4
Thermodynamic equilibrium
4.1
Chemical equilibrium in an adiabatic system at constant volume
Consider a closed adiabatic vessel originally containing only CO and O2 in proportion 2:1. Assume a spontaneous adiabatic reaction in the gas at constant
volume. The contents of the vessel at two different times t0 and t are
t0 :
t:
CO
1
1−α
O2
1/2
1
2 (1 − α)
CO2
0
α
⇒ [CO + 12 O2 ]t0 −−→ [(1 − α)CO + 12 O2 + αCO2 ]t
α is the fraction of CO that has reacted at time t. The enthalpy of formation
is less for CO2 than for CO and consequently the temperature will increase as
the reaction proceeds (α increases). However the system will not continue to
α = 1, as temperature increases the CO2 starts to dissociate and equilibrium is
established.
What about the entropy of the system? The entropy change of an ideal gas
depends on two state variables. If p and T are the independent variables we
have
1
1
ds̄ = c¯p dT − R dp
(63)
T
p
The entropy of species i when cpi is assumed constant is
s̄i (T, p) = s̄◦i (Tref ) + c̄pi ln
T
pi
− R ln ◦
Tref
p
(64)
The entropy of the mixture is thus:
1
S(T, p) = (1 − α)s̄CO + (1 − α)s̄O2 + αs̄CO2
2
T, S
S
T
0
.2
.4
.6
.8
1.0 α
Figure 8: Temperature and entropy as a function of α
23
(65)
The entropy of the system S=f(α) is plotted in fig.(8) The second law of thermodynamics gives that for an adiabtic closed system the spontaneous process
increases the system entropy, ds > 0. Equilibrium is found at Smax . For this
system maximum S is found at approximately α=0.65
4.2
Thermodynamic equilibrium in a general system
Let us obtain the general equilibrium conditions for a system before we proceed.
Consider the non adiabatic system shown in figure (9). It is a closed system in
contact with a heat-bath, the surroundings. The temperature (constant) of the
heat-bath is T . The heat flow Q is defined positive into the system. The system
Surrounding
System
Q
T
T
Figure 9: System in thermal equilibrium with a heat bath (surrounding)
plus surrounding is the universe. For a spontaneous process the total entropy
needs to increase
∆Stot > 0
where
∆Stot = ∆Ssys + ∆Ssurr > 0
(66)
The entropy change of the heat-bath is related to the heat flow Q and the
temperature T of the bath
Q
∆Ssurr = − .
T
With eq.(66) we get
Q
∆Ssys −
> 0.
(67)
T
Now, with the enthalpy change of the system
∆H = Q + V ∆p
we eliminate Q in eq.(67) and get
∆H − V ∆p − T ∆Ssys < 0.
(68)
∆H − T ∆Ssys < 0.
(69)
If p and T are constant
Thus for all spontaneous processes in a general system at constant p and T the
left hand side is negative. At equilibrium we have
∆H − T ∆Ssys = 0.
24
Let us introduce the Gibbs function or Gibbs free energy which is defined
as
G ≡ H(T, p) − T S(T, p)
(70)
and in differential form
∆G = ∆H(T, p) − ∆T S(T, p) − T ∆S(T, p).
(71)
At constant p and T
∆G = ∆H(T, p) − T ∆S(T, p)
(72)
Comparing eq.(72) with eq.(69) we find that for a spontaneous process the
Gibbs function of the system decreases,
∆G < 0
and at equilibrium it has its lowest value,
∆G = 0
.
4.3
Chemical equilibrium in an ideal gas mixture at constant temperature and pressure, the equilibrium constant
From above we have that for a system at equilibrium a change in composition
given by the reaction formula will give ∆G = 0. Now, let us consider a change
in an ideal gas mixture (h = h(T )). From eq.(57) and eq.(48) we have
!
!
Ni (h̄◦f,i (Tref ) + ∆h̄◦i (T ))
H=
Ni h̄◦i (T ) =
S=
!
Ni s̄i (T, pi ) =
!
"
Ni s̄◦f,i (Tref )
+
#
T
Tref
The Gibbs free energy per mole is
$
1 ◦
c̄pi dT − R ln(pi /p◦ ) .
T
ḡi (T, pi ) = h̄◦i (T ) − T s̄i (T, pi )
and with the expression for absolute enthalpy eq.(58) we get
ḡi (T, pi ) =
h̄◦f,i (Tref )
%
+
∆h̄◦i (T )
−T
"
s̄◦f,i (Tref )
&'
+
#
ḡi◦ (T )
where we have defined ḡi◦ (T ) as shown above.
T
Tref
$
1 ◦
c̄pi dT +RT ln(pi /p◦ )
T
(
ḡi (T, pi ) = ḡi◦ (T ) + RT ln(pi /p◦ ).
Now we differentiate Gibbs free energy,
!
G=
Ni ḡi (T, pi ),
25
remembering T and p constant.
!
!
dG =
dNi ḡi (T, pi ) +
Ni dḡi (T, pi )
$
"
$ ! "
!
=
dNi ḡi◦ (T ) + RT ln(pi /p◦ ) +
Ni dḡi◦ (T ) + RT d(ln(pi /p◦ ) .
&'
(
%
I
%
&'
(
II
Now dḡi◦ (T ) = 0 and
d(ln( pp◦i ))
=
d(ln pp◦
d(ln(pi /p◦ ) =
I:
+ ln( !NNi i )) ⇒
1
1 !
dNi −
dNi .
Ni
N
Now term II :
+
" 1
"!
$*
$
!)
1 !
Ni !
Ni RT
dNi −
dNi = RT
dNi −
dNi = 0
Ni
N
N
This means that
dG =
!,
"
$dNi ḡi◦ (T ) + RT ln(pi /p◦ )
(73)
which should be equal to 0 at equilibrium. For a general system reacting in the
following way
νA A + νB B + · · · −−→ νE E + νF F + · · ·
A small change in the system dNi = dξ · νi with
νi = −νreac ;
νi = νprod
This means that eq.(73)
0 = dξ
and
But
giving the final result
!
$!, "
νi ḡi◦ (T ) + RT ln(pi /p◦ )
νi ḡi◦ (T ) = −RT
!
!
!
νi ln(pi /p◦ ) = ln
νi ln(pi /p◦ ).
.
νi ḡi◦ (T ) = −RT ln
(pi /p◦ )νi
.
(pi /p◦ )νi
where νi is positive for products and negative for reactants.
The sum
!
νi ḡi◦ (T ) = ∆G◦T
is called the standard state Gibbs function change and
.
(pi /p◦ )νi ≡ Kp (T )
is the equilibrium constant. Thus the relation between equilibrium constant and
change in the Gibbs function is
∆G◦T = −RT ln Kp (T ).
26
(74)
Now in the same way as for enthalpy a Gibbs function of formation is
defined
!
◦
ḡf,i
(T ) ≡ ḡi◦ (T ) −
νj ḡj◦ (T )
(75)
j
where νj are the stoichiometric coefficients of the elements required to form
one mole of species of interest. For an element j in its most common form
at reference state (p◦ ) the Gibbs function of formation is assigned 0 for all
temperatures.
◦
ḡf,j
(T ) ≡ 0
◦
Gibbs function of formation, ḡf,i
(T ) is tabulated in JANAF (∆Ff◦ ) for different temperatures T . Thus the standard state Gibbs function change for
a reaction becomes
!
!
◦
∆G◦T (T ) =
νi ḡi◦ (T ) =
νi ḡf,i
(T )
(76)
(The term
!
j
νj ḡj◦ (T ) %= 0
for T %= Tref on reactant and product side but will be summoned to 0 as we
have the same amount of elements in reactants and products)
27