ellingham diagram (aftab ahmed laghari)
DESCRIPTION
A little overview of Ellingham Diagram.TRANSCRIPT
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