50
PART I
THERMODYNAMICS AND KINETICS
the system are exactly matched by corresponding opposite changes in the entropy of the surroundings.
For a process that occurs at constant pressure, the heat
flow is equal to the change in enthalpy (eqn 3.6), and
the entropy change of the surroundings can be written
in terms of enthalpy:
(3.9)
Thus, if the process is exothermic, ΔH is negati9e and ΔS sur
is positi9e, and the process is spontaneous. The entropy of
the surroundings increases when heat is released. This conclusion is
critical for understanding chemical equilibrium and will be used in the
formulation of the Gibbs energy, below.
If the change in the system is irreversible, the entropy change is not zero.
To determine the entropy change for an irreversible process, consider as an
example the isothermal expansion of an ideal gas against a fixed pressure,
P f (Figure 3.3). The value of fixed pressure is poised to be equal to the
final pressure of the gas after the expansion is completed. In this case,
the work performed is given by the product of the change in 9olume, ΔV,
and the pressure, which is expressed using the ideal gas law:
(3.10)
Ideal gas law:
The work performed in this irreversible process is different from what
was determined for a reversible process (eqn 3.3):
(3.11)
For example, if the final volume is twice the initial volume, V f = 2V i , then
the work is calculated to be −0.5 nRT and −0.693 nRT for the irreversible
and reversible, respectively. Since these are isothermal ideal gases, q = −w
and the entropy change can be related to the work:
w
nRT
V
V
f
i
ln
= −
P
nRT
V
=
w
P V
nRT
V
V
nRT
V
V
f
f
f
= −
= −
= −
Δ
Δ
Δ
Δ
Δ
S
H
T
sur = −
Δ x
Pex ϭ Pf
Irreversible
expansion
Vi
Vf
Figure 3.3
The irreversible
expansion for an
ideal gas with the
external pressure,
P ex , fixed to the
pressure, P f .
9781405124362_4_003.qxd 4/30/08 19:05 Page 50
PART I
THERMODYNAMICS AND KINETICS
the system are exactly matched by corresponding opposite changes in the entropy of the surroundings.
For a process that occurs at constant pressure, the heat
flow is equal to the change in enthalpy (eqn 3.6), and
the entropy change of the surroundings can be written
in terms of enthalpy:
(3.9)
Thus, if the process is exothermic, ΔH is negati9e and ΔS sur
is positi9e, and the process is spontaneous. The entropy of
the surroundings increases when heat is released. This conclusion is
critical for understanding chemical equilibrium and will be used in the
formulation of the Gibbs energy, below.
If the change in the system is irreversible, the entropy change is not zero.
To determine the entropy change for an irreversible process, consider as an
example the isothermal expansion of an ideal gas against a fixed pressure,
P f (Figure 3.3). The value of fixed pressure is poised to be equal to the
final pressure of the gas after the expansion is completed. In this case,
the work performed is given by the product of the change in 9olume, ΔV,
and the pressure, which is expressed using the ideal gas law:
(3.10)
Ideal gas law:
The work performed in this irreversible process is different from what
was determined for a reversible process (eqn 3.3):
(3.11)
For example, if the final volume is twice the initial volume, V f = 2V i , then
the work is calculated to be −0.5 nRT and −0.693 nRT for the irreversible
and reversible, respectively. Since these are isothermal ideal gases, q = −w
and the entropy change can be related to the work:
w
nRT
V
V
f
i
ln
= −
P
nRT
V
=
w
P V
nRT
V
V
nRT
V
V
f
f
f
= −
= −
= −
Δ
Δ
Δ
Δ
Δ
S
H
T
sur = −
Δ x
Pex ϭ Pf
Irreversible
expansion
Vi
Vf
Figure 3.3
The irreversible
expansion for an
ideal gas with the
external pressure,
P ex , fixed to the
pressure, P f .
9781405124362_4_003.qxd 4/30/08 19:05 Page 50