Standard Gibbs Energy of Formation
The standard Gibbs energy of formation is the standard reaction Gibbs
energy of formation of a compound from its elements in their reference
states.
Just like enthalpy, the standard Gibbs energy of a reaction can be found
using the standard
for products and reactants.
 G θ Gibbs

νenergy
 G θ  of formation
ν G θ
r

products
f

m
f
m
reactants
  r H θ  T r S θ
What is rG298 for the combustion of methane?
CH4 (g) +2O2 (g)  2H2O (l) + CO2 (g)
 r G θ   f G θ CO2 ( g )   2 f G θ H 2O (l ) 
  f G θ CH4 ( g )   2 f G θ O 2 ( g ) 
  394  2237    50.7  20  kJ mol-1
 817 kJ mol-1
© Dario Bressanini
The Gibbs Energy Change (cont’d)

For the methane combustion reaction
1 CH4(g) + 2 O2(g)  1 CO2(g) + 2 H2O(l)
rG =  np fG (products) -  nr fG (reactants)
= 2 fG [H2O(l)] + 1 fG [CO2(g)] - (7/2 fG [O2(g)]
+ 1 fG [CH4(g)] )
© Dario Bressanini
Free Energy Change, G,
and Spontaneity

Example 15-16: Calculate Go298 for the reaction in
Example 15-8. Use appendix K.
C 3 H 8 g  + 5 O 2 g   3 CO 2 g  + 4 H 2 O  l 
you do it
© Dario Bressanini
The Temperature
Dependence of Spontaneity

Example 15-17: Calculate So298 for the following reaction.
In example 15-8, we found that Ho298= -2219.9 kJ, and in
Example 15-16 we found that Go298= -2108.5 kJ.
C3H8g +5 O2g 3 CO2g +4 H2Ol
© Dario Bressanini
The Temperature
Dependence of Spontaneity

Example 15-17: Calculate So298 for the following reaction.
In example 15-8, we found that Ho298= -2219.9 kJ, and in
Example 15-16 we found that Go298= -2108.5 kJ.
C3H8g +5 O2g 3 CO2g +4 H2Ol
G  H TS
o
o
o
TS  H  G
o
© Dario Bressanini
o
o
The Temperature
Dependence of Spontaneity
C3H8g + 5 O2g  3 CO2g + 4 H2Ol 
G  H  TS
o
o
o
TS  H  G
o
o
H  G
S 
T
 22199
.  (21085
. ) kJ


298 K
 0374
. kJ K or - 374 J K
o
o
© Dario Bressanini
o
o
The Temperature
Dependence of Spontaneity


So298 = -374 J/K which indicates that the disorder of the
system decreases .
For the reverse reaction,
3 CO2(g) + 4 H2O(g) C3H8(g) + 5 O2(g)

So298 = +374 J/K which indicates that the disorder of the
system increases .
© Dario Bressanini
The Temperature
Dependence of Spontaneity

Example 15-18: Use thermodynamic data to estimate the
normal boiling point of water.
H 2 O l   H 2 O g 
equilibrium at BP  G = 0
G = H - TS or H = TS
H
T=
S
© Dario Bressanini
The Temperature
Dependence of Spontaneity
assume H@ BP 
H 
o
o
H H 2O(g)

o
H 298
o
H H 2O( l )
H   2418
.  ( 2858
. )
o
H  44.0 kJ@25 C
o
© Dario Bressanini
o
J
K
The Temperature
Dependence of Spontaneity
assume S@ BP 
S 
o
o
SH 2O(g)
o
S298
o
 S H 2O( l )
S  188.7  69.91
o
S
o
J
 118.8
© Dario Bressanini
K
or
J
K
kJ
- 0.1188
K
The Temperature
Dependence of Spontaneity
H H
44.0 kJ
T=  o

370
K
kJ
S S 0.1188 K
o
o
370 K-273 K=97 C
© Dario Bressanini
Calculating
Go from Enthalpy and Entropy Values–I
Problem: Potassium chlorate, one of the common oxidizing agents in
explosives, fireworks, and matchheads, undergoes a solid-state redox
reaction when heated, in which the oxidation number of Cl in the reactant
is higher in one of the products and lower in the other
(disproportionation):
4 KClO
3 KClO + KCl
3 (s)
4 (s)
(s)
Use Hof and So values to calculate Gosys ( Gorxn) at 25oC for this
reaction.
Plan: To solve for Go, use Hof values to calculate Horxn( Hosys),
use So values to calculate Sorxn( Sosys), and apply Equation 20.6.
Solution: Calculating Hosys from Hof values (with Equation 6.8):
Hosys =
Horxn = ∑m
Hof(products)
∑n
Hof(reactants)
= [3 mol KClO4 ( Hof of KClO4) + 1 mol KCl (
o of KClO )]
[4
mol
KClO
(
H
3
f
3
© Dario Bressanini
Hof of KCl)]
Calculating
Go from Enthalpy and Entropy Values–II
Hosys = [3 mol (-432.8 kJ/mol) + 1 mol (-436.7 kJ/mol)]
- [4 mol (-397.7 kJ/mol)]
= -144 kJ
Calculating
Sosys =
Sosys from So values (with Equation 20.3):
Sorxn = [3 mol KClO4 ( So of KClO4) + 1 mol KCl ( So of KCl)]
- [4 mol KClO3 (So of KClO3)]
= [3 mol (151.0 J/mol K) + 1 mol (82.6 J/mol K)]
- [4 mol (143.1 J/mol K)]
= - 36.8 J/K
.
Calculating
Gosys =
.
.
Gosys at 298 K:
Hosys - T Sosys = -144 kJ - [(298 K)(-36.8 J/K)(1kJ/1000 J)]
The reaction is spontaneous which is
o
G sys = -133 kJ
© Dario Bressanini
consistent with Go < 0
Effect of Temperature on Reaction Spontaneity
The temperature at which a reaction occurs influences the magnitude of
the T S term. By scrutinizing the signs of H and S, we can predict
the effect of temperature on the sign of G and thus on the spontaneity
of a process at any temperature.
Temperature-independent cases (opposite signs)
1. Reaction is spontaneous at all temperatures:
Ho < 0, So > 0
2 H2O2 (l)
2 H2O(l) + O2 (g)
Ho = -196 kJ and
So = 125 J/K
2. Reaction is nonspontaneous at all temperatures: Ho > 0, So < 0
3 O2 (g)
2 O3 (g)
Ho = 286 kJ and So = - 137 J/K
Temperature-dependent cases (same signs)
3. Reaction is spontaneous at higher temperature: Ho > 0 and So > 0
2 N2O(g) + O2 (g)
4 NO(g)
Ho = 197.1 kJ and
So = 198.2 J/K
4. Reaction is spontaneous at lower temperature: Ho < 0 and So < 0
o = - 822.2 kJ and
o = - 181.7 J/K
Dario Bressanini
2© Na
+
Cl
2
NaCl
H
S
(s)
2 (g)
(s)
Determining the Effect of Temperature on
Go–I
Problem: An important reaction in the production of sulfuric acid is the
oxidation of SO2 (g) to SO3 (g):
2 SO
+O
2 SO
2 (g)
2 (g)
3 (g)
At 298 K, Go = -141.6 kJ;
Ho = -198.4 kJ; and So = -187.9 J/K.
(a) Use the data to decide if this reaction is spontaneous at 25oC and how
Go will change with increasing T.
(b) Assuming Ho and So are constant with T, is the reaction
spontaneous at 900.oC?
Plan: (a) We examine the sign of Go to see if the reaction is
spontaneous and the signs of Ho and So to see the effect of T.
(b) We use Equation 20.6 to calculate Go from the given Ho and
So at the higher T (in K).
Solution: Continued on next slide.
© Dario Bressanini
Determining the Effect of Temperature on
Go–II
Solution: (a) Since Go < 0, the reaction is spontaneous at 298 K: a
mixture of SO2 (g), O2 (g), and SO3 (g) in their standard states (1 atm) will
spontaneously yield more SO3 (g). With So < 0, the term -T So > 0 and
becomes more positive at higher T. Therefore, Go will be less negative,
and the reaction less spontaneous, with increasing T.
(b) Calculating
Go =
Ho - T
= 22.0 kJ
Since
Go at 900.oC (T= 273 + 900. = 1173 K):
So = - 198.4 kJ - [(1173 K)(-187.9 J/K)(1 kJ/1000 J)]
Go > 0, the reaction is nonspontaneous at higher T.
© Dario Bressanini
Calculating
G at Nonstandard Conditions–I
Problem: The oxidation of SO2, which we discussed earlier, is too slow
2 SO2 (g) + O2 (g)
2 SO3 (g)
at 298 K to be useful in the manufacture of sulfuric acid. To overcome
this low rate, the process is conducted at an elevated temperature.
(a) Calculate K at 298 K and at 973 K.
Go298 = -141.6 kJ/mol of
reaction as written; using
Ho and So values at 973 K,
Go973 = -12.12 kJ/mol of reaction as written.
(b) In experiments to determine the effect of temperature on reaction
spontaneity, sealed containers are filled with 0.500 atm SO2,
0.0100 atm O2, and 0.100 atm SO3 and kept at 25oC and at 700oC. In
which direction, if any, will the reaction proceed to reach equilibrium at
each temperature?
(c) Calculate G for the system in part (b) at each temperature.
Plan: (a) We know
Go, T, and R, so we can calculate the K’s from
Equation 12.10.
Dario Bressanini
(b)©Calculate
Q, and compare it with each K from part (a).
Calculating
G at Nonstandard Conditions–II
Plan : Continued
(c) Since these are not standard-state pressures, we calculate G at each
T from Equation 20.11 with the values of Go (given) and Q [found in
part (b)].
Solution: (a) Calculating K at the two temperatures:
Go = -RT ln K
so K = e-( G/RT)
At 298 K, the exponent is
1000 J
-141.6 kJ/mol x
1 kJ
- ( Go/RT) = = 57.2
8.31 J/mol K x 298 K
K = e-( G/RT) = e57.2 = 7 x 1024
So At 973 K, the exponent is
-12.12 kJ/mol x 1000 J
1 kJ
- ( Go/RT) = = 1.50
8.31 J/mol K x 973 K
.
.
© Dario Bressanini
K = e-(
G/RT)
= e1.50 = 4.5
Calculating
G at Nonstandard Conditions-III
(b) Calculating the value of Q :
2
p2SO3
0.100
Q= 2
=
= 4.00
2
p SO2 x pO2 0.500 x 0.0100
Since Q < K at both temperatures, the denominator will decrease and the
numerator increase–more SO3 will form–until Q equals K. However, at
298 K, the reaction will go far to the right before reaching equilibrium,
whereas at 973 K, it will move only slightly to the right.
(c) Calculating
G, the nonstandard free energy change, at 298 K and
973 K
G298 = Go + RT ln Q
= -141.6 kJ/mol + (8.31 J/mol K x 1 kJ x 298 K x ln 4.00)
1000 J
= -138.2 kJ/mol
.
G973 = Go + RT ln Q
= - 12.12 kJ/mol + (8.31 J/mol K x 1 kJ x 973 K x ln 4.00
1000 J
© Dario Bressanini
= - 0.90 kJ/mol
.
© Dario Bressanini
Fig.
20.12
Calculating Gorxn for NH4NO3(s)
EXAMPLE 2:
NH4NO3(s)  NH4NO3(aq)
Is the dissolution of ammonium nitrate
product-favored?
If so, is it enthalpy- or entropy-driven?
© Dario Bressanini
9_amnit.mov
20 m07vd1.mov
Gorxn for NH4NO3(s)  NH4NO3(aq)
From tables of thermodynamic data we
find
Horxn = +25.7 kJ
Sorxn = +108.7 J/K or +0.1087 kJ/K
Gorxn = +25.7 kJ - (298 K)(+0.1087
kJ/K)
= -6.7 kJ
Reaction is product-favored
© Dario Bressanini
o
Calculating Gorxn
Gorxn =  Gfo (products) -  Gfo (reactants)
EXAMPLE 3: Combustion of carbon
C(graphite) + O2(g)  CO2(g)
Gorxn = Gfo(CO2) - [Gfo(graph) +
Gfo(O2)]
Gorxn = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an
element in its standard state is 0.
© Dario Bressanini
o
Go for COUPLED CHEMICAL REACTIONS
Reduction of iron oxide by CO is an example of
using TWO reactions coupled to each other in order
to drive a thermodynamically forbidden reaction:
Fe2O3(s)  4 Fe(s) + 3/2 O2(g)
Gorxn = +742 kJ
with a thermodynamically allowed reaction:
3/2 C(s) + 3/2 O2 (g)  3/2 CO2(g)
Overall :
Gorxn = -592 kJ
Fe2O3(s) + 3/2 C(s)  2 Fe(s) + 3/2 CO2(g)
Gorxn= +301 kJ @ 25oC
BUT
Gorxn < 0 kJ for T > 563oC
See Kotz, pp933-935 for analysis of the thermite reaction
© Dario Bressanini
Other examples of coupled reactions:
Copper smelting
Cu2S (s)  2 Cu (s) + S (s)
Couple this with:
S (s) + O2 (g)  SO2 (s)
Overall:
Gorxn= +86.2 kJ
(FORBIDDEN)
Gorxn= -300.1 kJ
Cu2S (s) + O2 (g)  2 Cu (s) + SO2 (s)
Gorxn= +86.2 kJ + -300.1 kJ = -213.9 kJ (ALLOWED)
Coupled reactions VERY COMMON in Biochemistry :
e.g. all bio-synthesis driven by
ATP  ADP for which Horxn = -20 kJ
Sorxn = +34 J/K
Gorxn = -30 kJ @ 37oC
© Dario Bressanini
Thermodynamics and Keq (2)
9_G_NO2.mov
20m09an1.mov
© Dario Bressanini
2 NO2  N2O4
Gorxn = -4.8 kJ
 pure NO2 has Grxn < 0.
 Reaction proceeds until Grxn = 0
- the minimum in G(reaction) see graph.
 At this point, both N2O4 and
NO2 are present, with more
N2O4.
 This is a product-favored
reaction.
Thermodynamics and Keq (3)
N2O4 2 NO2
Gorxn = +4.8 kJ



9_G_N2O4.mov
20m09an2.mov
© Dario Bressanini
pure N2O4 has Grxn < 0.
Reaction proceeds until Grxn
= 0 - the minimum in
G(reaction) - see graph.
At this point, both N2O4 and
NO2 are present, with more
NO2.
 This is a reactant-favored
reaction.
Thermodynamics and Keq (5)
Gorxn = - RT lnK
Calculate K for the reaction
N2O4  2 NO2
Gorxn = +4.8 kJ
Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K
4800 J
lnK = = - 1.94
(8.31 J/K)(298K)
K = 0.14
When Gorxn > 0, then K < 1 - reactant favoured
When Gorxn < 0, then K >1 - product favoured
© Dario Bressanini
Consider the synthesis of methanol:
CO(g) + H2(g)  CH3OH(l)
Calculate G at 25 oC for this reaction
where CO(g) at 5.0 atm and H2(g) at 3.0 atm
are converted to CH3OH(l) .
Given:
Gf
(CH3OH(l))
Gf
(CO(g))
Gf
(H2(g))
−166 kJ
−137 kJ
0
Cont’
We must use:
G = G + RT ln(Q)
G Gf
 Gf
 Gf
(CH3OH(l))
(CO(g))
(H2(g))
G 166137029kJ
Q=
1
PCO . P2H2
=
1
(5) . (3)2
=
2.2 x 10-2
G = 29x 103 + (8.3145)(298) ln( 2.2 x 10-2)
= − 38 kJ / mol
The second law is the greatest good and the
greatest bad on earth.
© Dario Bressanini
The good: Life is possible.
We eat concentrated energy in the form of food, process that
energy to synthesize complex biochemicals and run our organism,
excreting diffused energy as body heat and less concentrated
energy substances.
We use concentrated energy fuels to gather all kinds of materials
from all parts of the world and, without any energetic limitation,
arrange them in ways that please us.
We affect non-spontaneous reactions (pure metals from ores,
synthesizing curative drugs from simple compounds, altering
DNA).
We make machines that make other machines, machines that
© Dario Bressanini
mow
lawns, move mountains, and go to the moon.
The bad: Life is always threatened.
Every organic chemical of the 10,000 different kinds in our bodies
is metastable, synthesized by a nonspontaneous reaction and
kept from instant oxidation in air by activation energies.
Living creatures are energy processing systems that cannot
function unless biochemical cycles operate synchronically to use
energy to oppose second law predictions.
All of the biochemical systems that run our bodies are maintained
and regulated by feedback subsystems, composed of complex
substances that are synthesized internally by thermodynamically
nonspontaneous reactions, affected by utilizing energy ultimately
transferred from the metabolism of food. When these feedback
subsystems fail energy can no longer be processed to carry out the
many reactions we need for life that are contrary to the direction
© Dario Bressanini
predicted
by the second law.
GT,p
S
time
Changes in entropy of an isolated system with time.
with time until equilibrium is reached
Entropy increases
Change in the Gibbs free energy at constant temperature and pressure.
Under these conditions G decreases during the course of a spontaneous
change until equilibrium is reached.
© Dario Bressanini
Gibbs Free Energy
© Dario Bressanini
Chemical Potentials of the Ideal
Gas

Differentiating the chemical potential with
temperature
  J 

  Sm,J
 T  P
 P
Sm ,J T   S m ,J T   R ln P

© Dario Bressanini

The Standard Chemical Potential

For P1 = P = 1 bar, we define the standard
state chemical potential
°= (T, 1bar)
 P 
(T )   (T, 1 bar )  RT ln 

 1 bar 

© Dario Bressanini
The Chemical Potential for Real
Gases


The fugacity (f) represents the chemical
potential of a real gas.
Define the fugacity coefficient 
=f/P

For a real gas
 P
(T)   (T, 1 bar )  RT ln f

© Dario Bressanini

Obtaining Fugacity Coefficients

Comparing the chemical potential of the real
gas to the chemical potential of an ideal gas at
the same pressure
 P
   RT ln P 
P
 real   id    RT ln f



© Dario Bressanini

Calculating Fugacity Coefficients

The fugacity coefficients are obtained from
the compression factors (Z) as shown below
Z 1
RT ln   
dP
P
0
P
© Dario Bressanini
Partial molar (molal) quantities

If system is not simple, e.g. open system or
mixtures, then for e.g. substances A and B
 V 
dV (T , P, nA , nB )   
 T  P ,n
A ,nB
 V 
 V 


dP  
dnA  
dnB

n

n
 A  T , P ,n
 B  T , P ,n
A ,nB
 V 
dT   
 P  T ,n
B
A

(see lab manual, the five pages following error analysis)

For constant T,P and nA (e.g. 1 kg water in expt 2)
 V 

dV  
dn
 nB  T , P ,n B
A

Find V does not vary
linearly with nB
V
Slope
 V 
 V B  Vm,B
= 

n
 B  T , P ,n
A
partial molar volume
mB = moles solute / kg solvent
© Dario Bressanini
 V 
dV (T , P, nA , nB )   
 T  P ,n
A ,nB
 V 
 V 


dP  
dnA  
dnB

n

n
 A  T , P ,n
 B  T , P ,n
A ,nB
 V 
dT   
 P  T ,n
B
Note: at constant T and P, the total volume of the solution is:
nA
 dV  0  0   V dn
A
0


A
nB
  VBdnB
0
Vm,A, Vm,B are constant for a given composition
i.e. think of making up a solution at constant composition.
Then
V = Vm,A nA + Vm,B nB


A similar equation applies for any partial molar quantity.
© Dario Bressanini
A
DEFINE the chemical potential,
or partial molar free energy, of A

The exact differential for G is
 G 
dG(T , P, nA , nB )   
 T  P ,n
A , nB

 G 

A  
 nA  T , P ,n
B
 G 
 G 


dP  
dnA  
dnB .......

n

n




A T , P ,n
B T , P ,n
A , nB
 G 
dT   
 P  T ,n
Compare with combined law eqn (CL.G)
 G 
 G 
S  dw

, V 


dG + SdT - VdP = dwrev
 T 
 P 
 dG = - SdT + VdP + idni
G=A+PV
dA+PdV+VdP = - SdT + VdP + idni
 dA = -PdV - SdT + idni
and
 A 
 G 
 U 
 H 




 A  
 
 
 
 nA T ,V ,nB  nA T , P ,nB  nA  S ,V ,nB  nA  S , P ,nB
this the most useful defn., as it is the only one written in
terms of intensive variables.
B
A
P , nA , nB





© Dario Bressanini
T , nA , nB
recall:


Criteria for equilibrium:
dG = - SdT + VdP + idni
(dG) T , P   i dni  dw'
i




For a reversible, equilibrium process (we’ll take dw0
such as a phase change or chemical reaction (at const. T, P)
 dn

(

i
ice
i
i
0
liquid )dn = 0
i.e. equilibrium requires
ice  liquid
Hence criteria for equilibrium

equal T

equal P mechanical equil

equal 
© Dario Bressanini
thermal equil
chemical equil
ice
liquid
dnice = dnliquid