CHEE3003 2012 Weeks06-07 Lecture Slides Solutions

1

Key concepts from this section

1. What is a solution

2. Partial and Integral molar properties

3. Chemical potential of ideal gas in solution

4. Chemical potential of condensed phases

(liquids, solids) in solution

5. Thermodynamic activity

6. Complex reactions with solutions, conditions for

spontaneous reactions

Thermodynamics of Solutions

(Lectures Week 6, 2012)

2

Key concept

1. What is a solution


3

The logical development of ideas means that we have concentrated on single component (or pseudo-single component) phases to bed down many thermodynamic principles. Unfortunately (or fortunately) real life phases always consist of more than one component, i.e. they are multi-component systems.

A solution refers to any type of phase (gas, liquid or solid) that contains more than one chemical species of variable concentration such that the species cannot be physically distinguished or separated. Importantly, this is distinct from a mixture which is less well defined. You can have a mixture of any two things (e.g. apples and oranges, male and female) but you can only have a solution of chemical species in a single phase.


Mixtures

Solution of water and KMnO 4

is in factK+ + MnO4

-

Air: gas solution

Metal Alloy: Solid solution

4

Dissolution of Oxygen in water is essential for fish / aquatic life

O2 (gas, air) ↔ O2 ( aqueous) –

change from gaseous to aqueous solution

Low solubility of O2 in water is a problem for systems requiring oxygen

Examples of Solutions

It is important to know effects of various factors on the dissolution of Oxygen in water

TemperaturePressureComposition of water – how other elements effect the solubility of O in water?

Will river water have the same solubility of O as ocean water?

Water

Air:21 vol % O2

↕

O2(aq)dissolvedin water

5

Dissolution of CO2 in water

CO2 (gas) ↔ CO2 (aqueous)

Applications:• CO2 sequestration in deep saline waters (high pressure)- temporary solution to global warming and use of carbon as a fuel

* Carbonated drinks


Deep WaterIt is important to know effect of Temperatureand Pressure on maximum solubility of CO2

in water

Earth’s surface CO2

∆∆∆∆P

6

Osmosis – diffusion of solvent (e.g. water) across semipermiablemembrane from area of high to area of low solvent concentration


7

Solubility product –maximum solubility


Salts precipitate out of the Dead Sea as water vaporises

8

Dissolution of Hydrogen in metals

H2 (gas) ↔ 2 H (solution in metal)

Gas species H2 break down and dissolve in the metal phase as atoms

Applications:• Hydrogen storage – some alloys can dissolve significant concentrations of H• Defects during casting of some alloys – for example, in humid, wet environment H is absorbed in molten aluminium. Then, on solidification, H is rejected by solids producing gas bubbles in castings, forming defects and affecting mechanical properties in the final product


Alloy H (solution)

It is important to know effects of Temperature, Pressure and alloy composition on solubility of Hydrogen

H2 (gas)

9

1) Gas: Air solution of O2, N2, ...

2) Liquid: Gas in liquid:Oxygen in waterCarbonated water

Liquid in LiquidAlcoholic drinksPetroleum – solutions of hydrocarbons

Solid in LiquidSugar in waterSalt in water, Ocean water

3) Solid: Gas in SolidHydrogen in metal (H2 storage)

Liquid in SolidMercury in gold (amalgam)Hexane in paraffin wax

Solid in SolidHydrocarbon based Polymers with plasticizersMetal alloys, Steel Fe-C, bronze Cu-Sn

SUMMARY: Examples of different states (gas, liquid, solid)

in pure and in solution

10

DEMO 1 – formation of a solution

The question in focus – how to predict

chemical behaviour of species in solution?

DEMO 2 – partitioning of solute A between two immiscible liquids B and C

CuSO4

solutionin H2O

0%A solution

in C

50%A solution

in B

?%A solution

in C

?%A solution

in B

Mineral Turps (~C10H16)+Extractant (ACORGA M5774-Salicylaldoxime: C7H7NO2)

CuSO4 (30g/L) Aqueos Solution

CuSO4

CuSO4

11

In the following slides several examples are given where knowledge on the behaviour of solutes and solvents in solutions is important to predict the outcomes of the processes.


chemical behaviour of species in solution?

12

50%A solution

in B

50%A solution

in C

A? A?

30%A solution

in B

50%A solution

in B

BA

Example of different chemical behaviour of species in solutions

←S

epar

atio

nis

rem

oved

←S

epar

atio

nis

rem

oved

Case 1) The same solvent and solute, two separated solutions initially with different concentrations of A. .In this case the answer is clear - after separation is removed – A will diffuse from right to left, and B will diffuse from left to right compartments before the same concentration of A in both compartments is reached

Question – what is the direction of diffusion after separation between two separate compartments is removed?

Case 2) Two different solvents B and C are immiscible liquidswith the same concentration of A initially separated.

This case is not obvious – the same solute A but different solvents, so -different chemical environments–additional information on chemical behaviour of A in solvents B and C is needed to answer the question.

1)

2)

13

Example - phenol in water.

Phenol is present in different processes such as production of paints, polymers, oil refineries and coal.

It is soluble in water up to 8 g/l but is toxic to marine life in ppm concentrations.

One option for cleaning water from phenol before it is discharged from a process into environment is to contact the stream with Methyl Isobutyl Ketone (MIBK) since phenol is far more soluble in MIBK leaving very little phenol in water.


Liquid – liquid separation is used to clean water before discharge into environment?

ARD Extractor used for removal of phenol from water

14

30%A solution

in B

50%A solution

in B

A

50%A solution

in B

50%A solution

in C

A?

3)

4)

gas

gas

Example of different chemical behaviour of species in solutionsQuestion – what is the direction of mass transfer of volatile solute A after two solutions are placed in closed compartment?

Case 3) The same solvent B and solute A, two separated solutions initially with different concentrations of solute A.

In this case the answer is clear - A will move from right to left solution before the same concentration of A is reached in both volumes(similar to Case 1 above).

Case 4) Two different solvents B and C with the same concentration of solute A.

In this case the answer again is not obvious –additional information on chemical behaviour of solute A in solvents B and C is needed to answer the question(similar to Case 2 above).

?

Closed container

Closed container

15

5)

6)

30%A solution

in B

50%A solution

in B

SolidD

A + D ? A + D ?

SolidD

50%A solution

in B

50%A solution

in C

A + D ? A + D ?

SolidD

SolidD


Question – if reaction A+D can proceed only at certain concentration of A – will A in solution react with D?

Case 5) The same solvent B and solute A, two solutions with different concentrations of A.

In this case the answer is clear –that the reaction A+D will proceed ahead for higher concentration of A

Case 6) Two different solvents B and C with the same concentration of solute A.

In this case the answer is not obvious – additional information on the chemical behaviour of solute A in solvents B and C is needed.

16

DEMO – Ethanol and water change volume in solution

Properties of chemical species in solution are different from pure properties

At a given T and P a property of a chemical species in solution depends on

- solvent- concentration


Eth

anol

Wat

er

W+

Eth

solu

tion

17

Consider the Following...

A local barman attempts to make 70 ml of his “Rocket Fuel Special” by adding 50 ml of pure ethanol to 20 ml of pure water:

a) What would happen?

b) What would the final volume of the drink be?

c) What is the ethanol fraction in the vapour of the drink?

18

Answer

a) The two liquids mix spontaneously to form a liquid solution.

b) The final volume of the drink is 67 ml!

c) The mole fraction of ethanol in the liquid is 0.43 but the mole fraction in the vapour is less than this!

??××××→→→→

67 ml Solution of ethanol and water

20 ml Pure water50 ml Pure ethanol

19

Lessons

1. Don’t argue with the barman (let alone the bouncer), they do not understand thermodynamics.

2. The volume difference depends on the ratio of ethanol to water.

3. It does not matter if we added ethanol to water or vice-versa, the result would be the same. It is a state property.

4. If you have a very sensitive nose, you may be able to measure the vapour pressure / fugacity of the ethanol in your drink.

5. Do not drink anything that has 71 v/v% alcohol and stop asking the barman for their recommendation.

20

A Gap to Fill...

1. How do we describe chemical phenomena in solutions?

2. How do we predict whether a reaction will occur spontaneously when some/all of the reactants and products are in solution and no longer in pure state (remember we have only considered pure substances at 1 bar) so far in this course?

This example provokes two important questions:

21

Key concept

2. Partial and Integral molar properties


22

INTRODUCING TERMINOLOGY: State Property Y can be:

- Intensive or - Extensive

- Ideal Y* or - Real Y

- Molar Partial - property of a component “i” in the solution

- Molar Integral Y - property of the solution

- Molar mixing property– change of a property due to the mixing of pure component into solution, including

- Integral molar mixing

- Partial molar mixing

iY

23

Solution Properties1) For a pure component, the extensivestate property of the system is a function of three variables (using V as an example):

(((( )))), ,V V T P N====

2) The intensive, molar properties of a pure component are defined by any two other intensive variables:

(((( )))), etc.V V T P====

3) For a solution, containing m different chemical components, extensive properties then become a function of the number of moles of each species as well as two other variables (i.e. we need m + 2 independent variables to define the system):

(((( )))) (((( ))))1 2 3, , , , , ....., , ......, , ,

where is the set of amounts for all species

i mV V T P N N N N N V T P N

N m

′′′′= == == == =

′′′′

24

Partial Molar Property

The final term in the equation above is the sum of partial molar volumes. A partial molar property (using volume as a general example) is defined as:

j j m

mP N T N mT P N T P N

P N T N

V V V VdV dT dP dN dN

T P N N

V V VdV dT dP

T P

1

1, , 1 , , , ,

, ,

can then be written aThe total differential of a property in soluti s:

.

on

....≠≠≠≠ ≠≠≠≠

′ ′′ ′′ ′′ ′

′ ′′ ′′ ′′ ′

∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂ = + + + += + + + += + + + += + + + + ∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂

∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂ ⇒⇒⇒⇒ = + += + += + += + + ∂ ∂∂ ∂∂ ∂∂ ∂

j i

m

ii i T P N

dNN

i

1 , ,

The condition here is that the amounts of all other components except are kept constant≠≠≠≠

====

∂∂∂∂

∑∑∑∑

j i

m

i iiP N

P N

N

i

T

i T

V VdV dT dP V dN

T P

VV

N, ,

1,

3

,

(m /mol), the overbar denotes a partial molar property≠≠≠≠

====′ ′′ ′′ ′′ ′

∂ ∂∂ ∂∂ ∂∂ ∂ ⇒⇒⇒⇒ = + += + += + += + + ∂ ∂∂ ∂∂ ∂∂ ∂

∂∂∂∂==== ∂∂∂∂

∑∑∑∑

Partial Molar Property definition:- increment in that property when 1 mole of a component is added to a large amount of a solution so that the bulk composition of the solution stays constant.

25

Partial Molar Properties

Take careful note of the properties of a partial molar volume:

j i

ii T P N

VV

N≠≠≠≠

∂∂∂∂==== ∂∂∂∂ , ,

Always write Ni

never xi

Always at constant T and P

Amounts of all other species kept constant

Partial Property of component i in

solution

Differential of systemproperty

You can have a partial molar quantity of anyextensive variable:

, , , , , ,

etcj i j i j i

i i ii i iT P N T P N T P N

H S GH S G

N N N≠ ≠ ≠≠ ≠ ≠≠ ≠ ≠≠ ≠ ≠

∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂= = == = == = == = = ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂

26

Partial Molar and Integral Molar Properties

The total number of moles in the system, N, is defined as:

(((( ))))j i

m

ii

ii T P N

N

VV

N

N

NVV

N

1

, ,

- Integra

Any extensive property can be converted to an intensive molar property:

e.g.

Knowing this, we can also write a partial molar property a

l Molar Pr

s

per y

:

o t

≠≠≠≠

====

====

====

∂∂∂∂==== ∂∂∂∂

∑∑∑∑

j i

ii i

i T P N

i

NVV x i

x N

N N

, ,

Note that , where is the mole fraction of

since changes when changes.≠≠≠≠

∂∂∂∂≠ =≠ =≠ =≠ = ∂∂∂∂

27

Integral Molar Property

Since the partial molar volume is the volume occupied by each mole of each species, the total volume is given by:

m

i

m

i ii

mi

ii

m

i ii

ii

V N V

NVV V

N N

V x V x

1

1

1 1

The average molar volume of the solution can be calculated as:

- Integral m

Integral molar property = Patial m

olar

olar

volu

prop

me

Again this ca

e ty

be

r

n

====

====

====

====

====

= == == == =

⇒⇒⇒⇒ ==== ∑∑∑∑

∑∑∑∑

∑∑∑∑

∑∑∑∑

m m m

i i i i i ii i i

H x H S x S G x G1 1 1

extended to any variable:

etc= = == = == = == = =

= = == = == = == = =∑ ∑ ∑∑ ∑ ∑∑ ∑ ∑∑ ∑ ∑

The average molar property of a solution is also called an integral molar property of the solution.

28

Example: Partial Molar VolumeAn example of determining the partial volume shows the addition of a small amount of water into a water-ethanol solution:

- the change in volume of the solution is divided by the small amount of water added to give the partial volume of water at the specified T, P and amounts of water and ethanol in solution (ref. Koretsky1).

29

Example: Partial and Integral Molar Volume

The experimental results show that the partial molar volume can be a complex function of the composition of the solution (ref Atkins 2).

Depending on the mole fraction of water and ethanol, the molecules pack differently and occupy different volumes.

30

10

15

20

25

30

35

40

45

50

55

60

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Volume Fraction of Ethanol (C2H5OH) [cm3pure/cm3pure]

Mo

lar

Vo

lum

e [

cm3

mo

l-1

]

-0.035

-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

V m

ix [

cm3

/cm

3 o

f p

ure

]

Water Part.Vol. Ethanol Part.Vol.

Int.mol.Vol. Relative Vol.Change due to mixing

Example: Partial and Integral Molar Volume

∆∆∆∆mixV

Refer to DEMO 60vol% Eth + 40vol% W

ethanolV

waterV

solutionV

solutionmixV∆∆∆∆

31

Question: Why partial ≠ pure property? What changed?

The Partial Property of a species (e.g. water) in the solution (e.g. with ethanol) is different due to the interatomic / intermolecular interactions with surrounding atoms / molecules (e.g. of ethanol). A similar statement is true to for the ethanol.

Pure Water H2O

H2O H2O H2O H2O H2O

H2O

H2O

Solution of Water H2O and Ethanol Eth

Eth

Eth

Eth

Eth

Eth H2O

H2O

H2O

H2O

H2O

H2O

Eth Eth

Eth

H2O H2OEth Eth

Eth

EthH2O

H2O H2O H2O H2O

H2O H2O H2O H2O

H2O H2O H2O H2O

H2O H2O H2O H2O H2O

H2O

32

INTRODUCING TERMINOLOGY: State Property Y can be:

- Intensive or - Extensive

- Ideal Y* or - Real Y

- Molar Partial - property of a component “i” in the solution

- Molar Integral - property of the solution

- Molar mixing property– change of a property due to the mixing of pure component into solution, including

- Integral molar mixing

- Partial molar mixing

iYm

iii

Y x Y1====

====∑∑∑∑

33

Key concept

3. Chemical potential

of ideal gas in solution


34


chemical behaviour of species in solution

For example - refer to DEMO – partitioning of solute A between two immiscible liquids B and C- the question is how much A will be in the final equilibrium state?

Pure C(0%A )

50%A solution

in B

?%A solution

in C

?%A solution

in B

Initial State Final State

35

Recall from Section 5

Three Criteria of Equilibrium

j i

i ii T P N

GG J mol chemical potential

N

partial molar Gibbs Free Energy

µµµµ≠≠≠≠

∂∂∂∂= = −= = −= = −= = − ∂∂∂∂ , ,

[ / ]

-

Chemical Potential

Thermal Equilibrium

Mechanical Equilibrium

Chemical Equilibriumi i

T T

P P

G G

α βα βα βα β



============

36

RECALL: “MASTER EQUATION”

dG = VdP – SdT

describes the direction of change in a system as a function of two important practical variables: temperature, T, and pressure, P.

INSERTfor additional study

PRELIMINARIES: EFFECT OF TEMPERATURE

dG = VdP-SdT

1) At P = const: dG = -SdTand (dG/dT)P= -S or (d∆∆∆∆G/dT)P= -∆∆∆∆ S

2) ∆∆∆∆G = ∆∆∆∆H-T∆∆∆∆S so ∆∆∆∆G/T = ∆∆∆∆H/T-∆∆∆∆S so d(∆∆∆∆G/T) / d(1/T) = ∆∆∆∆H

2

1

2 22

11 1

T

P

1 2 P

*

P P P* 2PP P

1

dG VdP SdT ; at T=const: dG=VdP and (dG/dP)V

for a change from P to P G VdP

for 1 mol of ideal gas (V =RT/P)

PRTG V dP ( )dP =RT [ln P ] RT ln( )

P P

∆∆∆∆

∆∆∆∆

= − == − == − == − =

====

= = == = == = == = =

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

PRELIMINARIES: EFFECT OF PRESSURE:

37

Chemical Potential of Pure Ideal Gas

j i

i

i T ,P ,N

Actual State

iStandard State

Actual State

iStandard S

i

tate

0G

Partial Molar Gibbs free energy

dG

Difference d

Standard partial Gibb

G(chemical potential) G

G

s

N≠≠≠≠

++++ ∂∂∂∂= = →= = →= = →= = →

→→→→

∂∂∂∂ ∫∫∫∫

∫∫∫∫

0ifree energy G →→→→

To derive Partial Molar Gibbs Free Energy (Chemical Potential) of ideal gas at pressure P:

1. Reference Standard State is introduced, and 2. Difference between the values at Standard State and actual state at

pressure P is calculated3. The Partial Molar Gibbs Free Energy (Chemical Potential) equals to(Standard) + (Difference) :

38


at pressure P

0i i i

0 0i

Chemical potential (Partial Molar Gibbs

1. Pure gas at P 1atm is frequently selected a

Free Energy) of pure ideal gas is

s Standard State for gas: G ,

t

G G RT ln

hen

P ,

P in

µµµµ = = += = += = += = +

====

0 0

0

P P0

i i i i 0P P

0

2. Integration from Standard State (pure P )

to actual state at p

atm, and P

ressure P for an ideal gas

RT Pgives dG = dP , so that G G RT l

P

t

P

1a m

nµµµµ = = += = += = += = +

====

∫ ∫∫ ∫∫ ∫∫ ∫

39


at pressure P

0

P

0

0i ii

iP

0i

For Pure Standard State


Standard partial Gibbs free energy

(P =1atm) dG RT ln P

( P in atm

(chemical potential) G G RT ln P

)

G

µµµµ

= →= →= →= →

= + →= + →= + →= + →

→→→→

====

∫∫∫∫

Partial Molar Gibbs Free Energy (Chemical Potential) of ideal gas at pressure P equal to (Value at Standard State) + (Difference) :

40

Chemical Potential of Real Gas and fugacity

i

0 i ii i i 0 P 0

ii

0i

Fugacity f (T, P, x') is introduced to retain simplicity of expressions

f (T ,P , x') f(T,P,x')= G (T ,P , x') G RT ln with lim 1

Pf

f is the value of fugacity of gas in The Standard State

is

µµµµ→→→→

= + == + == + == + =

(((( ))))

0i

0i i i i i

commonly selected to be pure ideal gas at 1 atm and f 1

then (T,P)= G (T ,P ) G RT ln f for f in atm

is a measure of a chemical potential

in the form of ' adjusted

F

pressure

ugacity

.'

µµµµ

====

= += += += +

41

Chemical Potential of Real Gas and fugacity

(((( ))))0i i i i

For moderate pressures f/P 1 ,

and (T,P)= G (T ,P ) G RT ln Pµµµµ

≈≈≈≈

= += += += +

42

Total pressure and partial pressures Pi of

Several Gases in Solution behaving ideally

- Dalton’s lawNA mols of ideal gas A - in a container with volume V

will produce pressure PA=NART/V

If we add NB mols of ideal gas B into the same container with volume V – it will produce additional pressure PB=NBRT/V

What is the total pressure?

Dalton’s Law: The total pressure exerted by a solution ofgases behaving ideally is the sum of the pressures exerted by the individual gases occupying the same volume

Ptotal=PA+PB=(NA+NB)RT/V ; PA=xAPtotal and PB=xBPtotal

43

Total pressure and partial pressures Pi of

Several Gases in Solution behaving ideally

Ptotal

0

PA=x A

*P total

1.0Mol Fraction, xA

Pre

ssur

e

PB =x

B *Ptotal

Dalton’s Law: Ptotal= PA+PB = (NA+NB)RT/V

44

Chemical Potential of Ideal Gas

with partial pressure Pi in solution

For the Pure Standard State:Partial Molar Gibbs Free Energy (Chemical Potential) of ideal gas at partial pressure Pi in solution equals to (Molar Gibbs free energy in Standard State – pure gas at 1 atm) + (Difference between pure gas at 1 atm and at partial pressure Pi) :

0

P

0

0i ii

i

P

0

i i

i

For Pure Standard State (


P =1atm) dG RT ln P

(

(chemical poten

Standard

tial) G G R

partial

P in

Gibbs

T ln

fr

P

ee energy G

atm)

µµµµ

= →= →= →= →

= = →= = →= = →= = →

→→→→

++++

∫∫∫∫

45

Pure gas iat 1 atm

Chemical Potential of Ideal Gas

with partial pressure Pi in solution

0iG

For the Pure Standard State:Partial Molar Gibbs Free Energy (Chemical Potential) of ideal gas at partial pressure Pi in solution equals to (Molar Gibbs free energy in Standard State – pure gas at 1 atm) + (Difference between pure gas at 1 atm and at partial pressure Pi) :

Gas i in the solution

0

P

i iPdG RT ln P====∫∫∫∫

0i i iG G RT ln P= += += += +

46

Solution made of NA mols of A and NB mols of Bwith total pressure P = PA+PB

Calculating Integral Molar Gibbs Free Energy of Ideal Gas Solution

A

B

Let's calculate the change of Gibbs Free Energy of the system due to the formation of an ideal gas solution from N mols of pure A at pressure P and N mols of pure B at pressure P(initially not in solution) at a given T

NA mols of pure A at P

where Pi =xi*Ptotal is Partial Pressure of component “i”xi – mol fraction of component “i”xi = Ni / Ntotal = Ni / (NA+NB)Pi = xi*Ptotal , i=A or B

NB mols of pure B at P

INSERTINSERTfor additional study

47

0i i

0 0A B A BTotal A B A B

0ATotal to

A

tal A B

B1 ) Initial State: N mols of pure A at P and N mols of pure B at P

(not in solution) at a given T G G RT ln P

G N * G + N * G N (G RT ln P ) +N (G RT ln P )

G G / N x (G

RT ln P )

x

= += += += +

= = + += = + += = + += = + +

= = + += = + += = + += = + +0B

0 0 0 0A B A BA B A B A B

A B

(G RT ln P )

x G x G ( x x )RT ln P x G x G RT ln P

(note x x 1 )

+ =+ =+ =+ =

= + + + = + += + + + = + += + + + = + += + + + = + ++ =+ =+ =+ =

NA mols of pure A at P NB mols of pure B at P



48

(((( )))) (((( ))))0 0A B A BA B A A B B

0 0A Btotal A A B B Total

0 0A BA A B B

A B

G N G N G N G RT ln P N G RT ln P ,

G G / N ( N (G RT ln P ) N (G RT ln P

2 ) Final State: Solution of N and N mols of A and B at

) )/N

x (G RT l

P

n P ) x (G RT ln P ) (substi

and

t

ute P

T

= + = + + += + = + + += + = + + += + = + + +

= = + + + == = + + + == = + + + == = + + + =

= + + + == + + + == + + + == + + + = i i

0 0A BA A A B B B

0 0A BA B A A A B B B

0 0A BA B A A B B

A B

x P )

x G x RT ln( x P ) x G x RT ln( x P )

x G x G x RT ln x x RT ln P x RT ln x x RT ln P

x G x G x RT ln x x RT ln x RT ln P

(note that x x =1)

====

= + + + == + + + == + + + == + + + =

= + + + + + == + + + + + == + + + + + == + + + + + =

= + + + += + + + += + + + += + + + +++++




49

0

A

0

A BB

A B

A

B

G x G x G

1) In

R

itial State: N mols of pure A at P and N mols o

T ln P

f pure B at P

(not in solution) at a given T

2) Final State: Solution of N and N mols of A

− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −= + += + += + += + +

0 0

A BA B A A B B

A A B B

G x G x G RT ln P x RT ln x x RT ln x

Note that the term x RT ln x x RT ln x - difference between

G for separate pure gas and G for solutio

and B a

n - is a

t P and

lways neg ve

T

ati

= + + + += + + + += + + + += + + + +

++++


NA mols of pure A at P NB mols of pure B at P



50

A B

total A A B B

0 0A A B BA B

Mi

1) for solution of N mols of A and N mols of B at temperature T

for P 1atm, P = x , P = x

Partial Gibbs Free Energies G G RT ln x and G G RT ln x

for ideal gas: G

====

= + = += + = += + = += + = +

==== i

total

m 0 0

i A Bi A A B Bsolutioni 1

RT ln x

___________________________________________________________

2) It can be shown (see Insert above) that Integral Gibbs Free Energy (P =1 atm)

G x G x (G RT ln x ) x (G RT ln x ) =====

= = + + += = + + += = + + += = + + +∑∑∑∑0 0A BA B A A B B x G x G x RT ln x x RT lnx

___________________________________________________________

3) Gibbs free energy change due to formation of solution from pure ideal ga

= + + += + + += + + += + + +

0 0

A BMixing A B A A B Bsolution

ses

(Integral molar Gibbs Free Energy of Mixing):

G G ( x G x G ) = x RT ln x x RT ln x 0∆∆∆∆ = − + += − + += − + += − + + ≤≤≤≤

Integral Molar Gibbs Free Energy ofIdeal Gas Solution

51Pure A Pure B

0 mol fraction of B xB6666 1between xB=0 and xB=1 (A and B solution)

T=const0AG

0 0

A BA B A A B BSolution

Total

G ( x G x G ) x RT ln x x RT ln x

at P =1atm

= + + += + + += + + += + + +

total

0 0A BA B

Two separate pure gases at P = 1atm

x G x G++++0BG

Mix

A A B B

G

x RT ln x x RT ln x

∆∆∆∆ ====++++

Integral Molar Gibbs Free Energy of Ideal Gas Solution

52

Ideal Gas Solution

0 0A BMixing A Bsolution

A A B B

G G ( x G x G )

x RT ln x x RT ln x

∆∆∆∆ = − += − += − += − +

= += += += +

Pure A Pure B

∆∆ ∆∆ Mix

ingG

T=const

MixNegative values of G indicate that

1. the solution of ideal gases is more stable

than separate pure A and pure B, and

2. the formation of solution of ideal gases

∆∆∆∆

is spontaneous

0 mol fraction of B xB6666 1between xB=0 and xB=1 (A and B solution)

mix

A B

G - change due to formation of solution of ideal gases

N and N mols of A and B at total P = 1 atm and T

∆∆∆∆

06666

53

DEMO – formation of solution

Ideal Gas Solution

O2+N2 CO2+H2O+N2O2+N2

Spontaneous diffusion of O2

54

Key concept

4. Chemical potential of condensed phases (liquids, solids)

in solution


55

Equilibrium Partial Pressure over condensed phase

H2O

Pure water Liquid Water- Ethanol

Solution

Gas Equilibrium partial pressure PH2O < P0

H2O

PH2O depends on xH2O :higher xH2O –

- higher PH2O

Eth Eth

Eth

Eth

Eth

GasPH2O = P0

H2O

P0H2O = saturation

vapor pressure- standard state

H2O H2O H2O H2O

H2O H2O H2O H2O

H2O H2O H2O H2O H2O

H2O H2O H2O H2O H2O

H2O

H2O H2O

H2O

H2O H2O

H2O

Questions:

What is the steam (H2O) partial pressure above the water- ethanol solution compared to the H2O partial pressure above pure water?

Is it proportional to the H2O concentration?

56

Equilibrium Partial Pressure over condensed phase

A A

A

A

A

A

A

A

A

A

Pure Liquid A

A

A

A

A

Liquid SolutionA in B

GasPartial pressure above solution:PA < P0

A

PA ~ xA – mol fraction

B B

B

B

B A

AAAA A

GasPA = P0

A

P0A = saturation

vapor pressure- Standard state

Note

for low

pressures

f P

P will further

be used

in place of

fugacity f

≃≃≃≃

Saturation P0i is a measure of chemical interactions if “i” in a solution

57

Relationship between partial pressure PA and solution concentration in an ideal solution

EXAMPLE at T = const

Mol fraction A

PA

, (at

m)

01.00

Ideal Solution

(Raoult’slaw)

If A-A interactions .... A-B (the same), then Partial pressure PA is proportional to the concentration

of A in the solution – such behaviour is called “Ideal” or “Raoultian”

A A

A

A

A

A

A

A

A

B B

B

P0A P0

A - saturation vapour pressure

58

Relationship between partial pressure PA and solution concentration in a non-ideal system


Mol fraction A

PA

, (at

m)

01.00

Ideal Solution

Positive deviation

-Real Solution

A A

A

A

A

A

A

A

A

B B

B

P0A P0


If A-B interactions are weaker than A-A, then Partial pressure PA is higher

than over an ideal solution


59



Mol fraction A

PA

, (at

m)

01.00

Ideal Solution

Negative devia

tion

-Real SolutionA A

A

A

A

A

A

A

A

B B

B

P0A P0


If A-B interactions are stronger than A-A, then Partial pressure PA is lower

than over an ideal solution


60

Chemical potential of components in condensed (liquid or solid) solution phases

0

is frequently used:

component "i" (stable as gas at 298.15K and 1 atm)

Pure Standard

G

State - ideal gas at

aseo

P 1at

s

m

u

Recall: Pure Standard State

====

0i i i i i

(stable as liquid at 298.15K and 1 atm)

L

- Chemical potential of pure ideal gas "i" is

component "i"

Pure Sta

iquid

nd

G G RT ln P [ P in atm]µµµµ = = += = += = += = +

0i

0i

(stable as solid at 298.15K and 1 atm)

ard State - pure liquid G

component "i"

Soli

Pure Standard State - pure solid G

d

Chemical potential of condensed component "i"

" i" in solution

i i , pure solutionPure "i"

0 0i i i idG G G G Gµ ∆µ ∆µ ∆µ ∆ →→→→===== = + += = + += = + += = + +∫∫∫∫

Sol

iStd

i

0i

ch

G

ange

d

G

G∫∫∫∫

61

A,Solution

Liquid Solution

A in B: GA,PureA,Pure

Pure Liquid A

G G====

i , pure solution0

i i i G

Calculating Chemical potential of condensed component "i"

in solution G Gµ ∆µ ∆µ ∆µ ∆ →→→→= = += = += = += = +

A, pure A in solutionG∆∆∆∆ →→→→

A, pure A in solution - the Gibbs Free Energy change

between pure species A and A in sol n

G

utio

∆∆∆∆ →→→→

62

∆G2

EquilibriumSaturatedGas, PA,pureA=P0

A

∆G3∆G1

A,Solution

Liquid Solution


Pure Liquid A

G G==== 0

1 2

3

A, pure A,solution A,solution A, pure 1 2 3

Since G is a State Property (not dependent on the path)

to calculate G of change (pure A) (A in solution)

we may follow the path 0 1 2 3, so th

G G

a

G G

t

G G∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

∆∆∆∆

∆∆∆∆→→→→ = − = += − = += − = += − = +

→→→→

++++

→ → →→ → →→ → →→ → →

EquilibriumSaturated

Gas, PA,soland PB,sol

A,Solution,gasGA,Solution,gasG

i , pure solution0

i i i G




63

∆G2

Gas, PA,pureA

∆G3=0

Gas, PA,sol, PB,sol

∆G1=0

A,Solution

Liquid Solution


Pure Liquid A

G G====

A,liquid A,gas A,liquid A,gas 1 3

A, pure A,solution A,solution A, pure 2

For any liquid A in equilibrium with species A in the gas phase

G G G 0, then G 0 and G 0,

then G G G G

∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

→→→→

→→→→

= − = = == − = = == − = = == − = = =

= − == − == − == − =

0

1 2

3


i , pure solution0

i i i G




64

Solution Solution

A A2 Pure Pure

A above solution A above pure

i i

A abo

from dG=VdP-SdT :

G VdP ( RT / P )dP

RT ln( f / f )

for low pressures when gas behaves ideally and f P

RT ln( P

∆∆∆∆ = = == = == = == = =

====

========

∫ ∫∫ ∫∫ ∫∫ ∫

ve solution A above pure

A, pure A,solution A,solution A, pure 1 2 3

A, pure A,solution A,s

/ P )

Gibbs free energy change

between A in pure and A in solution

G G G G G G

G RT ln( P

∆ ∆ ∆ ∆∆ ∆ ∆ ∆∆ ∆ ∆ ∆∆ ∆ ∆ ∆

∆∆∆∆

→→→→

→→→→

= − = + + == − = + + == − = + + == − = + + =

==== olution A, pure/ P )

i , pure solution0

i i i G



65

A, pure A,solution A,solution A, pure

A,solution A, pure

A,solution A, pure A,solution A, pure

Rearranging: G G G

RT ln( P / P )

G G RT ln( P / P )

if pure is selected as Standard Stat

∆∆∆∆ →→→→ = −= −= −= −====

= += += += +

0 0A A, pure A A, pure A A

0 0A,solution A A,solution A


Chemical Potential

e,

then G G and P P (if A ideal and f P )

G G RT ln( P / P )

in general form G G RT ln( f / f )

−−−−

= = == = == = == = =

= += += += +

= += += += +

i , pure solution0

i i i G



66

Gas, PA,pureA

∆G3=0

Gas, PA,sol, PB,sol

∆G1=0

0A

A,Solution A 0A

Liquid Solution A in B

PG G RT ln( )

P= += += += +

0A,Pure AA,Pure

Pure Liquid A

G G G= == == == =

i , pure solution0

i i i G

SUMMARY: Calculating Chemical potential of condensed component "i"

in solution G Gµ ∆µ ∆µ ∆µ ∆ →→→→= = += = += = += = +0 0A A, pure A A, pure


0A,solution A

Chemical Potential

- G of 1 mol of A in solution = Partial Molar Free Energy

-

Pure Standard State, G G and P P

G G RT ln( P / P ) -

RT ln( P / P )

= == == == =

= += += += +

correction term

AA, pure A,solution 0

A

PG RT ln( )

P∆∆∆∆ →→→→ ====

Solution

2 Pure

Sol .A

A 0PureA A

G VdP

PRT( )dP RT ln( )

P P

∆∆∆∆ = == == == =

====

∫∫∫∫

∫∫∫∫


67

Key concept

5. Thermodynamic activity


68

ii 0

i i

i i

i

i

i i

Definition :

This leads to a simplified equation for partial Gibbs Free Energy:

G G RT lna

For the special case when the gas is

f a Activity of component

ideal, f P and f P on

" i

e can

" f

====

⇒⇒⇒⇒ = + ⇐= + ⇐= + ⇐= + ⇐

= == == == =

ο

ο ο

i ii ii

i i

i

use:

PG G RT ln

P

Fugacity f (T,P,x') is a function of concentration of component in soluti

a

o

P

P

n

→ = +→ = +→ = +→ = +

==== ο

οο

Activity of a component in solution

69

ACTIVITY AS A FUNCTION OF CONCENTRATION

The activity of a species (or “effective concentration”) is commonly expressed as asfunction of concentration:

aA = γAxAwhere γA is the activity coefficient of A in a solution of a given composition, and

xA is the molar fraction of A in the solution.

70

Pure Standard State is often used: . For gaseous components - Pure Standard State is 1 atm pure gas,

i.e. aA = PA/1atm, where PA = the partialpressure of the gas species A in atm.

. For a component in condensed phases(in solid or liquid solution), the pure element or compound is taken as the Pure Standard State,and aA = γAxA relative to the pure component

Note activity a is dimensionless value as it is always relative to the standard state

0 0A A A A

AA 0

A

A

Also, , for the special case when A gas above pure A and solution

behave as ideal (P

Pa

=f and P =f )

where P - equilibrium p a

P

arti

====

0A

l pressure of A above solution

and P - equilibrium partial pressure of A above pure A

in the corresponding state (liquid or solid)

71

0 A,solA,solution A 0

A

0A,solution A, pureA A

AA

For pure component standard state

PG G RT ln( )

P

G G RT lna

this is a general equation valid for solids, liquids and gases

G - refers to only part of solution

µµµµ

µµµµ

= += += += +

≡ = +≡ = +≡ = +≡ = +

≡≡≡≡ Partial Molar Gibbs Free Energy

Chemical Potential

≡≡≡≡≡≡≡≡

72

0 0 0 1 2TT T P

T T T0 0 1 2 2 1 TT 298.15 P P 298.15298.15 298.15 298.15

T T0 0 1P PT 298.15

298.15 298.15

at temperature T G H T S ; for C A BT CT

BH H C dT ; C dT ( A BT CT )dT [ AT T CT ]

2C C

S S dT ; dT ( AT B CTT T

−−−−

− −− −− −− −

− −− −− −− −

= − = + += − = + += − = + += − = + +

= + = + + = + −= + = + + = + −= + = + + = + −= + = + + = + −

= + = + += + = + += + = + += + = + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫T 3 2 T

298.15298.15

0

A,solution A, pureA A

C)dT [ Aln(T ) BT T ]

2

For pure component standard state

G G RT lnaµµµµ

−−−−= + −= + −= + −= + −

≡ = +≡ = +≡ = +≡ = +

∫∫∫∫

Chemical Potential as a function of temperature


73



Mol fraction A

PA

, (at

m)

01.00

Ideal Solution

Positive deviation

-Real Solution

A A

A

A

A

A

A

A

A

B B

B

A

0A

A A

Dividing P - equilibrium Partial Pressure of A above solution -

by P - equilibrium Partial Pressure of A above pure A (solid or liquid respectively)

we can obtain "normalised" Partial Pressure P /P0 given in the next slide

P0A P0


74

“Normalised” partial pressure = activity as a function of concentration aA = γAxA

γA = 1 - ideal” behaviourγA > 1 - positive deviation from ideality.

1.0

Act

ivity

aA

= P

A/ P

0 pure

A,s

at

1.0

0

Ideal

A-B = A-A

(γγγγ A=1)

(Rao

ult’slaw

)+vedeviation fro

m ideal

A-B weaker than A-A (γγγγ A

>1)

Mol Fraction, xA

A

A

A

A


Gas

PA < P0A


B B

B

B

B A

A

0

A A

A

A

A

A

75

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x,Ethanol

P(E

than

ol)/

P,V

ap(E

than

ol)

Ideal 2 constant Marg Exp Data

Example - Water – Ethanol- important solution for biofuels - as the result of any fermentation reaction will contain water and ethanol. A major cost in biofuel production is the separation of ethanol from water

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x, H2OP

(H2O

)/P

vap

(H2O

)

Ideal 2 constant Marg Exp Data

761.0

Act

ivity

aA

= P

A/ P

0 pure

A,s

at

1.0

0

Ideal

A-B = A-A

(γγγγ A=1)

(Rao

ult’slaw

)

-vede

viatio

n fro

m idea

l

A-B st

rong

er th

an A

-A

(γγγγ A<1

)

Mol Fraction, xA

A

A

A

A


Gas

PA < P0A


B B

B

B

B A

A

0

“Normalised” partial pressure = activity as a function of concentration aA = γAxA

γA = 1 - ideal” behaviourγA < 1 - negative deviation from ideality.

A A

A

A

A

A

77

Feature of activity function vs concentration:Activity in dilute solutions -Henry’s Law

aA = hAxAdilute concentrations always exists such that have hA=const (γγγγ0

A)

A A

A


GasPA < P0

A


B B

B

B

B B B B

Act

ivity

, aA

= P

A/ P

0 pure

A,s

at

1.0

0

ideal-ve

deviation from ideal

1.0Mol Fraction, xA

hA

0Henry’s Law

78

Feature of activity function vs concentration:Activity in dilute solutions -Henry’s Law

aA = hAxAdilute concentrations always exists such that have hA=const (γγγγ0

A)

A A

A


GasPA < P0

A


B B

B

B

B B B B

Act

ivity

, aA

= P

A/ P

0 pure

A,s

at

1.0

0

ideal

+ve deviation from ideal

1.0Mol Fraction, xA

hA

0

Hen

ry’s

Law

79

Description of measurement of Vapour pressureAlloy (Au-Cu) sample sheet is placed on a hot plate in a vacuum chamber and heated up to desired temperature. After the system has been stablised, a molybdem plate is inserted into the vacuum chamber above the sample and the gaseous metal deposits onto the Mo plate. The amount of deposition and time are then used to calculate the vapourpressure of the metal.

Vapour Pressure & Activity Au-Cu alloy [Hall, 1951] T = 1052oC

Au in Cu (mol

fraction)

Log(PAu) [atm] PAu[atm] aAu γγγγAu

0 0.00E+00 0.050 -12.6 2.51E-13 0.125 -11.85 1.41E-12 0.250 -11.5 3.16E-12 0.500 -11.23 5.89E-12 0.750 -10.92 1.20E-11 1.000 -10.71 1.95E-11

Example – vapour pressure and activity

Au Au

pure in solution

1) ??? what is at x = 0.5 ?

2) what is G of reaction Au Au ?

µµµµ∆∆∆∆ →→→→

80

Example – vapour pressure and activity

0.00E+00

2.00E-12

4.00E-12

6.00E-12

8.00E-12

1.00E-11

1.20E-11

1.40E-11

1.60E-11

1.80E-11

2.00E-11

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Mol Fraction of Au

Vap

our

Pre

ssur

e [a

tm]

Au Vapour Pressure Raoult's behaiviour Poly. (Au Vapour Pressure)

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Mol Fraction of Au

Au

Avt

ivity

0.000

0.200

0.400

0.600

0.800

1.000

Act

ivity

Coe

ffici

ent

Au activity Raoults's (ideal) l ine

Au Activity coefficient Poly. (Au activity)

Vapour Pressure & Activity Au-Cu alloy [Hall, 1951] T = 1052oC

Au in Cu (mol

fraction)

Log(PAu) [atm] PAu[atm] aAu γγγγAu

0 0.00E+00 0.000 0.050 -12.6 2.51E-13 0.013 0.258 0.125 -11.85 1.41E-12 0.072 0.580 0.250 -11.5 3.16E-12 0.162 0.649 0.500 -11.23 5.89E-12 0.302 0.604 0.750 -10.92 1.20E-11 0.617 0.822 1.000 -10.71 1.95E-11 1.000 1.000

Solution

81

Example – vapour pressure and activity measurement with Knudsen Cell


82

Example – vapour pressure and activity measurement with Knudsen Cell

1) The Mass Spectrometer with Knudsen cell in high vacuum is used to directly measure vapour pressures of various species evaporated into the chamber from the Knudsen cell

2) Another method – a quartz crystal microbalance (QCM) is used to measure the mass loss from the Knudsen cell into the ultra high vacuum (UHV – pressure below 1*10-10 mbar) and therefore the equilibrium vapor pressure within the cell Knudsen cell.


83

Key concept

6. Complex reactions with solutions,

conditions for spontaneous reactions


84

30%A solution

in B

50%A solution

in B

BA1 2

A in solution 1 A in solution 2

A in solution 1 A in solution 20 0

A AA in solution 1 A in solution 2


Criteria for equilibrium

1 ) G G 2 )

3 ) G RT lna G RT lna4 ) a a

µ µµ µµ µµ µ========

+ = ++ = ++ = ++ = +====

============================================================

B in solution 1 B in solution 2

same for all other components a a

========================================================================

====

Direction of reactions with solutions- diffusion will proceed till there is no difference in chemical potentials of individual species between phases

These arguments apply to any system at equilibrium –solid, liquid, gas.

Ope

ned

part

ition

85

Direction of reactions with solutions- diffusion will proceed till there is no difference in chemical potentials of individual species between phases

DEMO – partitioning of solute A between two immiscible liquids B and C

Pure C (0%A)

50%A solution

in B

?%A solution

in C

?%A solution

in B

These arguments apply to any system at equilibrium –solid, liquid, gas.


A in solution 1 A in solution 20 0

A AA in solution 1 A in solution 2


Criteria for equilibrium

1 ) G G 2 )

3 ) G RT lna G RT lna4 ) a a

µ µµ µµ µµ µ========

+ = ++ = ++ = ++ = +====

============================================================

B in solution 1 B in solution 2

same for all other components a a

========================================================================

====

86

REACTION EQUILIBRIA INVOLVING SOLUTIONS

Consider the general chemical reaction occurring at constant temperature and pressure:

bB + cC +…. = mM + nN +….

where b, c, ..m, n.. are the stoichiometric coefficients indicating the number of moles of the species B, C, .., M, N…. respectively.

87

(((( )))) (((( ))))

R R e action

Pr oducts Re ac tan ts M N A CR

The Gibbs Free Energy change associated with the general reaction

bB + cC + ... = mM + nN + ... G

G G G mG nG ... - bG cG ...

substituting chemical potentials G

∆∆∆∆

∆∆∆∆

−−−−

= − = + + + += − = + + + += − = + + + += − = + + + +∑ ∑∑ ∑∑ ∑∑ ∑0

i i i

0 0

M NR M N

0 0

B CB C

0 0 0 0

M N B C

G RT lna , then obtain

G m(G RT lna ) n(G RT lna ) ...

b(G RT lna ) c(G RT lna ) ...

after rearranging we obtain :

( mG nG ... bG cG ...)

∆∆∆∆

= += += += +

= + + + + += + + + + += + + + + += + + + + +

− + − + − =− + − + − =− + − + − =− + − + − =

+ + + − − − ++ + + − − − ++ + + − − − ++ + + − − − +

M N B C

m n0 M N

R R b cB C

RT(m lna n lna ... b lna c lna ...)

a a ... products G G RT ln

reac tan tsa a ...∆ ∆∆ ∆∆ ∆∆ ∆

+ + − − − →+ + − − − →+ + − − − →+ + − − − →

= += += += +

88

R

m no M N

R R b cB C

products

reac tan ts

The condition for spontaneous change

for non-standard conditions

for any general reaction G 0

a a ....G G RT ln i.e.

a a ....

term for cor

standard

conditions

∆∆∆∆

∆ ∆∆ ∆∆ ∆∆ ∆

<<<<

= += += += +

= += += += +rection term

for non-standard

conditions

89

(((( ))))

R

m no M N

R R b cB C

oR eq

o m nR M N

eq b cB C

For the special case

when the system is at equilibrium, G 0

a a ....0 G G RT ln

a a ....

G RT ln K

G a a .... productsK exp i.e.

RT reac tan tsa a ....

∆∆∆∆

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆

∆∆∆∆

====

= = + == = + == = + == = + =

= += += += +

−−−− = == == == =

eq

,

where K is the equilibrium constant

(usually K is used without "eq" subscript)

dG> 0

dG< 0 dG= 0

90

DEMO – Reaction equilibria

CH4

O2

CO2

H2ON2

CnH2n+2 - paraffin

CH4 + 2O2+N2↔CO2+2H2O+N2

o 1 2R CO2 H 2O

eq 1 2CH 4 O2

G a a K exp

RT a a

∆∆∆∆ −−−−= == == == =

91

30%A solution

in B

50%A solution

in B

SolidD

A + D ? A + D ?

Direction of reactions with solutions

50%A solution

in B

50%A solution

in C

A + D ? A + D ??Will A in solution

react with D?What is the extent

of reaction?

5)

6)

SolidD

SolidD

SolidD

reaction

0A A A

reaction A + D ...

will proceed

if G 0

with G G RT lna

∆∆∆∆

→→→→

<<<<

= += += += +

92

Example:Consider the general chemical reaction occurring at constant temperature and pressure:

B(s) + cC(g)+…. = mM(l) + nN(g)

if C and N are gaseous then (for pure component standard state) ac = PC and aN = PN

mo M

R R bB

nNc

C

a .P

P

...G G RT ln

a ....∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

93

Example:Consider the general chemical reaction occurring at constant temperature

and pressure:

bB(s)+ cC(g) +…. = mM(l) + nN(g)

if B and M are pure then aB = 1 and aM = 1

no N

R R cC eq

P ....G G RT ln

P ....

1

1∆ ∆∆ ∆∆ ∆∆ ∆

××××= += += += +

××××

94

Consider the general chemical reaction occurring at constant temperature and pressure:

bB(s)+ cC(g) +…. = mM(l) + nN(g)

if B and M behave ideally, then (for pure component standard state)

aB =xB and aM = xM

no

mMbB

NR R c

C eq

P ....G G RT ln

P ..

x

x ..∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

95

Example: Will PbO be reduced in CO/CO2 gas?PbO(s)+CO(g)↔Pb(l)+CO2(g)

1) for gaseous: aCO = PCO and aCO2 = PCO2

.2) if PbO and Pb are pure then aPbO = 1 and aPb = 1

1CO2( g

1Pb( l )o

R R)

1CO

1PbO ( g( s ) )

aG G RT ln

a

P

P∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

1CO2( g )o

R R 1CO( g )

1

1

PG G RT ln

P∆ ∆∆ ∆∆ ∆∆ ∆

××××= += += += +

××××

PbO(s)Pb(l)

96

Example: PbO(s)+CO(g)↔Pb(l)+CO2(g)

3) if PbO and Pb are in solutions and behave ideally (Raoult’s Law), then (for pure component standard state)

aPbO =xPbO and aPb = xPb

1CO2( g )o

R

1Pb( l )

1PbO( s )

R 1CO

x

x

PG G RT ln

P∆ ∆∆ ∆∆ ∆∆ ∆

××××= += += += +

××××

97

The approach is true for ALL chemical reactions

m no M N

R R b cB C

a a ....G G RT ln

a a ....∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

dG> 0

dG< 0 dG= 0

98

SUMMARY Direction of reactions with solutions

i i

j i

0i i ii i

i T ,P ,N

i ii i0 0

i i

i i i i

>1 - positive deviation, <0-negative d

G1 ) G chemical potentiaal; 2 ) G =G RT lna

N

f P3 ) a , for low pressures a

f P

4 ) a x ( is function of composition)

γ γγ γγ γγ γ

µµµµ


≠≠≠≠

∂∂∂∂= − = += − = += − = += − = + ∂∂∂∂

= == == == =

====

i

A in solution 1 A in solution 2 A in solution 1 A in solution 2

0

A A in sol

eviation,

=0 -ideal(Raoult's Law) - do not confuse with ideal gas

5 ) condition for equilibria: G G (or )

G RT lna

γγγγ

µ µµ µµ µµ µ= == == == =

++++0

Aution 1 A in solution 2


A in soltn.1 A in soltn.1 A in soltn.2 A in soltn.2

0 0 0i i ii i i i i

0i i

G RT lna a a x x

6 ) G RT lna G RT ln x G RT ln x RT ln

7 ) Henrian Law or h


γ γγ γγ γγ γγγγγ

= += += += +====

====

+ = + = + ++ = + = + ++ = + = + ++ = + = + +

m no M N

R R b cB C

a a ....8 ) G G RT ln

a a ....∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

99

Example: Mono-pressure nitric acid plant:

A key part of the process if the oxidation of ammonia to nitrous oxide in the NH3 converter:

NH3(g) + 1.25O2(g) → NO(g) + 1.5H2O(g) ∆∆∆∆rG0(1073.15) = -276,764 J/mol

Q.? Is reaction spontaneous at 800oC if the gas stream has the following partial pressures?

NH3: 1×10-6 MPaO2: 0.15 MPaNO: 0.1 MPaH2O: 0.15 MPa

100

Example: Mono-pressure nitric acid plant:

A key part of the process if the oxidation of ammonia to nitrous oxide in the NH3 converter:

NH3(g) + 1.25O2(g) → NO(g) + 1.5H2O(g) ∆∆∆∆rG0(1073.15) = -276,764 J/mol

Q.? Is reaction spontaneous at 800oC if the gas stream has the following partial pressures?

Solution:

NH3: 1×10-6 MPa =9.9×10-06 atmO2: 0.15 MPa = 1.480 atmNO: 0.1 MPa = 0.987 atmH2O: 0.15 MPa = 1.480 atm

1 1.5o NO H 2O

R R 1 1.25NH 3 O2

1 1.5

6 1.25

R

a aG G RT ln

a a

0.987 1.480276,764 8.314* 1073.15ln -173169 J/mol

9.9 10 1.480

G <0 - reaction will proceed spontaneously

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆

−−−−

××××= + == + == + == + =

××××

××××= − + == − + == − + == − + = × ×× ×× ×× ×

101

Example: Can Mn catalyst on Al2O3 substrate be used in a process at 900oC and syngas with CO/CO2 volume ratio =106?

1. Will Al2O3 and Mn react?2. Will Mn be oxidised to MnO? 3. Will Al2O3 be reduced to Al?

Mn

Al2O3

CO/CO2

103

Example: Step 1. Will Al2O3 and Mn react?Solution:

1. Formulate reaction 3Mn+Al2O3↔3MnO+2Al2. Assumptions: pure Mn, Al, Al2O3, MnO, no intermediate compounds or solutions3. Find ∆∆∆∆RG0

1173K

1) Mn(s) + 0.5O2(g) →MnO ∆∆∆∆R1G01173K=-388900 +76.3T = -299 KJ mol-1

2) 2Al(l) + 1.5O2(g) →Al2O3(s) ∆∆∆∆R2G01173K=-1682900 +323.2T = - 1303 KJ mol-1

3) 3Mn(s)+Al2O3(s) ↔3MnO(s)+2Al(l) (R3) = 3*R1 – R2 ∆∆∆∆R3G0

1173K=3*(-299)-(-1303) = +406 KJmol-1 > 04. Find ∆∆∆∆RG1173K

since ∆∆∆∆R3G1173K> 0, the reaction of reduction of Al2O3 and oxidation of Mnwill not proceed, so Mn catalyst will not react with Al2O3 ceramic substrate –the Al2O3 can be used as ceramic substrate for Mn catalyst (if no intermediate compounds or solutions are formed)

3 1o MnO Al

R3 R3 3 1Mn Al 2O3

3 1MnO Al

MnO Al Mn Al 2O3 3 1Mn Al 2O3

o 1R R

a aG G RT ln ,

a a

a afor pure phases a a =a a =1 , then ln 0,

a a

then G G 0 406KJmol

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆ −−−−

= += += += +

= = == = == = == = =

= + = += + = += + = += + = +

104

Step 2. Will Mn be oxidised to MnO?1. Formulate reaction 1) Mn(s) + CO2(g) →MnO + CO(g) 2. Assumptions: Mn is oxidised to MnO,

pure Mn & MnO, ideal CO &CO2

3. Find ∆∆∆∆R1G01173K= R2+R3-R4=-118 KJ/mol

R2) Mn(s) + 0.5O2(g) →MnO ∆∆∆∆RG01173K= -299 kJ/mol

R3) C + 0.5O2 →CO ∆∆∆∆RG01173K= -214 kJ/mol

R4) C+O2→CO2 ∆∆∆∆RG01173K= -396 kJ/mol

4. Find ∆∆∆∆R1G1173K

As ∆∆∆∆R1G1173K>0 – then Mn catalyst will not be oxidised in a process at 900oC and syngas with CO/CO2 volume ratio =10 6.

(((( ))))(((( ))))

1 1o MnO CO

R1 R1 MnO Mn1 1Mn CO2

oCO CO2 R1 R1 CO CO2

6

a aG G RT ln , assume pure phases a a =1,

a a

for P and P in atm G G RT ln P / P

117985 8.314* 1173.15* ln 10 117985 134750 16765J / mol

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

= + == + == + == + =

= + == + == + == + =

= − + = − + = += − + = − + = += − + = − + = += − + = − + = +

105

Step 3. Will Al2O3 be reduced to Al?1. Formulate reaction 1) Al2O3(s)+3CO→2Al(l) + 3CO2(g) 2.Assumptions: pure Al2O3(s) reacts to pure Al(l),

CO & CO2 are ideal

3. Find ∆∆∆∆R2G01173K = -R2-3R3+3R4=759KJ/mol

R2) 2Al(l) + 1.5O2(g)→Al2O3(s) ∆∆∆∆RG01173K= -1303 KJ/mol

R3) C + 0.5O2→CO ∆∆∆∆RG01173K= -214 KJ/mol

R4) C+O2 →CO2 ∆∆∆∆RG01173K= -396 KJ/mol

4. Find ∆∆∆∆R2G1173K

As ∆∆∆∆R1G1173K>0 – then Al2O3 substrate will not be reduced to Al in a process at 900oC and syngas with CO/CO2 volume ratio =106.

(((( ))))

2 3o Al CO2

R1 R1 Al 2O3 Al1 3Al 2O3 CO

o 3 3CO CO2 R1 R1 CO2 CO

6

a aG G RT ln , assume pure phases a a =1,

a a

for P and P in atm G G RT ln P / P

759528 3* 8.314* 1173.15* ln( 10 ) 759528 404252355276 J / mol

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆−−−−

= + == + == + == + =

= + == + == + == + =

= + = − == + = − == + = − == + = − =

106

Example: What temperature and syngas composition (CO/CO2 ) rangesare safe for Mn catalyst on Al2O3 substrate be used in a process ?

1) 3Mn(s)+Al2O3(s) → 3MnO(s)+2Al(l) ∆∆∆∆R1G0T=...

2) Mn(s) + CO2(g)→MnO + CO(g) ∆∆∆∆R2G0T=...

3) Al2O3(s) +3CO → 2Al(l) + 3CO2(g) ∆∆∆∆R3G0T=...

2 3o Al CO2

R3 R3 1 3Al 2O3 CO

a aG G RT ln ?

a a∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

1 1o MnO CO

R2 R2 1 1Mn CO2

a aG G RT ln ?

a a∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

3 1o MnO Al

R1 R1 3 1Mn Al 2O3

a aG G RT ln ?

a a∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

If in addition we would need to select a ceramic material for substrate and a metal for catalyst in addition to the question about the ranges of conditions – that would require collection of a lot of data and extensive calculations.

Further calculations are needed to calculate ranges of T and CO/CO2

107

APPENDIX –EXAMPLES OF Ni OXIDATION AT DIFFERENT CONDITIONS

108

m no M N

R b c

to

B C

solid , 2 ,gas sol ,

al P

i

t

d

a a ....bB cC ... m

Example 1: Will pure Ni oxidis

M nN ... G G RT lna a ....

Ni 0.5O NiO

E.g., for the reac

e at 1300K and P =1 atm?

tio

1a) using C expressions

n

β γβ γβ γβ γ

∆ ∆∆ ∆∆ ∆∆ ∆

+ + → + + = ++ + → + + = ++ + → + + = ++ + → + + = +

+ − >+ − >+ − >+ − >

2

0MOpure 2 ,gas pure F

0 0 00MO M OMeOF

0 0 0 1 2TT T P

T0 0T 298.15 P P298.15 298.15

of metal with oxygen M 0.5O MO G

Molar Gibbs Free Energy of Formation G G G 0.5G

G H T S ; for C A BT CT

H H C dT ; C dT

∆∆∆∆

∆∆∆∆−−−−

++++

= − −= − −= − −= − −

= − = + += − = + += − = + += − = + +

= += += += + ∫∫∫∫

�

(((( ))))2

T T 1 2 2 1 T298.15298.15

T T T0 0 1 3 2 TP PT 298.15 298.15298.15 298.15 298.15

0 0 00MO M OF ,MeO

T0298.15 ,MeO P ,MeO298.15

B( A BT CT )dT [ AT T CT ]

2C C C

S S dT ; dT ( AT B CT )dT [ Aln(T ) BT T ]T T 2

G G G 0.5G

H C dT

∆∆∆∆

− −− −− −− −

− − −− − −− − −− − −

= + + = + −= + + = + −= + + = + −= + + = + −

= + = + + = + −= + = + + = + −= + = + + = + −= + = + + = + −

= − − == − − == − − == − − =

= += += += +

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

(((( ))))(((( )))) 2

2 22

T0 P ,MeO298.15 ,MeO

298.15

T T0 0 P ,Me298.15 ,Me 298.15 ,MeP ,Me298.15 298.15

T T0 0 P ,O298.15 ,O 298.15 ,OP ,O298.15 298.15

F

CT * S dT

T

C H C dT T * S dT

T

C0.5* H C dT T * S dT

T

H∆∆∆∆

− + −− + −− + −− + −

− + − + −− + − + −− + − + −− + − + −

− + − + =− + − + =− + − + =− + − + =

====

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫0 0MeO ,T MeO ,TFT * S∆∆∆∆−−−−

109

0 0 01 2

TT T P

0 0 T

T 298.15 P P298.15 298.1

t

solid ,

ota

2 ,ga

l P

s solid ,


H H C dT ; C dT

Ni 0.5

Example 1a): Will pure Ni oxidise at 1300K and P =1 atm? (using C ) Continua

O NiO

tion...


−−−−= − = + += − = + += − = + += − = + +

= += += += +

+ − >+ − >+ − >+ − >

∫∫∫∫T T 1 2 2 1 T

298.155 298.15

0 0 T T T 1 3 2 TP PT 298.15 298.15298.15 298.15 298.15


2

C C CS S dT ; dT ( AT B CT )dT [ A ln(T ) BT T ]

T T 2

− −− −− −− −

− − −− − −− − −− − −

= + + = + −= + + = + −= + + = + −= + + = + −

= + = + + = + −= + = + + = + −= + = + + = + −= + = + + = + −

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

2

0 0 00NiO Ni OF ,NiOG G G 0.5G∆∆∆∆ = − −= − −= − −= − −

110

0 0 01 2

TT T P

0 0 T

T 298.15 P P298.15

solid , 2 ,g

total P

as solid ,


H H C dT ; C dT

Ni

Example 1a): Will pure Ni oxidise at 1300K and P =1 atm? ( 1a) using

0.5O Ni

C ) Continuat

.

O

ion ..


−−−−= − = + += − = + += − = + += − = + +

= += += += +

+ − >+ − >+ − >+ − >

∫∫∫∫T T 1 2 2 1 T

298.15298.15 298.15

0 0 T T T 1 3 2 TP PT 298.15 298.15298.15 298.15 298.15


2

C C CS S dT ; dT ( AT B CT )dT [ A ln(T ) BT T ]

T T 2

− −− −− −− −

− − −− − −− − −− − −

= + + = + −= + + = + −= + + = + −= + + = + −

= + = + + = + −= + = + + = + −= + = + + = + −= + = + + = + −

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

0298 ,15H

0298 ,15S

00-167000CP C*T-2

0.008460.007530.00418CP B*T1

46.80025.10029.960CP A*T0

38.08030.950205.000[J mol-1K-1]

-2406006300[J mol-1]

NiOsolid,γγγγNi solid,ββββO2,gasValue

01300H

01300S01300G -337051-66286-295209[J mol-1]

115.4775.45252.41[J mol-1K-1]

-1869413180532930[J mol-1]

NiOsolid,γγγγNi solid,ββββO2,gas

2

0 0 00 1NiO Ni ONiOF

0NiOF

G G G 0.5G 123161[ Jmol ]

G <0 - therefore pure Ni will oxidise at 1 atm and 1300K

∆∆∆∆

∆∆∆∆

−−−−= − − = −= − − = −= − − = −= − − = −

111

solid , 2 ,gas solid ,

0 0 0NiONiO NiOF F F

0 0NiO

0 0total F

iO

F

NF F

Ni 0.5O NiO

alternative way

Example 1b): Will pure Ni

G H T * S ,

where H and S are values

oxidise at 1300K and P =1 atm? (using

H

&

S )


∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆

∆∆∆∆

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆+ − >+ − >+ − >+ − >

= −= −= −= −

0 01 1 1NiO NiOF F

approximately valid over a range of temperatures

H 235800Jmol ; S 86.2 Jmol K∆ ∆∆ ∆∆ ∆∆ ∆− − −− − −− − −− − −= − = −= − = −= − = −= − = −

112

0 0total

so d ,

F F

li

Example 1b): Will pure Ni oxidise at 1300K and P =1 atm? (using H & S )

Continuati

N

on

i 0.ββββ

∆ ∆∆ ∆∆ ∆∆ ∆

++++ 2 ,gas solid ,

0 0 0NiONiO NiOF F F

0 0NiO NiO ,TF F

0 1NiOF

5O NiO

alternative way G H T * S ,

where H and S are values

approximately valid over a range of temperatures

H 235800Jmol

γγγγ

∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆ −−−−

− >− >− >− >

= −= −= −= −

= −= −= −= −

(((( ))))

0 1 1NiOF

0 1NiOF

0 NiO ,solid ,NiO NiOR F 0.5

Ni ,solid , O2 ,gas

; S 86.2 Jmol K

at 1300K G 235800 – 1300* 86.2 123740 Jmol

aG G RTln

a * P

Case

γγγγ

γγγγ

∆∆∆∆

∆∆∆∆

∆ ∆∆ ∆∆ ∆∆ ∆

− −− −− −− −

−−−−

= −= −= −= −

= − − = −= − − = −= − − = −= − − = −

= += += += +

============================================================================================================================================================================================================================

(((( ))))(((( ))))

total 2

NiO,solid Ni,solid O2,gas O2 2 O2

0.5NiOR

1. Will pure Ni oxidise at 1300K and P =1 atm in pure O ?

Pure NiO a =1; Pure Ni a =1, a =P ; pure O at 1atm: P =1

at1300K G =-123740+8.314*1300*ln 1/ 1*1 = -123740 J∆∆∆∆ /mol < 0

Ni will oxidise to NiO

113


0 0NiO ,solid , 1NiO NiO NiOR F F0.5

Ni ,solid , O2 ,gas

Ni 0.5O NiO

aG G R

Example Case 2: Will pure Ni oxidise in ai

Tln G 123740 Jmo

r

a

?

l* P


γγγγ

γγγγ

∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆ −−−−

+ − >+ − >+ − >+ − >

= + = −= + = −= + = −= + = −

====================================================================================================================

(((( ))))(((( ))))NiO,solid Ni,solid O2

0.5NiOR

Pure NiO a =1; Pure Ni a =1, air P =0.21

at1300K G =-123740+8.314*1300*ln 1/ 1*0.21 = -115306 J/mol < 0

∆∆∆∆

========================================================================================================


114


0 NiO ,solid ,NiO NiOR F 0.5

Ni ,solid , O2 ,

2 total

gas

Example Case 3: Will pure Ni oxidise

in Nitrogen gas with 1ppm O impurity at P =1 atm

Ni 0.5O NiO

aG G RTln

a * P

?


γγγγ

γγγγ

∆ ∆∆ ∆∆ ∆∆ ∆

+ − >+ − >+ − >+ − >

= += += += +

2

0 1NiOF

NiO,solid Ni,solid

62 total O

R

G 123740 Jmol

Pure NiO a =1; Pure Ni a =1,

Nitrogen gas with 1ppm O impurity at P =1 atm: P =10

at1300K

∆∆∆∆

∆∆∆∆

−−−−

−−−−

= −= −= −= −

============================================================================================================================================================================================================================

(((( ))))(((( ))))6 0.5NiOG =-123740+8.314*1300*ln 1/ 1*(10 ) = -49080 J/mol < 0


−−−−

============================================================================================================================================================================================================================

115

2

Ni,solid

solid , 2

62 t

,gas solid ,

R

otal O

Example: Will Ni oxidise?

Case 4: Will Ni in Cu alloy with a =0.01 oxidise at 1300K

in Nitrogen gas with

Ni 0.5O

1ppm O impurity at P =1 atm:

Ni

=

G

0

O

P 1 ?


∆∆∆∆

−−−−

+ − >+ − >+ − >+ − >

(((( ))))(((( ))))

0 0NiO ,solid , 1NiO NiO NiOF F0.5

Ni ,solid , O2 ,gas

6 0.5NiOR

aG RTln G 123740 Jmol

a * P

at1300K G =-123740+8.314*1300*ln 1/ 0.01*(10 ) = +694 J/

γγγγ

γγγγ

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆

−−−−

−−−−

= + = −= + = −= + = −= + = −

============================================================================================================================================================================================================================

mol > 0

Ni will not oxidise to pure NiO

116

2

Ni,solid

62 total O

Example: Will Ni oxidise?

Case 5: Will Ni in Cu alloy with a =0.01 oxidise at 1300K

in Nitrogen gas with 1ppm O impurity at P =1 atm: P =10

in contact with MgO-NiO crucibl

−−−−


0 0NiO ,solid , 1NiO NiO NiOR F F0.5

Ni ,solid , O2 ,

NiO

gas

,solid

Ni 0.5O NiO

aG G RTln G 123740 J

e having a 0.5

mo*

?

la P


γγγγ

γγγγ

∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆ −−−−

+ − >+ − >+ − >+ − >

= + = −= + = −= + = −= + = −

============================================================================================================================================================================================================

====

================

(((( ))))(((( ))))6 0.5NiOR

Ni,solid NiO

at1300K G =-123740+8.314*1300*ln 0.5/ 0.01*(10 ) = -6798 J/mol < 0

Ni from alloy (a =0.01) will oxidise to NiO in solution with MgO (a =0.5)

∆∆∆∆ −−−−

117

2 2

Ni,solid total

NiO ,solid

Example Case 6:

What maximum O impurity in N is allowed to avoid Ni oxidation

from Cu alloy with a =0.01 oxidise at 1300K, P =1atm,

in contact with MgO-NiO crucible having a 0.5?====


0 0NiO ,solid , 1R,NiO F ,NiO F ,NiO0.5

Ni ,solid , O2 ,gas

NiO ,solid ,

Ni ,solid

Ni 0.5O NiO

aG G RTln G 123740 Jmol

a * P

a

a


γγγγ

γγγγ

γγγγ

∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆ −−−−

+ − >+ − >+ − >+ − >

= + = −= + = −= + = −= + = −

============================================================================================================================================================================================================================

2

20 0F ,NiO F ,NiO NiO

O0.5, O2 ,gas Ni

2 7O2 ,gas

G G aexp P exp *

* P RT RT a

at 1300K P =[exp(-123740/8.314/1300)*0.5/0.01 ] =2.84*10 atm

γγγγ

∆ ∆∆ ∆∆ ∆∆ ∆

−−−−

−−−− = → == → == → == → =

2 2 0.284 ppm O in N is safe

============================================================================================================================================================================================================================

118

0 01 1 1NiO NiOs 2 ,gas s F F

2

t

g 2

o

,

tal

Example: Case 7 What minimum CO/CO ratio in gas is needed to protect

pure Ni from oxidation at 1300K

R1) Ni 0.5O NiO H 235800Jmol ; S 86.2 Jm

(P =1

ol K

R2) CO 0.5O

atm)?

∆ ∆∆ ∆∆ ∆∆ ∆− − −− − −− − −− − −+ → = − = −+ → = − = −+ → = − = −+ → = − = −

++++ 0 01 1 1g 2 ,g R2 R2

0 01 1 1s 2 ,gas s gas R3 R3

0R3

CO H 281900Jmol ; S 85.7 Jmol K

R3 R1 R2

R3 ) Ni CO NiO CO H 46100Jmol ; S 0.5 Jmol K

at 1300 K G = 46100-13

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆

− − −− − −− − −− − −

− − −− − −− − −− − −

→ = − = −→ = − = −→ = − = −→ = − = −

− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− −− −− −= −= −= −= −

+ → + = = −+ → + = = −+ → + = = −+ → + = = −1

0 NiO ,solid , CO ,gasR3 R3

Ni ,solid , CO2 ,gas

Ni,solid NiO,solid

0CO ,gas Ni ,R3

R3CO2 ,gas

00(-0.5)=+46750 Jmol

a * PG G RTln

a * P

pure Ni a =1 ; pure NiO a =1

P aGat equilibrium G 0,then exp *

P RT

γγγγ

γγγγ

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆∆∆∆∆

−−−−

= += += += +

−−−−= == == == =

solid

NiO ,solid

CO ,gas 2

CO2 ,gas

a

P 46750 1at 1300K exp * = 1.32*10

P 8.314* 1300 1−−−−−−−− ====

119

References:1. Koretsky, M.D., 2004. Engineering and Chemical Thermodynamics,

John Wiley & sons Inc., Hoboken, NJ.2. Lee, H.G., 1999. Chemical thermodynamics for metals and materials ,

Imperial College Press, London.3. Atkins, P. & de Paula, J., 2006, Atkins' Physical Chemistry, 8th edition,

Oxford University Press, Oxford, NY.

List of DEMOS:

1. Ethanol – water volume2. Solubility product – make the solution 3. Two immiscible liquid solvents and a solute –

CuSO4 partition can be affected by H2SO4

4. Reaction equilibria – candle in air

120

Example: 2Ag(s)+0.5O2(g)↔Ag2O(s)

1) for gaseous: aO2 = PO2

2) if Ag and Ag2O are pure then aAg=1 and aAg2O = 1

2

2

1Ag O( s )o

R R 2A

0.5O (s ) g )g(

aG G RT ln

Pa∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

2

2

1Ag O( s )

2A

oR R 0.5

O (( g )g s )

G1

R lnP1

G T∆ ∆∆ ∆∆ ∆∆ ∆ = += += += +

2

2

1Ag O( s )o

R R 2 0.5Ag( s ) O ( g )

aG G RT ln

a a∆ ∆∆ ∆∆ ∆∆ ∆

= += += += +

121

Example: 2Ag(s)+0.5O2(g)↔Ag2O(s)

3) if Ag and Ag2O behave ideally (Raoult’s Law), then (for pure component standard state)

aAg =xAg and aAg2O = xAg2O

2

2

1Ag O( s )

2A

oR R 0.5

O (( g )g s )

Gx

R lnPx

G T∆ ∆∆ ∆∆ ∆∆ ∆ = += += += +

1

Lecture week 7 (2012)CHEE3003

CALCULATION AND REPRESENTATION OF COMPLEX EQUILIBRIA

UNDER NON-STANDARD CONDITIONS

2

Key Concepts:

Graphical representation of complex equilibria

- Ellingham Diagrams

- Predominance Diagrams

3

For example:Will a catalyst oxidise in particular gas stream in a range of temperatures? Will production gas oxidise tubes or reduce ceramic (complex oxide) filters at a given temperature?Can you use Fe plate with Al2O3 refractory?Can you heat PbO in Ni crucible?Can you selectively reduce some of the metal oxides and keep others unchanged?Which PbO, Pb3O4 or PbO2 will be stable at what T and P?

Or more general:What range of operational conditions has to be used to achieve a certain target or to avoid a certain reaction? E.g. temperature, total pressure, composition

!!! Graphical and computer tools are used for complex systems

Real systems are not pure, complex, multi-component, multi-phase.

A number of seemingly simple important practical questions require extensive calculations to answer.

4

ELLINGHAM first constructed a very useful diagram

– now called ELLINGHAM DIAGRAM

– that shows stability of certain groups of compounds.

– can assist in fast answer to a wide range of questions

– are now widely used.

In the following slides we will construct several lines for the ELLINGHAM DIAGRAM.

5

2

0pure 2 ,gas pure F MO

0 00MO M OF MeO

e.g. METAL OXIDES

For the reaction of metal with oxygen M 0.5O MO G

where n is oxidation state of the metal oxide

Molar Gibbs Free Energy of Formation G G G 0.5G

∆∆∆∆

∆∆∆∆

−−−−

++++

= − −= − −= − −= − −

��

2

2 22

0

T T0 0 P ,MeO298.15 ,MeO 298.15 ,MeOP ,MeO298.15 298.15

T T0 0 P ,Me298.15 ,Me 298.15 ,MeP ,Me298.15 298.15

T0 0 P ,O298.15 ,O 298.15 ,OP ,O298.15 2

CH C dT T * ( S dT )

T

CH C dT T * ( S dT )

T

C0.5* H C dT T * ( S dT )

T

====

= + − + −= + − + −= + − + −= + − + −

− + − + −− + − + −− + − + −− + − + −

− + − +− + − +− + − +− + − +

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫T

98.15

0 0 0F F FMeOMeO MeO

0 0F FMeO MeO

G H T * S

the values H and S for metal oxides do not change a lot with temperature

- approximate "mean" values valid over a range of temperatures are frequent

∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

= −= −= −= −

∫∫∫∫

ly

tabulated and used - this is common for other groups of compounds (e.g. sulfides etc)

REPRESENTATION OF EQUILIBRIA ∆∆∆∆GF0 vs T

6

Although CP of compounds are complex functions of temperature, the terms ∆∆∆∆RH0 and ∆∆∆∆RS0 do not change significantly with T !!!

2

0 0 00 0MO M Opure 2 ,gas pure F FMO MeOe.g. METAL OXIDES 0.5M 0.5O MO G ; G G 0.5G 0.5G∆ ∆∆ ∆∆ ∆∆ ∆− + = − −− + = − −− + = − −− + = − −��

m n0 0 0 00 0M NM N B CR R Rb c

B C

0 0 0 1 2R R R PTT T

00298.15 ,iR T

i

REACTION EQUILIBRIA bB + cC + ... = mM + nN + ...

a a ...G G RT ln G ( mG nG ... bG cG ...)

a a ...

At temperature T G H T S ; for C A BT CT

H iH

∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

∆∆∆∆

−−−−

−−−−

= + = + + − − −= + = + + − − −= + = + + − − −= + = + + − − −

= − = + += − = + += − = + += − = + +

==== (((( ))))T T00 P ,i298.15 ,iP ,i R T298.15 298.15

i i i

Ci C dT and S iS i dT

T

where i - stoichiometric coefficients (products "+"; reactants"-")

∆∆∆∆

+ = ++ = ++ = ++ = +

∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑∫ ∫∫ ∫∫ ∫∫ ∫

7Temperature, K

∆G

F0 [k

J m

ol-1] 4Ag(s) + O2(g)↔2Ag2O(s)

∆∆∆∆ F’G0 =∆∆∆∆ F’

H0 -T∆∆∆∆ F’

S0 =-56.2+0.1212 J mol-1 O2

0

-60

-40

-20

20

40

0 100 200 300 400 500 600 700

1)2Ag(s) + 0.5O2(g)↔Ag2O(s) ∆∆∆∆FG0=-28.1+0.0606T [kJ per mol Ag2O]

2) 4Ag(s) + O2(g)↔2Ag2O(s) ∆∆∆∆F’G0=-56.2+0.1212T [kJ per mol O2]

! Per mole of O2 !

∆∆∆∆F’H0 =-56.2kJ mol-1

- intersection at T = 0oK

−−−−∆∆∆∆F’S0=+0.1212 kJ mol-1K-1

- slope

0 0 0F F FMeOMeO MeOFor metal oxides G H T * S∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆= −= −= −= −

REPRESENTATION OF EQUILIBRIA ∆∆∆∆FG0 vs T

Standard conditions:pure metal, pure oxide

8Temperature, K

∆G

F0 [k

J m

ol-1]

4Ag(s) + O2(g)↔2Ag2O(s)

∆∆∆∆ F’G

0 =∆∆∆∆ F’H

0 -T∆∆∆∆ F’S0 =-56.2+0.1212 J mol-1 O2

0

-60

-40

-20

20

40

0 100 200 300 400 500 600 700

4Ag(s) + O2(g)↔2Ag2O(s) ∆∆∆∆F’ G0=-56.2+0.1212T [kJ per mol O2]

−−−−∆∆∆∆F’S0

0 0 0F F FMeOMeO MeOFor metal oxides G H T * S∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆= −= −= −= −


From higher disorder -4 moles solid+1mol gas

to low disorder(higher order)-2 moles of solid

entropy (measureof disorder) decreased ∆∆∆∆S<0

So for metal oxides ∆∆∆∆F G

0=∆∆∆∆F H0– T∆∆∆∆F S

0

-slopes are positive and similar

9

2Ag2O0

R F' 4 1Ag O2

0R F ' O2

R

0F' O2

aG G RT ln

a * P

For pure metal and pure oxide G G RT ln( 1 / P)

At equilibrium

G 0, so

G RT ln P

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆

∆∆∆∆

= += += += +

= += += += +

====

====

Temperature, K

∆G

F0 [k

J m

ol-1]

4Ag(s) + O2(g)↔2Ag2O(s)

0

-60

-40

-20

20

40

0 100 200 300 400 500 600 700

4Ag(s) + O2(g)↔2Ag2O(s) ∆∆∆∆F’G0=-56.2+0.1212T [kJ per mol O2]REPRESENTATION OF EQUILIBRIA ∆∆∆∆FG0 vs T

∆∆∆∆F’ G0 – measure of chemical affinityof metalto oxygen

Decomposition T

∆∆∆∆F G0>0 Ag is stable

∆∆∆∆F G0<0AgO2 is stable

∆∆∆∆ F’G

0 =∆∆∆∆ F’H

0 -T∆∆∆∆ F’S0 =-56.2+0.1212 kJ mol-1 O2

10


If Gibbs Free energies of formation are presented per 1 mol of common gaseous species – O2 , then the relative stabilities of metal oxides can be readily compared

e.g.Ag1+ : 4Ag(s) + O2(g) ↔ 2Ag2O(s)

Fe2+ : 2Fe(s) + O2(g) ↔ 2FeO(s)

Al3+ : 4/3Al(s) + O2(g) ↔ 2/3Al2O3(s)

Ti4+ : 2Ti(s) + O2(g) ↔ TiO2(s)

11

Example: What temperature range is safe for Mn catalyst on Al2O3

substrate to be used in a process?

1) 2Mn(s) + O2(g) ->2MnO ∆∆∆∆R1G01173K=2*∆∆∆∆FG0

MnO=-777800-152.6T

2) 4/3Al(l)+O2(g)->2/3Al2O3(s) ∆∆∆∆R2G01173K=2/3∆∆∆∆FG0

Al2O3=-1121933-215.5T Jmol-1

3) 2Mn(s)+2/3Al2O3(s)↔2MnO(s)+4/3Al(l) (R3) = R1–R2=344133+62.9T >0

∆G

o , k

J m

ol-1

O2

4/3Al(l) + O2(g

) =2/3A2O3(s)2Mn + O2(g) = 2MnO(s)

∆∆∆∆R1G0=-777800+152.6T

Temperature, K

∆∆∆∆R3G0 =3/2(∆∆∆∆R1G

o-∆∆∆∆R2Go)

0 K 1173 K

-500-

-1000-∆∆∆∆R2G

0 =-1121933+215.5T Jmol-1

Presentation of ∆∆∆∆F’G0 per 1 mol of oxygen enables direct graphical analysis of relative stability of different metals and oxides

12

∆G

o , k

J m

ol-1

O2

R1) 2A + O2(g) =2AO(s)R2)

B + O 2(g)

= BO 2(s)

Temperature, K

TE

∆∆∆∆R3G0 =∆∆∆∆ R2G

oBO2-∆∆∆∆R1G

o AO

∆∆∆∆G0R3<0

BO2 is more stable than AO

Reaction 3 will proceed from left to right

B will reduce AO to A

∆∆∆∆G0R3>0

AO is more stable than BO2A will reduce BO2 to B

0F

0R 2

0 02 R1 F A

ELLINGHAM DIAGRAM RELATIVE STABILITIES OF GROUP OF COMPOUNDS G vs T

E.g. for metal oxides, series of reactions G are written for 1 mol O

R1 2A + O 2AO G 2 G

R2

∆∆∆∆

∆∆∆∆

∆ ∆∆ ∆∆ ∆∆ ∆

−−−−

→ =→ =→ =→ =0 0

2 2 R2 F B

0 0 0 0 0B AF F2 R3 R2 R1

B + O BO G G

R3=R2-R1: B+2AO BO +2A G G G G 2 G

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆ ∆ ∆ ∆∆ ∆ ∆ ∆ ∆∆ ∆ ∆ ∆ ∆∆ ∆ ∆ ∆ ∆

→ =→ =→ =→ =

→ = − = −→ = − = −→ = − = −→ = − = −

R3=R2-R1

13

ELLINGHAM:

Combined Gibbs Free energies of formation per 1 mol of common gaseous species for similar groups of compounds (oxides, sulphides, chlorides etc. ).

Now called ELLINGHAM DIAGRAMS

1) Are comprehensive sources of data2) Present relative stability of various compounds or species – analysing which one will be stable3) Make analysis and calculations very simple 4) Allows analysis of sensitivity to process variables

!!! Use faster than computers !!!

14

ELLINGHAM DIAGRAM – EXAMPLE:? Can pure metallic Fe be kept at 1000oC in contact with Cr2O3 without reduction of Cr2O3?

1) 2Fe + O2 ->2FeO ∆∆∆∆R1G0

1000C=-360 kJ mol O2-1

2) 4/3 Cr + O2->2/3Cr2O3

∆∆∆∆R2G01000C=- 525 kJ mol O2

-1

3)2/3Cr2O3 + 2Fe->2FeO+4/3Cr(R3) = (R1)–(R2)

∆∆∆∆R3G01000C=-360-(-525)= +165 kJmol-1 > 0

Reaction 3 will not proceed –Cr2O3 is more stable at 1000oCFe can contact Cr2O3 at 1000oC

The answer can be obtained directly from the graph - Cr2O3 line is lower than FeO

15

Example: Can other ceramic substrates (SiO2, TiO2, Cr2O3, MgO) be used for Mn catalyst and in what range of temperatures?

Cr2O3 can’t

SiO2 TiO2Al2O3MgO

can be used –lines are lower,

oxides are more stable

16

Example: Which ceramic (oxide) crucible (SiO2, TiO2, Cr2O3, MgO)can be used to melt Al?

Cr2O3 SiO2TiO2can’t

MgOcan be used –

the line is lower, MgO

is more stable

17

REPRESENTATION OF EQUILIBRIA2A(s,l)+O2=2AO(s,l) ∆∆∆∆F’G

0 = ∆∆∆∆F’H0 -T∆∆∆∆F’S

0

Slope of the line changes with the change of state

Sgas>Sliq>Ssolid

2A(s)+O2=2AO(s)2A(l)+

O2=2AO(s)

2A(l)+O2=

2AO(s)

2A(l)+O2=2AO(l)

AO

AO−−−−∆∆∆∆RS0

∆∆∆∆FS0=2S0AO-S0

O2-2S0A

18


Slope of the line changes at the change of state

19


Slope of the line changes at the change of state

20

REACTIONS AT NON STANDARD CONDITIONS IN O2 at fixed PO2

4Ag(s) + O2(g) -> 2Ag2O(s) Effect of PO2

1) if Ag and Ag2O are pure,

then ∆∆∆∆G=∆∆∆∆FG0+RTln(1/PO2) = (∆∆∆∆FH0-T∆∆∆∆FS0 ) - RTln(PO2)

2) ∆∆∆∆FG0 is negative for most of oxides and increases with temperature

3) For a given O2 partial pressure PO2 at Ptotal =1 atm, PO2 <1, lnPO2 <0

then the term [ Rln(PO2) ]<0, then the term [-RTln(PO2)] is positive,

and makes more and more positive contribution to ∆∆∆∆G as temperature increases

Richardson added nomographic scale for the [-RTln(PO2)] term to Ellingham diagrams

21

COMPOUNDS IN NON STANDARD CONDITIONS IN O2 at fixed PO2

Effect of PO24Ag(s) + O2(g) -> 2Ag2O(s) if Ag and Ag2O are pure, then ∆∆∆∆G=∆∆∆∆G0+RTln(1/PO2) =∆∆∆∆G0-RTln(PO2)

RTlnPO2

PO

2 sc

ale

[atm

]

10-X

∆RG=∆∆∆∆F’G

0-RTlnPO2

1

∆∆∆∆G<0 : Ag2O stable at fixed PO2 ∆∆∆∆G>0 : Ag stable at fixed PO2

∆∆∆∆G<0Ag2Ostableat T

∆∆∆∆G>0Ag stableat T

Temperature, K

∆G

F0 [K

J m

ol-1]

0

-60

-40

-20

20

40

0 100 200 300 400 500 600 700

4Ag(s) + O2

(g)↔2Ag2

O(s) ∆∆∆∆G

0At equilibrium0=∆∆∆∆G=∆∆∆∆F’G0-RTlnPO2

∆∆∆∆F’G0=RTlnPO2intersection

22

ELLINGHAM DIAGRAMRichardson’s nomographic scale for RTlnPO2

23

Superposition of Ellingham Diagram and RTlnPO2

2A+O2↔2AO ∆∆∆∆RG=∆∆∆∆RG0 +RTlnPO2

e.g. at T2

term ∆∆∆∆RG0 = (cd)

is compensated by term -RTln(PO2) =(cd)at PO2=10-8 atm

so that ∆∆∆∆RG=0

and reaction 2A+O2↔2AO is at equilibrium

at PO2>10-8 or T<T2

∆∆∆∆RG <0 – Me is oxidised

at PO2<10-8 or T>T2

∆∆∆∆RG >0 – Me is oxidised

∆∆∆∆ RG0

RTlnPO

2

24

? What is critical oxygen partial pressure PO2 in the gas to avoid Cu oxidation at 1000oC and 1 atm total pressure?Cu + O2 ->Cu2O – answer PO2 = 10-7 atm

25

Example: What critical CO/CO2 ratio in the gas to avoid Cu oxidation at 1000oC?0

2 2 R1

02 2 R2

2 2

2 20 Cu2O CO

R3 R3 4 2Cu CO2

0R1 R

R1) 4Cu(s) + O (g) 2Cu O G

R2) 2CO(g) + O (g) 2CO (g) G

R3) 4Cu(s) + 2CO (g) 2Cu O+2CO R3 = R1 - R2

a PG G RT ln

a * P

( G

∆∆∆∆

∆∆∆∆

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

↔↔↔↔

↔↔↔↔− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −

↔↔↔↔

= + == + == + == + =

= −= −= −= −2 2

0 Cu2O CO2 4 2

Cu CO2

20 0Cu2O CO

R1 R24Cu CO2

0 COR2

CO2

a PG ) RT ln RT ln

a P

a PG RT ln G 2RT ln

a P

PThe nomographic scale for the term G 2RT ln

P

is added to the El

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆

+ ++ ++ ++ +

= + − −= + − −= + − −= + − −

−−−−

lingham diagram

26

? What is critical CO/CO2 ratio in gas to avoid Cu oxidation at 1000oC and 1 atmtotal pressure?4Cu + 2CO2↔2Cu2O+2COanswer CO/CO2 = 10-3.5 atm

27

Example: What critical H2/H2O ratio in the gas to avoid Cu oxidation at 1000oC?0

2 2 R1

02 2 2 R2

2 2 2

2 20 Cu2O H 2

R3 R3 4 2Cu H 2O

0R1 R2

R1) 4Cu(s) + O (g) 2Cu O G

R2) 2H (g) + O (g) 2H O(g) G

R3) 4Cu(s) + 2H O(g) 2Cu O+2H R3=R1 - R2

a PG G RT ln

a * P

( G

∆∆∆∆

∆∆∆∆

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

↔↔↔↔

↔↔↔↔− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −

↔↔↔↔

= + == + == + == + =

= −= −= −= −2 2

0 Cu2O H 24 2Cu H 2O

20 0Cu2O H 2

R1 R24Cu H 2O

0 H 2R2

H 2O

a PG ) RT ln RT ln

a P

a PG RT ln G 2RT ln

a P

PThe nomographic scale for the term G 2RT ln

P

is added to the Ell

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆

+ ++ ++ ++ +

= + − −= + − −= + − −= + − −

−−−−

ingham diagram

28

? What is critical H2/H2O ratio in gas to avoid Cu oxidation at 1000oC and 1 atmtotal pressure?4Cu + 2H2O ↔2Cu2O+2H2

answer H2/H2O = 10-3.7 atm

29

Example: At what temperature SiO2 can be reduced to Si by pure carbon?

(C+O2↔2CO line on the Ellingham Diagram)

02 2 R1

02 R2

2

R1) Si(s) + O (g) SiO G

R2) 2C + O (g) 2CO(g) G

R3) SiO (s) + 2C Si+2CO(g) R3= - R1 + R2

∆∆∆∆

∆∆∆∆

↔↔↔↔

↔↔↔↔− − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − − − − − − − − − − − − − −

↔↔↔↔

30

? At what temperature SiO2 can be reduced to Si by pure carbon?

31

ELLINGHAM DIAGRAM –EXAMPLE:? Will Ni be oxidised at 1000oC at 1 atmtotal pressure in nitrogen gas N2 with 1ppm O2 impurity?

2Ni + O2 ->2NiO

RTlnPO2=-150 kJmolO2-1

∆∆∆∆F’G0 =-240 kJmolO2

-1

∆∆∆∆RG = ∆∆∆∆F’G0-RTlnPO2 =

= -240 – (-150) =-90 kJmolO2-1 <0

Ni will be oxidised

(critical PO2 is 10-10 atm)

32

∆∆∆∆FG0 vs T diagram for hydrocarbons shows that in general low molecular weight hydrocarbons are thermodynamically more stable than those of high molecular weight

This is used in Petroleum industry for hydrocarbon promoting cracking reactions by heating

e.g.

Cn+mH2(n+m)+2→CnH2n + CmH2m+2

33

OTHER ELLINGHAM DIAGRAMS:

Ref: Y.K.Rao, “Stoichiometry and thermodynamics of metallurgical processes”, Cambridge Univ. Press. 1985, p. 371:

Chlorides, Fluorides:- Kellogg,Trans.AIME, 1950, v188, pp.862-72; ibid.1951, 191, pp. 137-141

Oxides, Chlorides, Fluorides:–Glassner, ANL-5750, Argonne National Laboratory, Argonne, Ill., 1957

Oxides, Halides, Carbides:- Wicks & Block,Bulletin 605, Bureau of Mines, US Dept. of Interio, Washinton, DC, 1963

Oxides, Sulfides, Halides, Nitrides, Hydrides, Selenides, Tellurides: – Reed, MIT Press, Cambridge, Mass., 1971

34

PREDOMINANCE DIAGRAMS

We can now predict the conditiosn for a particular chemical reactions to proceed under non-standard conditions.

Question:

What can be done if gas contains several species and a number ofalternative products can be formed?

Answer:

The graphical representation concept can be extended to include other variables

35


● Graphical representation of multiple reaction equilibria as functions of process variables

● Two gas composition / activity variables + many condensed phases+ total pressure on one graph – analysis of which one is stable

● Analyse sensitivity to process variables

● Map over ranges of operating conditions

● More possibilities

● Easy for analysis

● Use of computer power

36

PREDOMINANCE DIAGRAMS- graphical representation of multiple reaction equilibria as functions of process variables

EXAMPLE: Si-C-O system forms the basis for 2 important industrial processes:1) Production of SiC from SiO2 using C (SiC is used in abrasive / grinding and as heat resistant refractory material)2) Production of Si metal from SiO2 using C (Si is used in photovoltaic cells for converting solar energy into electricity)

Conditions necessary to produce these different products from the same starting materials can be evaluated using :PREDOMINANCE DIAGRAMS

FeedSiO2(s) | C(s) | ?O2(gas)

SiC or Si

CO,gas

Electric Energy

High T

Molten Zone

CO2,g

SiOg

37

EXAMPLE: Production of SiC or Si metal from SiO2 using C

Feed SiO2(s) | C(s) | ?O2(gas)

SiC or Si

CO,gas

Electric Energy

High T

Molten Zone

CO2,g

SiOg

SiO2(s)

CO+SiO2(s)→Si liq

+CO2(gas)

C(s)

CO2(gas)

COgas

CO2,gas +C

(s)→2COgas

Siliq

C(s) Siliq2Cs+SiO2(s)→Si liq+2COgas

Possible Reactions

The question :what areconditions (e.g. PO2 and aC

at a given T)to produce SiC or Si

38


EXAMPLE: Si-C-O SYSTEM, Temp FIXED------------------------------------------------------------------------

Possible condensed phases/species: SiO2, SiC, Si, C, …

Possible gaseous species:O2, CO, CO2, SiO, …

------------------------------------------------------------------------Question: What temperature and reduction/oxidation conditions (PO2, aC, PCO, PCO2, … ) are needed to get stable Si or SiC ? ------------------------------------------------------------------------

Variables selected for diagram: log(aC) and log(PO2)

Condensed phases for analysis: SiO2, SiC, Si------------------------------------------------------------------------

39

-14 -13 -12 -11-3

-2

-1

0lo

g(a C

)

log(PO2)

SiCSi

Si SiO2

SiC SiO2

SiC(s)

SiO2(l)

Si(l)

(4) C(s)+0.5O2(g)->CO(g) P

CO=10-1atmPCO=10-2 atm

PCO=1 atm

C(s) stable

! ! ! ! Industrial conditions

for SiC at Ptotal =1atm

PREDOMINANCE DIAGRAMS are used to select operating conditions fuel/O2 ratio, Temperature, Ptotal , otherExample: Si-C-O system at T=2133.5K

40

PREDOMINANCE DIAGRAMS CONSTRUCTION

Procedure:

1. Fix temperature and select 2 variables describing gas composition or activity of an element (e.g. T=2133.5K, aC , PO2)

2. Identify all possible condensed phases (e.g.Siliq, SiO2,liq, SiCs)

3. Write down all possible reactions between combinations of these condensed phases with selected variables and derive equations for critical conditions (equilibria) corresponding to these reactions

4. Assess relative stabilities of the condensed phases and identify ranges of stabilities of the condensed phases in terms of selected 2 variables

41

PREDOMINANCE DIAGRAMS EXAMPLE Si-C-O SYSTEM, T=2133.5K

log(

a C)

log(PO2)

0

?SiC

? SiO2

?Si

Aim – to define operation conditions in terms of aC and PO2

at a given temperature to produce Si(l) or SiC(s) from SiO2(l)

Reactions to consider:Si(l) + O2(g) -> SiO2(l) (1)Si(l) + C(s) -> SiC(s) (2)SiC(s) + O2(g) -> SiO2(l) +C (3)

For pure Si, SiC, SiO2: aSi = 1, aSiC=1, aSiO2=1

42

(((( )))) (((( )))) (((( ))))

2 ,solid solid solid 2

2 2

possilbe phases are 1.pure SiO ; 2.pure Si ; 3.pure SiC ;4. C in solution and 5.gas with O

PREDOMINANCE DIAGRAMS example : Si-C-O system, T=2133.5 K

Si l + O g -> SiO l (((( ))))(((( )))) (((( )))) (((( )))) (((( ))))

(((( )))) (((( )))) (((( )))) (((( ))))

0R1

0R2

02 2 R3

1 G =?

Si l + C s ->SiC s 2 G =?

SiC s + O g -> SiO l +C 3 G =?

---------------------------------------------------------------------

Data taken from Fact

∆∆∆∆

∆∆∆∆

∆∆∆∆

2

2

-1

0 0

C ,2133.5 O ,2133.5

0 0 0

Si ,2133.5 SiC ,2133.5 SiO ,2133.5

Sage using View Data in Jmol :

G =-51566.0 G =-514127.7

G =-110659.5 G =-205727.7 G =-1150491.2

--------------------------------------------

(((( )))) (((( ))))

(((( ))))

2 2

0 0 00 -1SiO ,2133.5 Si ,2133.5 O ,2133.5R1

0 0 00SiC ,2133.5 Si ,2133.5 C ,2133.5R2

------------------------------------------

G =G -G -G =-1150491.2- -110659.5 - -514127.7 =-525704.0Jmol

G =G -G -G =-205727.7- -110659.5 - -5156

∆∆∆∆

∆∆∆∆ (((( ))))

(((( ))))

-1

0 0 0 -1R3 R1 R2

6.0 =-43502.2Jmol

G = G - G =-525704.0 – -43502.2 = -482201.8 Jmol∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

43


(note logX=lnX/2.303)


PREDOMINANCE DIAGRAMS e.g. Si-C-O system, T=2133.5 K

================================================================================

(((( )))) (((( )))) (((( )))) (((( ))))2

2

2

2 2

0 0 -12 2 R1 F SiO ,2133.5

0SiO0 0 R1

R1 R1 R1 OSi O O

R1 Si l + O g -> SiO l 1 G = G =-525704.0 Jmol

a G10 G= G RT ln G RT ln log P

a P P 2.3

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

============================================================================================================================================================================

= + = + → == + = + → == + = + → == + = + → =

2

2

O

O C

03RT

log P -525704.0 / ( 2.303* 8.314* 2133.5 ) 12.9

log P 12.9 does not depend on a

= − −= − −= − −= − −

= = −= = −= = −= = −

44


0O2 R1R1 log P G / ( 2.303RT ) 12.9∆∆∆∆→ = = −→ = = −→ = = −→ = = −

log(

a C)

log(PO2)

(1)

Si(l

)+O

2(g)↔

SiO

2(l)

Si← →SiO2-14 -13 -12 -11

-3

-2

-1

0Reaction 1 produced line logPO2=-12.9 which divides the range of conditions into two

to the right–more oxidising –SiO2 is stable

to the left –more reducing

– Si is stable

45





================================================================================

(((( )))) (((( )))) (((( )))) (((( )))) 0 0 -1R2 F SiC ,2133.5

00 0 R2SiC

R2 R2 R2 CSi C C

C

R2 Si l + C s ->SiC s 2 G = G =-43502.2 Jmol

Ga 10 G= G RT ln G RT ln log a

a a a 2.303RT

log a

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

============================================================================================================================================================================

= + = + → == + = + → == + = + → == + = + → =

C O2

-43502.2 / ( 2.303* 8.314* 2133.5 ) 1.1

log a 1.1 - d

oes not dep

e

nd on P

====

= −= −= −= −

= −= −= −= −

46


log(

a C)

log(PO2)

↑SiC↓Si

-14 -13 -12 -11-3

-2

-1

0

0C R2R2 log a G / ( 2.303RT ) 1.1∆∆∆∆→ = = −→ = = −→ = = −→ = = −

(2) Si(l) + C(s) -> SiC(s)

Reaction 2 produced line log(aC)=-1.1 which divides the range of conditions into two

Up from the line– higher aC –SiC is stable

Down from the line–- lower aC-– Si is stable

47





================================================================================

(((( )))) (((( )))) (((( )))) (((( ))))

2

2

2 2

0 -12 2 R3

0SiO C0 0 R3C

R3 R3 R3 C OSiC O O

C

R3 SiC s + O g -> SiO s +C 3 G = -482201.8 Jmol

a a Ga0 G= G RT ln G RT ln log a log P

a P P 2.303RT

log a

∆∆∆∆

∆∆∆∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

============================================================================================================================================================================

−−−− = + = + → = += + = + → = += + = + → = += + = + → = +

2 2

2

C O

O

C O

( 482201.8 ) / ( 2.303* 8.314* 2133.5 ) log P

log a 11.8 log P - depends on both

a and P

= − − += − − += − − += − − +

= += += += +

48


0C R3 O2 O2R3 log a G / ( 2.303RT ) log P 11.8 log P∆∆∆∆→ = − + = +→ = − + = +→ = − + = +→ = − + = +

log(

a C)

log(PO2)

(3) SiC← → SiO2

-14 -13 -12 -11-3

-2

-1

0

(3) S

iC(s) + O 2

(g)↔

SiO 2(l)

+ C(s)

Reaction 3 produced line log(aC)=11.8+logPO2

which divides the range of conditions into two

Up or Leftfrom the line– higher aC

and lower PO2–SiC is stable

Down or Rightfrom the line–- lower aC or

more oxidising-– SiO2 is stable

49





================================================================================

(((( )))) (((( )))) (((( )))) (((( ))))2

2

2

2 2

0 0 -12 2 R1 F SiO ,2133.5

0SiO0 0 R1

R1 R1 R1 OSi O O

R1 Si l + O g -> SiO l 1 G = G =-525704.0 Jmol

a G10 G= G RT ln G RT ln log P

a P P 2.3

∆ ∆∆ ∆∆ ∆∆ ∆

∆∆∆∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆

============================================================================================================================================================================

= + = + → == + = + → == + = + → == + = + → =

(((( )))) (((( )))) (((( )))) (((( ))))2 2O

0 0 -1R2 F SiC ,2133.5

0 SiCR2 R2

Si C

O

03RT

log P -525704.0 / ( 2.303* 8.314* 2133.5 ) 12.9

R2 Si l + C s ->SiC s 2 G = G =-43502.2 Jmol

a0 G= G RT ln

log P 12.9

a a

∆ ∆∆ ∆∆ ∆∆ ∆

∆ ∆∆ ∆∆ ∆∆ ∆

= = −= = −= = −= = −

= += += += +

= −= −= −= −

(((( )))) (((( )))) (((( )))) (((( ))))

00 R2

R2 CC

C

0 -12 2 R

C

3

R3

G1G RT ln log a

a 2.303RT

log a -43502.2 / ( 2.303* 8.314* 2133.5 ) 1.1

R3 SiC s + O g -> SiO s +C

log a 1.

3 G = -482201.8 Jm

G

1

ol

0

∆∆∆∆∆∆∆∆

∆∆∆∆

∆∆∆∆

= + →= + →= + →= + →

= −= −= −= −

====

= = −= = −= = −= = −

====2

2

2

2 2

2

0SiO C0 0 R3C

R3 R3 C OSiC O O

C O C O

a a Ga= G RT ln G RT ln log a log P

a P P 2.303RT

log a ( -482201.8 ) / ( 2.303* 8.314* 2133.5 ) log P log a 11 . 8 lo P g

∆∆∆∆∆ ∆∆ ∆∆ ∆∆ ∆

−−−− + = + → = ++ = + → = ++ = + → = ++ = + → = +

= − += − += − += − + = += += += +

50


0O2 R1R1 log P G / ( 2.303RT )∆∆∆∆→ =→ =→ =→ = 0

C R3 O2R3 log a G / ( 2.303RT ) log P∆∆∆∆→ = − +→ = − +→ = − +→ = − +lo

g(a C

)

log(PO2)

SiCSi (2)

(3) SiC SiO2

(1)

Si(l

)+O

2(g)

↔S

iO2(

l)

(2) Si(l) + C(s) -> SiC(s)

(3) S

iC(s) + O 2

(g)

↔SiO 2

(l) + C(s)

(1) Si SiO2

-14 -13 -12 -11-3

-2

-1

0

0C R2R2 log a G / ( 2.303RT )∆∆∆∆→ =→ =→ =→ =

51


0O2 R1R1 log P G / ( 2.303RT )∆∆∆∆→ =→ =→ =→ =

log(

a C)

log(PO2)

(1)

Si(l

)+O

2(g)

->S

iO2(

l)(3)

SiC(s)

+ O 2(g)

-> SiO 2(l)

+ C(s)

(1) Si(l)+O2(g)->SiO2(l)

(1) Si7 67 67 67 6SiO2

(2) Si(l) + C(s) -> SiC(s)

-14 -13 -12 -11-3

-2

-1

0 Reaction 1 Line divides the range of conditions into two: ●Si is not stable to the right●SiO2 is not stable to the left

- this eliminates redundant parts of the two other lines

52

-14 -13 -12 -11-3

-2

-1

0


log(

a C)

log(PO2)

(1)

Si(l

)+O

2(g)

->S

iO2(

l)

(2) Si(l) + C(s) -> SiC(s)

(3) S

iC(s) + O 2

(g) ->

SiO 2(l)

+ C(s)SiCSi (2)

(1) Si SiO2(3) SiC SiO2

Si

Reaction 2 Line divides the range of conditions into two: ●Si is not stable Up●SiC is not stable Down

- this eliminates redundant part of another line

53

-14 -13 -12 -11-3

-2

-1

0


log(

a C)

log(PO2)

SiCSi (2)

(1) Si SiO2

(3) SiC SiO2

Si

SiCSiO2

Finally, three lines derived from Reactions 1, 2 and 3 divide the range of conditions into three fields of stability of SiC, Si and SiO2

54

-14 -13 -12 -11-3

-2

-1

0


log(

a C)

log(PO2)

SiCSi (2)

(1) Si SiO2

(3) SiC SiO2

SiC(s)

SiO2(l)

Si(l)

C(s) stable

Oxidising→←Reducing

Car

buris

ing→

←D

ecar

buris

ing

55

-14 -13 -12 -11-3

-2

-1

0

55


log(

a C)

log(PO2)

SiCSi (2)

(1) Si SiO2

(3) SiC SiO2

SiC(s)SiO2(l)

Si(l)

C(s) stable

The diagrams show the ranges of stability of SiC, Si and SiO2 in terms of aC and PO2.The gas phase, however, contains a number of other species including CO, CO2, SiO etc. At these reducing conditions CO(g) is a predominant species.For the process run at Ptotal=1atm , the PCO -partial pressure of CO can only be less than 1atm. PCO isobars are important.

56

(((( )))) (((( )))) (((( ))))

CO

2

- other gaseous speciesPREDOMINANCE DIAGRAMS e.g. Si-C-O system, T=2133.5 K

To constract the lines with constant CO partial pressure P ,

the following reaction is analysed:

R4 C s + 0.5O g -> CO g (((( ))))

2

2

2

2

C CO O

0 -1R4

00 R4CO

R4 R4 C CO O0.5C O

C CO O

4 G = -297038.3 Jmol

GP0 G= G RT ln log a log P 0.5 log P

2.303RTa P

log a -297038.3 / ( 2.303* 8.314* 2133.5 ) log

log a 7

P 0.5 log P

.

3 log P 0.5 log P

∆∆∆∆

∆∆∆∆∆ ∆∆ ∆∆ ∆∆ ∆

= − += − += − += − +

= + → = + −= + → = + −= + → = + −= + → = + −

−−−−

= + −= + −= + −= + −

2

2

2

CO C O

CO C O

CO C O

for P =1 log a 7.3 0 0.5 log P

for P =0.1 log a 7.3 1 0.5 log P

for P =0.01 log a 7.3 2 0.5 log P

= − + −= − + −= − + −= − + −

= − − −= − − −= − − −= − − −

= − − −= − − −= − − −= − − −

57

(((( )))) (((( )))) (((( ))))2

2

C CO O

- other gaseous speciesPREDOMINANCE DIAGRAMS e.g. Si-C-O system, T=2133.5 K

R4 C s + 0.5O g -> CO g

log a 7.3 log P 0.5 log P= − + −= − + −= − + −= − + −

-14 -13 -12 -11-3

-2

-1

0lo

g(a C

)

log(PO2)

SiCSi (2)

(1) Si SiO2

(3) SiC SiO2

SiC(s)

SiO2(l)

Si(l)

(4) C(s)+0.5O2(g)->CO(g) P


PCO=1 atm

C(s) stable



PCO=const linesprovideadditionalinformationon gas composition

NOTE!Gas contains other species. These

diagramsare usedto selectoperating conditions:fuel/O2 ratioTemperaturePtotal

other

58

-14 -13 -12 -11-3

-2

-1

0lo

g(a C

)

log(PO2)

SiCSi (2)

(1) Si SiO2

(3) SiC SiO2

SiC(s)

SiO2(l)

Si(l)

(4) C(s)+0.5O2(g)->CO(g) P


PCO=1 atm

C(s) stable



Comments on PREDOMINANCE DIAGRAM for the Si-C-O system at 2133.5 K

1) Gas solution over condencedphases

2) Complex multi-component equilibria = equilibriaof simple subsystems

3) “Crossing” the SiC/SiO2 line along ~1atm CO isobar will produce SiC

4) CO in equilibrium with C has PCO >> 1atm – strong driving force

SiO2(s)C(s)

CO2(gas)

COgas

Siliq

59

In addition to thermodynamic stability, other factors should be taken into account. E.g. solid/solid reactions are slow- liquid phase is desirable to accelerate reactions.* Also, CO is predominant gaseous species to remove oxygen –PCO should approach 1 atm.* Other gaseous species (e.g. SiO) and other phases (e.g. SiO liquid) should be considered.

Gaskell “Introduction to the thermodynamics of materials”, pp 420-433.

Alternative axes - more convenient for analysis

60

PREDOMINANCE DIAGRAMS- Si-C-O SYSTEM- Other variables can also be selected for axes and - Other variables can also be fixed. E.g.:

Construction of the Predominancediagramusing logPCO and 1/T coordinatesat constant PCO2=10-12 atm

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PREDOMINANCE DIAGRAMS- Si-C-O SYSTEM- Other variables

PCO is far less than 1 atmin this rangeof temperatures

Si can be produced only at high tempertures

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PREDOMINANCE DIAGRAMS Example: selection of conditions for roasting of the Cu-Fe-sulphides to form more soluble CuSO4 and Cu oxides and insoluble Fe oxides (rather than soluble sulphates) Ref: Biswas and Davenport.

Shaded area-operatingconditionsat Ptotal=1atm

O2 is added to “burn” S to SO2

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Shaded area-operatingconditionsat Ptotal=1atm

Higher temperaturesfavour formationof Fe2O3

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Higher temperaturesfavour formationof Fe2O3

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PREDOMINANCE DIAGRAMS Example: PbS smelting (FactSage).

P(total)=0.1atm

PbO(l)

PbS(l)

PbSO4(s2)

PbOPbSO4(s)

(PbO)4(PbSO4)(s)

Pb(l)

P(total)=1atm

Pb-S-O, 1200 C'+' = 0.1 atm P(total) isobar

log10(P(S2)) (atm)

log 1

0(P

(O2)

) (a

tm)

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0

-14

-13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

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References:

Lee, H.G., 1999. Chemical thermodynamics for metals and materials , Imperial College Press, London.

D. Gaskell, 2008. Introduction to Thermodynamics of Materials, Taylor&Francis, USA.

Y.K.Rao, 1985. Stoichiometry and thermodynamics of metallurgical processes, Cambridge Univ. Press.

Biswas and Davenport, Extractive Metallurgy of Copper, Pergamon

CHEE3003 2012 Weeks06-07 Lecture Slides Solutions

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