Environmental Engineering Reference
In-Depth Information
which, for only mechanical work and for an isothermal process concerning an ideal
gas, becomes (V related to 1 mol of gas)
dG =Vd
p
=
R
u
T
p
d
p
ð
Eq
:
5
:
22
Þ
In order to estimate the finite value of the free energy variation
G that a mole of an
ideal gas needs to isothermally pass from an initial pressure
p
0
to a final pressure
p
,we
integrate Equation (5.22)
Δ
ð
G=
ð
2
p
R
u
T
d
p
p
=R
u
Tln
p
dG =G
2
−
G
1
=
Δ
ð
Eq
:
5
:
23
Þ
p
0
p
0
1
Considering a standard initial state as reference,
p
0
being 1 bar, Equation (5.23) can be
written as
G=G
0
+R
u
Tln
p
:
:
ð
Eq
5
24
Þ
where G is the free energy of 1 mol of ideal gas at pressure
p
(in bar!) and
temperature T, while G
0
is its free energy at the standard temperature and pressure
(298.15 K and 1 bar).
Considering the fact that the free energy is a function of state and given Equation
(5.24)), the change of the free energy for a general reaction (RX. 5.20) is expressed by
the Van
'
t Hoff equation:
G
0
+R
u
Tln
p
C
p
D
p
a
A
p
B
Δ
G=
Δ
ð
Eq
:
5
:
25
Þ
The argument of the logarithm in Equation (5.25) is formally analogous to the equi-
librium constant (
K
p
) but has the same value only when the reaction has reached
equilibrium, i.e.,
G
0
+R
u
Tln
K
p
=0
G
0
=
Δ
G=
Δ
)
Δ
−
R
u
Tln
K
p
ð
Eq
:
5
:
26
Þ
Example 5.3 Calculation of equilibrium constants,
K
c
and
K
p
In a container with a capacity of 5 L, 1 mol of SO
2
and 1 mol of O
2
react at 1000 K.
Once equilibrium has been reached, the amount of SO
3
in the container is 68 g
(0.85 mol). Find
K
c
and
K
p
.R
u
= 0.082 L.bar.K
−1
.mol
−1
Solution
The reaction observed is
2SO
2
+O
2
!
2SO
3
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