CHAPTER V
FREE ENERGY AND CHEMICAL
EQUILIBRIUM
Chapter Outline
(V-1)
(V-2)
(V-3)
(V-4)
(V-5)
Page
Introduction
Gibbs and Helmholtz Free Energies
Free Energy and Maximum Work
Standard Free Energies of Formation
The Properties of Gibbs Free Energy (G)
90
91
92
93
94
95
98
(V-5-1) Pressure dependence of the free energy
(V-5-2) Temperature dependence of the free energy
(Gibbs-Helmholtz equation)
(V-6)
(V-7)
Equilibrium Constant and Gibbs Free Energy
The Temperature Dependence of the
Equilibrium Constant
(V-1)
Introduction
100
101
In CHAPTER IV it was shown that for any process, the entropy change in a
system plus the entropy change in the thermal surroundings tell us whether that
process can proceed spontaneously. Now we introduce a function of the system
that by itself indicates the spontaneous direction of a reaction. This function,
the free energy , lends itself to the treatment of the equilibrium state toward
which the process moves. The interrelation of thermodynamic properties and
the equilibrium state of chemical reaction is, for chemistry, the most important
accomplishment of thermodynamics.
(V-2)
Gibbs and Helmholtz Free Energies
System is in thermal equilibrium with surroundings at a temperature T. There is
a change in the system accompanied by a transfer of energy dq between system
and surroundings. From the Classius Inequality we have:
90
dq
, i.e. Tds  dq or
T
dq  Tds  0
ds 
This inequality may be developed in two ways:
(A) Under the conditions of constant volume we have no p-V work, i.e. in
the absence of other forms of work:
dE Tds  0
(V-1)
(B) Under the conditions of constant pressure, in the absence of other
forms of work other than p-V, we have :
dH  Tds  0
(V-2)
The two inequalities above (i.e. dE Tds  0 and
) and are so
important (they determine whether a process is going to occur or not) that we
shall assign to them two new thermodynamic functions namely:
(a)
The Helmholtz free energy, A, given by:
dA  dE  Tds
(V-3)
(b) The Gibbs free energy, G, given by:
dG  dH  Tds
Thus the two inequalities above (i.e. dE Tds  0
(V-4)
and
)
governing the conditions for a spontaneous process may be summarized as:
d  AT ,V  0
(from dE Tds  0 )
d G T , P  0
(from dH  Tds  0 )
Note that d  AT ,V  0 and d G T , P  0 refer to conditions at equilibrium.
(V-3)
Free Energy and Maximum Work
Now let us see the significant of G for balanced processes in which the
system might do work over and above that of expansion against its confining
pressure. The defining equation, H  E  PV , is substituted in G  H  TS
to give
G  E  PV  TS
(V-5)
91
For an infinitesimal change in G one now has
dG  dE  PdV  VdP  TdS  SdT
(V-6)
For constant-temperature, constant-pressure processes, Eq. (V-6) reduces to
dG  dE  PdV  TdS
(V-7)
The first law expression dE  dq  dW can now be inserted to give
dG  dq  dW  PdV  TdS
(V-8)
Here let us think of dW , the work done by the system. Some of this
work of expansion, and this contribution is  PdV . The remaining component,
which we assume can be harnessed to do useful work, is dw . Thus the total
work that can be done by the system can be described by
dW   PdV  dw
(V-9)
Insertion of Eq. (V-8) in (V-7)
dG  dq  PdV  dw  PdV  TdS , then
dG  dq  TdS  dw
(V-10)
Multiplying Eq. (V-9) by (-1), gives
 dG  dq  TdS  dw
(V-11)
For a balanced, reversible process
dqrev
 dS , so at constant T and P and for reversible process,
T
 dG T , P   dw

  w
  W  PV
 G
 GT ,
P
(V-12)
T, P
Equation (V-11) indicates
 The decrease in Gibbs free energy for a process occurring in a balanced,
reversible way is equal to the work (over and above any P-V work) that
can be obtained.
 In a reversible process, there is no wasted derive, and the workobtained in
the maximum work.
 The decrease in Gibbs free energy for a reaction is the maximum work
that could be obtained if the reaction were carried out at constant
temperature and pressure.
92
 The negative value of G indicates that the reaction proceeds
spontaneously with significant deriving force.
On the other hand the decrease in the Helmholtz free energy A is given as
 dAT ,V  dW
(V-13)
 AT ,V  W
So the difference between the two Equations (V-12) and (V-13) can be
given as follows
 G T , P    AT ,V    W  PV    W 
 G T , P    AT ,V    PV
(V-14)
Accordingly the Helmholtz free energy is called also the Work Function.
(V-4) Standard Free Energies of Formation
The difference in the free energies of the reactants and the products of a
chemical reaction gives the free energy change of this reaction
G o 
 nG of 
products

mG of
(V-15)
reac tan ts
The subscript (o) indicates that All free energies were measured under the
standard conditions (1 atm and 25 oC). The standard free energy change of a
reaction indicates the direction in which the reaction can proceed. It is therefore
helpful to have a tabulation of the free energies of chemical compounds. Then
the free energy change of any reaction we want to consider can easily be
calculated.
By using the fundamental equation G  H  TS the following
equation is written
G of  H of  TS of
93
(V-16)
Example (V-1)
Given the following information, calculate the standard Gibbs free energy of
formation of H2O (g) at 398.15 K.
Substance
H 2O( g )
H 2( g )
O2( g )
H of (kJmol 1 )
-241.818
S of (JK -1mol 1 )
188.825
0
130.684
0
205.138
Solution
The standard Gibbs energy of formation is the standard reaction Gibbs energy for
the following reaction.
H 2( g )  1/ 2O2( g )  H 2O( g )
G of ( H 2 O( g ) )  H of ( H 2 O( g ) )  TS of ( H 2 O( g ) )
G of ( H 2 O( g ) )   241 .818   298 .15   0.188825  0.130684  0.5  0.205138 
G of ( H 2 O( g ) )  228 .572 kJmol 1
(V-5) The Properties of Gibbs Free Energy (G)
By using the fundamental Equations H  E  PV and G  H  TS , the
following equation is written G  E  PV  TS and for an infinitesimal
change in G one now has
dG  dE  PdV  VdP  TdS  SdT .
Since the combination of the first and the second laws of thermodynamics
gives dE  TdS  PdV  0 , the following equation can be written
dG  SdT  VdP
(V-17)
If, G  f (T , P) then,
 G 
 G 
dG  
 dT  
 dP
 T  P
 P T
Comparing between Eq. (V-14) and Eq. (V-15) shows that
94
(V-18)
 G 

  S
 T  P
(V-19)
 G 

 V
 P T
(V-20)
Because of the importance of the free energy, Eqs. (V-19) and (V-20)
contains two of the most important pecies of information in thermodynamics.
(ii) Since the entropy of any substances is positive, the minus sign in Eq.
(V-16) shows that increase in temperature decreases the free energy if
the pressure is constant. The rate of decrease is greater for gases
which have large entropies than for liquids or solids which have small
entropies.
(iii) The fact that V is always positive means through Eq. (V-20) that
increase in pressure increases the free energy at constant temperature.
The larger the volume of the system the greater is the increase in free
energy for a given increase in pressure. The comparatively large
volume of a gas implies that the free energy of a gas increases much
more rapidly with pressure than would that of liquid or solids.
(V-5-1) Pressure dependence of the free energy
By variables separation and integration of Eq. (V-20) at constant temperature,
P2
GP2  GP1   V dP
(A)
(V-21)
P1
For solids and liquid, at moderate pressures the volume is
approximately constant independent of pressure, and can be taken out
of the integral:
GP2  GP1  V P2  P1 
95
(V-22)
(B)
For a gas, we use the equation of state to write V as a function of P.
For an ideal gas, and approximately for any gas, we can substitute the
ideal gas equation ( V 
nRT
)in Eq. (V-21):
P
P2
dP
P1 P
GP2  GP1  nRT 
GP2  GP1  nRT ln
(C)
(V-23)
P2
P1
If the free energy of any pure materials is expressed by integrating
Eq. (V-21) from the standard pressure, 1 atm, to any other pressure
P, then
GP  G o  V P  1
(liquids and solids)
(V-24)
GP  G o  nRT ln P
(gases)
(V-25)
Importance
The quantity
GP2
n

G P1
n
G  RT ln
G
is called the molar Gibbs free energy (chemical potential), so
n
 RT ln
P2
P1
P2
P
o
G  G  RT ln P
96
Example (V-2)
An ideal gas is allowed to expand reversibly and isothermally (25 oC) from a pressure
of 1 atm to a pressure of 0.1 atm.
(a) What is the change in the molar Gibbs free energy?
(b) What would be the change in the molar Gibbs free energy if the process
occurred irreversibly?
Solution
(a) G  RT ln P2 , then
P1
0.1
G  8.314 Jmol 1K 1 298 K  ln
1
1
G  5.705 kJmol
(b) G  5.705 kJmol 1


Example (V-3)
Assuming that O2 is an ideal gas, what will be the molar Gibbs free energy of
formation at100 atm?
Solution
o
G  G  RT ln P
G  0  (8.314 Jmol 1K 1 )  298 .15K   ln 100
G  11.42 kJmol 1
(D)
To calculate the effect of pressure on the free energy of a chemical
reaction, we just apply Eqs. (V-22) and (V-23) to each product and
reactant. If all products and reactants are solids or liquid, the
following equation is written
GP2  GP1  V P2  P1 
V 
where V 
products
(V-26)
V
reac tan ts
If at least one gaseous product or reactant is involved, the volume of
solids or liquids is ignored as compared to that of the gases, the following
equation is written
GP2  GP1  n( g ) RT ln
where n( g ) 
 ng 
products

ng
reac tan ts
97
P2
P1
(V-27)
(V-5-2) Temperature dependence of the free energy (GibbsHelmholtz equation)
The dependence of the free energy on temperature is expressed in several
different ways for convenience of different problems. Rewriting Eq. (V-19)
 G 

  S
 T  P
(V-28)
From the definition G  H  TS , we obtain  S 
GH
and Eq. (V-28)
T
becomes
GH
 G 

 
T
 T  P
(V-29)
a form which is sometimes useful.
It is important to know how the function
G
depends on temperature. By
T
the ordinary rule of differentiation, we obtain
1  G 
G
  G / T  

  
  2
 T  P T  T  P T
(V-30)
Using Eq. (V-28), Eq. (V-30) becomes
S G
  G / T  

   2
T T
 T  P
  G / T  
 TS  G 

  

 T  P
 T2 
Introducing equation H  TS  G to Eq. (V-31) gives
H
  G / T  

  2
T
 T  P
(V-31)
(V-32)
This equation is the Gibbs-Helmholtz equation, which we used frequently.
Multiplying both sides of Eq. (V-32) by T 2 and recalling that
1


d
  , we obtain
T2
T 


  G / T  

 H
 1 



T P
dT
(P=constant)
(V-33)
For a Chemical reaction, we just apply Eq. (V-33) to each products and
reactants.
98
G T2
T2
G T2
T2


GT1
T1
GT1
T1
T2
1
  H d  
T 
T
1
(V-34)
1
1
 H   
T2 T 1 
Any of Eqs. (V-29), (V-31), (V-32) and (V-34) are simply different versions of
the fundamental equation, Eq. (V-28). We will refer to them as the first, second,
third and fourth form of the Gibbs-Helmholtz equation.
Example (V-3)
The standard enthalpy of the reaction N 2  3H 2  2 NH 3 is -92.2 kJ mol-1 and the
standard Gibbs function is -31.0 kJ mol-1, both at 298 K. Estimate the Gibbs function at
(a) 500 K and (b) 1000K. Is the reaction spontaneous at room temperature? Is the
formation of ammonia favoured or disfavoured by a rise in temperature?
Solution
o
G T2 G 298
1
1

 H   
T2
298
 T2 T 1 
G500   31.0 kJmol -1 
1 
 1
   92.2 kJmol -1 
 


500 
298.15
 500 298 

G500  10.4 kJmol -1
G1000   31.0 kJmol -1 
1 
 1
   92.2 kJmol -1 
 


1000 
298.15
1000 298 

G1000  113.2 kJmol -1
Fig. (V-1)
Comments
140
120 G=S
With increasing temperature the free energy
-1
100 kJ mol
Increases, indicating that the process is not
-1
80 S= -205.5 kJ mol
favoured with increasing temperature.



G (kJ mol-1)

Explain the results in Fig. (V-1).
60
40
20
0
-20
-40
0
200
400
600
800 1000 1200 1400
T (K)
Col 2
Col 1 vs Col 2
Plot 2 Regr
99
(V-6) Equilibrium Constant and Gibbs Free Energy
We shall now derive an important relation between G o , and K P , the
equilibrium constant in terms of partial pressure. For the general reaction
aA  bB  cC  dD ,
G  cGC  dG D  aG A  bG B
(V-35)
By inserting Eq. (V-25) for each component, Eq. (V-32) becomes
G  cGCo  cRT ln PC  dGDo  dRT ln P  aGAo  aRT ln PA  bGBo  bRT ln PB
On rearrangements, we have
G  G o  RT ln
PCc PDd
(V-36)
PAa PBb
G o  cGCo  dG Do  aG Ao  bG Bo
where
Since the equilibrium constant ( K P ) of the upper general reaction using the
partial pressure can be given
KP 
PCc PDd
(V-37)
PAa PBb
Introducing the result in Eq. (V-37) to Eq. (V-36)
G  G o  RT ln K P
(V-38)
At equilibrium G  0 , then
G o  RT ln K P  0
(V-39)
G   RT ln K P
o
 The derivation of Eq. (V-39) is a thermodynamic proof the existence of an
equilibrium constant.
 Since G o is a function of temperature alone, it can not depend on the partial
pressure in Eq. (V-37).
 As known K P at certain temperature is constant.
 Eq. (V-39) not only proves the existence of an equilibrium constant, K P , it also
gives us an explicit formula for calculating K P from G data.
o
100
Example (V-4)
Calculate the equilibrium constant at 25 oC for the reaction
Pyruvic acid (l )  Acetaldhyde( g )  CO2( g )
By using the following information
G of (kJmol 1 )
-463.38
Substance
Pyruvic acid (l )
-133.30
Acetaldhyde(g )
CO2( g )
-394.38
Solution
G o 
 nG of 
products

mG of
reac tan ts
G   133 .30    394 .38   463 .38
o
G o  64.30 kJmol 1
G o   RT ln K P


 64.30   8.314  10  3 kJK 1mol1  298.15K   ln K P
K P  1.8  1011
(V-7) The Temperature Dependence of the Equilibrium
Constant
Since the Equilibrium constant can be given as
G o
ln K P  
RT
(V-40)
Differentiation of Eq. (V-40) with respect of temperature at constant pressure
gives

1  d G o / T
 d ln K P 

  
R
dT
 dT  P


P
(V-41)
The following results was obtained from section (V-5-2),

  G o / T


T


H o
  T2
P
101
(V-42)
Introducing Eq. (V-42) in Eq. (V-41) gives
H
 d ln K P 

 
 dT  P RT 2
This is the Van ,t Hoff equation .
o
(V-43)
Multiplying both sides of Eq. (V-43) by T 2 and recalling that
1


d
 ,
T2
T 
dT
we obtain
d ln K P P
H o 1

d
R
T
(V-44)
The integrated form of this equation, on the assumption that H o is
temperature-independent, is
ln K P(T )  ln K P(T ) 
(V-45)
ln K P(T )  ln K P(T ) 
(V-46)
2
Or
2
1
1
H o

 constant
RT
H o  1
1

 

R  T2 T1 
When much larger temperature ranges are considered, the basis of the
dependence of the equilibrium constant on temperature can more clearly be
seen by returning to the expressions
G o  H o  TS o
G o   RT ln K P
and
Or
S o H o
ln K P 

R
RT
(V-47)
From this form of the van't Hoff equation, we
the y-axis and 1/ T on the x-axis has an intercept
S o
H o
of
and slope given by 
. This is the
R
R
origin of LeChatelier's Principle for the heat
absorbed or evolved during the course of a
chemical reaction. For an endothermic reaction,
102
lnKP
see that at constant pressure, a plot with K P on
slope = -H/R
1/T (T, K)
Fig. (V-2) lnKP-1/T relation for
endothermic reaction
the slope is negative and so as the
increases,
the
equilibrium
constant increases, as illustrated in Fig. (V-2).
For an exothermic reaction, the slope is
lnKP
temperature
slope = -H/R
positive and so as temperature increases, the
equilibrium constant decreases, as illustrated
in Fig. (V-3).
103
1/T (T, K)
Fig. (V-3) lnKP-1/T relation for
exothermic reaction