Chemistry Reference
In-Depth Information
last equation can be used to calculate the solvation Gibbs energy, which in our case
is (Ben-Naim 2006)
*
∆
µµ
=
(,,
Tp
ρ
)
−
kT
B
ln
ρ
Λ
(2.47)
AA
A
A
A
However, for calculating the excess Gibbs energy, we do not need to calculate the pres-
sure derivative in Equation 2.45. In our case, the mole fraction
x
A
is calculated from
ρ
ρρ
A
x
A
=
(2.48)
+
AB
having calculated ρ
A
(
T
,
p
, λ
A
) and ρ
B
(
T
,
p
, λ
A
), we also get
x
A
(
T
,
p
, λ
A
), which in our
case is
2
ΛΨ
ΛΨΨΛ ΨΨ
λ
AA AB
x
A
=
(2.49)
2
2
λ
(
2
−
λ
)
+
(
−
λ
)
AAAAA
B
A
AAABB
The last equation can be solved to obtain the absolute activity λ
A
as a function of
T
,
p
,
x
A
. The result is
2
Λ
(
12
−
x
)
Ψ
+
2
x
ΨΨ
±
Ψ
A
A
AB
A
A
B
AB
λ
A
=
(2.50)
2
2
x
ΨΨ
(
ΨΨ
−
)
AAAA
AA
B
AB
where
22
=−
(
1
x
)
Ψ
+
4
x
(
1
−
x
)
Ψ
Ψ
(2.51)
A
B
A
A
A
B
To determine the correct solution, we take the two limits; at
x
A
→ 0 and
x
A
= 1, we
find that the “+” solution gives the correct limiting quantities,
ΛΨ
Ψ
x
A
BA
2
()
λ
A
=
+
Ox
(2.52)
A
2
AB
and for
x
A
= 1, we get the absolute activity of the pure A,
Λ
Ψ
λ
o
A
=
(2.53)
AA
Thus, from the “+” solution in Equation 2.50, we can get the chemical potential of A.
Similarly, we get the chemical potential of B and from these we get the excess Gibbs
energy per particle as a function of (
T
,
p
,
x
A
),
E
o
o
Gx
=
(
µµ
−
−
kT
ln)(
x
+
1
−
x
)
µµ
−
−
kT
ln(
1 −
x
A
)
(2.54)
m
AA
A
B
A
A
BB
B
