Lecture 17 Ellingham

Thermodynamics in Materials Engineering

Mat E 212 - Course Notes

R. E. NapolitanoDepartment of Materials Science & Engineering

Iowa State University

Reaction Equilibriaand an introduction to Ellingham Diagrams

A simple chemical reaction

CBA 2

State I

1A 1B

State II

2CCnBnAn CBA

Intermediate state

III GGG

BAC GGG 2

Reaction coordinate

An

Bn

Cn

1

1

0

0

0

2

AB nn

BAC nnn 2An22

An 12

An

An

An12

CCBBAA nnnG

CABAAA nnn 12

For any intermediate state:

Consider only the mixing of A and B

State I

1A 1B

CBA 2

State II

2CCnBnAn CBA

Intermediate state

III GGG

BAC GGG 2

0

0

2

AB nn

BAC nnn 2An22

An 12

An

An

An12

CCBBAA nnnG

CABAAA nnn 12

Reaction coordinate

An

Bn

Cn

1

1

0

For any intermediate state:

XB

G

Consider only the mixing of A and B

State I

1A 1B

AG

BG

For 1 mole Aand 1 mole B

(MIXED)

A+B

IG

BBAA GXGX

(UNMIXED)

A B For 1 mole Aand 1 mole B

BBAABBAA XXXXRTGXGX lnln mixBBAA STGXGX

Gibbs free energy along reaction coordinate

State I State II

CBA 2

1A 1B 2CCnBnAn CBA

Intermediate state

III GGG

BAC GGG 2

Reaction coordinate

AG

BG

mixedA+B

IGA B CCBBAA GnGnGn

CCBBAA nnn

CII GG 2

Equilibrium State0

An

G

A simple chemical reaction

State I State II

CBA 2

1A 1B 2CCnBnAn CBA

Intermediate state

III GGG

BAC GGG 2

CABAAA nnn 12CCBBAA nnnG

Recall: iii aRTG ln

CCABBAAAA aRTGnaRTGnaRTGnG ln12lnln

CABAAACABAAA anananRTGnGnGn ln12lnln12

0ln2lnln2

CBACBAA

aaaRTGGGn

G

Find minimumalongreactioncoordinate: G

CBA aaaRTG ln2lnln

A condition of equilibrium

State I State II

CBA 2

1A 1B 2CCnBnAn CBA

Intermediate state

III GGG

BAC GGG 2

CBA aaaRTG ln2lnln

2ln

C

BA

a

aaRTG

BA

C

aa

aRTG

2

ln

KRTG ln K is defined as an “Equilibrium Constant”

Temperature dependence

KRTG ln

RT

GK

exp

STHG Note that:

ST

H

T

G

2T

H

T

G

T

2

lnT

HKR

T

2

ln

RT

H

T

K

HT

G

T

1

HKRT

ln1

R

HK

T

1

ln

Temperature dependence

R

HK

T

1

ln

Kln

T/1

Endothermic

Exotherm

icH > 0

H < 0

K increases with temperature.

K decreases with temperature.

A simple reaction

)(2)( 2

1ss MOOM

ArO 2

MO

M

Furnace – control T

2Op

2Op

T

A simple reaction

)(2)( 2

1ss MOOM

TG 11.6630540S

)(2)( 2

1ss AgOOAg

H

R

HK

T

1

ln

Recall:

Kln

T/1

Endothermic

Exotherm

icH > 0

H < 0

A simple reaction

Kln

T/1

Exotherm

ic

H < 0

Here: 30540 H

0 H (Exothermic)

K decreases as T increases.

)(2)( 2

1ss MOOM

yB

xA

zC

aa

a

RT

GK

exp

2/1

2

1

op

If K decreases with increasing T,pO2 must increase with increasing T.

2

1

2

1ln

OpRTG

2ln

2

1OpRTG

A simple reaction

)(2)( 2

1ss MOOM

STHG

2ln

2

1OpRTG

KRTG ln

STHG These are tabulated.

(See Table A-1, p.582.)

G

T

H

S

An example

)(2)( 2

1ss FeOOFe TG 35.64263700

H SG

T

Fe+1/2O2=FeO

0 Gat this temperature.Increasing K

Increasing pO2

2ln

2

1OpRTG

atmpO 12

0 G0

The pO2 scale

G

T

Fe+1/2O2=FeO

atmpO 12

0

For any constant pO2:2

ln2

1OpRTG

TpRG O

2ln

2

1

12Op

12Op

Combined reactionsFeOOFe 22 2 )(130.08.528 kJTG

22 22 COOCO )(174.08.564 kJTG

2COFeCOFeO )(022.00.18 kJTG

G

T

2Fe+O 2=2FeO

0

2CO+O 2=2CO 2

FeO+CO=Fe+CO2

C

The Ellingham-Richardson diagramFeOOFe 22 2 )(7.128400,527 JTG )(76.30052,126 calTG

The Ellingham-Richardson diagramFeOOFe 22 2 )(7.128400,527 JTG

)(76.30052,126 calTG

KRTG ln

RT

GK

exp

At T=1000ºC:

127376.30052,126 G

)/(895,86 molcal

RT

G

KpO

exp1

2

)1273)(/987.1(

/86895exp

KmolKcal

molcal

151021.1

2

2

2

OFe

FeO

aa

aK

2

1

OpK

The Ellingham-Richardson diagramFeOOFe 22 2 )(7.128400,527 JTG )(76.30052,126 calTG

Constant pO2

15102

Op

The Ellingham-Richardson diagram

Constant pO2

15102

Op

9102

Op

The EQ value of pO2 will increase, and the reaction is forced to the left. What if we establish EQ at 1000ºC and then raise the temperature to 1500ºC?

FeOOFe 22 2