Handbook of Solvents - George Wypych - ChemTech - Ventech!

9.5 The mixed solvent effect on the chemical equilibrium thermodynamics 559
Division of equation [9.126] into the constituents must be done with full assurance to
maintain influence of different solvents. For this reason, it is necessary to have a few isotherms
of equilibrium constant dependencies on permittivity (K=f(ε)T). We have developed
an equation of a process depicted by scheme [9.45] after approximating each isotherm as
lnK vs. 1/ε function and by following approximation of these equations for ε-1, ε-2, ε-3...ε-j
conditions. We then can calculate the integral value of process entropy by differentiating relationship
[9.55] versus T, e.g., ΔG= -RTlnK:
i= m
i= m
⎡j=
n
j= n
i i j
ΔSi = R⎢ ( 1−i)
aji( T ε ) − j( dln ε/ dT) / ε ∑a
ji / T
⎢∑
i = 0
i = 0
⎣⎢j=
0
j = 0
−1 i −1
⎤
⎥
⎥
⎦⎥
i
/ ε [9.127]
Here, the first term of the sum in brackets corresponds to ΔS T/R value and the second
term to ΔS/R. We can determine “van’t Hoff’s” (original) constituents of the process entropy
if all the terms containing dlnε/dT are equal to zero:
i= m
j= n
∑ 1
i j
( ) ( )
ΔS = R −i
a / T ε [9.128]
T ji
i=
0
j=
0
Van’t Hoff’s constituent of enthalpy is determined in analogous manner as ΔH=ΔG+TΔS
(ΔH T):
i= m
j= n
∑ /
ΔH =−R
a T
T ji
i=
0
j=
0
i−1j ( )
ε [9.129]
Such approach may by illustrated by dependence of formic acid association constant
on temperature and permittivity in mixed solvents: water-ethylene glycol, 72 which is approximated
from equation
where:
Hence
( )
2
ln K = a + a / T + a / T + a / ε+ a / εT
a00 a01 a02 a10 a11 00 01 02
= 40.90
= -2.17×10 4
= 3.11×10 6
= -425.4
= 2.52×10 5
10 11
[ 01 2 02 / 11 / ε ( ln ε/ )( 10 11)
/ ε]
ΔH =− R a + a T + a + T d dT a T + a
i
[ 00 02 / 10 / ε ( ln ε/ )( 10 11)
/ ε]
ΔS = R a − a T + a − d dT a T + a
i
and consequently,