look at Hess's law and do some example calculations. As in week one, there is a formative ten question multiple choice test in order for you to get feedback on your progress. Your assessment for this section of the course will comprise a final ten question multiple choice test, and the virtual practical. Good luck with week two, and I hope you enjoy it. [BLANK_AUDIO]

2The Second Law

2.1The Second Law of Thermodynamics and Entropy
0:06

[BLANK_AUDIO].

In this video, I shall present the Second Law of Thermodynamics.

0:12

I shall consider the spontaneous direction in which processes occur, and then try to

understand this through the definition and elucidation

of a new state function, termed entropy.

0:24

I will then show a clip from the BBC television series Wonders of the Universe,

in which Professor Brian Cox of the University

of Manchester gives an exquisite description of entropy.

0:38

The Second Law of Thermodynamics was first stated by Rudolph Clausius in 1854.

0:44

Clausius had been studying the direction in which processes occur spontaneously.

For example, a hot cup of coffee will

spontaneously lose heat to its surroundings and cool down.

0:56

No cold cup of coffee in the entire history of the Universe

has ever spontaneously sucked heat in from its surroundings and become hot again.

1:07

This led Clausius to postulate that heat can never pass from a colder to

a warmer body without some other change

connected therewith occurring at the same time.

1:18

And this led to the definition of a new

state function called entropy, which is given the symbol S.

1:25

Entropy is best thought of as a measure of disorder of a thermodynamic system.

1:31

It is defined as follows: change in entropy delta S equals a reversible heat

transferred to a system, divided by the

temperature at which the heat transfer occurs.

1:44

This important definition is given as equation ten.

1:50

The derivation of equation ten is long and complicated.

So we shall satisfy ourselves with an

example of how the equation works in practice.

1:59

Imagine two copper ingots.

A hot ingot, at 323 degrees centigrade, is placed on

top of a room temperature ingot, at 25 degrees centigrade.

2:10

Heat energy spontaneously passes from the hot ingot to the cooler one.

Image that 596 joules of energy are transferred.

2:22

Once again, we show the definition of entropy.

2:26

For the hot block, delta S equals q reversible over T, which

equals minus 596 joules divided by 596 kelvin.

Or minus 1 joule per kelvin.

2:45

The heat is negative because it is leaving the block.

596 kelvin is 323 degrees centigrade.

2:56

Now consider the room temperature block.

2:59

Delta S equals q reversible over T.

3:03

Which equals 596 joules divided by 298 Kelvin.

Or, plus 2 joules per Kelvin.

3:12

The heat is positive this time, because it is entering the block.

298 Kelvin is 25 degrees C.

3:22

The overall entropy change is minus 1 plus 2

joules per Kelvin, which equals plus 1 joule per Kelvin.

So entropy has increased.

3:36

In fact, entropy always increases for isolated systems.

3:42

So, now let's just consider a non-spontaneous process.

3:46

If a gas is released in a small volume of

a large container, it will spontaneously fill the entire container.

3:56

So, let us consider the possibility of a

gas actually contracting by only 1% of its volume.

Let's say for one microsecond over the entire history of the Universe.

4:09

Well, the probability for one atom of a gas to

be found in 99% of its original volume is 0.99.

4:20

The probability for one mole of gas atoms to all be found in 99% of their

original volume, is 0.99 to the power of the Avogadro number.

4:33

Now, I'm pretty much certain that this is the smallest finite

number you have ever considered in your life, as it is equal to

10 to the power of minus 26,284 followed by 17 zeros.

4:51

Even if we take the age of

the entire universe into account, 13.75 billion years.

Is only 4.336 times 10 to the 23 microseconds.

5:05

Multiplying the original probability by this number

doesn't even begin to make a dent.

And the effective probability of one mole of a gas

contracting by only `1% of its volume for only one

microsecond over the entire life of the universe, is so

small, as to be considered zero, ie, it has never happened.

Nor is it ever likely to happen over any

time scale that might have any meaning to us.

5:33

So, now I want to show you the clip of

The Wonders of the Universe produced by the BBC, in which

Brian Cox uses the analogy of a sandcastle to give

the best explanation I have ever heard of what entropy is.

5:48

When he says in the video that entropy always increases, why is that?

Because it is overwhelmingly more likely that it will.

You should now have a good idea of exactly what he means

as you have seen the calculation for the probability of a non-spontaneous process.

[BLANK_AUDIO].

If you want to understand how the

Second Law of ThermoDynamics determines the arrow of

time, then I strongly recommend that you watch

the rest of the Wonders of the Universe.

It really is a truly superb television series.

6:23

Okay so, now we have a basic grasp of entropy as a measure of disorder.

In the next lecture we will look into far

greater detail at the underlying causes that govern spontaneity.[BLANK_AUDIO]

2.2The Second Law of Thermodynamics and Spontaneity
0:00
[BLANK_AUDIO].
0:12
In this video, I shall endeavor to clarify various topics concerning the spontaneity of chemical reactions that confused me when I studied thermodynamics as an undergraduate. Firstly, what is actually meant when we say a chemical reaction is spontaneous, and secondly, what exactly governs the spontaneity of chemical reactions? I'm sure that my physical chemistry lecturers did a fine job of trying to explain these topics to me, and my failure to understand almost certainly derives from a lack of reading on my part. However, if I can make this lecture make sense to you, then it will have served great purpose.
0:51
I want to start by considering the word spontaneous, so, I turn to the authority on the English language and look the word up in the Oxford English dictionary. Part of the definition is given here.
1:03
Meanings of the word spontaneous that I find confusing from a thermodynamic point of view are sudden, and without apparent external cause.
1:16
The reason for potential confusion is that a spontaneous reaction need not be sudden, or without apparent external cause at all.
1:27
Take for example the reaction between hydrogen gas and oxygen gas. As far as thermodynamics is concerned, this reaction is spontaneous. One mole of hydrogen gas burns in half a mole of oxygen gas, to produce half a mole of water, usually with a bright orange flame.
1:45
However, with suitable care, hydrogen gas and oxygen gas can be stored in the same balloon together, and they will not react with one another.
1:54
The reason for this apparent lack of spontaneity lies in the domain of chemical kinetics, which determines that chemical reactions have an energy barrier called the activation energy.
2:07
You will learn about this in detail with Professor Underson later on in the second part of the course. So, despite the spontaneous nature of this reaction, it takes a small amount of energy, either in the form of a flame or possibly just static electricity to ignite the gaseous mixture.
2:26
Once a reaction starts, then its spontaneity becomes all too evident.
2:30
So, hopefully this gives you a clear picture in your mind of why a spontaneous reaction doesn't necessarily react spontaneously.
2:39
After much consideration, in my opinion, the best way way to think of the word spontaneous, with regard to thermodynamics, is to think of the word feasible.
2:52
Okay. So having cleared up one potential source of confusion, let's tackle an even bigger one. What makes a chemical process spontaneous?
3:03
The reason for confusion here is that, as far as I could determine as an undergraduate, there are actually two criteria for this one outcome.
3:13
Do both have to be true for a reaction to be spontaneous? What happens if one is true and the other is false? As an undergraduate, I had no idea.
3:23
So, let us consider what the criteria are.
3:28
First, in any spontaneous chemical process, the entropy of the Universe must increase.
3:37
Second, in any spontaneous chemical process, the Gibbs energy must decrease.
3:46
If you aren't sure what Gibbs energy is for the moment, don't worry, as I will explain it in this lecture.
3:53
Clearly, these two statements appear different. And so, once again, how can there be two distinct criteria for one phenomenon?
4:06
The answer to this problem, is that the two criteria are not independent. The second one is both defined and derived from the first one, and absolutely depends on it. So, in other words, if one criterion is true, then the other one must also be true. It is impossible for one to be true without the other.
4:35
So, in order to derive the second criterion, and in doing so to define Gibbs energy, we need to start with the first criterion.
4:45
The first criterion states that for any spontaneous chemical process the entropy of the universe must increase. So, this is where we begin.
4:56
The change in entropy of the universe, caused by a chemical process, is equal to the change in entropy of the system plus the change in entropy of its surroundings. Remember, from the second lecture on definitions, that the system plus its surroundings make up the entire universe.
5:17
We now rely on the definition of entropy itself, given as equation ten in the last lecture.
5:26
The change in entropy of the surroundings is equal to the reversible heat exchanged to the surroundings, divided by the temperature at which the transfer occurs. So, we can replace delta S surroundings by q surroundings divided by T.
5:45
Now, crucially, whatever heat energy flows into the surroundings as a consequence of mechanical process, must have flowed out of the system.
5:58
Thus, the heat flow into the surroundings equals the heat flow out of the system, which is the minus the heat flow into the system.
6:10
This step is vital because it enables us to describe the total entropy change of the universe based purely on parameters of the system.
6:21
Okay, so we can replace Q surroundings divided by temperature, by minus Q system divided by temperature. And the right hand side of our equation is now purely system based.
6:36
Next, we recall equation four, which states that for an isobaric process, the heat energy supplied to the system is equal to the enthalpy change of the system.
6:50
Replacing minus q system divided by t, behind minus delta H system divided by T, leads to the equation, delta S universe equals delta S system minus delta H system divided by T.
7:11
Multiplying the equation through by temperature, and negating it, leads to minus T delta S universe equals delta H system minus T delta S system.
7:23
We need to keep in mind the right hand side of this equation, as it is important for the next section, so you may wish to write this down. In 1873, the American Josiah Gibbs defined a new state function, G which is equal to H minus TS.
7:46
And this Gibbs energy, or historically Gibbs free energy, entered the thermodynamic vernacular for the first time.
7:56
Change in the Gibbs energy, delta G therefore, equals delta H, minus delta TS.
8:06
Remembering that delta TS is a product and needs to be differentiated accordingly, delta G equals delta H minus T delta S minus S delta T.
8:21
But if we consider an isothermal process, S delta T equals zero, and thus, delta G equals delta H minus T delta S.
8:35
This is an equation of fundamental importance, and we denote it equation 11.
8:42
So now, if delta G equals delta H minus T delta S, delta G system equals delta H system minus T delta S system.
8:53
And it should hopefully be apparent why Gibbs energy was defined in this way. Because the right hand side of this equation is exactly the same as the right hand side of the equation at the top of the slide.
9:07
Thus, we have now forged a link between the total entropy change of the universe, and the Gibbs energy of the system.
9:18
Minus T delta S universe equals delta G system.
9:24
The last part is now trivial, dividing both sides of the equation by minus T yields delta S universe, equals minus delta G system over T.
9:38
Since T is absolute, and hence positive, we can now see that for spontaneous chemical process necessitating an increase in the entropy of the universe, the Gibbs energy of the system must be negative.
9:55
Moreover, the changing Gibbs energy of the system is defined based on the change in enthalpy and entropy of this system, and the temperature which the reaction is carried out.
10:07
It was originally thought that only reactions that were exothermic i.e., that give out heat, could be spontaneous. Then, it became apparent that some endothermic reactions, i.e., that draw in heat, were also spontaneous. Three examples of such processes are: the dissolution of potassium chloride in water, which has an enthalpy change of plus 19 kilojoules per mole, the melting of ice, which has an enthalpy change of plus six kilojoules per mol, and the decomposition of ammonium carbonate, which as an enthalpy change of plus 68 kilojoules per mol.
10:49
As we have shown, delta G equals delta H minus T delta S. So, if delta H is positive, then delta S must be sufficiently positive to make minus T delta S outweigh delta H are the temperature of the reaction, thus making delta G negative and rendering the process spontaneous.
11:16
In all three of the reactions shown above, there is a significant increase in entropy or disorder.
11:23
Potassium chloride exists as a highly ordered crystalline ionic lattice of alternating potassium cations and chloride anions. Dissolution dismantles the structure, to leave hydrated ions moving around in solution.
11:39
The melting of ice again causes the dismantling of a highly ordered crystalline solid.
11:46
The decomposition of ammonium carbonate, sees one mole of highly regular ionic crystalline solid, transformed into multiple moles of gas and liquid. Gases have significantly higher entropies than solids as we shall see in lecture ten.
12:05
In actual fact there are four possible situations.
12:11
Reactions with negative enthalpy change and positive entropy change will be spontaneous regardless of temperature.
12:21
Reactions with positive enthalpy change and negative entropy change will never be spontaneous.
12:31
Reactions with negative enthalpy change and negative entropy change will switch from spontaneous to non-spontaneous as temperature rises, and reactions with positive enthalpy change and positive entropy change will switch from non-spontaneous to spontaneous as temperature rises.
12:54
In order calculate the point at which a reaction switches spontaneity, we are looking for the Gibbs energy to change sign, and consequently, we need the point where it passes through zero. At this point, T delta S equals delta H, and thus, T equals delta H over delta S.
13:17
On this slide, we're going to work an example calculation for a reaction with an enthalpy change of 19 kilojoules per mole and an entropy change of 45 joules per Kelvin per mole.
13:30
The simplest way to get such a calculation wrong is to forget the units. Remember, enthalpy changes are usually quoted in kilojoules per mole rather than joules per mole.
13:42
Also, remember the resulting temperature will be in Kelvin.
13:48
So, we divide 19,000 joules per mole by 45 joules per Kelvin per mole and we get 422 Kelvin.
14:00
Subtraction of 273 yields a temperature of 149 degrees Centigrade.
14:07
The reaction becomes spontaneous above 149 degrees Centigrade.
14:14
So, just to finish this lecture, I should like to make one final comment regarding spontaneity. That is that there is no such thing as a degree of spontaneity of a reaction. Reactions can not be highly spontaneous, or mildly spontaneous, or slightly spontaneous.
14:36
They simply are spontaneous or they are not.
14:41
Okay, so now hopefully we understand the somewhat enigmatic term spontaneous, and we understand that spontaneity is driven by an increase in universal entropy, which absolutely necessitates a decrease in the Gibbs energy of the system. The two criteria are effectively just one criterion.
15:03
In the next lecture, we shall look at how Gibbs energy controls chemical equilibrium. [BLANK_AUDIO]

2.3The Second Law of Thermodynamics and Equilibrium

0:10
In this video I shall present the concept of chemical potential energy and look at how Gibbs energy controls chemical equilibrium.
0:20
Consider chemical equilibrium which a moles of a and b moles of b are in equilibrium with c moles of c and d moles of d.
0:30
We need to introduce a new concept called the reaction gives energy symbol delta r g.
0:37
We shall get to the definition of reaction Gibbs energy shortly, but for now we will just accept that it can be interpreted as a difference in chemical potential between the products and the reactants at the composition of the reaction mixture.
0:53
Thus if mu i represents the chemical potential of species i, at the composition of the reaction mixture, then the reaction gives energy equals the chemical potential of the products multiplied by this documetric coefficients minus the chemical potential of the reactants multiplied by this documetric coefficients.
1:14
This important interpretation is given as equation 12. As a reaction progresses the chemical potentials of both its reactants and products change and consequently the reaction Gibbs energy also changes. The equilibrium composition of the reaction mixture will correspond to the minimum value of the Gibbs energy, G, and the reaction Gibbs energy at this point will be zero. [BLANK_AUDIO]. To work our way towards a definition of reaction Gibbs energy, we need to consider what happens if the reaction shifts towards the products by an infinitesimal amount d psi.
1:54
Sai is termed the extent of reaction or reaction coordinate. It varies from 0 corresponding to the reactants to 1 mol, since when 1 mol of the reactions is written has occurred then the reactants have been totally transformed into products.
2:11
So, if we now consider the effect of this shift towards products on the number of moles of each species present, the number of moles of each species has changed by d side multiply by the stoichiometric coefficient of that species.
2:27
The number of moles of reactant has decreased slightly. And the number of moles of product has increased slightly. Thus minus ad si for a.
2:39
Minus bd si for b.
2:43
Plus cd si for c.
2:47
And plus dd si for d. [BLANK_AUDIO]. Thus the change in the Gibbs energy delta G equals the chemical potentials of each species multiplied by the change in the number of molds of that species. In the equation shown, the products and reactants have been grouped together.
3:16
If we now take out the factor d si, the rest of the right hand side of the equation is equal to the reaction gives energy. Thus, delta g equals d si multiplied by delta r g. And hence by rearrangement delta r g equals delta g divided by d sine.
3:41
This equation forms the definition of delta r g and is given as equation 13. The reaction Gibbs energy. Is thus defined as the slope of the graph of Gibbs energy plotted against extent of reaction [BLANK_AUDIO]. The graph shows a typical variation of Gibbs energy with extent of reaction for a spontaneous chemical reaction.
4:12
We can see that the Gibbs energy of the products is less than the Gibbs energy of the reactants. And thus the standard reaction Gibbs energy, delta r G standard, is negative hence the spontaneous nature of the reaction.
4:29
So, the reaction Gibbs energy represents the slope of this graph.
4:36
As the reactants begin to form products, we notice that the slope of the graph and hence delta rg is negative.
4:45
This means that delta g is negative for increasing si and thus the forwards reaction is spontaneous and the reaction proceeds.
4:55
If we examine a composition closer to pure products, we can see that the slope of the graph enhanced delta rig, is positive.
5:05
This means that delta g is negative for decreasing psi and thus the backwards reaction is now spontaneous and the reaction occurs in reverse.
5:17
The reaction reaches equilibrium when the slope of the graph, enhance delta rG is 0. At this point, the chemical potential of the products and reactants are equal.
5:36
The chemical potential of a species may be written as the standard chemical potential of that species plus RT times a natural log of its activity.
5:47
Activities are dimensional quantities. It is the activity of the species that changes as the extent of reaction changes.
5:57
We can therefore write the reaction Gibbs energy, in terms of chemical potentials in this form.
6:04
If we then group all the standard potentials together. And take a factor of RT out of all the logarithms of activity terms, we get the following equation.
6:18
The standard potential terms are equal to the standard reaction gives energy.
6:25
We then use three successive logarithm identities to incorporate the stoichiometric coefficients as powers, to group the product and reactant terms together, and to express the result as a fraction.
6:45
I apologize for the small font here, but it was the only way to make this slide fit.
6:51
Okay, back to our usual sized font.
6:54
The fractional term is called the reaction quotient. At equilibrium, as we have seen, delta rG is 0, the activities of the various species will be the equilibrium activities and the reaction quotient is termed the thermodynamic equilibrium constant k. And thus our equation becomes 0 equals delta r g standard, plus r t ln k and thus, delta r g standard equals minus r t ln k.
7:30
This huge importantly equation is giving us equation 14. It enables the server dynamic equilibrium constant for a reaction, to be calculated from the standard reaction Gibbs energy. Thus ultimately, it is the standard change in Gibbs energy that determines the position of chemical equilibrium.
7:52
Activities as stated previously are dimensionless quantities. And thus so is thermodynamic equilibrium constant.
8:01
They have varying definitions depending on the type of chemical reaction being considered. The two of the most common are concentration divided by [UNKNOWN] cubed for solution chemistry, and pressure divided by one bar for gas phase reactions.
8:21
One molar, and one bar are considered standard states for the above types of reaction. [BLANK_AUDIO]. Finally in attempt to clarify all of this I have prepared an animation.
8:37
The animation initially shows a reaction with a large negative standard reaction Gibbs energy.
8:43
The equilibrium position of this reaction where the reaction gives energy and hence the slope of the curve is zero, lies effectively at completion.
8:54
Now watch what happens as the standard reaction Gibbs energy becomes less negative.
9:02
The position of equilibrium begins to shift away from pure products.
9:11
When the standard reaction gives energy zero the reaction reaches equilibrium at 50% extent of reaction. This is perfect equilibrium.
9:23
As the standard reaction Gibbs energy becomes more positive the equilibrium shifts further. Now lying closer to the pure reactants than to the pure products. The reaction is now not spontaneous.
9:41
When the standard reaction gives energy it has a large positive value, the position of equilibrium effectly lies at pure reactants. It is the backwards reaction that is spontaneous at all compositions and the reaction produces no discernible products.
10:01
Okay. So we've now covered the various aspects of the second law of thermal dynamics, which governs the spontaneity of chemical reactions based on the Gibbs energy.
10:13
In the next lecture, we shall examine the third law of thermodynamics, which concerns the entropy of perfect crystals and leads to determination of absolute entropies. [BLANK_AUDIO]

3.1The Third and Hess'Laws

The Third Law of Thermodynamics and Absolute Entropy

0:00
[BLANK_AUDIO]. In this video, I shall present the third law of thermodynamics, and show how it enables determination of absolute entropy.
0:20
The third law of thermodynamics states that the entropy of any perfect crystal at absolute zero is a well-defined constant.
0:28
This constant value at zero kelvin is defined as zero on the entropy scale.
0:35
Thus, the entropy of a perfect crystal at zero kelvin is zero.
0:42
From this definition of the zero point on the entropy scale, we can now determine absolute entropy, and hence standard molar entropies.
0:54
Imagine an infinitesimal amount of heat, delta Q reversible is supplied reversibly to an object.
1:03
The resulting entropy change, delta S, is calculated according to the definition of entropy, as delta S equals delta q rev divided by T. The heat capacity of the object is defined as delta q rev divided by delta T.
1:23
And thus, delta q rev equals the heat capacity, multiplied by delta t. Substituting this back in our equation for the change in the entropy gives delta s equals the heat capacity, multiplied by delta t divided by temperature.
1:42
If the temperature changes from Ti, the initial temperature, to Tf, the final temperature, then delta S equals the integral between Ti and Tf of CdT over T.
2:01
If we assume that the heat capacity is constant over the range of temperature, then we can bring the heat capacity outside the integral.
2:10
The integral of dx over x is a standard integral and is equal to log x plus c.
2:16
Thus, the integral of dT over T is log T.
2:22
Putting the limits in gives us delta S equals C multiplied by log Tf minus log Ti.
2:31
We now use the logarithm identity shown on the right, to get the final result, the delta s equals c log tf over ti, show as equation 15. Thus, entropy can be determined by measurement of the heat capacity as a function of temperature.
2:56
Once the graph of heat capacity versus temperature has been determined, it needs to be replotted as heat capacity divided by temperature, versus temperature, and then the area under the graph between two temperatures, calculated from equation 15, is the entropy change between those two temperatures.
3:17
To determine absolute entropies, this process needs to be carried out for all temperatures, from zero kelvin up to whatever the desired standard temperature happens to be.
3:28
Two problems present themselves: First, there will be discontinuities in the heat capacity graph at melting and boiling points. At these points it will be necessary to add the entropy diffusion and vaporization.
3:44
Also, measurement of heat capacity very close to zero Kelvin is extremely difficult. An example of how this problem can be overcome is the Debye T cube law, which enables us to approximate heat capacities of non-metallic substances at constant volume, by the molar heat capacity at constant volume equals a T cubed.
4:09
Slide eight shows tabulated standard molar entropies at 298 K. The substances have been grouped in the order, gases, liquids, then solids. What you should notice is that gasses generally have higher entropies than liquids, which have higher entropies than solids. This should hopefully make sense if you consider the molecular motion of atoms and molecules in solids, liquids and gasses.
4:35
Another point of interest is that even changes in solid structure impact on entropy. Graphite, which comprises honeycomb sheets of carbon atoms separated by layers of electrons, has a higher entropy than diamond, which is a three dimensional array of tetrahedrally bonded carbon atoms. In graphite, the layers can slide over one another, and this flexibility leads to the higher entropy value.
5:03
Okay, so now we have covered all four laws of thermodynamics, and defined a good number of state functions. In the next and final lecture we shall examine how an intrinsic property of state functions leads to Hess' law and see a series of examples of Hess' law calculations [BLANK_AUDIO].

3.2Hess' Law

0:00
[BLANK_AUDIO]. In this lecture I shall present Hess' Law, and we will work through a series of Hess' Law based example calculations.
0:20
A thermodynamic system is described by a series of state functions.
0:26
State functions are such that how the system came to be in that particular state, have no bearing whatsoever on their value.
0:36
This caused Germain Hess in 1840 to propose that the enthalpy change of a process is independent of the pathway between the initial and final states.
0:48
What this basically means is that if a chemical reaction can be performed in one step, or in a series of sequential steps.
0:57
The enthalpy change for the single step process will be equal to the sum of the enthalpy changes of the sequential steps.
1:11
Necessarily, by definition, this path independence holds for all state functions, and hence Hess' Law is applicable to standard enthalpies of reaction, standard internal energies of reaction, standard entropies of reaction, and standard gives energies of reaction.
1:29
So, just to reiterate, the state function that are most commonly questioned in Hess' Law questions are enthalpy, internal energy, entropy, and Gibbs energy. [BLANK_AUDIO]. A series of formal definitions of enthalpy changes for various processes exist. And these are listed on slides four and five. I do not propose to, to read them out or go through them particularly, but you should spend a few minutes to familiarize yourself with them.
2:06
It should be noted that each standard enthalpy has a symbol. And whilst there is some controversy over the symbols used, the ones that I have given here are the ones proposed by IUPAC.
2:22
Okay so all Hess' Law problems require you to calculate a change in a given state function. You are invariably given some ancillary data, and that data will always allow you to construct an alternative pathway to get from the initial state to the final one. The alternative route that you construct will then have the same change of state function as the requested one.
2:52
So let's see how the theory works as an actual calculation. So the task we are given is to calculate the enthalpy change for the combustion of propane and oxygen which is given as the equation at the top, given the following enthalpies of formation.
3:07
And we have enthalpies of formation of propane gas, oxygen gas, carbon dioxide gas, and water liquid.
3:19
You'll notice that the enthalpy formation of oxygen gas is 0 kilojoules per mole. This stems from the definition of enthalpy formation, which is the enthalpy change for the formation of a substance from its elements in their standard state. Oxygen gas already is an element in its standard state. So in essence, the enthalpy formation of oxygen, is for the conversion of oxygen gas into oxygen gas, hence the entropy formation is zero. So how do we go about solving this problem? We start with the equation that we've got to calculate the enthalpy change for. Underneath the equation, we write constituent elements in their standard states. So for propane and five moles of oxygen, the constituent elements in their standard states will be three moles of carbon solid, four moles of hydrogen gas, and five moles of oxygen gas. We then draw arrows from the constituent elements in their standard states, up to the species in the actual equation.
4:36
So going from constituent elements in their standard states to propane gas, is the enthalpy formation of propane gas.
4:46
Going from constituent elements in their standard states to 5 moles of oxygen gas is 5 times the enthalpy formation of oxygen gas.
4:57
Going from constituent elements in their standard state to 3 moles of carbon dioxide gas is 3 times the enthalpy of formation of carbon dioxide gas. And in a similar fashion the final arrow is 4 times the enthalpy of formation of liquid water. The enthalpy change that we're asked to calculate is the enthalpy further compression of propane which is given the symbol delta CH standard. We are asked to calculate the enthalpy change going from the reactants to products.
5:33
But instead of going directly from reactants to products, we can go from reactants, down to the constituent elements in their standard state. So in other words we can dismantle the reactants. And then we can go from constituent elements in their standard states back up to products. So we can dismantle the reactants, and reconstruct the products.
5:59
How do we know that the atoms that we get, or the molecules and atoms that we get when we deconstruct the reactants will make the products exactly?
6:09
The answer to that is that the equation is balanced, so that must be the case.
6:15
So, if we consider our alternative route, so the enthalpy of combustion of propane is going to be equal to minus the enthalpy of formation of propane. Minus because we are traveling in the opposite direction to the arrow.
6:35
Minus 5 times the enthalpy of formation of oxygen. Again, minus because we're traveling against the direction of the arrow.
6:45
Then, plus 3 times the enthalpy of formation of carbon dioxide, plus 4 times the enthalpy of formation of water liquid. And if we total all of this up and are careful with our minus signs, we should get the enthalpy of combustion of propane is minus 2,220 kilojoules per mole.
7:12
Whenever you do a calculation like this or indeed any calculation in science, always try to consider is the answer sensible. So first of all the enthalpy of combustion has turned out to be negative value. That suggests that the reaction is exothermic, which is what you would expect given that propane is actually a fuel. And 2,220 kilojoules per mole is quite a large enthalpy change, again which is what you would expect given that propane is actually a fuel. So the answer does seems sensible.
7:50
Another way to look at this kind of problem, is that if you are given a series of reactants from R1 through to Rn, with balancing numbers mu 1 through to mu n. Being converted to products P1 through to Pn with balancing numbers mu 1 through to mu n, then in order calculate the enthalpy change for such a reaction, it's basically the sum the enthalpies of formation of the products. Each one multiplied by its balancing number, minus the sum of the enthalpies of formation of the reactants, again each one multiplied by its balancing numbers. So this is exactly the same calculation as we had before, minus the enthalpy of formation of propane. Minus 5 times the enthalpy of formation of oxygen, plus 3 times the enthalpy of formation of carbon dioxide, plus 4 times the enthalpy of formation of water.
8:59
In actual fact both of these methods are identical, it's just a slightly different way of approaching the problem. Okay, in this second example calculation, we're asked to calculate the change in enthalpy for the reaction given at the top, given a series of reaction enthalpies for a series of other reactions. The basis of questions like this is that it must be possible to construct the reaction written at the top from the reactions that are given underneath.
9:32
So the way to do this is to look at the reactions and try to find a species that appears only once in those reactions.
9:45
So for example, if we look at OF2 gas. The only one of the three equations that actually contains OF2 gas is equation one.
9:57
And equation one converts one mole of OF2 gas into products. Clearly, in the equation we are trying to calculate the enthalpy change for, we need two moles of OF2 to be converted. So consequently we need two lots of equation one to happen.
10:18
If we look at equation two, we see that equation two converts one mole of SF4 gas. And it is the only equation to contain SF4 gas.
10:31
In the equation that we're trying to calculate the enthalpy for, SF4 gas actually appears as a product and not a reactant. So, we actually need to perform equation two backwards and consequently, we multiply it by minus 1.
10:48
For equation three, we convert one mole of sulfur gas to products, and again this is the only equation that contains sulfur. We need to convert two moles of sulfur gas to products, so consequently, we need two lots of equation three.
11:07
So in summary, we need two lots of equation one, minus one lot of equation two, and two lots of equation three.
11:17
Just to prove that this combination of these reactions does actually equal the overall reaction that we're trying to calculate the enthalpy change for, we need to do some reaction arithmetic.
11:33
So what I've done is I've written the three reactions out, but I've multiplied their coefficients by the numbers that we just calculated. So I've written out double reaction one.
11:46
I've written reaction two backwards. That basically simulates the minus 1, so I've written the products on the reactant side and the reactants on the product side.
11:58
And equation three I've also multiplied the coefficients by 2. What we then do is we add every thing on the left had side of the equations together so we actually get six species, and then we add every thing on the right hand side of the equations together.
12:18
The next thing to do is to see if anything can be cancelled on either side of the equation. And we should hopefully see that there are 4HF gas on the left, and 4HF gas on the right, so they cancel.
12:35
There are two waters on the left and two waters on the right so they also cancel.
12:41
There are two oxygens on the left and two oxygens on the right so they cancel. And there is one sulfur dioxide on the left which will cancel one of the two sulfur dioxides on the right.
12:56
Having performed all of this cancelling, we are left with two sulfur gas and two OF2 gas on the left hand side, and one SO2 gas and one SF4 gas on the right hand side. And this is the original equation that we have to work out the enthalpy change for. So we've now shown that our combination of the various reactions does actually indeed equate to the reaction that we need. And so now it's just a simple matter of multiplying the enthalpy changes for each reaction by the multiplying numbers that we calculated. So we need two lots of reaction one, minus one lot of reaction two, and two lots of reaction three. And again, if we add of all this together and are careful with our signs, we should get the delta RH equals minus 320 kilajoules.
14:00
Given these two examples, you should be quick to perform the very vast majority of Hess' Law calculations that should come your way. [BLANK_AUDIO]

4.3Thermodynamics Recap

0:00
[BLANK_AUDIO]. Hi everyone. So we've now reached the end of week two. We've covered all the laws of thermodynamics and the various concepts related to them. Hopefully the two multiple choice tests have provided you with good feedback, and you've been active in the discussions on the forum. All that remains now is to complete the final multiple choice test and the virtual practical in order to receive your assessment for this section of the course. Next week you will begin lectures on Kinetics with Mike Anderson. Good luck with the test and the practical. [BLANK_AUDIO]