Lecture 17 Ellingham
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Transcript of 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
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
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