Upload
others
View
7
Download
0
Embed Size (px)
Citation preview
09 - 11. TermodinamikaLarutan dan Aktivitas Ion
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. Termodinamika Larutan
10. Penggunaan Persamaan Gibbs – Duhem
11. Aktivitas Ion
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
Thermodynamic Activity (Module 04)
The change of molar Gibbs free energy of a single gas
with pressure at constant temperature
501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity (Module 04)
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 =
601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity (Module 04)
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.
701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity (Module 04)
Although the choice of standard state is arbitrary, in liquid solutions and solid
solutions, the pure materials at one atmosphere pressure and specified crystal
structure are usually taken as the standard state.
iio
ipurei,i a ln RT G-G G-G ==
Example:
Solutions of acetone and water, pure water as standard state for water and pure
acetone as standard state for acetone.
Solid solutions of iron and nickel, pure iron (with specified crystal structure, bcc or
fcc) as standard state for iron, and pure nickel as standard state for nickel.
801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity
a ln RT f
f ln RT G - G dG i
i
iii
G
G
i ===
if ln d RT Gd =
i
ii
f
fa
P ln d RT P
dP RT Gd ==
Gi = Gi° + RT ln ai
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
901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Activity
Δ ሜGA,pure→solution
A A
A
A
A
A
A
A
A
A
Pure A
A
A
A
A
Solution A in B
Gas
fA < fA
fA ~ xA – mol fraction
B B
B
B
B A
AAAA A
Gas
PA = P0A = fA
P0A = saturation
vapor pressure
A A
GA0
GA
ΔG1=0 ΔG3=0
GA,Solution = GA0+ RT ln(
fA
fA0)
GA,Pure = GA,Pure = GA0
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.
න
𝐺𝐴°
𝐺𝐴
𝑑𝐺𝐴 = GA − GA° = RT ln
fA
fA°ΔG2=
GA − GA° = RT ln aAΔG4=
S: E. Jak
ΔG4= ΔG1 + ΔG2 + ΔG3
1001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Aktivitas(Module 04)
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
1101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
IDEAL Solution
ai =fi
fi° =
Xi fip
fi° =
Xi fio
fi° = Xi
Gi (solution) = Goi (pure)
+ RT ln Xi
1.0
Activity a
A0
Mol Fraction, XA
0
1.0
Ideal: aA = XA
Murni
NelayanMurni
Petani
Aktivitas N
ela
yan
Aktivitas P
eta
ni
Jumlah Petani
en.wikipedia.org
A
A
AB B
B
B
B A
1201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
NON IDEAL Solution (Positive Deviation)
Gi (solution) = Goi (pure)
+ RT ln i + RT ln Xi
1.0
Activity a
A
0
Mol Fraction, XA0
1.0aA = A XA
Murni
NelayanMurni
Petani
Aktivitas N
ela
yan
Aktivitas P
eta
ni
Jumlah Petani
A
A
AB B
B
B
B A
1301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
NON IDEAL Solution (Negative Deviation)
Gi (solution) = Goi (pure)
+ RT ln i + RT ln Xi
1.0
Activity a
A
0
Mol Fraction, XA0
1.0aA = A XA
Murni
NelayanMurni
Petani
Aktivitas N
ela
yan
Aktivitas P
eta
ni
Jumlah Petani
A
A
AB B
B
B
B A
1401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Graphical Representation
1501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Mixing of Ethanol and Water
commons.wikimedia.org
0
10
20
30
40
50
60
70
80
90
100
110
1
Water
50 mL
Ethanol
50 mL
50 mL Ethanol +
50 mL water
1601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Change of Solution Volume by Addition of Moles B
Partial molar volume of a component in a
solution is volume change of solution when one
mole of particular component is added to it
if 50 cm3 water is added to 50 cm3 ethanol, total
volume will be ~ 97 cm3 (not 100 cm3).
The same principles applies to other
quantities: Enthalpy, Entropy, Gibbs free
energynB
V𝐒𝐥𝐨𝐩𝐞 =
𝛛𝐕
𝛛𝐧𝐁 𝐓,𝐏,𝐧𝐀
= ഥ𝐕
Volume of solution A-B as a a function of
moles of B added to the solution
1701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Partial Molar Quantities (Module 03)
,...n,nP,T,a
a
cb
n
VVSlope
==
na
V
Constant T and P
na
V
Volume of a solution (a + b + c + …) as a function of
moles a (na) addedVolume of a as a function of na in
pure a (slopes is V/na = Va)
Constant T and P
If a is added to pure a, partial molar volume would be equal to the molar volume aa VV =
1801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Partial Molar Quantities
Consider volume of solution consisting of materials A and B. Volume of this
solution, at constant pressure and temperature, is a function of the amount of A
(nA) and the amount of B (nB), n represents the number of moles.
P,TBA )n,n(VV =
B
n,P,TB
A
n,P,TA
dn n
V dn
n
VdV
AB
+
=
Bn,P,TA
A
n
VV
=
BBAA dn V dn V dV +=
For dV at constant T and P:
Partial molar volume of A:
Then:
1901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Partial Molar Quantities
BBAA dn V dn V dV +=
Integration equation above :
BBAA n V n V V +=
Dividing by total number of moles (nA + nB)
BA
BB
BA
AA
BA n n
n V
n n
n V
n n
V
++
+=
+
BBAA X V X V V +=
BBAA dX V dX V Vd +=
XA and XB are the mole
fractions of A and B in the
solution.
V is molar volume.
2001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Graphical RepresentationdV = VA dXA + VB dXB
Because XA + XB = 1,
then: dXA + dXB = 0,
dXA/dXB = -1
dV
dXB= VB − VA
At composition XB, tangential
line intersects A axis at , and
B axis atAV
Pure BPure A
XB
V
XB
V
B
AdX
VdX
BV
AV
XA
BV
1: XA𝑑𝑉
dXB T,P= XA𝑉𝐵 − XA𝑉𝐴
2: XB𝑑𝑉
dXB T,P= XB𝑉𝐵 − XB𝑉𝐴
3: 𝑉 = 𝑉𝐴 XA + 𝑉𝐵 XB
B
PT,B
A V dX
VdXV :31 =
++ A
PT,B
B V dX
VdXV :23 =
−−
: dXB
2101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Relative Partial Molar Quantities
By mixing of A and B to form a solution, volume changes:
V- V V V initialfinalMmixing ==
Vfinal is the volume of the solution
Vinitial is the volume of components
before mixing.
Volume change upon mixing (Vm):
V n V n - V n V n V BBAABBAAM −+=
( ) ( ) V - V n V-V n V BBBAAAM +=
A.pure of volume molar to relative solution in A volume molar Partial
solution. enters it as A pure of volume in change The
V A,of molar partial relative called is V-V ermTrel
AAA
rel
BB
rel
AAM V n V n V +=
2201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Relative Partial Molar Quantities
It can be written as:
rel
BB
rel
AAM V n V n V +=
For one mole of solution, this becomes:
In the above equations, volume is used as a representative of thermodynamic
property. The same forms of equations apply to Gibbs free energy, Enthalpy,
Entropy and similar quantities, example:
rel
BB
rel
AAM V X V X V +=
A
o
AA
rel
A a ln RT G G G =−=
For an ideal solution, ai = Xi
i
rel
i X ln RT G =
V n V n - V n V n V BBAABBAAM −+=
( ) ( ) V - V n V-V n V BBBAAAM +=
𝑇erm VA − VA is called relative partial molar of A, V𝐴𝑟𝑒𝑙
2301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Entropy of Mixing: IDEAL SOLUTION
dU = T dS - P dV
dH = T dS + V dP
dF = -S dT - P dV
dG = -S dT + V dPRelative partial entropy of component A:
( )B
B
n,PAn,P
rel
A
rel
A X ln RT T
-G T
- S
=
=
A
rel
A X ln R- S =
Molar entropy mixing for a solution A + B is:
rel
BB
rel
AAM S X S X S +=
BBAAM X ln X X ln X R- S += =i
iiM X ln X R- S
:componentsmany for general, in
dP V dT S dG +−=
dT S- Gd :ncompositio and pressure constant atrelrel
=
𝑇erm VA − VA is called relative partial molar of A, V𝐴𝑟𝑒𝑙
rel
BB
rel
AAM V X V X V +=
2401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Entropy of Mixing, for two gases (Module 03)
Va Vb
Entropy changes for two gases
Consider two gases on either side of a partition, a on the left
and b on the right, each at the same temperature and
pressure.
b12ba12a12 )SS( n)SS( nSS −+−=−
VT
Upon removal of the partition, the two gases will mix.
Ideal gases do not interact, total entropy change is the sum of two entropy changes of the
individual gases:
S2 − S1 = n R ln𝑉𝑡𝑉𝑎
Untuk 1 gas
S2 − S1 = na R ln𝑉𝑎+𝑉𝑏
𝑉𝑎+nb R ln
𝑉𝑎+𝑉𝑏
𝑉𝑏
2501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Entropy of Mixing, for two gases (Module 03)
S2−S1
𝑛𝑎+𝑛𝑏=
na
𝑛𝑎+𝑛𝑏R ln
𝑉𝑎+𝑉𝑏
𝑉𝑎+
nb
𝑛𝑎+𝑛𝑏R ln
𝑉𝑎+𝑉𝑏
𝑉𝑏
∆𝑆𝑚𝑖𝑥 = 𝑋𝑎 R ln𝑉𝑎+𝑉𝑏
𝑉𝑎+ 𝑋𝑏 R ln
𝑉𝑎+𝑉𝑏
𝑉𝑏
Va+Vb
Vb=
na+nb
nb=
1
Xb
∆𝑆𝑚𝑖𝑥 = −R 𝑋𝑎 ln 𝑋𝑎 + 𝑋𝑏 ln 𝑋𝑏
∆𝑆𝑚𝑖𝑥 = −R
𝑖=1
𝑖=𝑛
𝑋𝑖 ln 𝑋𝑖
2601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Enthalpy of Mixing: IDEAL SOLUTION
Gibbs free energy change for a reaction as a function of temperature:
( ) 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
=
Therefore:
rel
A
rel
A
H
T
1d
T
Gd
=
AA
rel
A X ln RT a ln RT G ==
( ) rel
AA H
T
1d
X ln Rd=
AA HH 0 −==
R ln XA is not a function of temperature
dU = T dS - P dV
dH = T dS + V dP
dF = -S dT - P dV
dG = -S dT + V dP
Conclusion: Enthalpy mixing of
an ideal solution is zero
solution) (ideal 0H XH X Hrel
BB
rel
AAM =+=
rel
BB
rel
AAM V X V X V +=
GM = XA RT ln XA + XBRT ln XB
2701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
IDEAL SOLUTION
Volume change upon mixing for ideal solutions:
V P
G
T
=
rel
A
T
rel
AV
P
G=
( )0V
dP
X ln RTd rel
AA ==
solution) (ideal 0V XV X Vrel
BB
rel
AAM =+=
=i
iiM X ln X R- S
=i
iiM X ln X T R G
0 HM =
0 VM =
dU = T dS - P dV
dH = T dS + V dP
dF = -S dT - P dV
dG = -S dT + V dP
2801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
IDEAL SOLUTION
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
VM
XB
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
HM
XB
0.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
S M
XB
-12,000
-10,000
-8,000
-6,000
-4,000
-2,000
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
GM
XB
1800 K
1500 K
1200 K
900 K
600 K
300 K
Materials A and B can spontaneously form a solution
because the change in Gibbs free energy is negative
upon mixing.
2901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions: REGULAR SOLUTIONS
In 1895, Margules suggested the activity coefficients, A and
B, of the components of a binary solution can be
represented by power series:
ln γA = α1XB + 12 α2XB
2 + 13 α3XB
3 + …
ln γB = β1XA + 12β2XA
2 + 13β3XA
3 + …
ln γA =Ω
RTXB2 =
Ω
RT1−XA
2
In 1929, Hildebrand suggested the following equation to
predict activity coefficients, A and B, in binary solution A-B:
ln γA =α′
RTXB2 =
α′
RT1−XA
2
Ω = temperature dependent interaction parameter.
https://en.wikipedia.org/
3001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions: REGULAR SOLUTIONS
Regular Solution: a non ideal solution whose thermodynamic properties can be described
using a simple agebraic function due to similarity between atoms A and B (no strong
interaction & both atoms randomly mixed
rel
BB
rel
AAM G X G X G +=
( )AAA
o
AA
rel
A X ln RTa ln RT G G G ==−=
( ) ( )BBBAAAM X ln RT XX ln RT X G +=
( ) ( ) ln X ln XRTX ln XX ln X RT G BBAABBAAM +++=
xs
M
ideal
MM G G G +=
𝑇erm VA − VA is called relative partial molar of A, V𝐴𝑟𝑒𝑙
GM = GMideal + GM
xs
3101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions: REGULAR SOLUTIONS
A Regular Solution of two components is defined as one in which the activity coefficient of A
has the form:
ln 𝛾A =Ω
RTXB2 =
Ω
RT1−XA
2
GM = RT XA ln XA + XB ln XB + 𝑅𝑇 XA ln 𝛾A + XB ln 𝛾B
GM = RT XA ln XA + XB ln XB + XAΩXB2 + XBΩXA
2
GM = RT XA ln XA + XB ln XB + XA + XB ΩXA XB
GM = RT XA ln XA + XB ln XB + ΩXA XB
3201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions: REGULAR SOLUTIONS
ln 𝛾A =Ω
RTXB2 =
Ω
RT1−XA
2
GM = RT XA ln XA + XB ln XB + ΩXA XB
GM = HM−T SM
HM = Ω XA XB
SM = −R XA ln XA + XB ln XB
Entropy mixing in regular solution is equivalent to entropy mixing in ideal solution.
xs
M
ideal
MM G G G +=
M
xs
M
xs
M HH G ==
0 Sxs
M =
SMideal = −R
i
Xi ln Xi
GMideal = R T
i
Xi ln Xi
HMideal = 0
HM = HMideal + HM
xs
SM = SMideal + SM
xs
GMxs = HM
xs − T SMxs
3301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Act
ivit
y (a
)
XB
Non Ideal Solutions: REGULAR SOLUTIONS (300)
0
0.2
0.4
0.6
0.8
1
1.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Act
ivit
y co
effi
cien
t (
)
XB
-3000
-2500
-2000
-1500
-1000
-500
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
HM
XB
-4,500
-4,000
-3,500
-3,000
-2,500
-2,000
-1,500
-1,000
-500
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
GM
XB
ln 𝛾A =Ω
RTXB2 =
Ω
RT1−XA
2
HM = Ω XA XB
GM = GMideal + GM
xs
GMxs = HM
xs = HM
Ω = -10.000 JΩ = -7.000 J
Ω = -4.000 J
IdealIdeal
3401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solution: Negative deviation from Raoult’s law
D.R. Gaskell, 2003
3501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
0
500
1000
1500
2000
2500
3000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
HM
XB
0
0.5
1
1.5
2
2.5
3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Act
ivit
y (a
)
XB
Non Ideal Solutions: REGULAR SOLUTIONS (300 K)
ln 𝛾A =Ω
RTXB2 =
Ω
RT1−XA
2
HM = ΩXA XB
xs
M
ideal
MM G G G +=
M
xs
M
xs
M HH G ==
Ω = 10.000 JΩ = 7.000 J
Ω = 4.000 J
0
10
20
30
40
50
60
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Act
ivit
y co
effi
cien
t (
)
XB
Ideal
-2,000
-1,500
-1,000
-500
0
500
1,000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
GM
XB
Ideal
3601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solution: Positive deviation from Raoult’s law
D.R. Gaskell, 2003
3701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions: REGULAR SOLUTIONS
M
xs
M
xs
M HH G ==
0 Sxs
M =
0S- T
G xs
M
xs
M ==
etemperatur of ntindenpende is Gxs
M
XS
BB
XS
AA
xs
M G X G X G +=
2
B)A(T2)A(T1
XS
A X ln RT ln RT G21
===
Practical use: converting activity data for regular solution at one temperature to activity data at
another temperature
2
1
1A
2A
T
T
T etemperatur at ln
T etemperatur at ln =
BBAA X V X V V +=
dU = T dS - P dV
dH = T dS + V dP
dF = -S dT - P dV
dG = -S dT + V dP
3801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions: REGULAR SOLUTIONS
etemperatur of ntindenpende is Gxs
M
XS
BB
XS
AA
xs
M G X G X G +=
2
B)A(T2)A(T1
XS
A X ln RT ln RT G21
===
2
1
1A
2A
T
T
T etemperatur at ln
T etemperatur at ln =
D.R. Gaskell, 2003
3901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 1
Copper and gold form complete ranges of solid solution at temperature between 410 and
889oC. At 600oC, the excess molar Gibbs free energy of formation of solid solution is given
by:
GMxs = - 28,280 XAu XCu Joule
Calculate the activities of Cu and Au in the solid solution of XCu =0.4 at 600oC. Calculate
partial pressures of Cu and Au!
ln 𝑝𝐶𝑢𝑜 (atm) = -
40,920
𝑇− 0.86 ln 𝑇 + 21.67
ln 𝑝𝑍𝑛𝑜 (atm) = -
45,650
𝑇− 0.306 ln 𝑇 + 10.81
4001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 2
At 473 oC the Pb-Sn system exhibits regular solution behavior, and the activity coefficient of
Pb is given by:
log (Pb) = -0.32 (1-XPb)2
Write the corresponding equation for the variation of Sn with composition at 573 oC!
4101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 3&4
3. In a liquid Cu-Sn alloy at 1400K and at XSn =0.4, the values of activitiy of Sn and Cu
are 0.333 and 0.362, respectively. Calculate Sn, Cu, , !
4. The variation with composition of Gxs for liquid Fe-Mn alloys at 1863 K is listed
below:
a. Does the system exhibit regular solution behavior?
b. Calculate GFexs
and GMn
xsat XMn = 0.6
c. Calculate GM at XMn = 0.4
d. Calculate the partial pressure of Mn and Fe exerted by the alloy of XMn = 0.2
XMn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Gxs, J 395 703 925 1054 1100 1054 925 703 395
xs
MM G ,G MM S ,H
4201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions
In non ideal solutions, the curve of specific Gibbs free
energy of the solution as function of composition may
have many forms.
In the region between XB1 and XB2, solution
minimizes its free energy by decomposing
into two solutions. A single solution in this
region is unstable relative to mixture of
solutions of XB1 and XB2.
4301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions
Non ideal behavior: two immiscible materials, materials A and B form no solutions.
Molar Gibbs free energy against the
composition is simply a straight line
between the two points on the A and B
axes.
4401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Non Ideal Solutions
Materials A and B form a compound AB, which does
not dissolve in either A or B
4501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Thermodynamic Models of Solutions
Empirical models - polynomials• Regular
• Margules (2 and 3 suffix)
• Interaction parameter model for dilute solutions
Structurally-based models
• Masson polymer model
• Quasi-chemical
• Cell
• Central atom
• Sub-lattice (for solids)
• Associate Solution Model
4601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Regular / Subregular Model
A
A
AB B
B
B
B A
Eo = ωAA EAA + ωBB EBB +ωAB EAB
𝜔𝐴𝐴 : number of bonds associated with the formation of AA
𝜔𝐵𝐵 : number of bonds associated with the formation of BB
𝜔𝐴𝐵 : number of bonds associated with the formation of AB
𝐸𝐴𝐴 : Energy associated with the formation of AA
𝐸𝐵𝐵 : Energy associated with the formation of BB
𝐸𝐴𝐵 : Energy associated with the formation of AB
If N atoms in solution and z coordination number,
number of bonds:
ωAA =1
2N z XA
2
ωBB =1
2N z XB
2
ωAB =12 N z XAXB
Eo =N z
2XA EAA + XB EBB + XA XB 2EAB − EAA − EBB
HM =N z
2XA XB 2EAB − EAA − EBB
HM = GMxs = ΩXA XB (REGULAR SOLUTION)
If the reference states are taken as pure A and B:
GMxs = XA XB Ω𝐴𝐵
𝐴 XA +Ω𝐴𝐵𝐵 XB
Kaufman & Bernstein, 1970, Subregular Model:GM = RT XA ln XA + XB ln XB + 𝑅𝑇 XA ln 𝛾A + XB ln 𝛾B
GM = GMideal + GM
xs
4701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Extrapolation of Gibbs Excess Energy
A
A
AB B
B
B
B A
GM = RT XA ln XA + XB ln XB + 𝑅𝑇 XA ln 𝛾A + XB ln 𝛾B
GM = GMideal + GM
xs
Muggianu equation, 1975:
GMxs = XA XB 𝐿𝐴𝐵
𝑜 + 𝐿𝐴𝐵1 XA − XB + XB XC 𝐿𝐵𝐶
𝑜 + 𝐿𝐵𝐶1 XB − XC + XA XC 𝐿𝐴𝐶
𝑜 + 𝐿𝐴𝐶1 XA − XC
Kohler equation, 1960:
GMxs = XA + XB
2XA
XA + XB
XB
XA + XB𝐿𝐴𝐵𝑜 + 𝐿𝐴𝐵
1 XA − XB
XA + XB
+ XB + XC2 XB
XB + XC
XC
XB + XC𝐿𝐵𝐶𝑜 + 𝐿𝐵𝐶
1 X𝐵 − XC
XB + XC
+ XA + XC2 XA
XA + XC
XC
XA + XC𝐿𝐴𝐶𝑜 + 𝐿𝐴𝐶
1 XA − XC
XA + XC
L : excess interaction parameter
Toop equation, 1975:
GMxs = XA XB 𝐿𝐴𝐵
𝑜 + 𝐿𝐴𝐵1 XA − XB− XC + XA XC 𝐿𝐴𝐶
𝑜 + 𝐿𝐴𝐶1 XA − XC − XB
+ XB XC 𝐿𝐵𝐶𝑜 + 𝐿𝐵𝐶
1 XA − XC +XB − XC
XB + XCXA
4801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Quasichemical Model
4901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
NON-IDEAL SOLUTIONS
A lot of substances formordered solutions with complex functionsof properties in terms of compositions
Appropriate models have to be used to adequately describethermodynamic properties
e.g.:molten salts slag – molten oxidesquasi-chemical solution model is used
Solution with
strong interatomic
interactions
resulting in very
negative MH
(-84) as well as in
strong ordering
(so that that MS
is close to 0)
Enthalpy MH
of mixing
Entropy MS
of mixing
5001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Gibbs-Duhem Relation
Molar Gibbs free energy of an A-B solution can be written as:
BBAA X G X G G +=
BBAABBAA Gd X GdX dX G dX G Gd +++=
Differentiation:
P,TBA )n,n(GG =
B
n,P,TB
A
n,P,TA
dn n
G dn
n
GdG
AB
+
=
Bn,P,TA
A
n
GG
=
Compare:
BBAA dX G dX G Gd +=
0Gd X GdX BBAA =+
BBAA X V X V V +=
XAdഥHA + XBdഥHB = 0
XAdതSA + XBdതSB = 0
en.wikipedia.org
5101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Gibbs-Duhem Relation: Activity
This equation is known as Gibbs-Duhem equation.
0Gd X GdX BBAA =+
( ) ( ) 0a ln RTd X a ln RTdX BBAA =+
( ) ( )B
A
BA a lnd
X
X- a lnd =
If the activity of one component of a binary solution is known as a
function of composition, then the activity of the other can be
determined.
aNi experimental
aFe ??
Gibbs Duhem Reation is used to find activity of a
component based on experimental activity of another
component in binary system if the solution is not a
regular solution
5201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
As XB → 1, aB → 1, log aB → 0
and XB / XA → ∞.
The curve exhibits a tail to infinity
as XB → 1
As XB → 0, aB → 0, log aB→
-∞ and XB / XA → 0.
The curve exhibits a tail to
minus infinity as XB → 0
Gibbs-Duhem Relation: Activity
( ) ( )B
A
BA a lnd
X
X- a lnd =
If the variation of aB with
composition is known, then
integration of above equation from
XA = 1 to XA gives the value of log
aA at XA:
( ) ( )B
A
BA a logd
X
X- a logd =
( )=
==
AB
AB
X at a log
1 X at a logB
A
BAA a logd
X
X - X at a log
5301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 1:
A and B are in a solution, material B behaves ideally.
Activity of B is equal to mole fraction of B (pure B as standard state)
( ) ( )B
A
BA a lnd
X
X- a lnd =
( ) ( )B
A
BA X lnd
X
X- a lnd =
( )B
B
A
BA
X
dX
X
X- a lnd =
1X X BA =+
BA dX- dX =
( )A
AA
X
dX a lnd =
AA X a =
If material B in a solution of A and B behaves ideally, then material A
behaves also ideally.
5401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Gibbs-Duhem Relation: Activity Coefficient
( ) ( ) 0X ln d X X ln d X BBBAAA =+
( ) ( ) ( ) ( ) 0X ln d XX ln d X ln d X ln d X BBAABBAA =+++
( ) AAA dXX ln d X =
( ) BBB dXX ln d X =
0dXdX BA =+
( ) ( ) 0 ln d X ln d X BBAA =+
( ) ( )B
A
BA lnd
X
X- lnd =
5501 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Activity and Activity Coefficient
At XB/XA=0 → aB = 0
ln aB= ∞ (infinite) At XB/XA=0 → γB = γBo
ln γB = ln γBo
(finite number)
lnγA = නXA=1
XA=XA
−XBXA
dlnγB = නXA=XA
XA=1 XBXA
dlnγBln𝑎A = නXA=1
XA=XA
−XBXA
dln𝑎B = නXA=XA
XA=1 XBXA
dln𝑎B
5601 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 2:
Copper zinc alloy. Activity of zinc is easy to measure because zinc is vollatile; it is possible to
measure zinc pressure in equilibrium with zinc containing liquids or solids. Activity of zinc is zinc
vapor pressure relative to vapor pressure of pure zinc. In temperature range 1400-1500K, an
expression that fits the data for activity coefficient of zinc is:
2
CuZn X -38,300 ln RT =
Applying Gibbs Duhem equation:
( ) ( )Zn
Cu
ZnCu lnd
X
X- lnd =
( )
−=
RT
X 300,38d
X
X- lnd
2
Cu
Cu
ZnCu
( ) CuZnCu dX RT
300,38 x 2X lnd =
ZnCu dX dX −=
( ) ZnZnCu dX XRT
300,38 x 2 lnd −=
2
ZnCu XRT
300,38 ln −=
5701 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 3:
lnγCd|XCd=0.1 = නXCd=0.1
XCd=1 XZnXCd
dlnγZn
The activity of zinc in liquid cadmium-zinc alloys at 708K is related to the alloy composition
by the following equation:
ln γ𝑍𝑛 = 0.87 𝑋𝐶𝑑2 − 0.3 𝑋𝐶𝑑
3
Calculate the activity of cadmium at XCd = 0.1!
5801 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 3:
lnγCd|XCd=0.1 = නXCd=0.1
XCd=1 XZnXCd
dlnγZn
XCd XZn ln γZn XZn/XCd
0.10 0.90 0.01 9.00
0.20 0.80 0.03 4.00
0.30 0.70 0.07 2.33
0.40 0.60 0.12 1.50
0.50 0.50 0.18 1.00
0.60 0.40 0.25 0.67
0.70 0.30 0.32 0.43
0.80 0.20 0.40 0.25
0.90 0.10 0.49 0.11
1.00 0.00 0.57 0.00
XCd XZn ln γZn XZn/XCd
0.10 0.90 0.01 9.00
0.20 0.80 0.03 4.00
0.30 0.70 0.07 2.33
0.40 0.60 0.12 1.50
0.50 0.50 0.18 1.00
0.60 0.40 0.25 0.67
0.70 0.30 0.32 0.43
0.80 0.20 0.40 0.25
0.90 0.10 0.49 0.11
1.00 0.00 0.57 0.00
XCd XZn ln γZn XZn/XCd
0.10 0.90 0.01 9.00
0.20 0.80 0.03 4.00
0.30 0.70 0.07 2.33
0.40 0.60 0.12 1.50
0.50 0.50 0.18 1.00
0.60 0.40 0.25 0.67
0.70 0.30 0.32 0.43
0.80 0.20 0.40 0.25
0.90 0.10 0.49 0.11
1.00 0.00 0.57 0.00
XCd XZn ln γZn XZn/XCd
0.10 0.90 0.01 9.00
0.20 0.80 0.03 4.00
0.30 0.70 0.07 2.33
0.40 0.60 0.12 1.50
0.50 0.50 0.18 1.00
0.60 0.40 0.25 0.67
0.70 0.30 0.32 0.43
0.80 0.20 0.40 0.25
0.90 0.10 0.49 0.11
1.00 0.00 0.57 0.000
1
2
3
4
5
6
7
8
9
10
0.00 0.10 0.20 0.30 0.40 0.50 0.60
XZnXCd
lnγZn
lnγCd = 0.59 → γCd = 1.81 → 𝑎Cd = 0.18
Area of trapezoid 1 =9 + 4
2(0.03 − 0.01)
5901 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 4:
From vapour pressure measurement, the following values have been determined for the
activity of mercury in liquid mercury bismuth alloys at 320oC. Calculate activity of bismuth in a
50 atom-percent Bi alloy at this temperature!
XHg 0.949 0.893 0.851 0.753 0.653 0.537 0.437 0.330 0.207 0.063
aHg 0.961 0.929 0.908 0.840 0.765 0.650 0.542 0.432 0.278 0.092
( ) ( )Hg
Bi
Hg
Bi a logd X
X- a logd =
( )=
==
BiHg
BiHg
X at a log
1 X at a logHg
Bi
Hg
BiBi a logd X
X - X at a log
6001 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Aqueous Solution
Solutions of electrolytes are non‐ideal at relatively low concentrations.
The greater the charge on the ions the larger the deviations from ideality.
For example, for one mole of CaCl2 dissolved the deviation from ideal behavior is larger than
from one mole NaCl due to the 2+ charge of calcium ion.
Perhatikan disosiasi dari molekul berikut:
V+ = koefisien stoikiometrik
Z+ = muatan positif
Z- = muatan negatif
Contoh: 2
4
2
42 OS H 2 OS H−+−+ +=
i
o
ii a ln TR +=
μi = μio + RT ln γi mi
ai = i mi
ai = aktivitas i
i = koefisien aktivitas i
mi = konsentrasi molal i
Untuk larutan ideal, i = 1 : ai = mi
z zz
v
z
v B v A v B A −
−
+
+
−
−
+
+ +=
6101 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Aqueous Solution
i
o
ii mln T (ideal) R+=
ln T mln T (real) ii
o
ii RR ++=
ln TR (ideal) -(real) iI-iii ==
Menurut teori Debye-Hueckel, perubahan potensial kimia yang
timbul dari interaksi ion-ion adalah sebagai berikut:
( )1-
2
0iAI-i
2
e z N -
=
( )1-
2
0iAi
2
e z N - ln TR
=
NA= bilangan Avogadro (6,022 x 1023 / mol)
zi = bilangan muatan ion π = 3,14
eo = muatan elektron (1,602 x 10-19 C)
= kostanta dielektrik (78 for water)
-1 = Debye Hueckel reciprocal length (m)
k = konstanta boltzmann (1,380 x 10-23 J/K)
ci = Konsentrasi dalam molal
https://en.wikipedia.org/
I = ionic strength
𝜅 =8 𝜋 𝑁𝐴 𝑒𝑜
2
1000 𝜀 𝑘 𝑇
I = 12σ ci zi
2
6201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Debye-Huckel Equation
https://en.wikipedia.org/
Limiting Law (for I < 10-2)
Individual ion: log 𝛾𝑖 = − 𝐴 𝑧𝑖2 𝐼
Mean activity coefficient: log 𝛾 = − 𝐴 𝑧+ 𝑧− 𝐼
A = 0.509 at 25 C
Extended DH (for I < 10-1)
Individual ion: log 𝛾𝑖 = −𝐴 𝑧𝑖
2 𝐼
1+ 𝑎𝑖 𝐵 𝐼
Mean activity coefficient: log 𝛾 =−−𝐴 𝑧+ 𝑧− 𝐼
1+𝑎 𝐵 𝐼
A = 0.509 at 25 C
B = 0.33
6301 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Davies Equation
Davies (for I < 0.5)
Individual ion: log 𝛾𝑖 = −𝐴 𝑧𝑖2 𝐼
1+ 𝐼− 0.2 𝐼
Mean activity coefficient:
log 𝛾 = −𝐴 𝑧+ 𝑧−𝐼
1 + 𝐼− 0.2 𝐼
A = 0.509 at 25 C
6401 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021
Case 1
Calculate the ionic strength and the mean activity coefficient of 2.0m mol kg-1 Ca(NO3)2 at 25 oC.
Solution:
I = ½(22*0.002 + (-1)2*(2*0.002)) = ½ *6*0.002 = 0.006
Debye-Huckel limiting equation:
log 𝛾 = − 𝐴 𝑧+ 𝑧− 𝐼
log 𝛾 = - |2*1|*A*(0.006)1/2
= - 2*0.509*0.0775= -0.0789
𝛾 = 0.834
I = 12σ ci zi
2
6501 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