Equilibrium for A Level Teachers and Students

Introduction

The purpose of this set of articles on Equilibrium is to introduce Teachers and Students to a molecular

interpretation of the Properties of Systems that determine the Equilibrium behaviour of Systems of

molecules at a level suitable for A Level Chemistry Students. Underlying the whole approach are simple

ways of showing that the Properties of Systems are the average values of the corresponding molecular

properties using Statistics in GCSE Mathematics Courses. These include Properties such as Pressure,

Volume Energy, Enthalpy and Work which are already the A Level Syllabus. In addition it means

providing a simple introduction to the Gibbs Free Energy, G, and it’s property that G must be a

minimum in any System at Equilibrium. These topics are not included in the A Level Syllabus but are

essential to any discussion of why Equilibrium occurs. To relate the expressions for G in this approach to

standard treatments of G also necessitates introducing Entropy and some A Level Statistics.

The Series of articles are numbered sequentially and listed below. Here we offer some ways on how

Readers can use this set of articles which was written using Microsoft Word.

Option 1

Print out the whole set of articles and read as a normal book.

Option 2

To make it easier for Readers to use on Screen without printing out the entire text we list the locations of

the individual articles by page number in this document in the list of articles that we give below.

1

Option 3

In addition to listing the page numbers of the articles there are provided a series of Bookmarks on the title

page of each of the articles to make access to easier. To use this option click Insert on the Word menu

bar followed by the Bookmark option and choose the name of the Article or Section required.

The series of Articles and the page numbers on which they start are as follows:

Section 1 Equilibrium for Teachers Page No. 3

1. Equilibrium for Teachers 4

2. Equilibrium and the First Law of Thermodynamics 13

3 Equilibrium the Gibbs Free Energy and the Second Law of Thermodynamics 25

Section 2 Equilibrium for A Level Chemistry Students 35

4 Equilibrium and Statistics 36

5 Equilibrium and Mechanisms 43

6 Equilibrium and Energy 55

7 Equilibrium and the Gibbs Free Energy 64

8 Exothermic Chemical Reactions 76

9 Endothermic Chemical Reactions 88

Section 3 Equilibrium for A Level Chemistry and Mathematics Students 97

10 Equilibrium and Entropy 98

11 Free Energies from Experimental Data 108

12 The Stirling Equation for Chemists 116

2

Section 1 Equilibrium for Teachers

These three articles are intended for Teachers and others who already know about standard

Thermodynamics Properties including the Energy, Enthalpy, Gibbs Free Energies and Entropy. In the

first article we introduce a new definition of Equilibrium using the Maxwell Boltzmann Distribution of

energies as a key part of it. It defines Thermodynamic, Kinetic controlled and Non Equilibrium Steady

States and shows how these types of Equilibrium are related to one another using simple Statistical ideas.

In the second article we show that the First Law of Thermodynamics implies that Work is always done by

a System in Equilibrium with its Surroundings in any change of conditions and that consequently any

function which can describe Equilibrium must take account of the interconversion of Energy and Work.

This leads to the Gibbs Free Energy.

In the third article we show how the Second Law enables us to set limits on the amount of Work a

System can do and this, combined with the Third Law of Thermodynamics, enables us to show that the

Gibbs Free Energy is a minimum at Equilibrium using arguments based solely on the amount of Enthalpy

that can be converted into Work at Equilibrium. The Third Law is used to prove that at Equilibrium the

amount of Enthalpy convertible to Work if the System were returned to a crystalline state at zero Kelvin

by a reversible path must be a maximum. This enables us to avoid using Entropy in the argument.

This treatment of the Gibbs Free Energy is simple enough to introduce early in an A Level Chemistry

Course. It then enables a Teacher to explain why Chemical Reactions occur and why Equilibrium is

reached. Using standard treatments it is not possible to explain why Reactions occur or Equilibrium is

reached. Traditionally in Chemistry Courses there is no discussion of why these things happen.

1 Equilibrium for Teachers2 Equilibrium and the First Law of Thermodynamics3 Equilibrium the Gibbs Free Energy and the Second Law of Thermodynamic

3

Equilibrium for Teachers

R M Gibbons

BSc PhD DIC

Proofs to: R M Gibbons

4 Little Acre

Beckenham

Kent BR3 3ST

Abstract:

Equilibrium causes problems in teaching A Level Chemistry. Simple definitions based long term

constant behaviour cannot distinguish between Thermodynamic Equilibrium, Kinetic controlled

Equilibrium, and Non Equilibrium Steady States. Thermodynamic definitions of Equilibrium do not

account for Kinetic controlled Equilibrium and use ideas that are thought too advanced for A Level

students The A Level Course already contains the information, in the form of the Maxwell Boltzmann

Distribution (MBD) of energies, required to provide a simple explanation of how these types of

Equilibrium arise but it is not used for this purpose in standard texts.

In this article we introduce new definitions of Equilibrium that includes all three of these types of

Equilibrium and recognise the role of Statistics, via the MBD. We go on to give a molecular

interpretation of how these types of Equilibrium arise and how to recognise when a switch from Kinetic

control to Thermodynamic Equilibrium will occur. We demonstrate this using the exothermic reaction of

Oxygen and Hydrogen and the endothermic reaction of Oxygen and Nitrogen as examples. The

discussion brings out the importance of the Mechanism needed to produce Equilibrium.

4

Section 1 Introduction

We start with some definitions of terms used in this article. In the discussion below we use the term

System to describe a system of molecules with properties per mole such as Energy and conditions such as

Temperature and Pressure. The System and all the properties are referred to using capitalised words so

Temperature and Energy refers to the temperature of the System and its Energy respectively.

Standard introductions to Equilibrium of a System (1) are usually based on a simple definition in terms of

constant behaviour of the properties of the System such as:

A System is at Equilibrium when it has constant Pressure, Temperature, Density and Composition that are

independent of time.

We recognise Equilibrium by observing that the state of the System does not vary over a long period of

time.

There are problems with such definitions because constant behaviour occurs in many Systems that are

plainly not at Equilibrium, such as the Ocean currents, where layers of water occur that have differences

of Temperature that persist for hundreds of years. Thermodynamic definitions of Equilibrium correctly

describe Thermodynamic Equilibrium but, when applied, frequently result in Kinetic controlled

Equilibrium, as do both the Systems we discuss as examples in the Section 3. It is this sort of

inconsistency that makes this material difficult for Teachers to present to Students and I hope the account

that follows provides a coherent account of these problems which will make them easier to teach. It is

rigorous because the MBD already includes the affect on Equilibrium of all the advanced

Thermodynamic Properties thought to be too difficult for A Level such as the Gibbs Free Energy.

Our new definitions of Equilibrium, which we give in Section 2, avoid these problems. They take

account of the fact that for all three types of Equilibrium, the molecules are distributed according to the

MBD of energies at constant temperature that are accessible to the molecules at the conditions of the

5

System. They also recognise that to obtain Thermodynamic Equilibrium the molecules will be distributed

according to the MBD for all conditions on (reversible) paths that a System may pass through in changing

from one Equilibrium state to another. Accessibility and Mechanisms by which changes occur are closely

related.

Our new definitions of Equilibrium are an improvement on existing definitions but they do not deal

with all aspects of Equilibrium. We conclude with a short discussion of what needs to be included to

provide a complete account of Equilibrium and how the use of simple Statistical methods can explain

many aspects of Equilibrium.

Section 2 Three Definitions of Equilibrium

The starting point for our definitions of equilibrium is that the molecules have a MBD of energies at the

conditions of the System which include constant Temperature. Our new definition for Thermodynamic

Equilibrium is:

A System is at Equilibrium when it has constant values of Pressure, Temperature, Density and

composition that are independent of time, the molecules have a MBD of energies and Mechanisms are

available for all energy changes to occur both at the conditions of the System and at all points on a

(reversible) path between the initial state of the System and its final Equilibrium condition.

When these conditions are satisfied a System will always reach its Thermodynamic Equilibrium. This is

the first definition of Equilibrium to ensure that a System will always reach Thermodynamic Equilibrium.

When mechanisms are not available for some changes of energy other types of Equilibrium result.

When Mechanisms are not available on a reversible path at conditions different from the conditions of

the System Kinetic controlled Equilibrium results. Just as for Thermodynamic Equilibrium the molecules

are distributed in the energy levels that can contribute to the Energy at the conditions of the System. This

observation is the justification for treating such Systems as if they were at (Thermodynamic) Equilibrium.

6

Typical examples are described in Section 3 where this sort of behaviour is found in mixtures of gases at

Temperatures where they do not react, even though they can react. The term” Kinetic controlled “ arises

from this behaviour where the System fails to reach Thermodynamic Equilibrium because the rate of the

reaction, which would enable the System to reach Thermodynamic Equilibrium, is zero. This leads to the

definition of Kinetic controlled Equilibrium as:

A System is at Kinetic controlled Equilibrium when it has constant values of Pressure, Temperature,

Density and composition that are independent of time, the molecules have a MBD of energies and

Mechanisms are available for all energy changes to occur at the conditions of the System but Mechanisms

are not available at points on a (reversible) path between the initial state of the System and its final

Equilibrium condition.

When Mechanisms are not available at the conditions of the System we have a Steady State non

Equilibrium condition where Systems have constant uniform Properties. Typical examples of this sort of

behaviour are found in high pressure gas mixtures where different compositions remain unchanged for

months and in the Ocean Currents, where warm surface currents with cold currents flowing in the reverse

direction underneath, have Temperature and composition differences that have been maintained for

hundreds of years.

This leads to the following definition of Non Equilibrium Steady States:

A System is in a Non Equilibrium Steady State condition when it has constant values of Pressure,

Temperature, Density and composition that are independent of time, the molecules have a MBD of

energies and Mechanisms are not available for all energy changes to occur at the conditions of the System

and at points on a (reversible) path between the initial state of the System and its Thermodynamic

Equilibrium condition.

The unifying concepts relating these three types of Equilibrium is the Statistical behaviour of the

molecules and the barriers that can prevent a System from attaining Thermodynamic Equilibrium. How

7

the barriers arise, and identifying the molecules that can surmount them, enable us to analyse the type of

Equilibrium we are observing.

These are abstract ideas and are most easily demonstrated by applying them to some typical cases. We

do that in the Section 3 for two chemical reactions for which there are established reaction mechanisms.

The idea of mechanisms is much wider than mechanisms for chemical reactions and we will discuss

mechanisms further in Section 4.

Section 3 Equilibrium for Exothermic and Endothermic Reactions.

The two Systems we take as examples to illustrate the ideas in our definitions of Equilibrium are the

exothermic reaction of Oxygen and Hydrogen at 100 C and the endothermic reaction of Oxygen and

Nitrogen, both at atmospheric Pressure. Both of these Systems are at Kinetic controlled Equilibrium at

these conditions. We discuss below how they reach Thermodynamic Equilibrium from their initial State

at 1 atm and 100C. To do this we must introduce a mechanism for each reaction.

We use the Activated Collision Energy Model to explain the mechanisms of the reactions. This is a well

established model for chemical reactions in which high energy collisions of molecules allow the change

or transfer of electrons to occur in the reactions. The number of molecules with high enough energies for

the reactions to take place are determined by the MBD and their number increases with Temperature.

Correspondingly the rate of reaction increases with Temperature and the usual rule of thumb given is the

rate doubles for every increase of 10 C. For each combustible gas there is a Temperature, the Ignition

Temperature, below which the reaction does not occur. There are problems with this model but is useful

to illustrate our approach in a simple way.

For the first example, the equation for the reaction is:

2H2O + O2 = 2H2O (1)

8

A System made up of gaseous Hydrogen and Oxygen at 1Atm and 100 C is in a stable Kinetic controlled

Equilibrium with the molecules having a MBD of energies. It will remain in that state indefinitely; it is

kinetic controlled because no molecules have sufficient energy to react at these conditions. Provide a

source of ignition, however, and the reaction will proceed rapidly to produce an equilibrium mixture of

Hydrogen, Oxygen and Water as indicated by Equation 1 while generating large amounts of heat. The

source of heat arises mainly from the changes in the energy states of the electrons in the molecules which

occur in the reaction.

The molecular changes explain what is going on. At the initial conditions the molecules have a MBD.

There are no high energy molecules with sufficient energy to react at 1 atm and 100 C. Introduce a

source of ignition, like a match, and a small volume of gas is heated sufficiently to have molecules with

sufficient energy to react so the reaction occurs in that small volume. This releases energy which heats

up adjacent volumes of the gas mixture above the Ignition Temperature which then contain molecules

with sufficient energy to react and the reaction spreads throughout the whole volume.

In terms of our new definition we have provided a Mechanism, which is in this case a reaction

mechanism. How quickly this will occur depends on how the molecules are involved in the Reaction. If

the gas mixture is supplied premixed to a burner the rate will obviously depend on the rate of flow to the

burner. If the gases are premixed in a large container, on ignition they all will explode instantly. Clearly

the rate depends on the number of molecules involved at the time of the reaction and the reaction occurs

most quickly when they are all involved at the same time.

Our second System is a mixture of Oxygen and Nitrogen. Initially the System is at ambient

Temperature and 1 atm and is at a Kinetic controlled Equilibrium; no reaction occurs at these conditions.

The reaction to form Nitric Oxide is given in equation 2:

N2 + O2 = 2 NO (2)

9

This reaction occurs in lightning strikes in the atmosphere and produces an equilibrium mixture of

Nitrogen, Oxygen and Nitric Oxide. Millions of tons of Nitric Oxide are produced this way every day. In

terms of our definition of Equilibrium at ambient conditions and 1 atm the System is at Kinetic controlled

Equilibrium. The molecules have a MBD and there is no Mechanism at these conditions for the reaction

to occur.

In the conditions of a lightning bolt, where Temperatures can reach 5500 C, there are a large number of

molecules with sufficient energy for the reaction to occur and so it does, even though as an endothermic

reaction, it absorbs heat. The energy from the lightning bolt provides the energetic molecules that enable

the reaction to occur.

It is natural to ask where has the energy gone to? The answer is it has been used in changing the energies

of the electrons between their initial energy state in the Reactants molecules and their final state in the

Product molecules. It is one example of how the requirement of a MBD always produces an Energy at

Equilibrium which is not at the maximum Energy of the System but a balance between the affects of

interactive energies and kinetic energies of the molecules.

Note the difference between this case and the exothermic reaction discussed previously. Endothermic

reactions that need high Temperatures to proceed require an external Mechanism to initiate them and

cease as soon as the external Mechanism is removed because endothermic reactions absorb energy and

tend to decrease the Temperature, and so cause the reaction to stop. By contrast exothermic reactions

often need an external Mechanism to initiate them but then proceed to react via internal mechanisms.

As for exothermic reactions, the rate at which Products are formed depends on the number of molecules

that are excited by the external Mechanism.

Section 4 Discussion

The new definitions of Equilibrium explain the connection between all three types of Equilibrium that

we have discussed. The introduction of the MBD allows us to both define the Equilibrium State and

10

identify the active molecules which enable the reactions to occur. We need mechanisms to describe all

chemical and physical changes; the idea is much wider than reaction mechanisms and we give a full

discussion of it in an article on the role of Mechanisms in Equilibrium.

Our definitions show how Equilibrium occurs and defines the Equilibrium States correctly. It does not

answer the questions of why reactions occur or why, at Equilibrium, we have a mixture of Products and

Reactants. To deal with these questions we must take account of the Energy of the System and its ability

to do Work. Energy and Work are interchangeable and it is their interconversion that lies at the heart of

Thermodynamic Equilibrium.

Statistics play a key role in explaining the interconverstion of Work and Energy. The inclusion of the

MBD in the Syllabus acknowledges the Statistical behaviour of Systems but the only use made of it is to

explain the occurrence of high energy molecules in reaction mechanisms. Traditionally Chemistry

courses do not include explanations of Chemical Properties using Statistics. This was understandable

when A Levels began in the nineteen eighties and Statistics was not taught in Mathematics Syllabuses.

Today, when Statistics are a popular option at A Level and most of the Statistical ideas that are needed

are included in the GCSE Mathematics Syllabus, it is an opportunity for cross curricular integration.

We showed in this article how the use of simple Statistical ideas can explain how Systems at Kinetic

controlled and Thermodynamic Equilibrium and Steady State non Equilibrium are related. In future

articles we show how simple Statistics and some physical insights lead to a simple definition of

Thermodynamic Equilibrium and expressions for Equilibrium Constants which show why Equilibrium

occurs. To do this we must introduce new State Functions which account for the interconversion of Work

and Energy in Systems and explain why and how it occurs.

In conclusion our new definitions of Equilibrium are an improvement on existing simple definitions and

the first to show how all three types of Equilibrium that we have discussed are related. Our definition of

Thermodynamic Equilibrium is as correct as the rigorous definition but does not indicate why a System

11

goes to Equilibrium. So it is only one step on the way to providing a complete account of Equilibrium.

The real advantage of this approach is that it leads to a simple introduction to the Gibbs Free Energy

which is the key property needed to describe Equilibrium at constant Pressure. This function does take

account of the affect of changes in Work and Energy for Systems at constant Pressure. We will take this

topic up in another article.

I hope this article will be useful to Teachers in the discussion of Equilibrium with their Students as it

stands.

References

1 eg E N Ramsden , “A Level Chemistry”, Stanley Thornes Ltd, 3rd edition, 1994

12

Equilibrium and The First Law of Thermodynamics

R M Gibbons

BSc PhD FRCS

Proofs to: R M Gibbons

4 Little Acre

Beckenham

Kent BR3 3ST

Abstract

We introduce a new definition of Equilibrium for Chemical Systems which makes clear the

statistical basis of the Laws of Thermodynamics. It is the first definition to cater for

Thermodynamic and Kinetic Equilibrium and Steady State non Equilibrium Systems. It

introduces the concept of a Mechanism needed to reach Equilibrium and ways of identifying

those molecules involved in the Mechanism.

In this note we look at the consequences of this definition for the First Law of Thermodynamics

and how it suggests new insights into why the First Law has the form it has. The main

conclusion is that heat applied to a System in Equilibrium with its Surroundings changes the

Pressure or Volume and this ensures that a System will always do some Work in reaching a new

Equilibrium. We discuss the question of how much Work such a System can do in and the

behaviour of Systems constrained to be at constant Volume in the next article as these types of

behaviour are determined by the Second Law of Thermodynamics.

13

Section 1 Introduction

We start with some definitions of terms used in this article. In the discussion below we use the

term System to describe a system of molecules with properties per mole such as Energy and

conditions such as Temperature and Pressure. The System and all the properties are referred to

using capitalised words so Temperature and Energy refers to the temperature of the System and

its energy respectively.

We base our approach (1) on a new rigorous definition of Equilibrium and then work out the

consequences for the behaviour of molecular Systems of this definition which is stated as

follows:

A System is at Equilibrium when it has values of Pressure, Temperature, Density and

composition independent of time, the molecules have a Maxwell Boltzmann Distribution (MBD)

of energies and Mechanisms are available for all energy changes to occur both at the conditions

of the System and at all points on a (reversible) path between the initial state of the System and

its final Equilibrium condition.

We recognise the System is at Equilibrium by observing that the Properties of the System do not

change over a long period of time.

This is a rigorously correct definition of Equilibrium that is applicable to all Chemical Systems

apart from the few (quantum) Systems that obey Fermi-Dirac or Bose-Einstein Statistics. At a

given condition of Temperature Pressure and Composition, a System will be in Thermodynamic

Equilibrium when all those energy states in the MBD which contribute to the properties of the

System are occupied by their equilibrium numbers of molecules. We discuss in Section 2 how

conditions can and do arise where not all of the energy states which can contribute to the

14

properties of a System are accessible to molecules in the System. When that happens, and the

condition persists, we have a Kinetic rather than a Thermodynamic Equilibrium.

This new definition of Equilibrium has implications for the First and Second Laws of

Thermodynamics too. Both these Laws were discovered before the molecular theory of matter

of was established and classical statements of them reflect that. The First law is usually given in

terms of a statement and an equation for Heat Energy and Work such as:

Heat applied to a System produces a change of Energy and Work is done by the System,

∆Q = ∆U + ∆W (1)

where Q U and W are Heat, Energy and Work.

Posed in these terms it is not possible to ask questions such as why must a System do Work?

The introduction of the Statistical behaviour of molecular Systems enables us to ask this question

and provide the answer which explains why Work has the leading role in any change of Energy.

We discuss this further in Section 3 and summarise our conclusions in Section 4.

Similarly the classical statement of the Second Law makes no acknowledgement of molecular

behaviour. It is commonly expressed in the form of a negative statement such as:

Not all Energy can be converted into Work. (2)

Again the absence of any recognition that the Law is based on the behaviour of molecular

Systems behaving according to Statistical Laws precludes questions about how much Energy

can be converted to Work or how much Work a System can do. Introducing Statistics and

molecular behaviour allows us to establish what governs the maximum amount of Work a

System can do and address the question of what governs the amount of Work a System will do

15

for a specific change of conditions. We discuss these topics relating to the Second Law in the

following article.

Section 2 The First Law of Thermodynamics for Molecular Systems

To see how the new definition of Equilibrium works in practice, we apply it to two mixtures of

gases at 100 C and 1atm. The first is an equimolar mixture of Hydrogen and Oxygen. The

molecules in this mixture follow a MBD of energies and are in a state of Kinetic controlled

Equilibrium because at these conditions not all energy states which can contribute to Equilibrium

are available at the conditions of the system. This is easily demonstrated by applying a lighted

match to the mixture when a spontaneous reaction will occur and the gases react to form water

and produce a Thermodynamic state of Equilibrium in which all energy states that contribute to

Equilibrium are occupied in a mixture of Hydrogen Oxygen and Water. The lighted match has

provided a Mechanism which produces molecules with sufficient energy for the Reaction to

proceed.

Similarly a mixture of Oxygen and Nitrogen at Room Temperature and 1atm is at a Kinetic

controlled Equilibrium at these conditions. These gases do not react at Room Temperature

because there are no molecules with sufficient energy to provide a Mechanism by which the

Reaction can occur at these conditions. At high Temperatures, such as occur in a lightning

strike, Oxygen and Nitrogen react readily to form Nitric Oxide

The Activated Energy Collision Model for Reactions describes a Mechanism for Reactions to

occur when molecules with high enough kinetic energies collide and it is clear the rate of

reaction increases with the number of molecules with enough kinetic energies to react. The

number of active molecules increases with Temperature and so therefore does the rate of

16

reaction. There is also a minimum Temperature below which a Reaction will not occur; for

combustible gases this is the Ignition Temperature. It clear therefore that mixtures of gases will

be in a state of Kinetic controlled Equilibrium at Temperatures below the Ignition Temperature,

or its equivalent for non combustible gases, and will reach Thermodynamic Equilibrium at

Temperatures above the Ignition Temperature. The deciding factor in each case is the number of

active molecules available to provide a mechanism for the reaction to occur. The most rapid rate

is when all the molecules are involved simultaneously. A good demonstration of that is what

happens when a uniform mixture of Oxygen and Hydrogen is ignited; it explodes.

These examples demonstrate clearly the need for active molecules in a Mechanism for a System

to reach Equilibrium but they do not show convincingly the role of Work in reaching

Equilibrium. Our next example demonstrates this. We look at the behaviour a mixture of

Helium and Argon at Room Temperature and high Pressure in a pressure cylinder. This is the

sort of mixture routinely made up for analysis for use in chromatographs in the analysis of gases.

For analysis the problem is to ensure the mixture has uniform composition. It is well known in

the Speciality Gas Industry that high Pressure Gas Mixtures require special treatment to produce

uniformity.

17

We illustrate the problem notionally in Figure 1 for a mixture in which Argon is first compressed into the

cylinder followed by Helium. Two layers are formed with the Helium above the Argon and these will

remain separate for months as shown in Figure 1. It is standard practice in the Speciality Gas Industry to

overcome this problem in one of two ways. The first way is to mix the gases at low pressure where the

molecules move freely and randomly throughout the Volume and then compress the mixture into the

cylinder. The second way is to put the Helium in first followed by the Argon. This will produce some

mixing due to gravity as the lighter Helium molecules migrate to the top of the cylinder. This is not

sufficient to produce a uniform mixture so it is standard practice to heat the bottom of the cylinder for a

number of days, as we show in Figure 2a, to produce convection currents which rise due to gravity and so

provide a Mechanism with a large number of molecules for mixing to occur

18

.

We show the situation we wish to discuss in Figure 2b and look at the response of the System when we

heat the Helium layer at the top of the cylinder. The Helium layer is heated and convection within the

Helium will quickly increase the Temperature and the Pressure of the Helium which promptly expands

and increases the Pressure of the Argon. Still layers of gases are good insulators so the Temperature of

the Argon changes very little. However its Pressure will increase to match that of Helium due to the

effect of Helium Pressure and correspondingly its Volume will decrease. The Helium layer will have

done Work on the Argon layer as a result of its expansion. How long it will take for the Argon layer to

reach the Temperature of the Helium layer is unclear but it is clearly going be much slower than the

change in Pressure which occurs in both layers as quickly as the Temperature change in the Helium layer.

This is a convincing demonstration of how Pressure changes always occur mostly rapidly which we

interpret in terms of a Mechanism that always involves all the molecules.

19

Section 3 Discussion

The number of Active Molecules for Reactions is determined by the Maxwell Boltzmann

Distribution of Energies (MBD) which has the form (2)

pi = n i /N = e (-u i/ k T) / Q (3)

Q = ∑ e (-u i/ k T) (4)

and pi is the MBD probability that the ith molecule has an energy ui, ni is the number of

molecules with energy ui and N is the total number of molecules.

It is the values of ui/kT that determine the number of molecules with energy u i. When this value

is too large the values of ni are too small to contribute to the properties of the System at the

current conditions.

The values of u i/ k T for the energy states involved in the kinetic energies of molecules are all

very small and the differences between u i / k T values for kinetic energies are so small that it is

standard in evaluating summations of these energy states to treat them as continuous. It follows

that the distribution of molecules in the kinetic energy states is always determined by MBD as

all kinetic energy states that can contribute to the State of the System are always available.

The Pressure is caused by change of average momentum that molecules in the System experience

at the walls of the container. The transfers of momentum by all the molecules in the System are

what generate that average momentum change. Because of the closely spaced nature of the

energy states involved in Kinetic Energy, these transfers are always available and the molecules

will re-arrange themselves into the appropriate MBD for the local Temperature in each part of

the System. As all the molecules are involved in the re-arrangements this will occur quickly. Not

20

only will the transfers of momentum be quick, because they are controlled by the statistical

distribution of the molecules between energy levels, nothing can stop these transfers occurring

and a change in Pressure must occur. This molecular behaviour explains why there are no

examples of Kinetic controlled Equilibrium where there are differences of Pressure, whereas

stable non Equilibrium Systems with different Temperatures occur widely in nature.

The discussion above is clearly true for gases where the translational energy levels are known.

The differences between these energy levels is always much less than kT(3) and this ensures that

all energy levels contributing to the Pressure at given conditions are freely available to all the

molecules. For liquids and gases the kinetic motions that generate the Pressure arise from the

vibrations of the molecules which give rise to the continuous spectrum of the liquid or solid. The

differences of these energy levels give rise to a set of closely spaced energy levels that are freely

available to all the molecules.

So the first effect of an increase in Temperature in any change for a System in Equilibrium with

its Surroundings will be an increase in Pressure and the System will do work against the Pressure

of the Surroundings. When a System is heated all the molecules are never heated

simultaneously; some are made hotter initially and then there is transfer of Energy by

mechanisms involving collisions. The transfer of energy always takes place more slowly than

the transfer of momentum simply because not all the molecules are involved in the transfers.

This explains why the First Law has the form it has. Work is always done and then the Energy

change occurs.

Not all changes involve an increase in Temperature. Phase changes, such as vaporizing water at

its boiling point, occur at constant Temperature. Such changes always involve an increase in

21

Volume as the liquid is vaporized and this increase in Volume causes the System to do Work

against the Pressure of the surroundings.

Systems can be constrained to have a fixed Volume and then input of Energy leads to changes

that do not involve Work. The responses of such Systems to the input of Energy involve the

Second law of Thermodynamics and we defer consideration of these Systems to the following

article on the Second Law.

This idea of a Mechanism can be extended to other processes dependent on molecular collisions

for Physical Equilibrium as opposed to Chemical Reaction. Examples include mixing in hot

water tanks where in the absence of stirring layers at different Temperatures remain separate for

long periods but which get to a uniform Temperature quickly when a Mechanism is provided

involving many molecules by stirring the water.

Mechanisms also explains the stable long term behaviour of the Ocean Currents in which warm

currents flow in the surface layers from the Tropics to the Arctic Ocean and cold denser Currents

flow below in the reverse direction. These Systems stay separate because the Mechanism for

mixing the two layers involves only the few molecules in the layers adjacent to the interface

between to the hot and cold layers and they have to overcome the affect of gravity which favours

keeping the denser colder layer below the warmer less dense layer on the surface.

The emphasis on reversible changes in standard explanations of Equilibrium explains why

Mechanisms are not discussed. It is assumed that a System is either at Equilibrium or will reach

Equilibrium along a reversible path. There are good reasons for adopting this approach. It does

allow quantitative calculations to be made of the maximum amount of Work or Energy to be

calculated in any change.

22

However the examples given above are enough to demonstrate that there is no guarantee that a

System will ever reach Equilibrium unless there is a Mechanism. In fact all experimental

techniques in laboratories, from stirring calorimeters to shaking titration flashes, are designed to

provide Mechanisms involving large numbers of molecules to produce a rapid approach to

Equilibrium. The need for such interventions is not addressed in standard explanations of

Equilibrium; they are implicitly assumed to occur.

Section 4 Conclusion

The new definition of Equilibrium recognizes the Statistical behaviour of molecules Systems. It

is the first definition of Equilibrium which caters for Thermodynamic Equilibrium and both

Kinetic controlled Equilibrium and Steady State non Equilibrium Systems. It also provides the

justification for the use of Statistical Mechanics for Systems at Kinetic controlled Equilibrium at

a specific condition because it recognizes that only energy levels occupied by sufficient numbers

of molecules can contribute to Equilibrium at the given condition.

This approach enables us to identify why the Pressure or Volume must increase when a System

is heated and that consequently Work must be done before the System changes to a new

Equilibrium state. It is obvious from this that any function that defines Thermodynamic

Equilibrium must include terms for both Energy and Work. It is also clear that, to ensure we

reach a Thermodynamic Equilibrium, we must specify that a Mechanism exists for a change to

occur, or a Kinetic controlled Equilibrium or non Equilibrium Steady State may result. Analysis

of the numbers of molecules active in the change process allows us to determine whether a

System is at Thermodynamic Equilibrium or not.

The question of how much Work a System can do involves the Second Law of Thermodynamics

23

and this topic is taken up in the following article.

References

1 R M Gibbons, Equilibrium and Statistics”, in press

2 E B Smith, “Basic Chemical Thermodynamics“, Imperial College Press, 1997 ch 9

List of Figures

Figure (1) A mixture of Helium above Argon at high Pressure.

Figure (2a) A mixture of Helium and Argon heated at the bottom.

Figure (2b) A mixture of Helium and Argon heated at the top.

24

Equilibrium The Gibbs Free Energy and the Second Law of Thermodynamics

R M Gibbons

BSc PhD FRCS

Proofs to: R M Gibbons

4 Little Acre

Beckenham

Kent BR 3 3ST

Abstract

We look at the consequences for the Second Law of Thermodynamics of Gibbons’s new definition of

Equilibrium for Chemical Systems which makes clear the statistical basis of the Laws of

Thermodynamics.

Previously we have explained why Systems must do Work in changes to their Equilibrium State. Here

we discuss the maximum Work a System can do in a change of its Equilibrium State, the actual Work

Done by a System and the amount of Enthalpy or Energy that could be converted into Work at its new

Equilibrium State. We show this amount of Work is equal to the amount of Energy or Enthalpy that

would be converted into Work on a reversible path between the State of the System’s current conditions

and the System as a perfect crystal at 0 K.

This leads to the Gibbs Free Energy and shows that it is a minimum at Equilibrium because the

maximum amount of Enthalpy is convertible to Work. We can then define Thermodynamic Equilibrium

in terms of the Gibbs Free Energy. The key advantage of this approach in teaching Chemistry is it

explains why Reactions occur early in the course.

25

Section 1. Introduction

Gibbons (1) gave a new definition of Equilibrium that emphasized the statistical nature of Equilibrium

and for convenience repeat it here:

A System is at Equilibrium when it has constant values of Pressure, Temperature, Density and

composition for a long period of time, the molecules have a Maxwell Boltzmann Distribution (MBD) of

energies (2) and Mechanisms are available for all energy changes to occur both at the conditions of the

System and at all points on a (reversible) path between the initial state of the System and its final

Equilibrium condition.

In reference 1 we examined the consequences of this definition for the First Law of Thermodynamics

which showed that Work must always be done when Heat is supplied to a system but did not indicate how

much Work is done as result. The Second Law of Thermodynamics enables us to set limits on the

amount of Work that will be done in any change of equilibrium states and we discuss these limits in this

article. We have previously given a simple proof (3) that the Energy of a System at fixed Temperature

and Volume is the average of the molecular interactions and that this Energy has a unique value because

the standard deviation of the Energy is too small to measure. At fixed Temperature and Pressure there is

a similar relationship for the Enthalpy and its standard deviation.

The Second Law states that some Energy or Enthalpy cannot be converted into Work. In Section 2 we

discuss what determines how much Enthalpy can be converted into Work in any change of conditions and

show that the maximum amount of Enthalpy that can be converted into Work is given by the Work Done

on a reversible path between the Equilibrium States. This discussion introduces the question of what

happens to the Energy or Enthalpy not converted into Work. We show an Enthalpy balance can always

be written relating the Enthalpy to Enthalpy convertible to Work and Enthalpy not convertible to Work.

The behaviour of all Systems at 0 K enables us to take this analysis further. At 0 K all substances have

zero Kinetic Energy and their maximum interactive energies between molecules as they are at their rest

26

positions in a crystalline state. This behaviour is the basis of the Third Law of Thermodynamics which

states (4):

A System at Equilibrium in a crystalline state at 0K has an Entropy of zero.

It is implicit in this Statement that the System has zero Kinetic Energy at 0 K so it is trivial to say that a

System at 0 K has no Energy/Enthalpy convertible to Work. It follows that the Enthalpy convertible to

Work at any condition is given by the Enthalpy convertible to Work that occurs on a reversible path

between 0 K and the state of the System; furthermore this Enthalpy convertible to Work must be a

maximum and the difference of the Enthalpy and the Enthalpy not convertible to Work must be a

minimum.

When we identify the Enthalpy not convertible to Work with TS, where T is the Temperature and S the

Entropy, we see that this difference, H-TS, is the Gibbs Free Energy which must therefore be a minimum

at Equilibrium. We can then express the Second Law in terms of the Energy convertible into Work and

conditions for Equilibrium in terms of either the Gibbs Free Energy or the amount of Energy convertible

to Work.

We discuss briefly the advantages of this approach and suggest that it is the simplest way to introduce the

Gibbs Free Energy which is essential to any discussion of Equilibrium.

Section 2 The Second Law for Molecular Systems at Constant Pressure

From the discussion in reference 1 it is clear that Systems cannot be prevented from doing Work once a

System is heated. The question we have to ask is how much Work can or will the System do?

We cannot address that question without considering the surroundings. Work is done when the System is

pushing against its surroundings. If there is nothing to push against the Pressure will increase but the

System cannot do any Work. If the surroundings are at 1atm and the System, in the course of a change,

generates a higher Pressure it is not clear how to calculate how much Work will be done. It is clear that

27

the maximum Work will be done on a reversible path where the Surroundings at each stage has a Pressure

marginally less than the Pressure of the System as it changes from its initial to its final state. If a System

has a Pressure of less than 1 atm , eg solid carbon, with a very small vapour pressure p, and increases in

volume on heating in an external Pressure of P , it will do a small amount of Work on its surroundings

given by- p∆V while the surroundings will do Work of P∆V on the Carbon. We are accustomed to

dealing with solids such as Carbon and liquids such as water at atmospheric pressure which is much

larger than the Equilibrium Vapour Pressure of either Carbon or water at Room Temperature. In defining

the Equilibrium of these Systems at Room Temperature and 1 atm we are including the Work Done by

the Pressure of 1atm on Carbon and Water; these differences are very small.

Now the Second Law says not all Energy/ Enthalpy can be converted into Work. It does not address the

question of how much Energy will be converted to Work or what happens to the Energy not converted to

Work. For a System at a given condition of Temperature and Volume some of the Energy can be

converted into Work and some cannot be converted so we can write:

Energy = Energy Convertible to Work + Energy not Convertible to Work (1)

If we wish to Work at constant Temperature and Pressure we must allow for the Work done at constant

Pressure for a change in Volume. So we must use the Enthalpy in place of the Energy as this function

automatically includes Work for a change of Volume at constant Pressure and have:

Enthalpy = Enthalpy Convertible to Work + Enthalpy not convertible to Work (2)

Returning to Equation 1 there is a condition where the System can do no Work. That is at 0 K because at

that condition there is no Kinetic Energy and consequently no Pressure. It follows that the amount of

Enthalpy Convertible to Work by a System in going from 0 K to a given T and Pressure will be a

maximum when the change between the selected conditions and 0 K occurs by a reversible path. We can

rewrite equation 2 as:

28

Enthalpy convertible to Work = Enthalpy - Enthalpy not Convertible to Work (3)

When written in this form it is clear that at Equilibrium the Enthalpy convertible to Work must be the

maximum amount of Work the System could do if it was returned to 0 K via a reversible path. The

Enthalpy at a selected condition is a State Function and so are the other two quantities in equation 3. It is

also clear that when the Enthalpy convertible to Work is a maximum that the difference of the other two

quantities must be a minimum.

Now at Equilibrium a System is in its most probable distribution of energies, ie MBD, and we can use the

Boltzmann Equation (5) and equation 3 to show that the Enthalpy not convertible to Work is equal to TS,

where T is the Temperature and S is the Entropy. Identifying the Enthalpy not convertible to Work with

TS and the Enthalpy Convertible to Work with G, the Gibbs Free Energy we can write equation 3 using

standard Thermodynamic Functions as

G = H – TS, (4)

a result first obtained by Gibbs by very different arguments. However the approach set out here makes

clear immediately that G is the maximum amount of Work that a System can do if it was returned by a

reversible path to 0 K. Because the amount of Work done by the System is negative (Enthalpy is leaving

the System) it follows that the value of G is at its most negative so G has a minimum value at

Equilibrium.

Section 4 The Second Law for Systems at Constant Volume

We now look at the Second Law for Systems at constant Volume. Typical examples include the

behaviour of Systems in high Pressure steel containers. These Systems cannot do Work when Energy is

supplied to the System; their Pressures are not maintained by their Surroundings. We still have our basic

Energy Balance from equation 1 which we can rearrange as,

Energy Convertible to Work = Energy - Energy not Convertible to Work (5)

29

When the Energy change occurs at constant Volume, at the new Equilibrium there is a new balance

between Energy convertible to Work and Energy not convertible. As for the case at constant Pressure if

the System is returned to 0 K at constant Volume by a reversible path the Energy convertible to Work

must be a maximum. Using the MBD and the Entropy S we can show by the same arguments as were

used above for the Gibbs Free Energy that TS is equal to the Energy not convertible to Work and the

Helmholtz Free Energy, A, is equal to:

A = U- TS (6)

Section 5 Discussion

The Gibbs Free Energy is the key property in any discussion of Thermodynamic Equilibrium. It includes

both Work and Enthalpy terms, allows for their interconversion and has a minimum which determines

Thermodynamic Equilibrium. We can identify Kinetic Equilibrium as a minimum in the Free Energy of

the System over the accessible energy levels at the conditions of the System. It is not possible to give a

sensible discussion of Equilibrium without introducing Free Energies. Standard treatments of the Free

Energy are difficult and so are normally not introduced until later in Undergraduate Chemistry and

Physics Courses,

As a result there is little or no discussion of Equilibrium in A Level Chemistry Courses which causes

difficulties for Students who are introduced to a number of topics in Physical Chemistry all involving

Equilibrium, without any explanation of how or why Equilibrium occurs. It is to remedy this situation

that the present approach was developed. It introduces the Gibbs Free Energy and shows it has a

minimum at Equilibrium in any System. In principle this approach can be developed to include all types

of Systems. To keep the Mathematics needed to a minimum, here we limit application of Free Energies

to Chemical Reactions of Ideal Gases (6,7 ). For Ideal Gases we can take this one step further because we

have an expression for the Work Done in reversible changes between two Pressures at a fixed

Temperature. At a fixed Temperature the work done by an Ideal Gas in going from P1 to Po is given by

30

Work Done= ∆G = ∫ P d V = - RT log (P1/P0) , (7)

On rearranging we have:

G (T,P) = G(T, Po) + RT log(P1/Po) (8)

In references (6) and (7) we show these expressions for the Gibbs Free Energies define Equilibrium for

Ideal Gas Reactions. This allows us to provide explanations for both Exothermic and Endothermic

Reactions which are unexplained in standard treatments.

Finally we note there is no difficulty, in principle, of extending this approach to more complex Systems.

It is just that the Mathematics becomes more complex as partial derivatives have to be introduced to

describe the effect of mutual interactions between molecules. These topics are best left to more

advanced courses.

Section 6 Conclusion

.In this article and the previous one (1) we have applied the First and Second Laws to molecular Systems

and present here the main insights that result from Gibbons’s definition of Equilibrium and recognizing

that these Laws are based on observations of the responses of molecular systems to changes of Energy or

Enthalpy. For Equilibrium the key result is:

1. A minimum in the value of G at given conditions defines Thermodynamic Equilibrium at those

conditions. This concept is the key to all discussion of Equilibrium. The other consequences are

listed below.

2. The Equilibrium Pressure of the System at a given Temperature will always be reached because all

the energy levels that contribute to the Pressure at the conditions are always accessible and the

molecules distribute themselves statistically between the MBD levels whatever else happens.

3. For Systems in Equilibrium with their Surroundings any increase in Temperature, which causes the

31

Pressure to increase or the Volume to change, makes the System do Work. So a System will always

do Work against its surroundings; the surroundings determine how much Work can be done in any

change between Equilibrium States.

4. The maximum amount of Work a System can do in any change of state is the amount of Work

required for that change to be reversible.

5. At Equilibrium at a given Temperature and Pressure a System always has the maximum Enthalpy

that is convertible to Work. This is equal to the amount of Work that would be obtained if the

System was returned to 0 K by a reversible path.

6. The fact that the Enthalpy convertible to Work at Equilibrium at given conditions is a maximum is

sufficient to ensure that the Gibbs Free Energy is a minimum at Equilibrium.

7. Our definition of Equilibrium includes Thermodynamic Kinetic and Steady State non Equilibrium

Systems and allows us to distinguish between them

8. The concept of Mechanisms for changes is essential for any definition to ensure Equilibrium can be

reached.

9. We can state the Second Law positively as: A System at constant Pressure changes as much

Enthalpy as it can into Work in any change between Equilibrium States. The maximum amount of

Enthalpy is changed into Work in a reversible change or

At any Equilibrium condition at constant Pressure, a System has the maximum amount Enthalpy

convertible into Work, which is equal to the Work obtained in changing the state of the System

reversibly to 0 K.

10. The simplest statement of the Third Law is that at 0K in a perfect crystalline state all molecules are

at their rest positions which maximize their interactive energies with the other molecules. It is an

32

immediate consequence that in this State the Entropy is also zero and that the amount of Work that

the System can do is zero too.

This simple approach to the Free Gibbs Energy is a key part of the basis of an Introduction to

Equilibrium and Thermodynamics. These two articles on the First and Second Laws developed

from a series of articles written for A Level Students and Teachers to give a simple account of

topics on Equilibrium in the A Level Syllabus but not addressed by Standard Textbooks. This

involved developing simple Mathematical methods for all proofs using methods that are in the

Syllabuses for A Level Statistics and Pure Mathematics. The original impetus for starting the

project was the recognition that most of the Mathematical tools needed to obtain expressions for

the Energy, Entropy and Work form part of the Mathematics Syllabus. However it did also

require the development of new proofs, using A Level Mathematical methods, for the Stirling

Equation and some development in Statistics to simplify the calculation of the standard deviation

for the Energy.

I hope this approach will be of particular interest to Teachers and will encourage them to discuss

Equilibrium with their Students using the concept of the Gibbs Free Energy confident that they

have explanations for Thermodynamic and Kinetic Equilibrium and ways of distinguishing

between them.

References

1. R M Gibbons, “Equilibrium and the First Law of Thermodynamics”, in press

2. E B Smith, “Basic Chemical Thermodynamics “, Imperial College Press, 1997

3. R M Gibbons, “Equilibrium and Energy”, in press

4. E B Smith, “Basic Chemical Thermodynamics “, Imperial College Press, 1997, p69

5. R M Gibbons, “Equilibrium and Entropy”, in press

33

6. R M Gibbons, “Exothermic Chemical Reactions”, in press

7. R M Gibbons, “Endothermic Chemical Reactions”, in press

8. G Attwood and G Skipwood, “Chemistry Data Book”, John Murray, London, 1992

34

Section 2 Equilibrium for A Level Chemistry Students

These articles are intended for A Level Students of Chemistry who have no background in

Thermodynamics but who have been introduced to the First and Second Laws of Thermodynamics.

It is assumed they will have done elementary Statistics in their GCSE course and will be able to calculate

a mean and a standard deviation for simple examples such as the results to be expected from tossing a six

sided dice which is the first example that Students encounter in Statistics in GCSE Mathematics.

In the first two articles of this Section we use our new definition of Equilibrium to show the importance

of Statistics and Mechanisms in the types of Equilibrium that occur in nature. We show how to analyse

the type of Equilibrium in each of a number of typical examples.

In the next article we show that the Energy of a System arises from the sum of the energies of the

molecules which have widely varying energies. We show the Energy has a unique value at fixed

Temperature and Volume using simple Statistics by applying the same formulae as those used to obtain

the expected values for a six sided dice to a mole of dice. This is the key calculation in understanding the

interconversion of Work and Energy and in how Entropy arises.

In the fourth article of this Section we introduce the Gibbs Free Energy and show it is a minimum at

Equilibrium using arguments based on the amount of Enthalpy convertible to Work at Equilibrium. We

then apply it to explain vapour liquid Equilibrium for water. In the last two articles of this Section we

show how the Gibbs Free Energies lead to Equilibrium for Exothermic and Endothermic Reactions.

4 Equilibrium and Statistics

5 Equilibrium and Mechanisms

6 Equilibrium and Energy

7 Equilibrium and the Gibbs Free Energy

8 Exothermic Reactions

9 Endothermic Reactions

35

Equilibrium and Statistics

R M Gibbons

BSc PhD DIC

 

Proofs to: 4 Little Acre

Beckenham

Kent BR3 3ST

 Abstract:

 The concept of Equilibrium causes problems in teaching A Level Chemistry. Simple definitions based

long term constant behaviour cannot distinguish between Thermodynamic Equilibrium and Kinetic

controlled Equilibrium. Thermodynamic definitions of Equilibrium do not account for Kinetic

Equilibrium and use ideas that are thought to be too advanced. The A Level Course already contains the

information, in the form of the Maxwell Boltzmann Distribution (MBD) of energies, required to explain

how Kinetic controlled and Thermodynamic Equilibrium arise.  

In this article we introduce a new definition of Equilibrium that includes both Thermodynamic and

Kinetic Equilibrium and recognises the role of Statistics, via the MBD, in Equilibrium. We go on to give

a molecular interpretation of how and why Kinetic controlled and Thermodynamic Equilibrium arise and

how to recognise when a switch from Kinetic controlled to Thermodynamic Equilibrium will occur. We

demonstrate this using the exothermic reaction of Oxygen and Hydrogen and the endothermic reaction of

Oxygen and Nitrogen as examples. The discussion brings out the importance of the Mechanism needed

to obtain Equilibrium.

 

36

Section 1 Introduction

We start with some definitions of terms used in this article. In the discussion below we use the term

System to describe a system of molecules with properties per mole such as Energy and conditions such as

Temperature and Pressure. The System and all the properties are referred to using capitalised words so

Temperature and Energy refers to the temperature of the System and its energy respectively.

Standard introductions to Equilibrium of a System (1) are usually based on a simple definition in terms of

constant behaviour of the properties of the System such as:

 A System is at Equilibrium when it has constant Pressure, Temperature, Density and Composition for a

long period of time.

There are problems with such definitions because constant behaviour occurs in many Systems that are

plainly not at Equilibrium, such as the Oceans, where layers of water occur at different Temperatures that

persist for hundreds of years.

Thermodynamic definitions of Equilibrium correctly describe Thermodynamic Equilibrium but then,

when applied, frequently produce Kinetic controlled Equilibrium, as do both the Systems we discuss as

examples in the Section 3. It is this sort of inconsistency that makes this material difficult for Teachers

to present to Students and I hope the account that follows provides a coherent explanation of these

problems which will make them easier to teach.

 Our new definition of Equilibrium, which we give in Section 2, avoids these problems. It takes account

of the fact that for both types Equilibrium, the molecules are distributed according to the MBD of

energies that are accessible to the molecules at the conditions of the System. It also recognises that to

obtain Thermodynamic Equilibrium the molecules will be distributed according to the MBD for all

conditions on the reversible paths that a System may pass through in changing from one Equilibrium state

to another. Accessibility and Mechanisms by which changes occur are closely related.  Our new

37

definition of Equilibrium is an improvement on existing definitions but it does not deal with all aspects of

Equilibrium. We conclude with a short discussion of what needs to be included to provide a complete

account of Equilibrium and how the use of simple Statistical methods can explain much many aspects of

Equilibrium.

 Section 2 An Improved Definition of Equilibrium

The starting point for our definition of equilibrium is that for both Kinetic controlled and Thermodynamic

Equilibrium the molecules have a MBD of energies at the conditions of the System. Our new definition

is:

 A System is at Equilibrium when it has constant values of Pressure, Temperature, Density and

composition independent of time, the molecules have a MBD of energies and Mechanisms are available

for all energy changes to occur both at the conditions of the System and at all points on a (reversible) path

between the initial state of the System and its final Equilibrium condition.

 When these conditions are satisfied a System will always reach its Thermodynamic Equilibrium. When

Mechanisms are not available for some changes of Energy, Kinetic controlled Equilibrium will result.

We recognise the Equilibrium by observing that the Properties of the System do not change over a long

period of time.

 These are abstract ideas and are most easily demonstrated by applying them to some typical cases. We

do that in the Section 3 for two Chemical Reactions for which there are established reaction mechanisms.

The idea of Mechanisms is much wider than mechanisms for chemical reactions and we will discuss

Mechanisms further in Section 4.

 Section 3 Equilibrium for Exothermic and Endothermic Reactions.

The two Systems we take as examples to illustrate the ideas in our definition of Equilibrium are the

Exothermic Reaction of Oxygen and Hydrogen at 100 C and the Endothermic Reaction of Oxygen and

38

Nitrogen, both at atmospheric Pressure. Both of these Systems are at Kinetic controlled Equilibrium at

these conditions. We discuss below how they reach Thermodynamic Equilibrium from their initial State

at 1 atm and 100C. To do that we must introduce mechanisms for each of the reactions.

 We use the Activated Collision Energy Model to explain the mechanisms of the reactions. This is a well

established model for chemical reactions in which high energy collisions of molecules allow the change

or transfer of electrons to occur in the reactions. The numbers of molecules with high enough energies

for the reactions to take place are determined by the MBD and their number increases with Temperature.

Correspondingly the rate of reaction increases with Temperature and the usual rule of thumb given is the

rate doubles for every increase of 10 C. For each combustible gas there is a Temperature, the Ignition

Temperature, below which the reaction does not occur. There are problems with this model but is useful

to illustrate our approach in a simple way.

For the first example, the equation for the reaction is:

 2H2O + O2 = 2H2O (1)

 A System made up of gaseous Hydrogen and Oxygen at 1Atm and 100 C is in a stable Kinetic

Equilibrium and the molecules have a MBD of energies. It will remain in that state indefinitely. Provide

a source of ignition, however, and the reaction will proceed rapidly to produce an equilibrium mixture of

Hydrogen, Oxygen and water as indicated by Equation 1 while generating large amounts of heat. The

source of the heat arises mainly from the changes in the energy states of the electrons in the molecules

which occur in the reaction.

The molecular changes explain what is going on. At the initial conditions the molecules have a MBD.

There are no high energy molecules with sufficient energy to react at 1Atm and 100 C. Introduce a

source of ignition, like a match, and a small volume of gas is heated sufficiently to have molecules with

sufficient energy to react so the reaction occurs in that small volume. This releases heat which heats up

adjacent volumes of the gas mixture above the Ignition Temperature and the reaction spreads throughout

39

the whole volume.  In terms of our new definition we have provided a Mechanism, which is in this case a

reaction mechanism. The initial reactions by heating up the gas have provided molecules that allow the

energy changes to occur for the reaction to proceed throughout the System.  

How quickly this will occur depends on how the molecules are involved in the Reaction. If the gas

mixture is supplied premixed to a burner the rate will obviously depend on the rate of flow to the burner.

If the gases are premixed in a large container, on ignition they all will explode instantly. Clearly the rate

depends on the number of molecules involved at the time of the reaction and the reaction occurs most

quickly when they are all involved at the same time.

  Our second System is a mixture of Oxygen and Nitrogen. Initially the System is at ambient

Temperature and 1 atm and is at a Kinetic controlled Equilibrium; no reaction occurs at these conditions.

The reaction to form Nitric Oxide is given in equation 2:

 N2 + O2 = 2 NO (2)

 This reaction occurs in lightning strikes in the atmosphere and produces an equilibrium mixture of

Nitrogen, Oxygen and Nitric Oxide. It produces millions of tons of Nitric Oxide every day. In terms of

our definition of Equilibrium at ambient conditions and 1 atm the System is at Kinetic Equilibrium. The

molecules have a MBD and there is no Mechanism at these conditions for the reaction to occur. In the

conditions of the lightning bolt, where the Temperature can reach 5500 C there are a large number of

molecules with sufficient energy for the reaction to occur and so it does, even though as an Endothermic

Reaction it absorbs heat. The energy from the lightning bolt provides the energetic molecules that enable

the Reaction to occur.

It is natural to ask where has the Energy gone? The answer is it has been used in changing the energies of

the electrons between their initial energy state in the Reactants molecules and their final state in the

Product molecules. It is one example of how the requirement of a MBD always produces an Energy at

40

Equilibrium which is not at the maximum Energy of the System but a balance between the affects of

interactive energies and kinetic energies of the molecules.

Note the difference between this case and the Exothermic Reaction discussed previously. Endothermic

Reactions that occur only at high Temperature require an external Mechanism to initiate them and cease

as soon as the external Mechanism is removed because the endothermic reaction causes the System to

cool below the minimum Temperature at which the reaction can occur. By contrast Exothermic

Reactions that often need an external Mechanism to initiate them then proceed to react via an internal

mechanism. As for Exothermic Reactions, the rate at which the Product is formed depends on the

number of molecules that are excited by the external Mechanism.

 Section 4 Discussion

The new definition of Equilibrium explains both Thermodynamic and Kinetic controlled Equilibrium.

The introduction of the MBD allows us to both define the Equilibrium State and identify the active

molecules which enable the reactions to occur. We need Mechanisms to describe all Chemical and

Physical changes; the idea is much wider than Reaction Mechanisms and we give a full discussion of it in

an article on the role of Mechanisms in Equilibrium

 Our definition shows how Equilibrium occurs and defines the Equilibrium States correctly. It does not

answer the questions of why Reactions occur or why, at Equilibrium, we have a mixture of Products and

Reactants. To deal with these questions we must take account of the Energy of the System and its ability

to do Work. Energy and Work are interchangeable and it is their interconversion that lies at the heart of

Thermodynamic Equilibrium.

 Statistics play a key role in explaining the interconverstion of Work and Energy. The inclusion of the

MBD in the Syllabus acknowledges the Statistical behaviour of Systems but the only use made of it is to

explain the occurrence of high energy molecules in reaction mechanisms. Traditionally Chemistry

courses do not include explanations of Chemical Properties using Statistics. This was understandable

41

when A Levels began in the nineteen eighties and Statistics was not taught in Mathematics Syllabuses.

Today, when Statistics are a popular option at A Level and most of the Statistical ideas that are needed

are included in the GCSE Mathematics Syllabus, it is an opportunity for cross curricular integration.

We showed in this article how the use of simple Statistical ideas can explain how Kinetic controlled and

Thermodynamic Equilibrium occur. In future articles we show how simple Statistics and some physical

insights lead to a simple definition of Thermodynamic Equilibrium and expressions for Equilibrium

Constants which show why Equilibrium occurs.

In conclusion our new definition of Equilibrium is an improvement on existing simple definitions and the

first to account for Kinetic controlled and Thermodynamic Equilibrium. It is as correct as the

Thermodynamic definition but it is only one step on the way to providing a complete account of

Equilibrium. I hope it will be useful to Teachers in the discussion of Kinetic and Thermodynamic

Equilibrium.

 References

1.eg E N Ramsden “A Level Chemistry”, Stanley Thornes Ltd, 3rd edition, 1994

42

Equilibrium and Mechanisms

Proofs to: R M Gibbons

BSc PhD DIC

 

4 Little Acre

Beckenham

Kent BR3 3ST

 

 Abstract:

 The availability of Mechanisms determines whether a System reaches a Kinetic controlled or

Thermodynamic Equilibrium. We start by defining what a Mechanism is and go on to introduce a model

for Mechanisms involving molecular collisions. This is a generalised model based the Activated Energy

Collision model for Chemical Reactions. It enables us to recognise when a System will reach

Thermodynamic Equilibrium and when it is remain in a stable Kinetic Equilibrium.

  Using the new definition, that includes both Kinetic controlled and Thermodynamic Equilibrium, we

demonstrate through a number of examples that the type of Equilibrium a System reaches depends on the

Mechanisms available for the changes to occur. This leads to the conclusion that, when a System has all

its molecules simultaneously involved in a process, that change not only occurs most rapidly but must

occur for statistical reasons to produce a MBD at Equilibrium. All the molecules are always involved in

producing the Pressure. As result the changes of Pressure always must occur. We consider the

implications of this for the role of the Work done as opposed to Energy changes in relation to producing

Equilibrium.

43

 

Section 1 Introduction

We start with some definitions of terms used in this article. In the discussion below we use the term

System to describe a system of molecules with properties per mole such as Energy and conditions such as

Temperature and Pressure. The System and all the properties are referred to using capitalised words so

Temperature and Energy refers to the temperature of the System and its energy respectively.

We have previously introduced a new definition of Equilibrium that includes both Kinetic controlled and

Thermodynamic Equilibrium and for convenience repeat it here:

 A System is at Equilibrium when it has constant values of Pressure, Temperature, Density and

composition, the molecules have a MBD of energies and Mechanisms are available for all energy

changes at the conditions of the System and at all points on a (reversible) path between the initial state of

the System and its final Equilibrium condition.

  We treat Equilibrium for a System of molecules in terms of Driving Forces which reduce to zero at

Equilibrium, a Mechanism by which the System reaches Equilibrium and the number of active molecules

the Mechanism uses. Kinetic controlled Equilibrium occurs when Mechanisms involving some energy

states are not available.

  Driving Forces arise from differences of Properties such as Temperature Pressure and composition. It is

easy to identify Driving Forces from differences of Temperature as differences of average kinetic energy

which can be transferred by molecular collisions until all the molecules have a common average energy.

Similarly differences of Pressure have Driving due to differences of average momentum which reduce to

zero at Equilibrium. The Driving Forces for chemical reactions and vapour liquid equilibrium are much

less clear and we postpone discussion of them until we have defined the properties that govern them.

44

  For this definition to be useful we need to be able to define what Mechanisms are and to have ways of

applying them to Systems so we recognise what types of Equilibrium are occurring in a System. We

discuss different types of Mechanisms in Section 2 and summarise their main characteristics. In Section 3

we use these Mechanisms to analyse behaviour of typical Systems and show which type of Equilibrium

will reached. One benefit of this approach is that it enables us to analyse Systems that have constant

Temperature and Pressure but are not at Equilibrium.

Finally we discuss briefly the implications of the role of Work occurring via the Pressure on the

properties that describe simultaneous changes of Work and Energy.

 Section 2 Mechanisms

  Mechanisms need further explanation. They are the processes that enable a System to reach

Equilibrium. It is worth distinguishing two classes of mechanisms. Internal Mechanisms occur naturally

in reaching Equilibrium; a simple example of a mechanism is a convection current in a tank of water

heated at the bottom. If the tank is stirred to produce Equilibrium this is an external mechanism applied

from outside the System.

  All experimental measurements in a laboratory use external mechanisms to produce Equilibrium;

simple examples include stirring to produce uniform Temperature in a calorimeter or shaking a titration

flask to ensure a true endpoint. The idea of a Mechanism is the missing link between the world of the

Theoretical Chemist where everything is assumed to be able to reach Equilibrium and the real world

where most Systems do not reach Equilibrium; some of the systems we will look at have been in stable

Non Equilibrium States for hundreds of years.

 For chemical reactions there is a neat division. Exothermic reactions always proceed via internal

Mechanisms once an external mechanism has been applied. Endothermic reactions that require a high

Temperature must be initiated by an external mechanism and cease as soon as the external mechanism is

removed because the endothermic reaction cools the System below the minimum Temperature needed for

45

the reaction to occur. Other endothermic reactions can proceed spontaneously cooling the System as the

Reaction takes place.

  We can describe chemical reactions using the Activation Energy Collision Model in which reactions

occur via sufficiently energetic collisions between molecules which form intermediate complexes where

electron transfers occur. The number of molecules with sufficient energy to react can be estimated from

the MBD. The usual rule of thumb is the rate doubles for every increase of 10 C. For combustible gases

there is also a minimum Temperature, the Ignition Temperature, below which the gas will not react.

  We can generalise this model for any process for energy transfers dependent on molecular collisions.

These include energy transfer by conduction and diffusion where an activation energy barrier has to be

overcome. Of course we have to change the terms we use to describe our model. For example the

Ignition Temperature determines the minimum number of active molecules needs for the reaction to

proceed. In our terms at the Ignition Temperature there is a quorum of active molecules and the reaction

can proceed. Note we do not need to know the actual numbers involved to use this idea. But it is clear

that the rate increases with the number of active molecules and the fastest rate of approach to Equilibrium

will be when all the molecules are involved in the process at the same time. To demonstrate this further

consider the controls of a gas burner and we see the rate of burning increases with the flow; when all the

gas burns at once, as happens when a combustible gas is uniformly mixed with air, it produces an

explosion which is as instant a reaction as can occur.

The common characteristic of all types of mechanisms is that a mechanism that involves all the molecules

in producing a change leads to a fast approach to Equilibrium. To clarify these ideas we look in Section 3

at some typical Systems and mechanisms.

 Section 3 Some Examples of Equilibrium and Mechanisms

We now discuss the behaviour of four Systems made up of gases and liquids. Conventional theory says

they will reach Thermodynamic Equilibrium. Using the ideas about mechanisms we can recognise which

46

Systems reach Thermodynamic Equilibrium and which remain in a Kinetic controlled Equilibrium or a

non-equilibrium Steady State.

  We show two liquid Systems in Figures 1 and 2. Figure 1a shows an insulated water tank

heated at the bottom which warms up quickly to produce a uniform Temperature throughout.

 The mechanism here is convection currents, in which gravity causes the hotter water (less dense) to rise

and mix with cooler water above. It involves most of the molecules in dispersing heat throughout the

volume through collisions and produces a uniform Temperature quickly. Figure 1b shows an insulated

tank heated at the top. The hotter less dense water is on top and gravity

which prevents convection occurring. Heat transfer occurs by conduction at the interface of warmer and

cooler water which involves the (few) molecules in the interface layers. Consequently the rate of heating

is very slow. However stirring the tank , involving all the molecules, produces a uniform Temperature

quickly.

  The example shown in Figure 2, of the cold and warm ocean currents shows how stable systems

involving liquid layers are when the less dense warm layer is above the colder layer. The density of salt

47

water increases with the amount of dissolved salt. The warm currents are saltier than the water of the

Polar Oceans. When the warm currents reach the Polar Oceans they encounter cold Polar Ocean water

and are cooled. Because they are saltier the waters of the warm currents become denser than the Polar

water at the same Temperature and they sink to the bottom and form separate layer. This then returns to

the Tropics as a deep cold current.

  There are two Driving Forces involved. The first is Gravity which opposes transfer between layers. The

second is the salt concentrations which favours transfer of salt between the layers to equalise

concentrations. Both Driving Forces involve the small number of molecules at the interface. Mixing

only occurs at the interface and involves only a few molecules, which must overcome the effect of

gravity. This will therefore be a slow process as we see in the oceans where separate warm and cold

currents have continued for hundreds of years.

  In Figures 3 and 4 we show two Systems for gases. Figure 3 shows a mixture of Argon and Helium

which initially have different Temperatures, and are separated by a membrane. They then mix after

removing the membrane. At low Pressures the molecules move freely throughout the volume and mixing

48

occurs quickly via random collisions between molecules which transfer energy between higher and lower

energy molecules.

Figure 3a shows the case of where the Argon layer is above the Helium layer. At low pressures when

the gases mix this System comes to Equilibrium quickly via two Mechanisms The first arises from the

large the mean free paths of the molecules which enables all the of molecules able to move unhindered

throughout the volume and produce random collisions which transfer energy between higher and lower

energy molecules.. In figure 3a there is a second mixing effect due to Gravity as the lighter Helium

molecules rise and mix with the Argon atoms above them. Both of these mechanisms involve large

number of molecules in the mechanism producing Equilibrium quickly. In Figure 3b gravity works

against mixing so this case will lead to slower mixing.

At high Pressures molecules cannot move freely throughout the volume and are confined by their

neighbours. They have mean free paths of less than a molecular diameter and mixing occurs by diffusion

and the effect of gravity on the lighter Helium molecules which causes the layers to mix. The Helium

molecules must diffuse through the Argon molecules via a series of collisions. Mixing only occurs at the

49

interface and involves only a few molecules. This will therefore be a slow process,

This problem of mixing gases at high pressure is well known in the Specialty Gas Industry where it is

common practice to introduce Helium initially into a cylinder to ensure some mixing with a heavier gas

such as Argon from the effect of Gravity as described above. To produce a uniform mixture the bottom

of the cylinder is then heated to produce mixing by convection currents. This involves a large number of

molecules in the mechanism and a uniform mixture is produced as indicated in Figure 3a

  Figure 4 is our last example and involves a mixture of Helium and Argon at the same Temperature and

Pressure. Initially they are separated by a membrane. The membrane is then removed and the gases left

free to mix. At low pressures they mix quickly to become uniform. At high pressures they stay unmixed

for months. The mechanisms involved in these examples are the same as in the cases shown in Figure 3.

The question is what is the Driving Force in this case?

The answer is the Probability of the initial condition occurring when the whole of the volume is

available to both Helium and Argon. With the membrane in place this is an Equilibrium condition. On

the removal of the membrane a state with the Helium in Va and Argon in Vb has become an improbable

state among the possible states of the System. At low Pressures the molecules move freely throughout the

Volume and quickly produce the uniform mixture which is the most Probable State. It is a short step to

saying we only ever see the most Probable state of the system at Equilibrium. An essential part of our

50

definition of an Equilibrium state is that it is the most probable state. For all Systems at room

Temperature and above this is the Maxwell Boltzman Distribution (MBD) and we discuss the molecular

basis of this briefly in Section 4.

Section 4 The Maxwell Boltzmann Distribution.

Most students know that in MBD (1, 2) that pi, the probability of molecule i having an energy ui

, is proportional to e (-u i// k T )

Since the probabilities sum to unity we have:

P i = e (-u i /k T ) / Q (1)

where

Q = Σ i e (-u i /k T) (2)

n i is the number of molecules with energy ui so we also have

pi = n i /N (3)

where

N = ∑ n i (4)

For the MBD the argument u I / k T is closely related to the ratio of the energy of the molecule divided by

the average kinetic energy of a molecule, 3/2kT. The MBD describes the balance between attractive

forces between molecules that try to maximize the attractive energies and the kinetic energy that allows to

the molecules to move freely. So it is natural to have an argument that reflects this ratio.

This leads on to consideration of what the Equilibrium State of a System is. We treat the Energy of a

System at a fixed Temperature and Pressure as a unique value. In fact the Temperature, Pressure and

51

Energy are averages of corresponding molecular values. We talk about them as fixed values because the

deviations from the averages are so small. There are Statistical reasons for this that we will discuss later.

The MBD is not the only distribution of energies that occurs in molecular systems but it is the most

common and the one we observe.

Section 5   Discussion

It is clear from the examples in the Section 3 that at Equilibrium molecules distribute themselves

according to the MBD in all the accessible energy levels at the conditions of the System; ‘accessible’ here

means there are enough molecules in the energy levels concerned to affect the Energy of the System or to

facilitate a change of Energy of the System. Kinetic controlled Equilibrium arises when some energy

levels are not accessible for a change to occur. Thermodynamic Equilibrium always results when all

levels needed for a change of Energy are accessible. Note that this means the energy levels needed for

changes to occur are at any conditions and not just at the conditions of the System.

Now changes involving Work always occur via the Pressure. All the molecules generate the Pressure

and, as the sizes of the differences of the energy levels for the molecular kinetic energies are very small,

they are always accessible. So changes involving Work always must occur and occur most rapidly; other

Energy changes involving fewer molecules will occur at slower rates. Even for changes that occur

rapidly, eg an explosion of a combustible gas and air mixture, the System can reach Equilibrium only

after it had done all the Work it could. One result of this is that while Non-Equilibrium Steady States

often occur in nature with differences of Temperature, Non-Equilibrium Steady States with differences of

Pressure never occur. This has consequences for the behaviour of the properties that describe changes

where simultaneous changes of Work and Energy occur. Such changes involve interconversion of Work

and Energy. We discuss this further in another article after we have introduced the properties in question.

However before we do that we must first show how the Energy of the System, U, is related to the energies

52

of the molecules in the System and why U has a unique value at the conditions of the System. This is the

key to understanding how Work and Energy can interchange.

References

1. E N Ramsden, “A Level Chemistry”, Stanley Thornes Ltd, 3rd edition

List of Figures

 Figure 1(a) Bottom Heated Tank (b) Top Heated Tank

  Figure 2 Warm and Cold Ocean Currents

 Figure 3a Gas Mixtures Helium above Argon

Figure 3b Gas Mixtures Argon above Helium

Figure 3c Heated Gas Mixture of Helium and Argon

Figure 4a Helium and Argon unmixed at the same Temperature and Pressure

Figure 4b Helium and Argon mixed at the same Temperature and Pressure

53

Equilibrium and Energy

Proofs to: R M Gibbons

BSc PhD DIC FRSC

 

4 Little Acre

Beckenham

Kent BR3 3ST

 

 Abstract:

 The Energy of the System is the sum of the energies of the molecules in the System. We show how the

Energy of the System at Equilibrium has a fixed value at a given Temperature and Pressure while the

energies of the molecules have a Maxwell Boltzmann Distribution of energies (MBD). We do this using

the properties of an average and the Standard Deviation, σ, and the relationship between them applied to a

System of a mole of molecules using some simple Statistics. We demonstrate the simplicity of this

approach applying this technique to a mole of six sided dice. We finish with a brief discussion of Energy

and Work and how the First and Second Laws of Thermodynamic determine Equilibrium.

 

 

 

 

 

54

1 Introduction

We start with some definitions of terms used in this article. In the discussion below we use the term

System to describe a system of molecules with properties per mole such as Energy and conditions such as

Temperature and Pressure. The System and all the properties are referred to using capitalised words so

Temperature and Energy refers to the temperature of the System and its energy respectively.

The Energy of a System is the sum of the energies of the molecules that form the System. The average

energy of the molecules multiplied by the number of molecules in the System, N, is equal to the total

Energy of the System. The energies of the molecules vary very widely but the Energy of the System at

given conditions has a fixed value. We give a simple proof showing why this apparent paradox is just a

consequence of what is often called “The Law of Large Numbers”.

 To do this we must use some simple Statistics involving the average and the Standard Deviation, σ, and

their ratio. For convenience we give a full account in Section 2 of the Statistics involved which form part

of the GCSE Mathematics course. The only exceptions to this are the ratio of σ to the average is always

less than one for symmetrical distributions and the corresponding result for asymmetrical distributions.

These topics are not discussed in standard Textbooks on Statistics but simple proofs of them are given in

Appendix 1. They are new results and were crucial developments in producing this simple approach.

 We first use the calculation of the expected value and σ obtained from throwing a six sided dice to

demonstrate the methods; this is almost invariably the first problem that GCSE Mathematics Students do

in Statistics – for a single dice. Then we apply the method to a mole of dice and calculate the average and

σ for a mole of dice and show that σ is very small. We then apply the same technique to a System of

molecules with a MBD and show that σ for this System is also negligible. We conclude with a short

discussion of what is involved in a change of Energy as the conditions of the System change. Our

purpose in doing this is to show how changes in Energy occur in molecular Systems as this is important in

changes involving both Work and Energy.

55

Section 2 Statistical Methods

 The Statistics we require involve the average, the standard deviation, σ, and the idea of a distribution of

probabilities; all these topics are taught in the GSCE Mathematics course though the Normal Distribution

rather than the MBD is taught. The key idea is that the spread of values about the mean is determined by

the value of σ. With a very small value of σ the average value will be the only one observed. We wish to

calculate the average for a set of values, ei, occurring with different probabilities, pi, and their standard

deviation, σ. To demonstrate the Statistics let us first consider the model of a six sided dice using

standard formulae (1) to calculate the mean and variance for the outcome of throwing the dice

E av = ∑ e i pi = (1+2+3+4+5+6)/6 = 3.5 (1)

Variance= Nσ2 = ∑e2ipi - E2

av= (12 + 22 + 32 + 42 + 52 +62)/6 - 3.52 = 2.75 (2)

Standard Deviation σ = (2.75)0.5 =1.66 (3)

Now let us apply the same formulae to a mole of dice i.e. 6x1023. E av is unchanged so the expected value

of a mole of dice E m, is

E m = N E av (4)

But look what happens to σ compared to E m

Variance = (1.66) N and (5)

σ = (1.66N)0.5 (6)

σ/E m = (1.66N)0.5/(3.5N) = 0.368/N0.5 = 0.613x10 -11 (7)

Note σ is less than the mean. A simple proof of this is given in Appendix 1 for finite distributions.

56

This is a consequence of large numbers. Students are taught to analyze their results in terms of σ and will

know, on the basis of the Normal Distribution Function, that they can expect to find two thirds of their

results within σ about their mean value.

Section 3 The Energy for a MBD of Energy

We can use exactly the same formula as in the previous section to obtain expressions for the average

Energy and ó for mole of molecules with a MBD with probabilities pi. We first calculate the average

Energy for an Ideal Gas where we already know the answer.  Simple kinetic theory (2) shows the Kinetic

Energy for a mole of a monatomic gas is

U = 3/2 RT (8)

where U is the Energy, R the Gas Constant and T the absolute Temperature.

Dividing R by N, Avogadro’s Number , gives us k, the Boltzmann constant.

k = R/N (9)

and dividing both sides of equation 8 by N leads to u the average energy per molecule,

u =3/2kT (10)

For the MBD we show in Appendix 1 that the ratio σ/u < 30.5, so we can write that σ is bkT where b is a

number less than 30.5. Put these values into Equations 4 and 6 and we have,

E mol = 3/2kNT (11)

σ= bkTN0.5 (12)

σ/E mol =(2b/3)/N0.5) ≤ N-0.5 ≤ 10-11 (13)

57

It is clear that the values of u and σ for individual molecules have no significant effect on the value of σ

for a mole; the factor N-0.5 always ensures the value of σ is negligible. The result is that the value of σ for

a mole of molecules is infinitesimal and the Energy has a fixed value at given conditions.

 For monatomic substances kinetic energy is the sole form of energy related to motion. Molecules have

other types of energies in addition to kinetic energy. They can spin and have rotational energy. They

vibrate with characteristic frequencies. All these types of energy are interchangeable. They also have

electronic energies and can absorb energy by promoting electrons to higher energy levels or emit energy

by electrons falling to lower energy levels. For the transfer of energy the key point is that in a molecular

collision kinetic energy is transferred to the cooler molecule with lower kinetic energy. They also have

attractive and repulsive interactions with other molecules via their electrons; these interactions are the

sources of the heats of vaporisation and sublimation of liquids and solids respectively

58

We can extend the results in equation 13 to other Ideal Gas molecules, all of which have Energies of the

form of nkT where n is a small number greater than 3/2. It is clear replacing the average energy in

equations 4 and 6 with nkT in place of 3/2kT will have no significant effect on the outcome. There also

no trouble in extending this argument to real gases liquids and solids as all their energies can also be

written in the form nkT.

 This is a simplified version of a method introduced by introduced by Hill (3). We just give his key

results here for the Energy and σ:

Um= ∑ u i pi = 3/2NkT (14)

σ = ( ( C v kT2)/N)0.5 = ca 10-11 (15)

This behaviour of a mole of molecules means the Energy at a fixed Temperature and Pressure has a

unique value; the most probable value. We can similarly show the Pressure or any other molar property is

a statistical average with negligible Standard Deviation. Note the facts, that the kinetic energy of

monatomic gases has a negligible Standard Deviations and the relationship between Kinetic Energy and

Temperature in equation 1, means that the Temperature too is similarly constant.

 Section 4 Discussion

Showing that the Energy has a fixed value at one set of conditions raises questions about how the System

changes its Energy as the conditions are altered. At each set of conditions the molecules distribute

themselves according to the MBD about the average Energy value for the conditions. As the Temperature

increases higher energy levels become accessible to enough molecules for those levels to contribute to the

average Energy. In principle we have to use the entire range of energies of the molecules to calculate the

average. In practice only those within three times the average energy can contribute. Levels outside this

range occur with too small a probability to affect the average. The precise level depends on how

accurately we wish to calculate the Energy but the level given would exclude levels that could alter the

59

calculated energy at most by < 0.01%. In the Appendix we use a multiple of four times the average

energy to ensure even greater accuracy for the Energy.

 If the only affect of altering the conditions was to change the Energy this would be all we had to

consider. But the First Law of Thermodynamics says that Heat supplied produces changes of both

Energy and Work:

∆H = ∆U + ∆W (16)

 The Second Law says that some of the Energy is converted into Work but does not say how much;

saying some of the Energy can be converted implies that not all of the Energy can be converted. This

interconversion of Energy and Work is at the heart of Equilibrium and how it is produced. It also raises

the question of what happens to the Energy that is not converted to Work. We take up these questions in

another article.

References

1. G Attwood and G Skipworth, “ Statistics 1”, Heinemann, 1994, p160

2. Eg R Muncaster, “ A Level Physics”, Stanley Thornes Ltd, 4th ed., p 255

3 T. L Hill, “Introduction to Statistical Mechanics”, McGraw Hill, 1959, p 100-102

  List List of Figures

Figure 1 Probability of average Energy with different standard deviations

 

60

 Appendix 1. The Ratio of the Standard Deviation and the Mean

We have a symmetrical set of value ui with probabilities pi for which we calculate average, U a

U a = ∑ u i p i = u (1.1)

Variance = ∑ ui2 pi - u2

(1.2)

The largest value of the variance is when the values of ui are at the extreme range of the values of ui. So

for an average of u (dropping subscripts) we have half with a u value of zero and half with values of 2u

giving an average of u

For this set

 Variance = N u2 (1.3)

Standard Deviation = u (1.4)

For any other distribution the individual terms in the summation are less than the corresponding terms in

equation 2 and hence the Standard Deviation will be smaller than u. This proves what we set out to do.

The above proof applies to distributions between zero and 2u about an average of u. For the MBD the

distribution is between zero and 4u; higher energies occur too infrequently to affect the average value of

u. The worst case distribution in this case to produce the same average u must be

u = p4 (4u) + p0 (0) where (1.5)

p4 and p0 are the probabilities of the molecules having u values of 4u or zero.

Clearly p4 must have the value of ¼ to produce an average of u. Using this value in the expression for the

Variance leads to

Variance = N (p4 (4u) 2 – u2) = 3Nu2 (1.6)

Worst case Standard Deviation = 30.5 u (1.7)

61

The general result is the Standard Deviation is less than u (R – 1)0.5 where R is the range of values of u

contributing to the average.

 

 

 

62

Equilibrium and the Gibbs Free Energy

Proofs to: R M Gibbons

BSc PhD DIC FRSC

 

4 Little Acre

Beckenham

Kent BR3 3ST

  Abstract:

  We discuss the interrelation of Energy and Work and obtain formulae for the Work done in

changes at constant Pressure and by an Ideal Gas. This leads to the introduction of the Enthalpy

as the convenient property for changes at constant Pressure. We go on to introduce the Gibbs

and Helmholtz Free Energies, G and F respectively, and use arguments involving Work and the

behaviour of all Systems at zero degrees Kelvin to show that they have minimum values at

Equilibrium.

  Using standard treatments these topics are too difficult to include in A Level Courses. The

simple approach proposed here establishes that G has a minimum value at Equilibrium using

arguments based solely on Work considerations. We finish with a definition of Equilibrium in

terms of G and an expression for G for an Ideal Gas that forms the basis for Equilibrium in the

Chemical Reactions of Ideal Gases.

 

63

Section 1 Introduction

We start with some definitions of terms used in this article. In the discussion below we use the

term System to describe a system of molecules with properties per mole such as Energy and

conditions such as Temperature and Pressure. The System and all the properties are referred to

using capitalised words so Temperature and Energy refers to the temperature of the System and

its energy respectively.

The interconversion of Energy and Work is at the heart of how Equilibrium is reached and the

properties that control it. The First and Second Laws of Thermodynamics control these changes.

We start by establishing the Work done for changes at constant Pressure and for the Work done

by an Ideal Gas in going from an initial Pressure to final Pressure at constant Temperature.

These are the only expressions for Work that apply generally to all Systems. We show that the

Work done at constant Pressure combined with the Energy leads to the definition of the Enthalpy

and that this property is the natural choice to describe the affect changes of Energy at constant

Pressure.

 We next consider simultaneous changes of Energy and Work. We show for the Energy to

conform to the first two laws of Thermodynamics it must come in two forms, one convertible to

Work and the other not convertible. The difference of these quantities is in equal to the amount

of Work the System can do. These differences of the two types of Energy at constant volume

are the Helmholtz Free Energy, F, and at constant Pressure, the Gibbs Free Energy, G. We then

use arguments based on Work and our new definition of Equilibrium to show that at Equilibrium

the Work that can be done by the System is a maximum and consequently F and G have

64

minimum values. We then apply the concept of G to Vapour Liquid Equilibrium for some

simple Systems.

  When we alter the Temperature of a System we change its Energy. If the only affect of altering

the conditions was to change the Energy this would be all we had to consider. But the First Law

of Thermodynamics says that Heat supplied produces changes of both Energy and Work:

 ∆H = ∆U + ∆W (1)

 Standard forms of the Second Law say:

In the course of a change between Energy States some of the Energy is converted

into Work (2)

A better definition would address the question of what happens to all the Energy in the course of

a change:

In the course of a change between Energy States some of the Energy is converted into Work and

some cannot be converted into Work (3)

In this article we show that at Equilibrium the amount of Energy convertible to Work is a

maximum and is equal to the amount of Work Done in returning the System to the crystalline

state at 0 K by a reversible path. This interconversion of Energy and Work is at the heart of

Equilibrium and how it is produced. It also raises the question of what happens to the Energy

that is not converted into Work. We take up these questions in the following Sections but first

obtain expressions for Work for two types of change of conditions and give a new definition of

Equilibrium.

65

 Section 2 Expressions for Work and the Enthalpy

We need expressions for work that a System can do and derive them here for work done at

constant Pressure and by an Ideal Gas against its equilibrium Pressure at constant Temperature.

By definition:

Work = Force x Distance (4)

For a gas at constant Pressure moving a piston with cross section, A, a distance l as shown in

Figure 1a,

Work Done = P x A x l = PV (5) 

When we measure the heat of vaporisation at constant pressure it includes both the difference of

energy for the liquid and vapour, U l and U g , and the work done so that

Heat of vaporization = U l – U g + P (V l – Vg) = U l +P V l – (U g + P V g) (6)

It is convenient to define the Enthalpy, H, for heat changes at constant Pressure to automatically

66

include the work that occurs in a change at constant Pressure by defining

Enthalpy = H = U + PV (7)

The heat of vaporization then is the difference of Enthalpy of the liquid and gas at equilibrium,

Hl – H g, as Enthalpies include the work term.

For an Ideal Gas as shown in Figure 1b with the Pressure reducing from P1 to P0

Work Done = ∫-P d V = -RT∫ d V/V = - RT log (V1/V0) = - RT log (P1/ P 0) (8)

When P0 is equal to the Atmospheric Pressure then equation 8 tells us how much work is done by

an Ideal Gas at a fixed Temperature as the Pressure reduces from P1 against an external Pressure

equal to its equilibrium Pressure, i.e. reversibly; the negative sign arises because Energy is

leaving the System. In changing from a Pressure P1 to P0 the system has done work and lost

Energy; to maintain the Temperature an equal amount of Energy has to be supplied to the

system.

 Section 3 Definition of Equilibrium

Simple definitions of Equilibrium are based on constant behaviour of a System eg

A System is at Equilibrium when it had constant uniform Temperature, Pressure, and

Composition over a long period of time.

Such definitions cannot distinguish between Kinetic controlled Equilibrium and true

Thermodynamic Equilibrium because they take no account of the functions that define

Equilibrium. The definitions of Equilibrium we give here do enable us to distinguish between

Kinetic controlled and Thermodynamic Equilibrium and to identify the key property of the

67

functions we need to give a short description of Equilibrium.

Because we lack these essential tools to describe Equilibrium we must start with a long

definition. In the following section we develop the quantities we need to describe Equilibrium

and finish that discussion with a short description of Equilibrium. Our new definition is:

A system is at Equilibrium when it has the maximum amount of Enthalpy convertible to Work,

has constant and uniform Temperature Pressure, each phase has uniform density, it is in its

most probable state, its molecules have MBD of energies and Mechanisms are available for all

energy transfers to occur.

The maximum Work referred to is the amount of reversible Work done by the System in the

changes involved in going from a State where the System can do no Work to the current

Equilibrium State. A System can do no work when it is in its crystalline state at 0K so this

amounts to the statement that all the Energy that can be converted into Work is available to be

converted back into Energy if we returned the System back to the crystalline state at 0K. Note

this is not the same as useful work that be obtained from the System which is governed by the

final State the System will reach in any change which can occur without any external

intervention. In reaching Equilibrium a System will do as much Work as it can; the amount of

Work actually done will depend on the Surroundings. The maximum amount of Work Done in

any change between two Equilibrium States will be that done on a reversible path between the

two States.

Work has this leading role in the definition of Equilibrium because Work occurs via the

Pressure; at 0 K the velocities of the molecules are zero and the Pressure is zero. The molecular

energy states involved in producing the Pressure arise from differences of the kinetic energies of

68

molecules which are very small and therefore always available. They are always involved in the

distribution of molecules between energy levels. So the molecules distribute themselves

according to the MBD and Work is always done. By contrast many Energy states for electrons

are available in significant numbers only at high Temperatures. They cannot contribute to the

significantly to the average properties of a System over a wide range of low Temperatures

because, due to the MBD, they are available to too small a number of molecules. 

Section 4 Energy Work and Equilibrium

The earlier discussion showed that some Energy cannot not be converted into Work. This is

one statement of the Second Law of Thermodynamics. Complications in obtaining an expression

for Equilibrium arise from this because we need to describe two types of Energy, one of which

can be converted into Work and the other cannot. The Energy that cannot not be converted into

to Work when changing from Equilibrium at (T1,P1) to (T2,P2) is Energy used to change the MBD

of molecules from the equilibrium distribution at (T1,P1) to the MBD at(T2,P2). The most

obvious example of this is the Energy of vaporisation when a liquid is converted into a gas.

  From our definition of Equilibrium it follows that the maximum amount of Energy is available

for conversion into Work. Hence the difference between the Energy, U, and the Energy not

convertible to Work is the amount of Energy available for Work so we can write:

U – U not convertible to Work = U convertible to Work (9)

 Either side of this equation defines the Free Energy at a fixed volume. This is the Helmholtz

Free Energy, F, first proposed by Helmholtz (1, 2) using different arguments. If we wish to work

at constant Pressure we need to allow for work done at constant Pressure and can write:

69

H – H not convertible to work = H convertible to work (10)

This is the Gibbs Free Energy, G, first proposed by Gibbs (1, 2) using different reasoning. It is

the key quantity we require to describe Equilibrium at constant Pressure. While either side of

equation 10 gives the value of the G we can only calculate it from the expression on the left hand

side of the equation. Values of G are tabulated for most common substances in Data Books

used in A Level Courses (eg 3). We will show how values of G are related to both measured

values of Heat, Pressure, Volume and Temperature and molecular properties in a later article. In

this article we only discuss the key property of G that we need to know to use it to determine

Equilibrium. As we will see you do not need to know actual values of G to use the idea to

explain equilibrium.

To help clarify what is involved let us use the definition of G in equation 8 to write an equation

for the difference of G for an Ideal Gas between two Pressures at constant Temperature using

equation 8 for the Work

 G1 –G0 = RT log (P1/P0) (11)

When P0 is set at one atmosphere the difference of the G values is the amount of work that can

be done reversibly at constant Temperature in going from Pressure P 1 to one atmosphere. The

discussion in Section 3 makes clear why the Work term is a maximum and G consequently a

minimum What governs how much energy is made available for conversion is the energy

changes that lead to the Most Probable Distribution as the molecules redistribute themselves in

going from one equilibrium MBD to another. They occur in the system and so by definition are

not available for work. Simple examples of this are the heat of vaporization of water as it is

converted into steam and the heat of reaction for endothermic reactions; in both cases the energy

70

involved is used within the System and so is not available to do Work externally.

We finish this discussion by using G to give short definitions of Thermodynamic Equilibrium

and Kinetic controlled Equilibrium that includes all the conditions given in the long definition of

Equilibrium above:

 A System is at Equilibrium when the Gibbs Free Energy, G, has a minimum value, Mechanisms

are available to the System for all energy changes and the molecular energies have a MBD.

This definition applies to Thermodynamic Equilibrium. The Mechanisms referred to include all

conditions involved in the Energy changes in going from one State of the System to another. For

example, in the Reaction of Hydrogen and Oxygen to form water, it means that the energy states

needed for the reaction to occur are available.

We can modify this definition to include Kinetic controlled Equilibrium as follows:

A System is at Equilibrium when the Gibbs Free Energy, G, has a minimum value, Mechanisms

are not available to the System for all energy changes and the molecular energies have a MBD at

conditions different to that of the System.

We can extend this further to include Non Equilibrium Steady States as follows:

A System is at Equilibrium when the Gibbs Free Energy, G, has a minimum value, Mechanisms

are not available to the System for all energy changes and the molecular energies have a MBD at

conditions of the System.

At any Equilibrium condition at constant Temperature the molecules distribute themselves in a

MBD among accessible energy levels. This in itself will ensure there are well defined Properties

such as Energy and Enthalpy at those conditions. This is the justification for treating Systems at

71

Kinetic controlled Equilibrium as if they were at Thermodynamic Equilibrium as is routinely

done in practice. It explains how we can treat Equilibrium between Hydrogen and Oxygen at

100 C as an Equilibrium System even though the mixture of gases is not at Thermodynamic

Equilibrium.

Section 5 Th e Gibbs Free Energy and Vapour Liquid Equilibrium

 We now discuss an example of Vapour Liquid Equilibrium in terms of G. This example

illustrates how Equilibrium is a balance between maximizing the Energy and the most probable

state as required by a MBD. This case shows how the requirements of the MBD always produce

less than maximum energy of interaction and will move a system to a lower energy of interaction

in order to get to produce the Most Probable State.

We show in Figure 2 the equilibrium of a typical substance (e g water) at different Temperatures

and Pressures. Each form of water has its own Gibbs Free Energy, denoted by G g , G l and G s for

72

gas liquid and solid respectively, each with a MBD distribution and the overall distribution is

also a MBD. The water distributes itself between solid liquid and gas so that overall G is a

minimum. In each region where one phase is indicated that phase has the lowest Free Energy.

The Free energies, Gl and Gg, have to be equal on the boundaries where the phases meet;

otherwise the Free Energy of the total System would be reduced by increasing the amount in the

phase with the lower G. So on the Vapour Pressure boundary the Free Energies of the liquid and

the gas are equal. Similarly on the boundary between the solid and the gas the Free Energies of the

solid and the gas will be equal.

Section 6 C onclusion  

The key property in describing Equilibrium is the Gibbs Free Energy. It enables us to explain

why we get solids liquids and gases and how we get equilibrium between them. We have

introduced the key attribute of the Gibbs Free Energy, namely that it is a minimum at

Equilibrium, as a result of the conditions in our definition of Equilibrium that the maximum

amount of reversible Work is done in reaching Equilibrium and the MBD. The discussion of the

molecular cause of this, because the energy levels in the MBD that produce Pressure are always

available, make clear the physical reason why this occurs. Systems cannot be stopped from doing

Work because the energy levels involved in producing Pressure are always accessible.

In non reversible changes the System will do as much Work as it can and at the final Equilibrium

conditions, whatever they are, will have the maximum amount of Enthalpy convertible to Work

if we returned the System to 0 K and the crystalline state by a reversible path. In the next two

articles we show how the Gibbs Free Energies explain why exothermic and endothermic

reactions occur and why some go to completion and some do not. It remains to provide ways of

73

calculating values of G from experimental data and that topic is taken up in the last article of this

series.

References

1) E B Smith, “Basic Chemical Thermodynamics”, Imperial College Press, 5th edition, 2004,

Chapter 4

2) K S Pitzer and L Brewer, “Thermodynamics”, McGraw Hill, 1961, Chapter 13

3) G Attwood and G Skipwood, “Chemistry Data Book, John Murray, London, 1992

List of Figures

Figure 1a Work Done at Constant Pressure.

Figure 1b Reversible Work Done by an Ideal Gas as the Pressure changes from P1 to P0 at

constant Temperature.

Figure 2 Vapour Liquid Equilibrium in terms of the Gibbs Free Energy

Exothermic Chemical Reactions

74

R M Gibbons

BSc PhD FRSC

Proofs to: R M Gibbons

BSc PhD FRCS

4 Little Acre

Beckenham

Kent BR3 3ST

Abstract

We show that the differences of the Gibbs Free Energies are the Driving Forces producing

Equilibrium in Chemical Reactions for Ideal Gases and lead to expressions for the Equilibrium

Constant. We use the standard Activation Energy model for reactions to discuss when reactions

will occur and whether we will obtain a Thermodynamic Equilibrium or a Kinetic controlled

Equilibrium.

The reactions of Methane and Hydrogen with Oxygen illustrate the application of this treatment

as examples of exothermic reactions. The heat of reaction between Oxygen and Hydrogen is

the heat of formation for water while Oxygen reacting with Methane produces the heat of

combustion. We discuss how the Equilibrium for Ideal Gas reactions respond to changes of the

Equilibrium conditions in terms of the Le Chatelier’s Principle.

 

 

Section 1 Introduction

75

We start with some definitions of terms used in this article. In the discussion below we use the

term System to describe a system of molecules with properties per mole such as Energy and

conditions such as Temperature and Pressure. The System and all the properties are referred to

using capitalised words so Temperature and Energy refers to the temperature of the System and

its Energy respectively.

We discuss how Chemical Reactions for Ideal Gases reach Equilibrium largely following the

approach of Pitzer and Brewer (1). We use the Standard Activation Energy Model for the

Mechanism of Chemical Reactions (2) and discuss it briefly in Section 2. In section 3 we show

that the difference of the combined Gibbs Free Energies of the Reactants and the Products is the

Driving Force producing the Equilibrium for Reactions of Ideal Gases. When this Driving

Forces becomes zero the Reaction is at Equilibrium and this leads to the familiar expression for

the Equilibrium Constant in terms of the partial Pressures. We apply this approach to

Exothermic Reactions between Oxygen and the combustible gases, Methane and Hydrogen. We

introduce Le Chatelier’s Principle (3) and use it to show how the position of Equilibrium

changes with alterations to the Equilibrium conditions.

Section 2 The Activation Energy Model

Students will be familiar with the Activation Energy Model for Chemical Reactions from

earlier courses but we summarize it briefly here. We show it graphically in Figure 1 where

Reactants at one Energy level form an intermediate complex at a higher Energy level and the

Products, for exothermic reactions, at a lower energy level.  The intermediate complex is said to

be formed by “sufficiently energetic” kinetic collisions between reacting molecules which

overcome the repulsive forces enabling transfer or exchange of electrons to occur. Generally

76

these collisions involve the small number of the most energetic molecules in the MBD and, as

the number of these increases with Temperature, the rate of reaction increases with the

Temperature. The Activated Energy arises from the mutual repulsion of the electrons involved

in the Chemical Reaction. There are well known problems with this model but it is widely

accepted as a useful introduction to reaction mechanisms. The model is basically sound and I

hope to discuss some of the problems and suggest improvements to it in a later article.

For a reaction to occur it is not enough for a “sufficiently energetic” collision to occur, there

must also enough of these collisions for the reaction to continue. The number of collisions of a

given energy increases with the Temperature so there will be a minimum Temperature below

which the reaction will not occur. For combustible gases this is the Ignition Temperature. It is

clear that at lower Temperatures gases that could react, such as mixtures of Oxygen and

Hydrogen, will not react. In terms of our treatment of Equilibrium there is no Mechanism to

produce Equilibrium and the System is at a Kinetically controlled Equilibrium. The Ignition

Temperature has to be obtained experimentally for each reaction. The effect of a catalyst on a

reaction is schematically by the dotted line in Figure 1. Catalysts work by providing an

alternative reactive complex with a lower Activation Energy enabling more molecules to react

effectively at given conditions and so increasing the rate of reaction.

Section 3 Equilibrium in Chemical Reactions

We want to use our knowledge that the Gibbs Free Energy, G, is a minimum at Equilibrium to

describe Chemical Reactions. For Exothermic Reactions the value of G for the Products is

generally less than G for the Reactants so we might expect these reactions to go to completion to

minimize the G of the combined System of Products and Reactants. Usually the System comes to

77

an Equilibrium State involving both Reactants and Products. We explain why this occurs

below.

To apply G to Chemical Reactions we need to be able to obtain values of G for mixtures. There

ways of doing this generally but they involve more complicated Mathematics to deal with the

affects of mutual interactions between the components of the mixture. To avoid these problems

we restrict ourselves here to Ideal Gases. Most gases at Room Temperature and Atmospheric

Pressure are nearly Ideal so this simple model represents the behavior of real gases very well at

these conditions. The approach presented below uses concepts very closely related to those that

describe real systems. For mixtures of Ideal Gases however there are no mutual interactions so

we can write the G,, for such mixtures as the sum of the G values for the components. It is for

that reason we can write for a binary Ideal Gas mixture such as Oxygen and Hydrogen:

Gm = GO2 + GH2 (1)

 We use the forms for G derived in the previous article for Hydrogen and Oxygen:

G H2 = G0H2 + RT log (PH2) (2)

GO2 = G0O2 + RT log (PO2) (3)

Hydrogen and Oxygen can react to form water following the well known equation:

2H2 + O2 = 2H2O (4)

There is a similar equation for G of water:

GH2O = G0H2O +RT log (PH2O) (5)

78

 Whether the reaction will take place or not depends on the conditions. At 100C and a Pressure

of one Atmosphere Hydrogen and Oxygen can be mixed in a container and will not react.

However introduce a Platinum catalyst and the reaction proceeds readily (4). Alternatively

apply a match, from a safe distance with the mixture in a high pressure container, and the

reaction will proceed explosively.

In both cases we have provided a Mechanism with sufficient number of molecules able to react.

The catalyst allows the Reaction to proceed by reducing the Activation Energy so that enough

molecular collisions occur with sufficient energy for the reaction to proceed. The match

provides an initial volume of high Temperature in the mixture for high energy reactive collisions

to occur that release heat which warms up adjacent local volumes and continues the reaction

throughout the entire volume.

We can discuss the reaction in terms of Free Energies as shown in Figure 2. The Free Energy

of Reactant Mixture is shown schematically as the composition varies.

79

As GH2O is much lower this reaction will occur if there is a Mechanism to allow it to happen as we

have seen above. This difference of the combined values of G is the Driving Force of the

reaction and reduces to zero as the reaction proceeds. We can plot the variation of G for the

mixture as the reaction proceeds as the pressure of each of Hydrogen, Oxygen and water (as

steam) changes. A schematic plot of G is show in Figure 2 showing a minimum in the G value at

equilibrium where

G Products = G Reactants (6)

Substituting from equations 2, 3 and 5 for the G values in equation 6 leads to:

2(G 0H2O + RT log(PH2O)) = G0

H2 + RT log(PH2) +G0O2 + RT log(PO2) (7)

Gathering all the G0 values on one side and all the log terms of the other we have

-2G0H2O + G0

H2 + G 0O2 = 2 R T log (PH2O) – RT log (PH2) - RT log (PO2 ) (8)

Recalling negative logs are equivalent to dividing, we can rewrite this as:

-∆G = RT log (P2H2O/ (PH2PO2)) (9)

 

K = P2H2O/ (PH2PO2) (10)

Relating the Equilibrium Constant to ∆G is beyond the scope of the A level course but is

discussed briefly in Appendix 1 together with some properties of logs which it is useful to recall

for students as the properties of logs have much less prominence in Mathematical curricula than

previously.

80

  The other reaction we wish to discuss as an example of Exothermic Reactions is the

combustion of Methane. This follows the well known equation:

CH4 + 2 O 2 = CO 2 + 2H2O (11)

We can again equate the Gibb Free Energy values at equilibrium and collect the log terms as

before to arrive at an expression for the Equilibrium Constant:

K = (PCO2P2H2O) / (PCH4PO2

2) (12)

The treatment given above explains Equilibrium Constants which are usually introduced with no

explanation as to how they arise. A common extension of questions on Equilibrium Constants is

to ask how the Equilibrium will respond to changes in conditions and this topic is discussed in

the next section.

Section 4 Le Chatelier’s Principle

The discussion of the affect of changes in the conditions on the position of Equilibrium in a

Chemical Reaction is best done in terms of Le Chatelier’s Principle.

This can be stated as:

When the conditions for a System at Equilibrium change, the System reacts to resist the change.

To see how this works let us apply it to the two reactions discussed in the previous section for

changes of Temperature and Pressure. Both reactions are exothermic so the formation of more of

the products would release more heat which warms up the system. Conversely the reverse

reaction, forming less of the products is endothermic, which cools the system. So if we increase

the Temperature the system will react by cooling the system by moving in the endothermic

81

direction forming more of the Reactants. Reducing the Temperature causes the system to

respond by releasing heat by making more of the Products. The most favourable conditions for

making the Products will then be the lowest Temperature at which the reaction will occur at an

acceptable rate. The rate of course increases with Temperature because the number of

sufficiently energetic molecules increases with Temperature.

The responses to changes in Pressure are equally easy to predict once it is realized that for Ideal

Gases the Pressure increases with the number of molecules present irrespective of type. If we

increase the Pressure, the system reacts to decrease the Pressure by reducing the number of

molecules present so it will move the Equilibrium towards the side of the equation with the

fewest molecules and vice versa. If there are equal numbers of molecules of Reactants and

Products the Equilibrium will not change with the Pressure. For the reaction to form water, three

molecules of Hydrogen and Oxygen form two molecules of water. For the combustion of

Methane three molecules of reactants form three molecules of Products. So in the first case

increasing the Pressures causes the system to react by decreasing the number of molecules and

hence the Pressure. This favours increasing the amount of Products and moves Equilibrium to

the right. For the methane reaction there are equal numbers of molecules on both sides of the

equation so the Equilibrium Constant is independent of Pressure.

Section 5 Discussion

The discussion in the previous section suggests that all reactions come to equilibrium whereas

we know that many reactions go to completion so it is worth discussing how a reaction can

terminate. With large differences of G between Products and Reactants in a reaction, equilibrium

will so favour the Products that negligible amounts of Reactants will be left. Alternatively the

82

partial pressures of the reactants will be so small at Equilibrium that it is difficult to detect their

concentrations. Finally the concentrations of the Reactants may fall so low that the high

Temperature needed to sustain a reaction can no longer be maintained and the Equilibrium

becomes frozen with very small amounts of Reactants remaining.

The above discussion is restricted to Ideal Gas reactions and students may conjecture how this

treatment applies to non- ideal gases, solids and liquids. The answer is the basic approach

applies but with the Gibbs Free Energies replaced by equivalent properties that do take account

of the mutual interactions of the molecules in mixtures. Discussion of these requires more

advanced mathematics and so are not included here.

The discussion above suggests that Reactions in which the Products have higher values of G

than the Reactants will not occur. It is true they will not normally occur. However, because of

the dependence of G for each substance on its partial Pressure, it is possible for a combination of

Products and Reactants to have a lower value of G than either the Products or the Reactants. In

such cases a Reaction will occur if a Mechanism is available for the Reaction to occur. These

Reactions are Endothermic in which heat is absorbed by the Reaction and will be discussed in

another article.

Appendix 1 Properties of Logs and the Equilibrium Constant K

Logs occur in many expressions describing physical systems. We give a summary here of key

points about logs and use them to show how ∆G and K are related.

A log of a number is the power to which a base is raised to give the number. Any number may

be taken as the base but the two most commonly used are ten and the exponential number, e,

2.7138. Logs involving e occur in the Ideal Gas expressions for G. The same arguments apply

83

whether e or ten is used as the base and the explanations given below are in terms of base ten as

they are easier to follow.

The basic equation is:

y = 10x (1.1)

x is the log of y

log(y) = x (1.2)

Now raise the base 10 to the power on each side of the equation and we have:

10log(y) = y = 10x (1.3)

Equation 1.3 often causes students difficulty and a simple example clarifies the steps involved.

The log of 100 is 2: 100 =102 (1.4)

Log (100) = 2. (1.5)

Raise 10 to the powers on both sides of equation 1.5 and:

10 log (100) = 102 = 100 (1.6)

Raising a base to a power log(x) produces x. We can use equation 1.3 to relate K to ∆G.

Recalling from equations 9 and 10 in the text

-∆G = RT log (K) (1.7)

We can rearrange this as:

Log (K) = -∆G/RT (1.8)

84

, raising both sides to the base e and we have the relationship for K and ∆G:

K = e (-∆G/ k T) / (1.9)

References:

1. K S Pitzer and L Brewer, “Thermodynamics”, McGraw- Hill, London, 1961 chapter 15.

2. E N Ramsden, “A Level Chemistry, Stanley Thornes Ltd, 1994, p313-317

3. E N Ramsden, ibid, p231-3

4. K S Pitzer and L Brewer, ibid, p142

List of Figures

Figure 1 Activation Energy Model for Reactions

Figure 2 Free Energy Diagram for the Reaction of Hydrogen and Oxygen

Line A is the Free Energy of a mole of mixtures of Hydrogen and Oxygen

Line B is the Free Energy of a mole of a mixture of Hydrogen Oxygen and Water

85

Endothermic Chemical Reactions

R M Gibbons

BSc PhD FRSC

 

Proofs to: R M Gibbons

4 Little Acre

Beckenham

Kent BR3 3ST

Abstract

Endothermic reactions are a problem for standard A level textbooks; they are mentioned but

never explained. Endothermic reactions absorb heat. We show how endothermic reactions can

occur in terms of the Free Energies of the Products and Reactants. We consider the conditions

that are needed before they can occur and use two important examples of endothermic reactions

for Oxygen with Nitrogen and Carbon and Hydrogen to demonstrate key points about

Endothermic Reactions. We discuss the response of endothermic reactions to changes in

conditions of the reactions in terms of Le Chatelier’s Principle in the same way as for

exothermic reactions.

 

 

 

 

86

Section1. Introduction

We start with some definitions of terms used in this article. In the discussion below we use the

term System to describe a system of molecules with properties per mole such as Energy and

conditions such as Temperature and Pressure. The System and all the properties are referred to

using capitalised words so Temperature and Energy refers to the temperature of the System and

its energy respectively.

A level texts mention endothermic reactions but do not attempt to explain them. Endothermic

reactions play a very important role in the chemical systems that form our environment and our

fuel supplies. The atmosphere is 99% a mixture of Nitrogen and Oxygen that can react

endothermically to form Nitric Oxide:

N2 + O2 = 2NO (1)

Fortunately for us this reaction does not occur at Room Temperature. Yet every day this reaction

produces millions of tons of Nitric Oxide in two major ways all over the world. As to our fuel

supplies the key reaction in the formation of hydrocarbons between Hydrogen and Carbon is

endothermic:

C + 2H2 = CH4 (2)

This reaction does not occur readily and has to be done indirectly; we discuss this further in

Section 2 in terms of the gasification of coal.

More generally the C-H bond, which is found in all hydrocarbons, is Endothermic, but to

simplify things we will discuss this in terms of the C-H bond in Methane. The endothermicity

87

of the C-H bond is the key to making hydrocarbons such good fuels. It is also the key in

enabling the energy of the sun millions of years ago to be locked into fuels for our use today. In

the section 2 we show how endothermic reactions can occur even though they involve energy

transfers from a lower Free Energy at the start to an apparently higher Free Energy level which

would not normally occur.

In Section 3 we discuss the change in Equilibrium for endothermic reactions to changes in the

conditions at Equilibrium in terms of Le Chatelier’s Principle. In Section 4 we introduce the

remaining task in this series of articles, of how to obtain values of G from experimental data and

explain why we have deferred this topic to the end. The first five articles are essentially non

Mathematical and accessible to any A Level Chemistry Student. The calculation of Free

Energies does involve A Level Mathematical techniques and using standard methods would be

thought too difficult for A level Students. We indicate why we think the simplified methods we

have developed should be suitable for A Level Students doing A Level Mathematics and discuss

them in full in the next article

 Section 2 Free Energies and Endothermic Reactions

Endothermic reactions (1) can be discussed in exactly the same way as exothermic reactions

using Free Energy Diagrams as shown in Figure 1.

88

 Generally endothermic reactions involve transferring energy between molecules by putting

electrons into to the Products in such a way that energy is locked into the bond energies of the

Product molecules. The Products formed by endothermic reactions always have exothermic

reactions with Oxygen so most of them are them good fuels.

We use the heat of formation of Nitric Oxide by the reaction of Nitrogen and Oxygen as an

example. The Reactants, Oxygen and Nitrogen, have a lower Free Energy than the Product,

Nitric Oxide, so normally one might expect that no reaction would occur. However if we plot

the Free Energy of the combined Reactants and Products as the reaction converts some of the

Oxygen and Nitrogen to Nitric Oxide we obtain the sort of plot shown by the line B in Figure 1,

where there is a minimum in the curve. At the minimum the Free Energies of Products and

Reactants are equal and we obtain the usual expression for the Equilibrium Constant, K in terms

of the partial Pressures:

K = PNO2/ (PN2PO2) (3)

89

As the Free Energy for the system will go to minimum when it can, this minimum represents the

equilibrium between Oxygen, Nitrogen and Nitric Oxide. Mathematically it occurs because the G

value for each substance depends on the log of the partial Pressure for each substance which

varies with the amount of each substance in the mixture. This explains how we are able to form

Nitric Oxide even though it is energetically unfavourable to do so.

Let us now look at the conditions needed for the Reaction to occur. Nitrogen molecules have

strong triple bonds and Oxygen molecules have strong double bonds. These bonds must be

broken for the reaction to occur. In terms of our Activation Energy Collision Model this will

require high Temperatures to overcome the Bond Energies and this energy must be put into the

reaction by an external source. This is indeed shown by the two common situations on Earth

where this reaction occurs daily.

   This reaction occurs in car engines at ca 800C; car exhausts contain ca 0.3% NO (2) and

occurs as a by-product from the combustion of the fuel to propel the car. It occurs in copious

quantities in lightning strikes, which generate Temperatures of about 5000C in the upper

atmosphere. The Nitric Oxide formed in this way, after reacting with Oxygen in the atmosphere

is eventually responsible for about 10% of all nitrogenous fertilizers (3) used on agricultural

land.

Note the contrast with exothermic reactions where an external trigger may be required, as in the

ignition of Methane or Hydrogen, but after that the reaction is self propagating. For

endothermic reactions that require high Temperatures, the energy for the reaction to occur must

be supplied to keep the reaction going. After the supply of external energy is stopped, the

endothermic heat of reaction cools the System and so causes the reaction to stop. This is not

90

true for all endothermic reactions as many occur spontaneously at ambient temperatures and in

the process cause instant cooling of the System as the reaction draws the energy needed from the

System itself.

The other question we must address is where did the Energy go? In one sense it is easy to

explain it is locked in the Bond Energies of the Nitric Oxide. In Energy terms the change in

Enthalpy has gone into the bonds and is not available to do work. We need a new property to

describe this effect and take this up in the next article.

The other reaction we will discuss is that of Carbon and Hydrogen. This is strongly

endothermic. Though it is the key reaction in nature in the formation of all oil and gas for

simplicity we will discuss it here in terms of coal gasification for the production of synthetic

natural gas.

Coal can be gasified in a gasifier, at 1000C and a Pressure of 70 bars. Some of the coal reacts

with Oxygen to provide the heat to generate the 1000C Temperature needed for the reaction to

occur and to produce Carbon Monoxide:

2C + O2 = 2CO (4)

 The Carbon Monoxide then reacts with injected steam to form Hydrogen:

CO + H2O = CO2 + H2 (5)

 The rest of the carbon in the coal reacts with steam to form Hydrogen:

C + H2O = CO + H2 (6)

91

The final step is to convert the Carbon Monoxide and Hydrogen to Methane and other

hydrocarbons using a catalyst:

CO + 2H2 = CO2 + CH4 (7)

Despite this complexity, approximately 95% of the energy in the coal appears in the product gas.

This occurs because of the endothermicity of the C-H bond and is a much higher level of

conversion of energy than is found in other processes involving chemical reactions. The overall

efficiency of the gasification process is much lower after taking account of the purification

processes that must be included to deal with the byproducts that occur because of the materials

other than carbon that are in coal.

Section 3 Endothermic Reactions and Le Chatelier’s Principle.

We now look at how our two Endothermic Reactions respond to changes in conditions for the

reactions in terms of le Chateliers Principle (4) in the same way as we did previously for

Exothermic Reactions. According to his Principle the system opposes the change. So if we

increase the Temperature the system tries to reduce the Temperature by moving in the

endothermic direction, producing more of the Endothermic Product in both our sample cases;

Equilibrium moves to the right of the equation.

  If we increase the Pressure the system responds by trying to decrease the Pressure moving the

Equilibrium in favour of the side of the equation with fewer molecules in the gas phase. The

reaction of Nitrogen and Oxygen has equal numbers of molecules on both sides of the equation

so the Equilibrium is not affected by changes in Pressure. For the reaction of Carbon Monoxide

and Hydrogen in equation 7 there are three moles on the left hand side of the equation and two

on the right 2. Increasing the Pressure then will move the equilibrium in favour of producing

92

more Methane as the system responds to try and reduce the Pressure. The optimum conditions

for the gasification reaction will be the highest Temperature and Pressure that is practicable.

These conditions will be decided on grounds of cost and safety rather the behaviour of the

reaction itself.

Section 4. Discussion

In this series of articles we have set out the basic ideas needed to describe Equilibrium at a level

suitable for A Level Students. The simple treatment of the Gibbs Free Energy, G, in an earlier

article provided the key to explanations of Vapour Liquid Equilibria and for chemical reactions.

While we gave a physical interpretation of G in terms of a System’s ability to do work it is

clearly necessary to provide a way of calculating G from experimental measurements. We have

delayed this task until now because, to obtain expressions for G, requires a much more

mathematical treatment than has been necessary previously in this series and the introduction of

a new concept, the Entropy. In fact the theories needed for this purpose are generally thought to

be too difficult for A Level courses. And using the traditional treatments of these topics they are

too difficult. They all involve Mathematical methods too advanced to use in an introduction to

these topics.

   The treatments we use in the rest of this series are suitable for A level Students. All the

mathematical techniques we use are in the Mathematics A Level syllabuses, mostly in the first

modules for Pure Mathematics and Statistics. Several of them are worked examples in

Textbooks that we then apply to a mole of molecules. In several instances we use novel results

obtained using standard A level Mathematical techniques to provide simple ways of obtaining

the desired results in a much simpler way than other approaches. Discussion of these methods

93

and the relation of the Free Energy to Entropy are taken up in the next article.

References:

1 E N Ramsden, “A Level Chemistry”, Stanley Thornes Ltd, 1994, p313-317

2. E N Ramsden, ibid, p231-3

3. G Burden J Holman G Pilling D Waddington, Salters Advanced Chemistry “Chemical

Storylines”, Heinemann, 1994

4. E N Ramsden, ibid, p142

List of Figures

Figure 1 Schematic Diagram of the Free Energy for the Reaction of Nitrogen and Oxygen

Line A is the Free Energy of a mole of a mixture of Nitrogen and Oxygen

Line B is the Free Energy for a mole of mixtures of Nitrogen Oxygen and Nitric Oxide.

94

Section 3 Equilibrium for A Level Chemistry and Mathematics Students

The articles in Sections 1 and 2 involve little more than GCSE mathematics. The articles in this Section

involve some A Level Mathematics from the First Module of Statistics and some Calculus. We give full

details of the Mathematical Methods in appendices.

The arguments involved in the Entropy based on the MBD would normally be considered too advanced

for A Level Students using standard proofs. The simplified methods used here utilise standard A Level

Mathematical methods for all steps. There are four key stages in relating the Boltzmann Equation to an

expression for Entropy. The first, the number of permutations of molecules in the MBD, is a standard

theorem in the First Module of Statistics courses; a proof involving three equations is given for those not

doing A Level Statistics. The second is the use of Stirling’s Equation to convert expressions involving

log (N!) and log (n i !) to expressions involving the probabilities, p i , we give a derivation of Stirling’s

Equation using an integration by parts. The third transformation uses the relationship of exponential and

log functions and leads to the same summations as were encountered in obtaining the Energy as the

average of the energies of the molecules in a mole which leads to the standard expression relating the

change in Entropy to the change in Enthalpy divided by the Temperature.

The final article uses the Third Law and differentiation of the expression for the MBD to obtain

relationships between the absolute values of Enthalpy and Enthalpy and the measured Heat Capacities at

constant Pressure.

10 Equilibrium and Entropy

11 Free Energies from Experimental Data

12 The Stirling Equation for Chemists

95

Equilibrium and Entropy

R M Gibbons

BSc PhD FRCS

Proofs to: R M Gibbons

4 Little Acre

Beckenham

Kent BR3 3ST

Abstract

We start from our definition of Equilibrium and the expression for the Free Energy in terms of the

difference between Energy not convertible to work and Energy convertible to work, which we introduced

previously. The requirement that the System is at its the Most Probable Distribution at Equilibrium leads

to the Boltzmann equation and both introduces the Entropy and relates it to the Energy not convertible to

Work and the Energy convertible to Work. We show that the same expression for obtaining the Energy

of the System occurs in the definition of Entropy changes and that, combined with the expression for the

Work leads to a definition of the change of Entropy that occurs between two Equilibrium States.

We go on to show how differences of the Entropy relate to experimental measurements for changes at

constant Pressure and for Ideal Gases. We discuss the problems of obtaining absolute values of Entropy

and Enthalpy and Free Energy. This involves the Third Law of Thermodynamics which we introduce and

define using our expression for Entropy in terms of molecular probabilities.

96

Section 1 Introduction

We start with some definitions of terms used in this article. In the discussion below we use the term

System to describe a system of molecules with properties per mole such as Energy and conditions such as

Temperature and Pressure. The System and all the properties are referred to using capitalised words so

Temperature and Energy refers to the temperature of the System and its Energy respectively.

Next we give our definition of Equilibrium (1):

A System is at Equilibrium when it has constant values of Pressure, Temperature, Density and

composition for a long period of time, the molecules have a Maxwell Boltzmann Distribution (MBD) of

energies and Mechanisms are available for all energy changes to occur both at the conditions of the

System and at all points on a (reversible) path between the initial state of the System and its final

Equilibrium condition.

The behaviour of Systems of molecules at 0 K is the key to understanding how Systems behave at higher

Temperatures. The Interactive Energy of a System is at a maximum, with every molecule at its optimum

position in the crystalline state at 0 K. At all other Temperatures the Kinetic Energy of the molecules

ensures that the Energy of the System is a minimum subject to the amount of Kinetic Energy that the

molecules have at the Temperature of the System. It is the Kinetic Energy of the molecules that ensures

the maximum Interactive Energy is not obtained at Equilibrium at any other Temperature because it

makes it impossible for molecules to stay in the positions that maximize their interactions with other

molecules.

Our basic equation is an energy balance dividing the Energy of the System between Kinetic Energy that

can do Work and interactive energies between the molecules which are not convertible to Work. For any

System

Energy = Energy Convertible to Work + Energy not Convertible to Work (1)

97

There are two generally applicable expressions for Work Done in changing for one Equilibrium condition

to another. For changes at constant Pressure we have:

Work Done =P∆V (2)

For an Ideal Gas for change at constant Temperature between P1 and P0

Work Done = RT log (P1 / P0) (3)

For the Energy we showed in reference 1 that it is the most probable energy. It is the sum of all the

energies of the molecules in the System where ui is the energy of the ith molecule

U = N (∑ u i p i ) (4)

and pi is the MBD probability for the ith molecule has an energy u i

pi = n i /N = e (-u i/ k T) / Q (5)

Q = ∑ e (-u i/k T) (6)

 

Using Equation 1, these expressions for Work, and the expression for the Most Probable Distribution at

Equilibrium (MBD) we can relate the change in Energy not available to do Work to measured physical

values. In the process we introduce and define the Entropy and show how it is related to the Energy not

convertible to Work.

We summarise the Mathematical methods we use in Section 2. They are all included in A Level Pure

Mathematics (2) and Statistics courses (3). We give a full account of three key topics in Appendices for

the benefit of Students who are not studying Statistics.

The relationships we develop in this article enable us to obtain expressions for changes of Energy,

Enthalpy, Entropy and the Free Energy of a System. We go on to discuss what is required to enable us to

98

obtain absolute values of these quantities and introduce the Third Law of Thermodynamics as part of this.

This is the key step in obtaining absolute values of the Free Energies for different substances on a

common scale. We defer to the next and final article of this series the remaining steps to complete this

task.

Section 2 Mathematical Methods

We need to use some Pure Mathematical Methods involving integration and differentiation and one key

Statistics result that forms part of a Module 1 Statistics course(3). The first method involves the

relationship of natural logs and the exponential function:

Log (ex) = x (7)

We use this with the Boltzmann factor:

Log (e (-u i/k T)) = -u i / k T, (8)

The Statistical expressions we need all involve the factorial function N! The key result we require is for

the number of ways of arranging a set of N distinguishable objects of which ni are identical and i varies

from 1 to n. This is a standard problem in the first module of a Statistics Course (3) and we quote the

equation in Section 3 and give a proof of the result in Appendix 1. We then need to use this result to

obtain an expression for the probabilities, p i, which we also show in Section 3, and give the derivation in

Appendix 2.

The third technique we need is the Stirling Equation for log (N!):

Log (N!) = N log (N) –N (9)

We have published elsewhere (4) a method of proving this relationship using integration by parts and give

a brief account of this in Appendix 2.

99

Section 3 The Most Probable Distribution at Equilibrium and Entropy

We can use our definition of Equilibrium to obtain an expression for the Most Probable Distribution (5).

This is the one that occurs in more distinguishable ways than any other. So our problem is to find the

number of permutations that can occur in the MBD. We give a complete account in Appendix 2 of a

simplified version of an explanation from Margenau and Murphy (6) and we just quote the result here:

Most Probable Distribution, W = N! / (n1|! n2!……..) (10)

Boltzmann (7) found the way of relating this equation to other properties by writing:

S/k = log (W) (11)

It is clear that, as the values of ni change from one Equilibrium condition to another, that S will vary too.

So S is related to the Energy that stays within the System as it too varies with changes in values of n i.

This becomes much clearer when we use Stirling’s expression for log (N!), defined in equation 10, in

equation 11 and introduce the probabilities pi defined in equations 5 and 6.

We give the details in Appendix 2 and arrive at the second form of the Boltzmann equation:

S/ R = - ∑pi log (pi) (12)

The connection with the Energy becomes much clearer when we use, in equation 13, the relationship

between log and exponential functions: in equation 8 to arrive at a third form for Boltzmann’s equation:

S/R = ∑- u i p i//k T - log (Q) pi (13)

Carrying out the summations leads to the result, noting the probabilities sum to unity and the first

summation is the same as that in equation 5:

S/R =U/ RT – log (Q) (14)

We rearrange this into a form directly comparable with equation 1.

100

U= TS + RT log (Q) (15)

We can identify the corresponding terms in the two equations and see that TS is the energy not

convertible to work, U the Energy and –RT log (Q) is the energy convertible to work at constant volume.

This is the Helmholtz Free Energy, which Helmholtz discovered using very different reasoning. For

changes at constant Pressure we must add the work term for changes at constant Pressure and get the

Gibbs Free Energy, which Gibbs invented using other arguments.

U+PV –TS = H-TS = G (16)

We can see that TS is the Enthalpy not convertible to work at an equilibrium condition and -G is the

Energy convertible to Work at constant Pressure. When we obtain a value for a change of Entropy in

going from one equilibrium state to another, the change will follow the second law of thermodynamics

and some Energy will convert into work. The only universal expression we have for work in a change is

equation 2 at constant Pressure such as occurs in vapour liquid equilibrium. Since the G values are equal

we can write

Hl –T S l = Hg – TS g (17)

Rearranging,

∆S = ∆H/T (18)

This is the equation used in most textbooks to introduce the Entropy without any indications of its

connection with molecular behaviour.

For Ideal Gases at a given Temperature H is independent of Pressure so for reversible changes between

P1 and P0 the Work Done is equal to the change in Entropy times the Temperature:

∆G = T∆S =RT log (P1/P0) (19)

101

It is usual to select P0 as one atmosphere and Data Books such as reference 8 gives tables of values for G

for a variety of substances at 1atm.

Section 4 Discussion

The relationships obtained in the previous Section enable us to calculate changes in the Enthalpy,

Entropy and Free Energy between two Equilibrium States in terms of quantities we can measure. It is

natural to ask is there way in which we can calculate absolute values for these quantities for different

substances on a common basis ? This would allow us to compare the Free Energies of, for example of

Oxygen and Nitrogen at the same conditions, such as are quoted in Data Books e g (8). This can be done

but to do so we require the Third Law of Thermodynamics. We defer consideration of this to another

article.

References

1 R M Gibbons, in press

2 G. Attwood and G. Skipworth, “ Pure Mathematics 2”, Heinemann, 1994, p277-313

3. G. Attwood and G. Skipworth, “Statistics 1”, Heinemann, 1994, p160

4. R M Gibbons, in press

5. E B Smith “Basic Chemical Thermodynamics “Imperial College Press, 1997, ch 9

6. H Margenau and G M Murphy “The Mathematics of Physics and Chemistry”, Van Nostrand,

New York, 2nd ed., 1961

7. E B Smith ibid, p141

8. G Attwood and G Skipwood “Chemistry Data Book, John Murray, London, 1992

9. E B Smith ibid, p 69

102

Appendix 1 The Most Probable Distribution for MBD

The problem is to obtain the number of permutations we can get from a set of n i molecules with energies

u i which sum to a total energy U. This is a worked example in many first Statistics textbooks (e g 2); the

explanation I give here follows that of Margenau and Murphy (6). We assume we can distinguish the

molecules. Then we have N ways of choosing the first molecule, N-1 for second and so on leading to:

Number of ways of choosing = N (N-1)(N-2)…… = N! (1.1)

For each of the set of n i there are n i! ways of arranging them that we cannot distinguish because they

have the same energy. Multiply all these indistinguishable ways of choosing the molecules together and

we have:

Total number of indistinguishable ways of choosing the set of n i molecules = П n i! (1.2)

We also have:

Distinguishable permutations x Indistinguishable permutations = Total permutations

Substituting our expressions above into this and rearranging leads to:

Distinguishable arrangements, W = N! / (П n i!) (1.3)

Appendix 2 Probabilities and Stirling’s Equation.

Taking logarithms on both sides of equation 11 gives

Log (W) = Log (N!) -∑log (n i!) (2.1)

Using Stirling’s equation, which we derive in Appendix 3, in equation 2.1 gives,

Log (W) = N Log (N) –N – (∑n i log (n i )–n i ) (2.2)

103

Doing the summation the sum over the n i terms cancels with N and noting a subtraction for a log is the

same as dividing we are left with

Log (W) = - ∑ n i/N Log (n i/N) (2.3)

The probability pi = n i/N (2.4)

So Log (W) = - ∑ pi log (pi) (2.5)

This is what we require for the second form of the Boltzmann equation.

Appendix 3 Derivation of Stirlings Equation

Stirlings equation is:

Log (N!) = N log (N) –N (3.1)

We can obtain this result easily starting from the definition

N! = N (N-1) (N-2) ….. = П i (N-i) (3.2)

Factorise this by taking N out of each bracket and writing

xi = i /N we have (3.3)

N! = NN П (1-xi) (3.4)

for values of xi from 0 to 1/N.

Taking the natural logs of this we have

Log (N!) = N log (N) + ∑ log (1 – xi) (3.5)

We can approximate the summation by an integral

Log (N!) = N log (N) + N ∫ log (1-x) d x (3.6)

104

The integral can be evaluated by an integration by parts following a change of the variable between the

limits of 1/N and 1 leading to:

Log (N!) = N log (N) - N + log (N) (3.7)

For large N, log (N) is negligible compared to N, and we have Stirling’s equation given above – his result

also contains a term1/2 log (N) that is ignored because for large values log (N) is negligible compared to

N.

105

Free Energies from Experimental Data

R M Gibbons

BSc PhD FRSC

Proofs to: R M Gibbons

4 Little |Acre

Beckenham

Kent BR 3 3ST

Abstract

The Gibbs Free Energies, G, are the key properties in defining and producing Equilibrium. To use values

of G we need to have their values for each substance on a common scale so we can compare their values

with one another. In this article we show how absolute values for G and other properties can be

obtained from experimental measurements. The key to doing this is the Third Law of Thermodynamics

and the crystalline state at 0 K.

We briefly discuss the Mathematical methods and the experimental measurements of heat capacity and

the heat of Sublimation at 0 k that enable us to obtain the absolute values of Energy Enthalpy and Entropy

that determine the values of G.

We conclude with a short summary of the aims and purpose of this series of articles.

106

Section 1 Introduction

We start with some definitions of terms used in this article. In the discussion below we use the

term System to describe a system of molecules with properties per mole such as Energy and

conditions such as Temperature and Pressure. The System and all the properties are referred to

using capitalised words so Temperature and Energy refers to the temperature of the System and

its energy respectively.

The Free Gibbs Energy values determine Equilibrium. They cannot be measured directly and we must

calculate them from measured values of properties we can measure. This means heat capacities and heats

of Sublimation at 0 K from which we can calculate Energies, Enthalpies and Entropies which in turn

allow us to obtain values of G. We start in Section 2 by setting out the relationships between the heat

capacities and the Energies, Enthalpies and Entropies.

For their values to be useful they must be measured on a common scale and the key to that is the Third

Law of Thermodynamics which we discuss in Section 3. We discuss briefly in section 4 the aims and

purposes of this series of articles. In conclusion I set out why I thought it worthwhile to develop this

approach.

Section 2 Absolute Values of U and H from Experimental Data

The experimental properties that we can measure for changes in Energy or Enthalpy are the heat

capacities. Cv is the increase in Energy for an increase in Temperature of 1 K at constant volume. Cp is

the increase in Enthalpy for a Temperature increase of 1 K (1) at constant Pressure. We set out here how

Energy and Enthalpy and C v and C p respectively are related. C v is the rate of increase in Energy with

Temperature at constant volume

C v =d U/d T (1)

107

at constant Volume. To obtain the increase in U as the Temperature changes from 0 K to a Temperature

T we must integrate this expression between 0 and T.

∆U = ∫ C v d T (2)

At 0K molecules have no kinetic energy and are all at their rest positions in a crystalline solid with an

interaction energy, U0. By adding the value of U0 to the values of U and H from equations 1 and 2 we

obtain the total Energy at each Temperature. This value of U0 is the heat of sublimation at 0 K so the total

Energy at a Temperature T is:

U =U0 + ∫ C v d T (3)

 There is a similar expression for changes at constant Pressure relating Cp to H:

H = H0 + ∫ Cp d T (4)

Equations 1 to 4 show absolute values of U and H can be obtained from a series of measured values of C v

and Cp over a range of Temperatures between 0 and T combined with plotting the results and doing the

integration graphically from the area under the curve of a plot of heat capacity values versus

Temperature.

The absolute values of the Entropies pose additional problems and we deal with them in the next Section.

Section 3 The Third Law of Thermodynamics and Entropy

The Third Law of Thermodynamics states (2)

The Entropy of a substance in a crystalline state at 0 K is zero.

It follows as a direct consequence of equation 5 which defines Entropy

S/R = - ∑pi log (pi) (5)

108

where pi is the probability that a molecule has an energy ui and pi is given by the MBD :

pi = n i /N = e (-u i/k T) / Q and (6)

Q = ∑ e (-u i / k T) (7)

At 0 K the molecules are all at their positions that maximize their interactions. Their probabilities are

unity at those positions and zero elsewhere. Substituting values of unity for all p i, noting log (1) is zero,

in equation 5 leads to:

S/R = - ∑1ilog (1i) = 0 (8)

To obtain values of S from experimental data at constant Volume we start from equation 5 and

differentiate with respect to T:

D S /d T = (d U/d T)/T +U/T2 - 1/Q (∑ u i / (k T)e(-u i/k T) (9)

= C v /T + (U-U)/T2

So the final expression is:

S = ∫ C v /T d T (10)

and the limits are zero and T.

Similarly at constant Pressure we have;

S = ∫ Cp/T d T (11)

Values of S can be obtained graphically from plotting Cp/T versus Temperature and the integral

evaluated from the area under the graph. It is then a simple matter to combine values of T, H and S to

obtain G from the basic equation:

G = H-TS (12)

109

Since all substances behave as Ideal Gases at low enough Pressure the usual practice is to obtain values

of G0 and H0 for each substances in their Ideal Gas state at 1atm as a function of Temperature either from

measured values of the Heat Capacities or from equations which accurately represent the data. The

difference in the value of G for a substance at a given Temperature and Pressure, P, can then be calculated

from the Work Done in going from 1 atm to P at that Temperature

G = G0 + ∫ P d V (13)

For an Ideal Gas we can evaluate the integral by substituting RT/V for P and integrating between the limit

of P and P0 to give the usual expression:

G = G0 + RT log (P /P 0) (14)

To evaluate the integral for a non ideal gas we must have an empirical equation for the substance in

question. Such equations are available for a wide range of substances but we will not discuss this further

here. It is common practice in Data Books (3) to provide tables of values of G H and S for Ideal Gases at

1atm in their standard state of 25 C and 1atm Pressure.

Section 4 Discussion

With the development of expressions to obtain H U G and S from experimental data we can now review

and comment on the use of data for these properties. It will be apparent that to explain equilibrium

conditions we have no need of the actual values of G or S. When we have values of G and S we can use

them to predict equilibrium.

In this series of articles we have set out the basic ideas needed to describe Equilibrium. There are four

keys concepts involved in the whole approach. The first is the extended definition of Equilibrium

including the MBD and most probable state as well as constant Temperature and Pressure and

composition. The second is the simple proof that the Gibbs Free Energy is a minimum at Equilibrium at

fixed Temperature and Pressure solely on the basis that the amount of Enthalpy that can be converted to

110

work at must be a maximum. It is the simplicity of this argument that enables us to include this topic in

an A Level Chemistry Course. The Gibbs Free Energy is central to any explanation of Equilibrium and

the only reason it is has not been used in A Level Chemistry Courses prior to this is the difficulty of the

arguments leading to it. I hope the arguments I have presented here will overcome this obstacle.

The third key aspect is the simplified proofs, using just four equations, that show that values of the

Energy Temperature and Pressure have negligible Standard Deviations at fixed Temperature and

Pressure. The techniques used in the summations are simple and use the same equations that Students use

in GSCE Mathematics. The earlier proofs by T L Hill(4), from which I developed these simplified

methods, are far too complex for A Level Students and no doubt that is one reason why no one has

previously attempted to use this approach for an A Level course.

This enables us to demonstrate the need for two forms of energy, one convertible to work and the other

not convertible to work as a system changes from one equilibrium state with its MBD to another

equilibrium state with a different MBD.

The fourth key development is the use of the most probable distribution and the MBD to introduce the

Entropy and relate it to the Enthalpy not Convertible to Work. Because the equation for the number of

arrangements of N molecules divided into sets of n i of energy u i is a standard example in A Level

Statistics courses its use here amounts to cross curricular integration. For Students not studying A Level

Statistics the explanation adapted from Margenau and Murphy (5) provides a clear proof of the required

result using just three equations.

The Entropy then appears naturally as a consequence of the condition for the most probable distribution

at Equilibrium once we have carried out the summations involved. Our approach makes clear, in a way

not addressed by any other treatment, as to why we need Entropy at all. It also explains the Second Law

of Thermodynamics because it shows that some of the Energy in going from one Equilibrium state to

another is always used in going from the MBD at the initial conditions to the MBD at the final conditions

111

and therefore is not available for Work. The discussion of the interconversion of Work and Energy

introduced the Gibbs Free Energy as the convenient way to apply these concepts to Equilibrium. We give

physical interpretations for both the Entropy and the Gibbs Free Energies and showed how they apply to

systems of Ideal Gases.

The natural question is how easily can this approach be extended to real Gases, Liquids and Solids and

their mixtures? The answer is the basic formalism can be extended almost without change for pure

substances. The reason we restricted it to the Reactions of Ideal Gases is that this is the only model for

which we have a closed expression relating Pressure Volume and Temperature allowing us to obtain an

explicit formula for Work. The concepts apply to pure Solids and Liquids; we just can’t write closed

formulae for them using universal equations.

In practice, to apply these ideas in the real world, empirical equations fitted to experimental PVT data for

real gases and Ideal Gas heat capacities are used in these relationships. This has become the standard

method of designing oil, gas, and chemical plants in the past forty years.

When it comes to mixtures a further development is needed because interactions between components in

a mixture have to be allowed for. This can be done but introduces more complex mathematical ideas and

these are best left to a more advanced course. When this is done formulae similar to those discussed here

describe Equilibrium for mixtures.

Section 5 Conclusion

I started writing these articles in response to a remark by a colleague who lectured in Chemistry at one of

our Universities and said of his new Students: “They come to us knowing all the buzz words but they

don’t know what they mean”. I hope that this series will help introduce Students to the concepts needed

to describe Equilibrium in a way that they can understand and relate to. I trust that this series of articles

has helped them to understand some of the buzz words and to have a sound basic understanding of

Equilibrium which they can develop in more advanced courses.

112

References

1. E B Smith, “Basic Chemical Thermodynamics “, Imperial College Press, 1997, p 19-20

2. E B Smith ibid, p69

3 J G Stark and H G Wallace, “Chemistry Data Book, John Murray, London, 1992

 4 T. L Hill, “Introduction to Statistical Mechanics”, McGraw Hill, 1959, p 100-102

5 G. Attwood and G. Skipworth, “Statistics 1”, Heinemann, 1994, p160

 6 H Margenau and G M Murphy, “The Mathematics of Physics and Chemistry”, Van Nostrand, New

York, 2nd ed., 1961

113

The Stirling Equation for Chemists

R M Gibbons

BSc PhD FRCS

Proofs to: 4 Little Acre

Beckenham

Kent BR3 3ST

Abstract A simple derivation of the Stirling Equation using A Level Mathematics

114

Section 1 Introduction

The Stirling equation for the log of factorial N, log (N!), is widely used in chemical theories. Stirling’s

derivation of this is mathematically difficult and it is generally introduced in Chemistry courses without

explanation.

It would be preferable for students to have a simple explanation using standard mathematical methods. In

this note we show how to obtain Stirling’s result using integration by parts.

Section 2 Derivation of Stirlings Equation

Stirling’s equation is:

Log (N!) = N log (N) –N (1)

We can obtain this result easily starting from the definition

N! = N (N-1)(N-2) ….. = П i (N-i) (2)

Factorise this by taking N out of each bracket and writing

x i = i /N we have

N! = NN П (1-xi) (3)

for values of xi from 0 to 1/N.

Taking the natural logs of this we have

Log (N!) = N log (N) + ∑ log (1 – xi) (4)

for the range xi from 0 to 1/N as i goes from 0 to N-1

The sum in equation 4 cannot be evaluated directly but we can approximate it by replacing x i with a

continuous variable x so that

115

∑ log (1-xi) = ∫ log (1-x) d x (5)

Noting that N d i = d x then leads to our approximate expression:

Log (N!) = N log (N) + N ∫ log (1-x) d x (6)

where the limits are 1/N and 1.

The integral can be evaluated by an integration by parts following a change of the variable to leading to:

Log (N!) = N log (N) - N + log (N) (7)

For large N, log (N) is negligible compared to N, and we have Stirling’s equation given above.

It is natural to ask how our expression compares with Stirling’s series expansion (1):

Log (N!) = N log (N) – N +0.5log (2πN) +1/ (12N) (8)

Comparison of the values of the extra terms in equations 7 and 8 shows they are both negligible (≤

0.01%) for N ≥ 100000. The ratio of the values of the extra terms in equation 7 to those in equation 8 is

always less than 2. This simple derivation of Stirling’s formula for log (N!) provides an easy way to

introduce it to students who have to use it in modern chemical theories.

References:

1. H Margenau and G M Murphy, “The Mathematics of Physics and Chemistry”, pub. Van

Nostrand, New York, 1961, p 97

116