Transcript Solution thermodynamics theory

Solution thermodynamics
theory—Part IV
Chapter 11
When we deal with mixtures of liquids or
solids
• We define the ideal solution model
• Compare it to the ideal gas mixture, analyze
its similarities and differences
i  i (T )  RT ln(yi P)
ig
Component i in a mixture
of ideal gases
This eqn. is obtained by combining
Gi  i (T )  RT ln P
ig
  Gi (T , P)  RT ln yi
ig
i
ig
Now we define
  Gi (T , P)  RT ln xi
id
i
Ideal solution model
Other thermodynamic properties
for the ideal solution: partial molar volume
Gi  Gi (T , P)  RT ln xi
id
 Gi
 
 P
id
Vi
id
V
id

 Gi
  
T , x  P
  xiVi   xiVi
id
i
i

  Vi
T
partial molar entropy in the ideal solution
Gi  Gi (T , P)  RT ln xi
id
 Gi
 
 T
id
S
id
i
S   xi S
id
i
id
i

 Gi
  
 P, x
 T

  R ln xi  Si  R ln xi
P
  xi Si  R  xi ln xi
i
i
partial molar enthalpy in the ideal solution
Gi  Gi (T , P)  RT ln xi
id
H iid  Gi id  TSiid  Gi  RT ln xi  TSi  RT ln xi  H i
H id   xi H iid   xi H i
i
i
Chemical potential ideal solution
i  i (T )  RT ln fˆi
Chemical potential component i
in a Real solution
Gi  i (T )  RT ln fi
Chemical potential
Pure component i
Subtracting:
For the ideal solution
fˆi
i  Gi  RT ln
fi
i
id
ˆf id
 Gi  RT ln i
fi
Lewis-Randall rule
i  Gi  RT ln xi
id
i
id
ˆf id
i
 Gi  RT ln
fi
ˆf id  x f
i
i i
id
ˆ
i  i
Lewis-Randall rule
(Dividing by Pxi each side of the equation)
When is the ideal solution valid?
• Mixtures of molecules of similar size and
similar chemical nature
• Mixtures of isomers
• Adjacent members of homologous series
Virial EOS applied to mixtures
BP
Z 1 
RT
B   yi y j Bij
i
j
How to obtain the cross coefficients
Bij
0
1
ˆ
Bij  B  ij B
Bˆ ij 
Bij Pcij
Tcij
Mixing rules for Pcij, Tcij, ij, 11-70 to 11.73
ˆ
Fugacity coefficient  i
from virial EOS
P
2
ˆ
B11  y2 12 
ln1 
RT
12  2B12  B11  B22
For a multicomponent mixture, see eqn. 11.64
problem
• For the system methane (1)/ethane (2)/propane (3)
as a gas, estimate
fˆ , fˆ , fˆ and ˆ ,ˆ ,ˆ
1
2
3
1
2
3
at T = 100oC, P = 35 bar, y1 =0.21, and y2 =0.43
Assume that the mixture is an ideal solution
Obtain reduced pressures, reduced temperatures, and calculate
k
id
 Prk 0
1 
 exp ( Bkk  Bkk )
 Trk

id
fˆkid  k yk P
Results: methane (1) ethane (2) propane (3)
ˆ1  1.02;ˆ2  0.88;ˆ3  0.78
Virial model
fˆ1  7.49; fˆ2  13.25; fˆ3  9.76bar
1  0.98;2  0.88;3  0.76
Ideal solution
ˆf id  7.18; fˆ id  13.25; fˆ id  9.57bar
1
2
3
Now we want to define a new type of
residual properties
• Instead of using the ideal gas as the reference,
we use the ideal solution
Excess properties
M  M M
E
id
The most important excess function is
the excess Gibbs free energy GE
Excess entropy can be calculated
from the derivative of GE wrt T
Excess volume can be calculated
from the derivative of GE wrt P
And we also define partial molar excess properties
ˆ
Gi  i (T )  RT ln f i
Gi  i (T )  RT ln xi f i
id
subtracting :
ˆf
i
i 
xi f i
Definition of activity coefficient
Summary
Gi  RT ln  i
R
ˆ
G  RT ln 
E
i
i
Summary
  G  RT ln yi
ig
i
ig
i
  Gi  RT ln xi
id
i
i  Gi  RT ln  i xi
Note that:
  G  RT ln yi  i (T )  RT ln yi P
ig
i
ig
i
  Gi  RT ln xi  i (T )  RT ln xi f i 
id
i
id
ˆ
 i (T )  RT ln f i
i  Gi  RT ln  i xi  i (T )  RT ln  i xi f i 
 i (T )  RT ln fˆi
problem
• The excess Gibbs energy of a binary liquid mixture
at T and P is given by
G / RT  (2.6x1 1.8x2 ) x1 x2
E
a) Find expressions for ln 1 and ln 2 at T and P
Using x2 =1 – x1
GE/RT= x12 -1.8 x1 +0.8 x13
Since i comes from
Gi  RT ln  i
E
We can use eqns 11.15 and 11.16
E
dG
G  G  x2
 RT ln  1
dx1
E
1
E
E
dG
G  G  x1
 RT ln  2
dx1
E
2
E
then
E
dG / RT
2
 1.8  2 x1  2.4 x1
dx1
And we obtain
ln  1  1.8  2 x1  1.4 x  1.6 x
2
1
ln  2   x  1.6 x
2
1
3
1
3
1
If we apply the additivity rule and the
Gibbs-Duhem equation
G
E
RT
  xi ln  i
i
 x d ln 
i
i
At T and P
0
i
(b and c) Show that the ln i expressions satisfy
these equations
Note: to apply GibbsDuhem, divide the equation
by dx1 first
Plot the functions and show their
values
0
0
0.2
0.4
0.6
-0.5
-1
-2.5
-3
1
GE/RT
ln 1
-1.5
-2
0.8
ln 2
GE/RT
ln g2
ln g1
x1