Environmental Engineering Reference
In-Depth Information
E
XAMPLE
2.4 F
REE
E
NERGY AND
T
EMPERATURE
Problem
: The free energy change for a reaction is given by
Δ
G
0
=−
RT
ln
K
eq
,
(2.43)
where
K
eq
is the equilibrium constant for a reaction (Chapter 5). If
K
eq
for the reac-
tion H
2
CO
3
(
aq
)
→
H
+
(
aq
)
+
HCO
3
(aq) is 5
×
10
−
7
mol/L at 298 K, what is the
equilibrium constant at 310 K?
Δ
H
0
is 7.6 kJ/mol.
Solution
: Using the Gibbs-Helmholtz relationship, we have
=
Δ
H
0
∂
ln
K
eq
∂T
RT
2
,
(2.44)
and hence
ln
K
eq
(T
2
)
K
eq
(T
1
)
1
T
2
−
.
=−
Δ
H
0
1
T
1
R
·
(2.45)
Substituting, we obtain
K
eq
(at 310 K
)
=
5.63
×
10
−
7
mol/L.
Other properties such as vapor pressure, aqueous solubility, air/water partition con-
stant, and soil/water partition constant show similar changes depending on the sign and
magnitude of the corresponding standard enthalpy of the process. These are discussed
in Chapter 4.
2.4 CONCEPT OF MAXIMUM WORK
We have the following equation for the total free energy change for any process:
Δ
=
−
w
tot
+
Δ
−
Δ
=−
w
tot
+
Δ
G
q
P
V
T
S
P
V
(2.46)
since
T
Δ
S
=
q
. Thus for any
permissible
process at constant
T
and
P
it follows that
−Δ
G
=
w
tot
−
P
Δ
V
=
w
useful
.
(2.47)
Since
P
V
is the expansion work done or otherwise called “wasted work,” the dif-
ference between the total work and
P
Δ
V
is the “useful work” that can be done by
the system.
At constant temperature and pressure, the useful work that can be done
by a system on the surroundings is equivalent to the negative change in Gibbs free
energy of a spontaneous process within the system
. If the system is at equilibrium,
then
Δ
Δ
G
=
0 and hence no work is possible by the system.
E
XAMPLE
2.5 C
ALCULATION OF
M
AXIMUM
U
SEFUL
W
ORK
Problem
: Humans derive energy by processing organic polysaccharides. For example,
glucose can be oxidized in the human body according to the reaction C
6
H
12
O
6
(
s
)
+
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