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06. Kesetimbangan Kimia dan Diagram Ellingham
Zulfiadi Zulhan
Teknik MetalurgiFakultas Teknik Pertambangan dan PerminyakanInstitut Teknologi BandungINDONESIA
Termodinamika Metalurgi (MG-2112)
201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Jangan Mengunggah
Materi Kuliah ini di
INTERNET!
301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
1. Pendahuluan, istilah-istilah dan notasi
2. Hukum I Termodinamika
3. Hukum II Termodinamika
4. Hubungan Besaran-Besaran Termodinamika
5. Kesetimbangan
6. Kesetimbangan Kimia dan Diagram Ellingham
7. Proses Elektrokimia dan Diagram Potensial - pH (Pourbaix)
8. Ujian Tengah Semester
9. Aktivitas Ion
10. Termodinamika Larutan
11. Penggunaan Persamaan Gibbs - Duhem
12. Penggunaan Metoda Elektrokimia untuk menentukan Sifat-Sifat / Besaran-Besaran
Termodinamika
13. Keadaan Standar Alternatif
14. Koefisien Aktivitas dalam Larutan Encer Multi-Komponen
15. Diagram Fasa
16. Ujian Akhir Semester
Materi Perkuliahan
401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
dU = T dS - P dV
dH = T dS + V dP
dF = -S dT - P dV
dG = -S dT + V dP
dG = − S dT + V dP
At consntant temperature: dT = 0
dG = V dP
P ln d RT P
dP RT Gd ==
For ideal gas: V = RT/P
0 P as 1, f/P lim
f ln d RT Gd
→=
=At low pressure (pressure approaches zero), the
gas behaves ideally and its fugacity equal to its
pressure.
Thermodynamic Activity
For real gas, by analogy to pressure:
The change of molar Gibbs free energy of a single gas
with pressure at constant temperature
501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Fugasitas
https://id.wikipedia.org/
601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity
For mixtures of gases, molar Gibbs free energy (G) is replaced by partial molar Gibbs free
energy, .
For component i:
if ln d RT Gd =
G
0 P as 1, f/P lim
f ln d RT Gd
→=
= At low pressure (pressure approaches zero), the gas
behaves ideally and its fugacity equal to its pressure.
For real gas, by analogy to pressure:
Fugacity is property of a gas ~ pressure of non ideal gas.
701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity
Fugacity of a condensed phase (solid or liquid) is equal to the fugacity of the vapor in equilibrium
with condensed phase.
At equilibrium, partial molar Gibbs free energies of vapor and condensed phase are equal.
For the difference in partial molar Gibbs free energy between two states at constant
temperature (in term of fugacity):
1
2
f
f
12
f
f ln RT f ln d RT G - G G
2
1
===
Sometimes, it is written as:
1
212
f
f ln RT - =
,...n,nP,T,a
a
cb
n
VV
=
i
nnS,V,innS,P,innV,T,innP,T,iijijijij
n
U
n
H
n
F
n
G=
=
=
=
if ln d RT Gd =
801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity
Thermodynamic activity of a material is defined as the ratio of fugacity
of material to the fugacity in its standard state:
i
ii
f
fa
1
2
f
f
12
f
f ln RT f ln d RT G - G G
2
1
=== 1
212
f
f ln RT - =
Thermodynamic Activity is evaluated relative to a standard state at the same temperature.
Activity is set equal to one at the standard state.
In terms of a quantity, Activity is called “fugacity”
Fugacity is property of a gas ~ pressure of non ideal gas.
Activity is the ratio of the fugacity of a material to its fugacity in the standard state.
901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity
Standard state of a material can be arbitrarily defined, certain “conventions” generally are followed:
▪ Standard state of a gas is “pure gas at one atmosphere pressure”
▪ Standard states for liquid and solids are “pure liquid or solid at one atmosphere pressure”
i
ii
f
fa
Standard state vs. reference state
Reference state defines the condition of material in term of temperature, pressure, and physical
form: gas, liquid and crystal structure (for solid). Example: reference state for enthalpy=0 for pure
element at 298K.
Standard state is defined only in terms of pressure and physical form. Thermodynamic activity is
an isothermal concept.
Thermodynamic Activity is evaluated relative to a standard state at the same temperature.
Activity is set equal to one at the standard state.
1001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity
So, the difference in Gibbs free energy between a material under an arbitrary set of conditions and a
standard state at the same temperature is:
Example: activity of an ideal gas at pressure of 2 atm is 2 (because the fugacity of the standard state
is one atmosphere).
Pure nitrogen at a pressure of 0.1 atm has an activity of 0.1, if pure nitrogen at one atmosphere is
taken as standard state.
Note: thermodynamic activity has no units.
a ln RT f
f ln RT G - G dG i
i
iii
G
G
i ===
Fugacity of liquid or solid is equal to fugacity of vapor in equilibrium with it. If equilibrium vapor
pressure of liquid is 0.01 atm at a specified temperature, then its fugacity is 0.01
if ln d RT Gd =
i
ii
f
fa
P ln d RT P
dP RT Gd ==
dG = -S dT + V dP
Gi = Gi° + RT ln ai
1101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity
Value of thermodynamic activity changes not only with pressure but also with composition.
Fugacity of a material in an ideal solution varies linearly with mole fraction.
Pure A Pure B
Fugacity o
f B
, f B fB
p
Activity o
f B
, a
B
0
1
fB = XB fBp
Where fBp is the fugacity of pure B,then
If fBp = fB°, then aB = XB (ideal solution)
If in liquid A-B consisting of 80 mol % B, then aB
= 0.8, aA = 0.2 (ideal solution)
==
B
p
BB
B
BB
f
f X
f
fa
aB = B XB
aB = activity of B
B = activity coefficient of B
XB = mol fraction of B
1201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Aktivitas
https://www.pikiran-rakyat.com/
https://lokadata.id/
Murni
Nelayan
Murni
Petani
https://dialeksis.com/
https://idxchannel.okezone.com/
Aktivitas N
ela
yan
Aktivitas P
eta
ni
Jumlah Petani
Pure A Pure B
Activity o
f A
, a
A
Activity o
f B
, a
B
0
1
XB
aB =fB
fB° =
fBpXB
fB°
aA =fA
fA° =
fApXA
fA°
ideal solution:
fAp = fa°, then aA = XA
ideal solution:
fBp = fB°, then aB = XB
1301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Condition of Equilibrium
Partial molar Gibbs free energy of a material,
If G2 – G1 is negative, then reversible work (Wrev) is negative. It means
that material change spontaneously from state 1 to state 2, because no
work needs to be done to force the change.
δWrev = ഥG2 − ഥG2 dn = ΔഥG dn = ΔGAt equilibrim: δWrev = ΔG = 0
G
If G2 – G1 is positive, then reversible work (Wrev) is positive. Work would
have to be done on the system to force it to change from state 1 to state
2 (the change would not occur spontaneously).
https://www.tokopedia.com/
SISTEM
Q+ W+Lingkungan
G1
G2
i,1
i,2
i,1 > i,2
G1
G2
i,1
i,2i,1 < i,2
Gi = Gi° + RT ln ai
1401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Chemical Equilibrium
Reaction: b B + c C = d D + e E
Gc -G b-G e G d GW CBEDrev +==
B
o
BB a ln RT G G +=
( ) ( ) ( ) ( )C
o
CB
o
BE
o
ED
o
D a ln RT Gc a ln RT G b-a ln RT G ea ln RT G d ΔG +−++++=
c
C
b
B
e
E
d
Do
C
o
B
o
E
o
Da a
a a ln RT G cG bG eG d ΔG +−−+=
K ln RT G a a
a a ln RT G ΔG
c
C
b
B
e
E
d
D +=+=
K = equilibrium constant for a reaction, expressed in terms of activities
Gi = Gi° + RT ln ai
δWrev = ΔG
1501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Gaseous Equilibria
Case 1: a tank contains pure oxygen at total pressure of 1 atmosphere. Oxygen exists primarily in
diatomic form, O2. However, oxygen can also exist in monoatomic state. Calculate the equilibrium
concentration of monoatomic oxygen in the tank at 1000K.
Standard Gibbs free energy of formation of monoatomic oxygen is 187800 J per mole of
monoatomic oxygen at 1000K.
J 800 187 0 - 800 187 G - G G o
2O,f21o
O,f ===Reaction:
½ O2 = O
1/2
O
O
1/2
O
O
22p
p ln RT
a
a ln RT G −=−=
22.58- 1000 8.314
187800
RT
G-
p
p ln
1/2
O
O
2
=−=
= 10 x 55.1 p
p 10-
1/2
O
O
2
=
-10
OOO 10 x 5.1p 1,pp2
==+ Activity of diatomic oxygen in the tank is very nearly one. O2 is
predominant at 1000 K
1601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Gaseous Equilibria
205,1)*T - 0,5(0 - 161) * T - (249.000 G - G 2,21
, == o
Of
o
OfG
Reaction:
½ O2 = O
J/mol 205,1 * T - 0 2= o
OG
J/mol 161 * T - 249.200 = o
OG
) T * 58,45 - (249.000 G - G 2,21
, == o
Of
o
OfG
J/mol 190.550 G - G 2,21
,1000 == o
Of
o
Of
o
KG
J/mol 335.500- G - G 2,21
,000.10 == o
Of
o
Of
o
KG atm 99968,0p 1,pp OOO 2==+
http://hebasoffar.blogspot.co.id
Reaction:
b B + c C = d D + e E
ΔG = ΔG° + RT lnaDd aE
e
aBb aC
c
1701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Gaseous Equilibria
Case 2: consider the equilibrium in water-hydrogen-oxygen system. At 2000K, a system will be
operated so that the partial pressure of oxygen is 10-10 atm. What should be the ratio of
hydrogen to water vapor to achieve this condition?
2/1
OH
OH
22
2
p p
p 3652 K
8.20 RT
G- K nl K, 2000 at
==
=
=
Reaction: H2 + ½ O2 = H2O G° = -246 000 + 54.84 T
atm 1ppp
10 x 3652p x 3652 p
p
222
2
2
2
OHOH
5-1/2
O
H
OH
=++
==
At one atmosphere, a hydrogen water mixture in which water content of 3.52% and hydrogen
content of 96.48% will have an oxygen partial pressure of 10-10 atm at 2000 K.
This illustrates that it is possible to obtain low oxygen pressures by chemical means.
1801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Cu-O2 Equilibria
Consider the equilibrium among Cu2O, Cu and O2:
4 Cu (s) + O2 (g) = 2 Cu2O (s)
If pressure of oxygen exeeds equilibrium pressure, reaction proceeds to right (Cu2O will be formed).
If pressure of oxygen less than equilibrium pressure, reaction proceeds to left (Cu2O should
decompose into Cu and O2)
Cu2O
http://www.baofull.com
+ =
http://www.pometon.com
1901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Source of Information on G°
For chemical reaction, Gibbs free energy change at temperature T, G°T, is the sum of Gibbs free
energies of formation of products less the sum of the Gibbs free energies of formation of the reactants
at the same temperature
=reactants
o
Tf,r
products
o
Tf,p
o
T G n-G n G
o
T
o
T
o
T S T-H G =
G°T can also be calculated from values of H°T and S°T as follows (assuming that there are no phase
transformation between 298 K and T):
Td T
CS S and
Td CH H :where
T
298
o
po
298
o
T
T
298
o
p
o
298
o
T
+=
+=
4 Cu (s) + O2 (g) = 2 Cu2O (s)
Δ𝐺° = − RT lnaCu2O2
aCu4 pO2
Δ𝐺° = −339000 − 14.24 T ln T + 247 T Joule
2001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Source of Information on G°
T ln T C T B A Go
T,f ++=
Above form of equation arises because Cp° is treated as indenpendent on
temperature for the purpose of calculating G°.
( )298TC H H o
p
o
298
o
T −+=
+=
298
Tln C S S o
p
o
298
o
T
( ) ( )298 ln - T ln C T S T298T C H G o
p
o
298
o
p
o
298
o
T −−−+=
( ) T ln C T 298 ln CCS TC 298 H G o
p
o
p
o
p
o
298
o
p
o
298
o
T −−−−−=
T ln T C T B A Go
T,f ++=
4 Cu (s) + O2 (g) = 2 Cu2O (s)
Δ𝐺° = − RT lnaCu2O2
aCu4 pO2
Δ𝐺° = −339000 − 14.24 T ln T + 247 T J
2101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
T B T ln T C A Go
T,f ++=
4 Cu (s) + O2 (g) = 2 Cu2O (s)
Δ𝐺° = − RT lnaCu2O2
aCu4 pO2
Δ𝐺° = −339000 − 14.24 T ln T + 247 T J
2201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
4 Cu (s) + O2 (g) = 2 Cu2O (s)
Calculate oxygen pressure where Cu2O dissociates into Cu and O2 at 1000K.
Joule T 247 T ln T 24.14339000G
p a
a ln RT G
2
2
O
4
Cu
2
OCu
+−−=
−=
Activity of Cu2O is 1 because during the reaction, Cu2O exists as pure solid and
standard state is pure Cu2O.
Cu2O
http://www.baofull.com
+ =
http://www.pometon.com
Cu-O2 Equilibria
Reaction:
b B + c C = d D + e E
ΔG = ΔG° + RT lnaDd aE
e
aBb aC
c
2301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Copper
Cu2O
http://www.baofull.comhttp://www.pometon.com
https://dir.indiamart.com
CuO Cu(OH)2
http://www.instructables.com
http://www.nyctransitforums.com
http://www.reade.com
http://2il.orghttps://www.healingcrystals.com
CuFeS2
2401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Solid Vapor Equilibria
atm 10 x 14.1p 22.90,- p ln
p ln 1000 * 8.314 360 190 G K, 1000 at
10-
OO
O
22
2
==
=−=
At higher pressure of oxygen, activitiy of copper will be less than 1, copper will not exist.
At oxygen pressure of 1.14 x 10-10 atm, metallic copper and cuprous oxide can exist in
equilibrium.
At lower oxygen pressure, metallic copper will exist
Activity of Cu will become 1, if the pressure of oxygen is in equilibrium with Cu2O and
Cu.
Cu2O+→
2501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Gaseous Equilibria
Case 2: consider the equilibrium in water-hydrogen-oxygen system. At 2000K, a system will be
operated so that the partial pressure of oxygen is 10-10 atm. What should be the ratio of
hydrogen to water vapor to achieve this condition?
2/1
OH
OH
22
2
p p
p 3652 K
8.20 RT
G- K nl K, 2000 at
==
=
=
Reaction: H2 + ½ O2 = H2O G° = -246 000 + 54.84 T
atm 1ppp
10 x 3652p x 3652 p
p
222
2
2
2
OHOH
5-1/2
O
H
OH
=++
==
At one atmosphere, a hydrogen water mixture in which water content of 3.52% and hydrogen
content of 96.48% will have an oxygen partial pressure of 10-10 atm at 2000 K.
This illustrates that it is possible to obtain low oxygen pressures by chemical means.
Reaction:
b B + c C = d D + e E
ΔG = ΔG° + RT lnaDd aE
e
aBb aC
c
2601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Cu stable
Cu2O stable
CuO stable
Cu-O2 Equilibria0 = ΔG° + RT ln
aCuO4
aCu2O2 pO2
Reaction: b B + c C = d D + e E
ΔG = ΔG° + RT lnaDd aE
e
aBb aC
c
ΔG = ΔG° + RT lnaCuO4
aCu2O2 pO2
2 Cu2O + O2 = 4 CuO
At equilibrium, ΔG = 0
ΔG° = − RT lnaCuO4
aCu2O2 pO2
For pure CuO, aCuO = 1
For pure Cu2O, aCu2O = 1
ΔG° = RT ln pO2
0 = ΔG° + RT lnaCu2O4
aCu4 pO2
Δ𝐺° = −339000 − 14.24 T ln T + 247 T Joule
2701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Source of Information on G°
T ln T C T B A Go
T,f ++=
Above form of equation arises because Cp° is treated as indenpendent on
temperature for the purpose of calculating G°.
( )298TC H H o
p
o
298
o
T −+=
+=
298
Tln C S S o
p
o
298
o
T
( ) ( )298 ln - T ln C T S T298T C H G o
p
o
298
o
p
o
298
o
T −−−+=
( ) T ln C T 298 ln CCS TC 298 H G o
p
o
p
o
p
o
298
o
p
o
298
o
T −−−−−=
T ln T C T B A Go
T,f ++=
4 Cu (s) + O2 (g) = 2 Cu2O (s)
Δ𝐺° = − RT lnaCu2O2
aCu4 pO2
Δ𝐺° = −339000 − 14.24 T ln T + 247 T J
2801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
-350,000
-300,000
-250,000
-200,000
-150,000
-100,000
0 200 400 600 800 1000 1200 1400 1600
G
o=
RT
ln P
O2
Temperatur [K]
dG = A + C*T ln T + BT
dG = dH - T*dS
ΔGf,To = ΔHo - T ΔSo
ΔG = ΔG° + RT lnaCu2O2
aCu4 pO2
4 Cu + O2 = 2 Cu2O
At equilibrium, ΔG = 0
ΔG° = − RT lnaCu2O4
aCu4 pO2
For pure Cu, aCu = 1
For pure Cu2O, aCu2O = 1
ΔG° = RT ln pO2
2901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
M + O2 = MO2
If the difference in heat capacities between
reactants and products is small, then G°T
equation assumes to form A - BT:
G°T =H° - T S°
ΔG = ΔG° + RT lnaMO2
aM pO2
At equilibrium, ΔG = 0
ΔG° = − RT lnaMO24
aM pO2
For pure MO2, aMO2 = 1
For pure M, a𝑀 = 1
ΔG° = RT ln pO2
G
° T=
H°
-T
S
°=
RT
ln P
O2
0 200 400 600 800 1000
H°
G°T =H° - T S°
3001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Ellingham
Diagram
twitter.com/kachra_peti
Ellingham diagram is a graph showing the
temperature dependence of the stability of
compounds. This analysis is usually used
to evaluate the ease of reduction of
metal oxides and sulfides. These diagrams
were first constructed by Harold
Ellingham in 1944.en.wikipedia.org
Harold Johann Thomas Ellingham, (1897–1975), British physical chemist, best known for Ellingham diagrams, which summarize a large amount of information concerning extractive metallurgy www.wikiwand.com
Increasing
Stability
3101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Ellingham DiagramΔG°=
RTlnpO2
TTm,A
ΔG°=
RTlnpO2
TTm,AO2
3301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
ΔG°=
RTlnpO2
(kJ)
RT ln PO2 vs. T
O
At 1050°K, the
oxygen pressure
in equilibrium with
Cu2O and Cu is
10-9 atm
3401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Ellingham Diagram
-273
ΔG° = RT ln pO2ΔG° = 2.303 RT log pO2
ΔG° = 2.303 RT log (10𝟎) Oxygen Potential (PO2)
ΔG° = 2.303 RT log (10−𝟑)
ΔG° = 2.303 RT log (10−𝟓)
ΔG° = 2.303 RT log (10−𝟏𝟒)
O
3501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Ellingham Diagram
ΔG° = RT ln pO2ΔG° = 2.303 RT log pO2
ΔG° = RT ln (𝟏)
Oxygen Potential (PO2)
p O2 = 1 atm
ΔG° = RT ln (10−𝟏𝟎)
K
ΔG° = RT ln (10−𝟐𝟎)
O
3601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Ellingham Diagram
-273
ΔG°
2.303= RT l𝑜𝑔 pO2 + RT log
1
1
2
Equilibrium pressure of
oxygen in a system by
controlling the ration of
H2 and H2O in vapor:
2H2 + O2 = 2H2O
At equilibrium, ΔG = 0
ΔG° = − RT lnpH2O2
pH22 pO2
ΔG° = RT ln pO2 + RT lnpH2pH2O
2
ΔG°
2.303= RT l𝑜𝑔 pO2 + RT log
103
1
2
ΔG°
2.303= RT l𝑜𝑔 pO2 + RT log
107
1
2
ΔG°
2.303= RT l𝑜𝑔 pO2 + RT log
1012
1
2
H2
H
OH
O p
p
K
1p
2
2
2
=
3701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Ellingham Diagram
K
Equilibrium pressure of
oxygen in a system by
controlling the ration of
H2 and H2O in vapor:
2H2 + O2 = 2H2O
At equilibrium, ΔG = 0
ΔG° = − RT lnpH2O2
pH22 pO2
ΔG° = RT ln pO2 + RT lnpH2pH2O
2
ΔG° = RT ln pO2 + RT ln1
1
2
ΔG° = RT ln pO2 + RT ln10−3
1
2
ΔG° = RT ln pO2 + RT ln107
1
2
ΔG° = RT ln pO2 + RT ln1012
1
2
H2
H
OH
O p
p
K
1p
2
2
2
=
3801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Ellingham Diagram
-273
ΔG°
2.303= RT l𝑜𝑔 pO2 + RT log
1
1
2
ΔG°
2.303= RT l𝑜𝑔 pO2 + RT log
103
1
2
ΔG°
2.303= RT l𝑜𝑔 pO2 + RT log
107
1
2
ΔG°
2.303= RT l𝑜𝑔 pO2 + RT log
1012
1
2
C
Equilibrium pressure of
oxygen in a system by
controlling the ration of
H2 and H2O in vapor:
2CO + O2 = 2CO2
At equilibrium, ΔG = 0
ΔG° = − RT lnpCO22
pCO2 pO2
ΔG° = RT ln pO2 + RT lnpCOpCO2
2
2
CO
CO
O p
p
K
1p 2
2
=
3901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Ellingham Diagram
K
Equilibrium pressure of
oxygen in a system by
controlling the ration of
H2 and H2O in vapor:
2CO + O2 = 2CO2
At equilibrium, ΔG = 0
ΔG° = − RT lnpCO22
pCO2 pO2
ΔG° = RT ln pO2 + RT lnpCOpCO2
2
ΔG° = RT l𝑜𝑔 pO2 + RT log1
1
2
ΔG° = RT l𝑜𝑔 pO2 + RT log10−3
1
2
ΔG° = RT ln pO2 + RT ln107
1
2
ΔG° = RT l𝑜𝑔 pO2 + RT log1012
1
2
C
2
CO
CO
O p
p
K
1p 2
2
=
4001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Mengapa kemiringan
kebanyakan garis (slope)
pada diagram tersebut
positif?
Apa maksudnya jika slopenya
negatif?
Diagram Ellingham
Decreasing
Oxygen
Potential
Increasing
Stability
AlAl2O3
FeO
Fe
4101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Diagram Baur – Glaessner / Diagram Chaudron
0
10
20
30
40
50
60
70
80
90
100
0 200 400 600 800 1000 1200 1400
Temperature [°C]
CO
/ (
CO
+ C
O2)
[%]
Fe3O4
FeO
Fe
Ptot = 0.2 atm
Ptot = 1.0 atm
Ptot = 5.0 atm
Reaction:
b B + c C = d D + e E
ΔG = ΔG° + RT lnaDd aE
e
aBb aC
c
Fe3O4 + 4CO = 3 Fe + 4CO2
Fe3O4 + CO = 3 FeO + CO2
FeO + CO = Fe + CO2
3 Fe2O3 + CO = 2 Fe3O4 + CO2
CO2+C = 2CO
AISE, 1999Fe2O3
4201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Diagram KelloggReaction:
b B + c C = d D + e EΔG = ΔG° + RT ln
aDd aE
e
aBb aC
c = ΔG° + RT ln K
Log pO2
Log p
SO
2
MeS
Me
MeO
MeSO4
4. 2MeO + 2SO2 + O2 = 2MeSO42 log pSO2
+ log pO2 = −log K4
2. 2Me + O2 = 2 MeOlog pO2 = − log K2
3. 2MeS + 3O2 = 2MeO + 2SO22 log pSO2
− 3 log pO2 = log K3
1. Me + SO2 = MeS + O2log pO2 − log pSO2
= log K1
5. MeS + 2O2 = 2 MeSO42 log pO2 = − log K5
6. S2 + 2O2 = 2 SO22 log pSO2
− 2 log pO2 = log K6 + log pS2
7. 2SO2 + O2 = 2SO32 log pSO2
+ log pSO2= −log K7 + 2 log pSO3
At equilibrium, ΔG = 0 ΔG° = -RT ln K
1 2
3
4
5
4301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
http://www.aimehq.org/
4601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Gases Dissolved in Metals (Sievert‘s Law)
Gases, such as hydrogen, oxygen and nitrogen, dissolve in liquid and solid metals.
In gaseous state, these gases exist in diatomic form, H2, O2 and N2.
If these gases dissolve in metals, they are found in monoatomic form, H, O and N.
For hydrogen dissolving in copper:
H2 (g) = 2H ( in copper solution)
Equilibrium constant:
2H
2
H
p
a K =
Activity of diatomic gases dissolved as dillute
solution is linear to their concentration in metal.
aH = [H]
2H
2
p
H K = 2/1
H
2/1
2p K H =
Sievert‘s Law: the amount of diatomic gases dissolved in metals at a given temperature
is proportional to sequare root of diatomic gas partial pressure.
en.wikipedia.org
4701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Property Relations Derived from U, H, F, G
dU = T dS - P dV
dH = T dS + V dP
dF = -S dT - P dV
dG = -S dT + V dP
( ) dT S Gd −=
−= S T H G
( ) dT S Gd −=
−= S T H G
T
G H S
−=
( ) dTT
H dT
T
G-Gd
−=
Multiply by 1/T
( ) dT
T
H dT
T
G-
T
Gd22
−=
T
1dH
T
Gd
=
( ) T
1d
R
H K lnd
−=
( ) T
1d
R
H K lnd
2
1
2
1
T
T
K
K
−=
T
1
T
1
R
H
K
K nl
121
2
−
−=
Van‘t Hoff relationship
Gibbs free energy change for a reaction as a function of temperature:
en.wikipedia.org
At equilibrium, ΔG = 0 ΔG° = -RT ln K
4801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Variation of Equilibrium Constant with Temperature
Example: decomposition of calcium carbonate (lime stone) into calcium oxide
and carbon dioxide
CaCO3 = CaO + CO2
Following data are available for pressure of carbon dioxide in equilibrium with
CaO and CaCO3:
Temperature (K) Pressure (atm)
1030 0.10
921 0.01
a. Calculate change in enthalpy
T
1
T
1
R
H
K
K nl
121
2
−
−= H° = 166000 J per
mol CaCO3
b. Estimate temperature, at which the equilibrium
partial pressure of CO2 will be one atmosphere T3 = 1168 K
T
1
T
1
R
H
K
K nl
121
2
−
−=
K =aCaO PCO2aCaCO3
4901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Terima kasih!
Zulfiadi Zulhan
Program Studi Teknik Metalurgi
Fakultas Teknik Pertambangan dan Perminyakan
Institut Teknologi Bandung
Jl. Ganesa No. 10
Bandung 40132
INDONESIA
www.metallurgy.itb.ac.id