Chemical Equilibrium
• In general the combustion products consist of more than just
CO2, H2O, O2, and N2
• For rich mixtures CO also exists in the products and at high temperatures
the molecules dissociate to form H, O, OH, NO via the following reactions:
H 2  2H
O2  2O
H 2  O2  2OH
O2  N 2  2 NO
• The opposite direction reactions are also possible
2H  H 2
2O  O2
2OH  H 2  O2
2 NO  O2  N 2
• At equilibrium the rate of the forward reaction equals the rate of the
backward reaction.
H 2  2H
O2  2O
H 2  O2  2OH
O2  N 2  2 NO
1
Chemical Equilibrium
• At equilibrium the relative proportion of the species mole fraction is fixed
• For the general equilibrium reaction
nA A  nB B  ncC  nD D
• The equilibrium composition for species A, B, C, D is given by:
X
K (T ) 
X
nC
C
nA
A
X
X
nD
D
nB
B
 P

P
 ref




nC  nD  n A  nB
where K is the equilibrium constant which is tabulated as a function of
temperature for different equilibrium reactions, Pref is 1 atm and P is in
units of atmospheres.
Note X A 
nA
n A  nB  nC  nD
2
Chemical Equilibrium
Recall that for a rich mixture (<1) the reaction equation could not be
balanced (5 unknowns a, b, d, e, f and only 4 atom balance equations for
C,H,O,N ) even if we neglect dissociation (i.e., low product temperature)

C H    (  )(O2  3.76 N 2 )  aCO2  bH 2O  dN2  eCO  fH 2
4
If the product species CO2, H2O, CO and H2 are at equilibrium and described
by the water-gas reaction:
CO2  H 2  CO  H 2O
The equilibrium constant for this reaction provides the fifth equation :
K (T ) 
X CO  X H 2O
X CO2  X H 2
Note K is tabulated as a function of T

eb
a f
P  1 atm
Chemical Equilibrium (example)
1 kmol of CO2, ½ kmol of O2 and ½ kmol of N2 reacts to form a mixture
consisting of CO2, CO, O2, N2 and NO at 3000K and 1 atm. Determine the
equilibrium composition of the product mixture.
CO2  1 / 2O2  1 / 2 N 2  aCO  bNO  cCO2  dO2  eN 2
3000K and 1 atm
C
O
N
1 = a+c
3 = a+b+2c+2d
1 = b+2e
c=1-a
d = 1/2(1 + a - b)
e = 1/2(1 - b)
Have 2 unknowns a, b so need 2 equilibrium equations
1.
CO2  CO  1 / 2O2
K1( 3000 K)
2.
1 / 2O2  1 / 2 N 2  NO
K 2( 3000 K)
Chemical Equilibrium (example)
From the equilibrium constant expression
K1  0.3273 
X CO  X O1 /22
X CO2
ntot= a+b+c+d+e = a+b+(1-a)+1/2(1+a-b)+1/2(1-b) = (4+a)/2
X CO 
a
(4  a) / 2
X O2 
1 / 2(1  a  b)
(4  a) / 2
X CO2 
1 a
(4  a) / 2
Substituting yields:
K1  0.3273 
X CO  X O1 /22
X CO2
a 1  a  b 



1 a  4  a 
1/ 2
(1 )
Chemical Equilibrium (example)
Similarly for the second equilibrium reaction
K 2  0.1222 
X NO
2b

X O1 / 2  X 1N/ 2 (1  a  b)(1  b)1 / 2
2
(2)
2
Solving equations 1 and 2 yields:
a= 0.3745 b= 0.0675
From the atom balance equations get:
c= 0.6255 d= 0.6535
e= 0.4663
Role of Equilibrium Solvers
• If the products are at high temperature (>2000K) minor species will be
present due to the dissociation of the major species CO2, H2O, N2 and O2.

C H   (  )(O2  3.76 N 2 )  aCO2  bH 2O  cN 2  dO2  eCO  fH 2
4
 gH  hO  iOH  jNO  kN  
• Hand calculations are not practical when many species are considered,
one uses a computer program to calculate product equilibrium composition.
Composition of Octane-air Mixtures at Equilibrium
(stoich)
Mole fraction of each species
(lean)
Temperature (K)
(rich)