EQUILIBRIUM COMPOSITION
Statement
Consider the isobaric decomposition of CO2 to CO and O2, having initially 1 mol of CO2 at 25 ºC and 100
kPa.
a) Find the equilibrium composition.
b) Find the temperature at which the equilibrium molar fraction of oxygen is 0.25.
c) Plot the equilibrium composition versus temperature.
Considérese la descomposición isobárica del CO2 en CO y O2, teniendo inicialmente 1 mol de CO2 a 25
ºC y 100 kPa. Se pide:
a)
Determinar la composición de equilibrio.
b)
c)
Calcular a qué temperatura la fracción molar de oxígeno es del 25%.
Representar la composición de equilibrio en función de la temperatura.
Solution.
a)
Find the equilibrium composition.
The stoichiometric reaction to be considered is:
1
CO 2 =CO+ O 2
2
and the mixture ratio:
aCO2 =xCO2 CO2 +xCO CO+xO2 O2
The 4 unknowns {a, xCO2 , xCO , xO2 ) are found by solving the 4 equations:
normalisation: 1  xCO2  xCO  xO2
balance of C:
a  xCO2  xCO
balance of O: 2a  2 xCO2  xCO  2 xO2
equilibrium:
C
xCO xO2
i 1
xCO2
 xi i 
 p 
 
p 

 i
K (T , p  )
The equilibrium constant is:
 g
h  T 
K (T , p  )  exp  r  r  1 
RT 
T
 RT



Here,
i=1+1/21=1/2, but p=p;
hr=(111)+(1/2)0(394)=283 kJ/mol, thus:
Equilibrium composition
gr=(137)+(1/2)0(394)=257
kJ/mol
and
1
34000 K 

K (T , p  )  exp 10.34 

T


Solving the set of 4 equations for the 4 unknowns is not difficult, but it is better to convert it to
one equation with one unknown, i.e. to use the three linear equations to solve for the others in
terms of xO2:
1
2
 p 
34000 K 
3

2 xO22    exp  10.34 
  2 xO2  xO2  1
T


p 
For several given temperatures one gets:
b)
c)
T [K]
a
xCO2
xCO
xO 2
2000
2500
3000
0.99
0.94
0.81
0.98
0.81
0.43
0.015
0.13
0.38
0.008
0.065
0.19
Find the temperature at which the equilibrium molar fraction of oxygen is 0.25.
Similarly, for a given xO2 =0.25 we find:
T [K]
a
xCO2
xCO
xO 2
3300
0.75
0.25
0.50
0.25
Plot the equilibrium composition versus temperature.
Fig. 1. The equilibrium composition vs. temperature.
Comments.
The variance for this system at equilibrium is V=2+CFR=2+EF=3, since C=#{CO2,CO,O2}=3, F=1
(gas phase), R=1 (CO2=CO+½O2) and E=#{C,O}=2. If we fix temperature and pressure, there is still a
free parameter, e.g. xO2, the other two molar fractions being determined by xi=1 and xii=K(T,p). The
atom conservation equations (C-balance and O-balance) serve to relate the equilibrium composition to the
initial composition. In this problem, the C/O ratio was fixed to C/O=1/2 and thence the free parameter
Equilibrium composition
2
(xO2) becomes fixed for given temperature and pressure, but a different C/O ratio might be fixed as initial
conditions, e.g. stating a set of amounts for{CO2,CO,O2}; the mixture ratio then becomes:
aCO2 +a
nCO,0
nCO2 ,0
CO+a
nO2 ,0
nCO2 ,0
O2 =xCO2 CO2 +xCOCO+xO2 O2
The atom balance equations would change, and so the solution, but the formulation is the same: a set of 4
equations for the 4 unknowns.
Instead of the equilibrium condition above used (  xi i  K (T , p  )  p p  
i 1
G(T,p,)=min (A(T,p,)=0) may have been used, as in Exercise 1, where we got:
C
G ( )   ni i   ni i (T , p  )   ni RT ln

 i
), the condition
p
  ni RT ln xi 
p
 




  (1   ) CO





(1

) xi ( ) RT ln xi ( )
CO
O

2
2 2
2

here
C
A( )    i i    i i (T , p  )   i RT ln
i 1
 

    CO
 CO
2

here
p
  i RT ln xi 
p
 



1 
1 

1
 O2   RT   ln
 ln
 ln 2 

 2

2


1
1
1 
2
2
2

But
1
34000 K 
 



   CO
 CO
 O2   RT ln K  RT 10.34 

2
2
T




so that
 


34000 K 
1 

1

2 
A(T ,  )  RT 10.34 

RT

ln

ln

ln



 2

T



1
1
1 
2
2
2


and in a similar manner we can get G(T,)G0, with G0  CO
, and plot them for some temperature
2
values (R=8.314 J/(mol·K)) to yield (see P-9.6):
Fig. 2. The equilibrium states at given T and p occur at G()=min or A()=0.
Back
Equilibrium composition
3