Termodinamika Larutan: Teori
(Solution Thermodynamics: Theory) (Solution Thermodynamics: Theory)
Lecturer: Bregas S T Sembodo
Pendahuluan
• Sistem nyata biasanya berupa campuran fluida
• Komposisi sering menjadi peubah (variabel) utama
• Dikembangkan teori dasar aplikasi termodinamika untuk campuran gas dan larutan
• Akan diperkenalkan:
• Akan diperkenalkan:
– Potensial kimia (chemical potential) – Sifat parsial (partial properties)
– Fugasitas (fugacity)
– Sifat ekses (excess properties)
– Larutan ideal (ideal solution)
Hubungan sifat-sifat fundamental
• Hubungan antara Energi Gibbs dengan suhu dan tekanan pada sistem tertutup:
dT nS dP
nV T dT
dP nG P
nG nG d
n P n
T
) ( )
) ( (
) ) (
(
, ,
−
=
∂ + ∂
∂
= ∂
– Diterapkan untuk fluida fase tunggal dalam sistem tertutup dimana tidak terjadi reaksi kimia.
• Untuk fase tunggal, sistem terbuka (a single-phase, open system):
∑
∂ + ∂
∂ + ∂
∂
= ∂
i
i n
T i P
n P n
T
n dn dT nG
T dP nG
P nG nG
d
, j
, , ,
) (
) (
) ) (
(
nj
T i P
i n
nG
, ,
)
(
∂
≡ ∂ µ Definisi potensial kimia:
∑
+
−
=
i
i i
dn dT
nS dP
nV nG
d ( ) ( ) ( ) µ
Hubungan sifat fundamental untuk sistem fluida fase tunggal dengan komposisi konstan maupun sebagai variabel:
Jika n = 1,
= − + ∑
i
i i
dx SdT
VdP
dG µ ( , , , ,..., ,...)
2
1
x x
ix T P G G =
x
T P
S G
,
∂
− ∂
=
x
P T
V G
,
∂
= ∂
Energi Gibbs dinyatakan sbg fungsi dari variabel-variabel kanonis
Sifat larutan, M Sifat parsial,
Sifat spesies murni, Mi
M
iPotensial kimia dan kesetimbangan fase
• Untuk sistem tertutup terdiri dari 2 fase dalam kesetimbangan:
∑
+
−
=
i
i i dn dT
nS dP
nV nG
d( )α ( )α ( )α µα α = − +
∑
i
i i dn dT
nS dP
nV nG
d( )β ( )β ( )β µβ β
i i
β
α
( )
)
( nM nM
nM = +
∑
∑
++
−
=
i
i i i
i
i dn dn
dT nS dP
nV nG
d( ) ( ) ( ) µα α µβ β
β
α
µ
µ
i=
iβ α
i
i dn
dn = −
Neraca massa:
Multi fase pada T dan P yg sama adalah dalam kesetimbangan jika potensial kimia setiap spesies pada semua fase sama.
Sifat Parsial (Partial properties)
• Definisi sifat molar parsial (partial molar property) spesies i :
arti : perubahan besaran M terhadap perubahan n (jml mol i) jika
nj
T i P
i
n
M nM
, ,
)
(
∂
≡ ∂
arti : perubahan besaran M terhadap perubahan n
i(jml mol i) jika P, T dan n
j(jml mol spesies lain) tetap
– Potensial kimia adalah Energi Gibbs molar parsial:
– Untuk sifat termodinamika M:
i i
≡ G
µ
,...) ,...,
, , ,
( P T n
1n
2n
iM
nM =
∑
+
∂ + ∂
∂
= ∂
i
i i n
P n
T
dn M T dT
n M P dP
n M nM
d
, ,
)
(
∑
+
∂ + ∂
∂
= ∂
i
i i n
P n
T
dn M T dT
n M P dP
n M nM
d
, ,
) (
∑
+ +
∂ + ∂
∂
= ∂ +
i
i i
i n
P n
T
ndx dn
x M T dT
n M P dP
n M Mdn
ndM ( )
, ,
0
, ,
=
−
+
−
∂
− ∂
∂
− ∂ dT
∑
M dx n M∑
x M dnT dP M
P dM M
i
i i i
i i n
P n
T
0
, ,
=
−
∂
− ∂
∂
− ∂
∑
i
i i n
P n
T
dx M T dT
dP M P
dM M dan −
∑
= 0i
i iM x M
Perhitungan sifat-sifat campuran dari sifat- sifat parsialnya
= 0
−
∑
i
i iM n nM
0
, ,
=
−
∂ + ∂
∂
∂ ∑
i
i i
n P n
T
M d x T dT
dP M P
M
∑
∑
+=
i
i i i
i
idM M dx
x dM
Persamaan Gibbs/Duhem
Sifat Parsial pada Larutan Biner
• Untuk sistem biner
2 2 1
1M x M
x
M = +
2 2 2
2 1
1 1
1dM M dx x dM M dx
x
dM = x1dM1 + M1dx1 + x2dM2 + M2dx2
dM = + + +
P and T tetap, menggunakan pers. Gibbs/Duhem
2 2 1
1dx M dx
M
dM = +
2 1
1 + x =
x
2 1
1
M dx M
dM = −
1 2
1 dx
x dM M
M = +
1 1
2 dx
x dM M
M = −
Constant T, P
Constant T, P
Pada suatu Lab dibutuhkan 2000 cm3 antifreeze solution (larutan antibeku)
mengandung 30 mol-% metanol dalam air. Berapa volume metanol murni dan air murni pada 25°C harus dicampur untuk memperoleh 2000 cm3 antifreeze pada 25°C?
Volume molar parsial dan volume murni masing-masing spesies diketahui.
2 2 1
1V x V
x
V = + V = (0.3)(38.632) +(0.7)(17.765) = 24.025 cm3 / mol Vt 2000
=
=
= n1 = (0.3)(83.246) = 24.974 mol V mol
n V
t
246 . 025 83
. 24
2000 =
=
= n1 = (0.3)(83.246) = 24.974 mol mol n2 = (0.7)(83.246) = 58.272
3 1
1
1 nV (24.974)(40.727) 1017 cm
V t = = = V2t = n2V2 = (58.272)(18.068) =1053 cm3
Entalpi sistem cairan biner dari spesies 1 dan 2 pada T dan P tertentu adalah:
Tentukan dan sbg fungsi x1, nilai numerik entalpi spesies murni H1 and H2, nilai numerik entalpi parsial pada pengenceran tak berhingga dan
H
1H
2∞
H
1 H2∞ )20 40
( 600
400x1 x2 x1x2 x1 x2
H = + + +
) 20 40
( 600
400x1 x2 x1x2 x1 x2
H = + + +
2 1
1 + x =
x
20 3
180
600 x x
H = − − x1 + x2 =1 H = 420−60x2 + 40x3
40 3
600 x
H = +
3 1
1 20
180
600 x x
H = − −
1 2
1 dx
x dH H
H = +
3 1 2
1
1 420 60x 40x
H = − + H2 = 600+ 40x13
1 = 0 x
mol H1∞ = 420 J
1 =1 x
mol H2∞ = 640 J
Hubungan antar Sifat Parsial
• Relasi Maxwell :
∑
+
−
=
i
i i
dn G dT
nS dP
nV nG
d ( ) ( ) ( )
S
V
∂
− ∂
=
∂
∂ Gi (nS)
∂
− ∂
=
∂
∂ Gi (nV)
∂
− ∂
=
∂
∂
n T n
P P
T , ,
∂
−
=
∂
nj
T i P
n
P n
T , , ,
∂
−
=
∂
nj
T i P
n
T n
P , , ,
∂
−
=
∂
i x
P
i S
T
G = −
∂
∂
,
i x
T
i V
P
G = −
∂
∂
,
PV U
H = +
dT S dP V G
d
i=
i−
ii i
i
U P V
H = +
Campuran Gas Ideal
• Teorema Gibbs’s
– Sifat molar parsial spesies (kecuali volume) dalam camp. gas ideal SAMA DENGAN sifat molarnya sbg gas ideal murni pada suhu campuran tetapi pada
gas ideal murni pada suhu campuran tetapi pada tekanan yg sama dengan tekanan parsialnya dalam campuran.
ig i ig
i
V
M ≠ M
iig( T , P ) = M
iig( T , p
i)
Untuk yg tergantung tekanan, mis. dS
iig= − Rd ln P const . T
i i
i i
ig i ig
i R y
P y R P
p R P
p T S P
T
S ( , ) − ( , ) = − ln = − ln = ln
Untuk yg tidak tergantung tekanan, mis.
) , ( )
, ( )
,
( T P H T p H T P
H
iig=
iig i=
iig =∑
i
ig i i
ig y H
H =
∑
i
ig i i
ig yU
U
) , ( )
,
( iig i
ig
i T P M T p
M =
i ig
i ig
i T P S T P R y
S ( , )− ( , ) = ln
) , ( )
,
( i i
i T P M T p
M =
∑
∑
−=
i
i i
i
ig i i
ig y S R y y
S ln
ig i ig
i ig
i H TS
G = − H (T,P) H (T,P)
G
iig= H
iig− TS
iig+ RT ln y
iig i ig
i =
i ig
i ig
i T P S T P R y
S ( , ) − ( , ) = ln
i ig
i ig
i ig
i
≡ G = G + RT ln y
µ P y RT
T
ii ig
i
= Γ ( ) + ln
µ
or P
y y
RT T
y G
i
i i
i
i i
ig =
∑
Γ ( ) +∑
lnG T RT P
i ig
i
= Γ ( ) + ln
∑
∑ −
=
i
i i
i
ig i i
ig
y S R y y
S ln
y S y
R y
y R
S y S
i i
i i
i i
i
ig i i
ig
− ∑ = − ∑ ln = ∑ ln 1 = ∆ >0
Perubahan entropi pencampuran gas ideal selalu positif ATAU entropi campuran gas ideal lebih besar dari pada entropi spesies murni konstituennya
Fugasitas dan Koefisien Fugasitas
• Potensial kimia:
– Menyediakan kriteria dasar kesetimbangan fase
nj
T i P
i n
nG
, ,
)
(
∂
≡ ∂ µ
– Namun, Energi Gibbs juga µ
ididefinisikan
berhubungan dengan energi dalam dan entropi (nilai absolut tidak diketahui, sulit ditentukan nilainya dari T dan P).
• Fugasitas:
– Suatu kunatitas yg menggantikan µ
ii i
i
T RT f
G ≡ Γ ( ) + ln
Dengan satuan tekanan
i i
i
T RT f
G ≡ Γ ( ) + ln P RT
T
G
iig= Γ
i( ) + ln
P RT f
G
G
i−
iig= ln
iG
iR= RT ln φ
iP f
ii
=
φ
Residual Gibbs energy Fugacity coefficient
) . (
) 1 (
ln dP const T
P
Z
∫ −
φ = ( 1 ) ( . )
ln
0const T
P Z dP
P i
i
= ∫ −
φ
Kesetimbangan Uap-Cair (VLE) untuk Spesies Murni
• Uap Jenuh:
• Cairan Jenuh:
v i i
v
i
T RT f
G = Γ ( ) + ln
l i i
l
i
T RT f
G = Γ ( ) + ln
f
v l iv l i
i v
i
f
RT f G
G − = ln
VLE
0
ln =
=
− l
i v l i
i v
i f
RT f G
G
sat i l
i v
i f f
f = =
sat i l
i v
i
φ φ
φ = =
For a pure species coexisting liquid and vapor phases are in equilibrium when they have the same temperature, pressure, fugacity and fugacity coefficient.
Fugasitas Cairan Murni
• Fugasitas spesies murni i sebagai cairan tertekan (compressed liquid):
sat i i i
RT f G
G − = ln
satG − G
sat= ∫
PV dP ( isothermal process )
i i
i
G RT f
G − = ln G G V dP ( isothermal process )
P i
sat i
i sat
∫
i=
−
∫
=
PP i
sat i
i
sat i
dP RT V
f
f 1
ln
Karena Vi sedikit dipengaruhi P (weak function)
RT P P
V f
f
il isatsat i
i
( )
ln = −
sat i sat i sat
i
P
f = φ RT
P P
P V f
sat i l
sat i i sat i i
)
exp ( −
= φ
For H2O at a temperature of 300°C and for pressures up to 10,000 kPa (100 bar) calculate values of fi and φi from data in the steam tables and plot them vs. P.
For a state at P:
G
i= Γ
i( T ) + RT ln f
iFor a low pressure reference state:
G
i*= Γ
i( T ) + RT ln f
i*) 1 (
ln
* i i*i
i
G G
RT f
f = −
i i
i
H TS
G = −
− − −
= 1 ( )
ln
**
* i i
i i
i
i
S S
T H H
R f
f
The low pressure (say 1 kPa) at 300°C:
J gK Si* =10.3450 kPa
P
fi* = * =1 J g Hi* = 3076.8
Si =10.3450 gK kPa
P
fi = =1
For different values of P up to the saturated pressure at 300°C, one obtains the values of fi ,and hence φi . Note, values of fi and φi at 8592.7 kPa are obtained
Fig 11.3 Values of fi andφi at higher pressure:
RT P P
P V f
sat i l
sat i i sat i i
)
exp ( −
= φ
Fugasitas dan Koefisien Fugasitas:
spesies dalam larutan
• Untuk spesies i dalam campuran gas nyata atau dalam larutan cair:
i i
i
≡ Γ ( T ) + RT ln f ˆ
µ
• Multi fase pada T dan P yg sama adalah dalam
kesetimbangan jika fugasitas masing-masing spesies konstituen sama pada semua fase:
Fugasitas spesies i dalam larutan (mengganti tekanan parsial yiP)
π β
α
i i
i
f f
f ˆ = ˆ = ... = ˆ
ig
R
M M
M = −
Sifat Residual:
ig i i
R
i
M M
M = −
Sifat residual parsial
ig i i
R
i
G G
G = −
i i
i
≡ Γ ( T ) + RT ln f ˆ
µ
P y RT
T
ii ig
i
= Γ ( ) + ln
µ
P y RT f
i ig i
i i
ˆ
= ln
− µ µ
i R
i
RT
G = ln φ ˆ
i n
T i P
i G
n nG
j
=
∂
≡ ∂
, ,
) µ (
P y
f
i i i
ˆ ≡ ˆ φ
Koefisen fugasitas spesies i dalam larutan
Untuk gas ideal,
= 0
R
G
i1 ˆ = ˆ =
P y
f
i i
φi fˆi = yiP
Fundamental residual-property relation
RT dT nG nG
RT d RT
d nG 1 ( ) 2
−
=
∑
+
−
=
i
i idn dT
nS dP
nV nG
d( ) ( ) ( ) µ
TS H
G = H −TS G = −
∑
+
−
=
i
i i dn RT dT G
RT dP nH
RT nV RT
d nG 2 f (P,T,ni)
RT nG =
G/RT as a function of its canonical variables allows evaluation of all other
thermodynamic properties, and implicitly contains complete property information.
The residual properties:
∑
+
−
=
i
i R i R
R R
RT dn dT G
RT dP nH
RT nV RT
d nG 2 = − +
∑
i
i i R
R R
dn RT dT
dP nH RT
nV RT
d nG 2 lnφˆ
or
∑
+
−
=
i
i R i R
R R
RT dn dT G
RT dP nH
RT nV RT
d nG 2 = − +
∑
i
i i R
R R
dn RT dT
dP nH RT
nV RT
d nG 2 lnφˆ
Fix T and composition:
x T R
R
P RT G
RT V
,
) /
(
∂
= ∂
Fix P and composition:
x P R
R
T RT T G
RT H
,
) /
(
∂
− ∂
=
Fix T and P:
nj
T i P
R
i n
RT nG
, ,
) /
ˆ (
ln
∂
= ∂ φ
Develop a general equation for calculation of values form compressibility- factor data.
φˆi
ln
nj
T i P
R
i n
RT nG
, ,
) /
ˆ (
ln
∂
= ∂ φ
∫
−= P
R
P n dP RT nZ
nG
0 ( )
∫
∂ ∂ − = P
i P
dP n
n nZ ) ˆ (
lnφ
∫
∂
=
n T i P
i n P
j
0
, ,
lnφ
i i
n Z nZ =
∂
∂( )
=1
∂
∂ ni
n
∫ −
=
P ii
P
Z dP
0
( 1 )
ln φ ˆ
Integration at constant temperature and composition
Fugacity coefficient from the virial E.O.S
• The virial equation:
– the mixture second virial coefficient B:
– for a binary mixture:
RT Z = 1 + BP
∑∑
=
i j
ij j
i
y B
y B
– for a binary mixture:
• n mol of gas mixture:
22 2 2 21
1 2 12
2 1 11
1
1
y B y y B y y B y y B
y
B = + + +
RT n nBP nZ = +
2
2 1 ,
, 1 ,
1
) 1 (
) (
n T n
T
P n
nB RT
P n
Z nZ
∂ + ∂
=
∂
= ∂
2
2 1 ,
0
1 , 1
) ( )
( ˆ 1
ln
n T P
n
T n
nB RT
dP P n
nB
RT
∂
= ∂
∂
=
∫
∂ φ2
2 1 ,
0
1 , 1
) ( )
( ˆ 1
ln
n T P
n
T n
nB RT
dP P n
nB
RT
∂
= ∂
∂
=
∫
∂ φ22 2 2 21
1 2 12
2 1 11
1
1
y B y y B y y B y y B
y
B = + + +
22 11
12
12
≡ 2 B − B − B
δ y
i= n
i/ n
(
11 22 12)
ˆ1
lnφ B y δ
RT
P +
= Similarly: lnφˆ2
(
B22 y12δ12)
RT
P +
=
For multicomponent gas mixture, the general form:
+ −
=
∑∑
(2 )2 ˆ 1
ln ik ij
i j
j i kk
k B y y
RT
P δ δ
φ where
kk ii
ik
ik
≡ 2 B − B − B
δ
Determine the fugacity coefficients for nitrogen and methane in N2(1)/CH4(2) mixture at 200K and 30 bar if the mixture contains 40 mol-% N2.
cm mol B
B
B
12 11 22 312
≡ 2 − − = 2 ( − 59 . 8 ) + 35 . 2 + 105 . 0 = 20 . 6 δ
( ) [
35.2 (0.6) (20.6)]
0.0501) 200 )(
14 . 83 ( ˆ 30
lnφ1 = B11 + y22δ12 = − + 2 = −
RT P
9511 .
ˆ 0
1 = φ
( ) [
105.0 (0.4) (20.6)]
0.1835) 200 )(
14 . 83 ( ˆ 30
lnφ2 = B22 + y12δ12 = − + 2 = −
RT P
8324 .
ˆ 0
2 = φ
Generalized correlations for the fugacity coefficient
∫ −
=
Pr rr r i
i
const T
P Z dP
0
( 1 ) ( . )
ln φ
1
0
Z
Z
Z = + ω
) (
1 B0 B1
T Z P
r
r +ω
=
−
) (
ln B
0B
1T P
r
r
ω
φ = +
1
0
Z
Z
Z = + ω
∫
− +∫
= Pr Pr r
r r r
r const T
P Z dP P
Z dP
0 0
1
0 1) ( . )
(
lnφ ω
or
1
0
ln
ln
ln φ = φ + ω φ
∫
−≡ Pr
r r
P Z dP
0
0
0 ( 1)
lnφ ≡
∫
Prr r
P Z dP
0 1
lnφ1
with
Table E1:E4 or Table E13:E16
) ,
,
( TR PR OMEGA
= PHIB φ
For pure gas
For pure gas
Estimate a value for the fugacity of 1-butene vapor at 200°C and 70 bar.
127 .
=1 Tr
731 .
=1 Pr
191 .
= 0 ω
627 .
0 = 0
φ and φ1 =1.096
1
0
ln
ln
ln φ = φ + ω φ
Table E15 and E16
638 .
= 0
φ f = φ P = ( 0 . 638 )( 70 ) = 44 . 7 bar
For gas mixture:
For gas mixture:
+ −
=
∑∑
(2 )2 ˆ 1
ln ik ij
i j
j i kk
k B y y
RT
P
δ δ
φ
) ( B
0B
1P
B RT
ijcij cij
ij
= + ω
2
j i
ij
ω
ω = ω + T
cij= ( T
ciT
cj) ( 1 − k
ij)
Prausnitz et al. 1986
cij cij cij
cij
V
RT P = Z
2
cj ci
cij
Z Z Z +
=
1/3 1/3 32
+
=
ci cjcij
V V V
Empirical interaction parameter
Estimate and for an equimolar mixture of methyl ethyl ketone (1) / toluene (2) at 50°C and 25 kPa. Set all kij = 0.
ˆ
1φ φ ˆ
22
j i
ij
ω ω = ω +
) 1
( )
(
ci cj ijcij
T T k
T = −
cij cij cij
cij
V
RT P = Z
2
cj ci
cij
Z Z Z +
=
1/3 1/3 32
+
=
ci cjcij
V V V
) ( B
0B
1P
B RT
ijcij cij
ij
= + ω δ
12≡ 2 B
12− B
11− B
22( )
0.0128lnφˆˆ = P
(
B + y2δ)
= −0.0128 φˆ = 0.987 lnφ1 = B11 + y22δ12 = −RT P
( )
0.0172lnφˆ2 = B22 + y12δ12 = − RT
P
987 . ˆ 0
1 = φ
983 . ˆ 0
2 = φ
The ideal solution
• Serves as a standard to be compared:
i i
id
i
G RT x
G = + ln = ∑
i i+ ∑
i iid
x G RT x x
G ln
∑
=
i
id i i
id
x M
M
i i
i
G RT x
G = + ln
cf.
G
iig= G
iig+ RT ln y
ii P
i x
P id id i
i R x
T G T
S G ln
,
−
∂
− ∂
=
∂
− ∂
=
S
iid= S
i− R ln x
iT i x
T id id i
i
P
G P
V G
∂
− ∂
=
∂
− ∂
=
,
i id
i
V
V =
i i
i i
id i id
i id
i
G T S G RT x TS RT x
H = + = + ln + − ln H
iid= H
i∑
∑
i
i i
i
i i
i i
i i
i i
id x S R x x
S =
∑
−∑
lni i
i
id
x V
V = ∑
i i
i
id
x H
H = ∑
i i
id
i
G RT x
G = + ln
i i
i
≡ Γ ( T ) + RT ln f ˆ
µ
The Lewis/Randall Rule
• For a special case of species i in an ideal solution:
id i i
id i id
i
= G = Γ ( T ) + RT ln f ˆ
µ
i i
i
G RT x
G = + ln
i i
i
T RT f
G ≡ Γ ( ) + ln
i i id
i
x f
f ˆ =
The Lewis/Randall rulei id
i
φ
φ ˆ =
The fugacity coefficient of species i in an ideal solution is equal to the fugacity coefficient of pure species i in the same physical state as the solution and at the same T and P.Excess properties
• The mathematical formalism of excess properties is analogous to that of the residual properties:
id
E
M M
M ≡ −
– where M represents the molar (or unit-mass) value of any extensive thermodynamic property (e.g., V, U, H, S, G, etc.)
– Similarly, we have:
∑
+
−
=
i
i E i E
E E
RT dn dT G
RT dP nH
RT nV RT
d nG 2
The fundamental excess-property relation
(1) If CEP is a constant, independent of T, find expression for GE, SE, and HE as
functions of T. (2) From the equations developed in part (1), fine values for GE , SE, and HE for an equilmolar solution of benzene(1) / n-hexane(2) at 323.15K, given the following excess-property values for equilmolar solution at 298.15K:
CEP=-2.86 J/mol-K, HE = 897.9 J/mol, and GE = 384.5 J/mol From Table 11.1:
x P E E
P T
T G C
, 2 2
∂
− ∂
=
. const a
CPE = = T
a T
G
x P
E = −
∂
∂
, 2 2
b T T a
GE
+
−
=
∂
∂ ln
integration From Table 11.1:
E E
T
S G
∂
− ∂
= a T b
T P x
+
−
=
∂ ln
,
c bT T
T T a
GE = − ( ln − )+ +
integration From Table 11.1:
x
T P
S
,
∂
−
=
b T a
SE = ln −
c aT TS
G
H E = E + E = +
integration
86 .
−2
=
= a CPE
c
a +
= (298.15) 9
. 897
c b
a − + +
−
= ((298.15)ln(298.15) (298.15) (198.15) 5
. 384
We have values of a, b, c and hence the excess- properties at 323.15K
The excess Gibbs energy and the activity coefficient
• The excess Gibbs energy is of particular interest:
id
E
G G
G ≡ −
i i
i
T RT f
G = Γ ( ) + ln ˆ
i i i
id
i
T RT x f
G = Γ ( ) + ln
i i E i
i
x f
RT f G
ˆ
= ln
i i
i
i
x f
≡ fˆ γ
i E
i
RT
G = ln γ<