LECTURE 6
AOSC 434
AIR POLLUTION
RUSSELL R. DICKERSON
THERMODYNAMICS (continued)
Wark and Warner Chapter 7,8
Seinfeld & Pandis, Chapter 3
From the last lecture, you will remember the concepts of enthalpy (heat),
entropy (disorder) and Gibbs Free Energy (the criterion feasibility).
G  H  T  S
ΔG   RT ln (Keq )
K eq  exp(  ΔG/RT)
How much CO is in equilibrium with CO₂ in auto exhaust? Again,
consider the combustion of isooctane (C₈H₁₈)
C8H18  12.5O2 ...(47N2 )  8CO2  9H2O...(47N2 )
The dissociation of CO₂ produces CO:
CO2  CO  1 / 2O2
H 0 rxn  283 kJ / m ole
G 0 rxn  258 kJ / m ole
R  8.31 J / m oleK
What is the concentration of CO₂ if there is complete combustion?
[CO₂] = 8 / (8 + 9 + 47) = 0.125
If P = 1.00 atm, then P(CO₂) = 0.125 atm. What is the equilibrium constant?
  258103 

K eq  exp(G / RT )  exp
 8.31 298 
 1045
Equilibrium lies far to the left!
10 45 
 
PCO PO2
PCO2
1/ 2
If we assume that all the CO and O₂ come from the dissociation of CO₂ then:
[CO₂] >> [CO] = 2[O₂]
P = [CO₂].
If total pressure is 1.00 atm then
CO 2
PCO ( PO2 ) 0.5 
10
PCO 

1
( PCO ) 3 / 2  PCO2 K eq
2
2 PCO2 K eq

2/3

0.125 2   3 1031 atm
This is not much to worry about, but combustion does not occur at room
temperature, it occurs at about 2000 K. What are ΔG and K eq at 2000 K?
Assume that ΔH and ΔS are independent of temperature. We want G2000 K
 45
2/3
H 2000 K  H 298 K  283 kJ / m ole
H 298  G298
S 2000 k  S 298 K 
298
 0.0839kJ / m oleK
A positive ΔS for the reaction means that the products are favored by the
entropy term in the Gibbs Free Energy, and the equilibrium will be pushed to
the right as temperature increases.
G2000  283 (2000)(0.0839)  115.2 kJ / m ole
K eq  9.8 10 4
PCO  3.1103 atm
At P = 1.00 atm
[CO] = 0.31% = 3100 ppm
It takes only 600 ppm of CO to cause death in humans; see Seinfeld, p.55.
The CO is produced quickly and does not reform CO₂ fast enough to be
destroyed as the exhaust gases cool.
But a car’s engine does not run at 1.00 atm pressure, the typical compression
ratio is 8:1 (120 psi). If P = 8 atm and we maintain a stoichiometric air/fuel
mixture,
PCO2  8[CO2 ]  1.0 atm
PCO 

2 PCO2 K eq

2/3
 1.25x 102 atm
This may look like more than we had at 1.0 atm, but at 8.0 atm total
pressure:
[CO] ≈ 1.6 x 10⁻³ = 1600 ppm
Only half as much! Higher compression reduces the amount of CO
produced.
What happens if we adjust the carburetor to produce a lean burn? Let there
be 10% excess oxygen, and leave the total pressure at 8.0 atm.
C8 H18  13.75O2  52N 2  8CO2  9 H 2O  52N 2  1.25O2
8
 0.114
8  9  1.25  52
 8  (0.114)  0.91 atm
[CO2 ] 
PCO2
[O2 ]XS 
1.25
 0.018
8  9  1.25 52
PO2  0.144atm
K eq 
Rearranging:
PCO  (PO2 )0.5
PCO 2


P CO K 
P(CO ) 
2
eq
PO2 
 ((0.91)9.8  104 ) /(0.144)0.5
 2.6  103 atm  m ixing_ ratio  3.3  104
[CO] = 330 ppm
This is much less than the 3100 ppm we calculated for a stoichiometric
mixture!
CONCLUSIONS
For minimal CO production:
1.
2.
3.
Burn as cool as possible.
Burn as lean as possible.
Burn as high compression as possible.
EXTRA EXAMPLE
How much molecular hydrogen will be formed by the decomposition of
water in the exhaust gas? This example uses water recombination instead
of the water dissociation reaction, whole number coefficients instead of ½,
and units of kcal mole⁻¹ instead of kJ mole⁻¹.
2H2  O2  2H2O
P (H2O)2
K eq 
(P (H2 )2 P (O2 ))
ΔG0  2(54.63) 109.26kcal mole1
ΔH0  2(57.80) 115.60kcal mole1
P (H2 )  P (H2O)/(P (O2 )Keq )0.5
Assume that the reaction takes place at one atmosphere pressure, 2000 K
and slightly lean, 1% excess oxygen. (The concentration of H₂ would be
somewhat more for a stoichiometric mixture.) Let P(O₂) ~ 1%, P(H₂O) ~ 10%
and T = 298 K,
Keq  exp(G / RT)  1079
Equilibrium lies far to the right and H₂O is strongly favored over H₂. But
at a combustion temperature of 2000 K what happens to ΔG?
G2000  H 0  TS 0
(H 0  G 0 )
S 
T
G2000  73 kcal / m ole
0
This difference of 36 kcal (about 30%) makes a large difference in the
calculated equilibrium concentration of H₂.
K eq  108
PH 2  104  100 ppm
In general, for the Gibbs free energy at some temperature other than 298 K.
0
0



H


G

GT  H 0  T 
298

