Taking the natural log of both sides of the above we can see the relation between ln PO2
and 1 .
P
G 1
ln PO2 = 2 ln PCO2 + 2R
T
CO
P
In this form the slope is 2 rxn and the intercept 2 ln PCO2 . The slope is negative as
CO
G < 0.
T
rxn
G
R
rxn
2 ln PCO2
PC O
(
PO2
)
2 ∆G rxn/R
ln PO2
2
2∆G rxn
(PPCO2
)
(
exp RT )
CO
1
T
PT
c) Assuming that the gases act as an ideal gas mixture we can write the equilibrium
reaction constant in terms of the mole fractions of each gas. But rst we have to nd out
the equilibrium amounts of each substance. The following table shows the knowns to begin
with.
reaction
CO 21 O2 *
) CO2
0
0
initial
nCO nO2
n0CO2
change
equilibrium x
The change in number of moles of oxygen is x ; n0O2 . From this we can calculate the
change in the number of moles of CO and CO2 as, dn0CO = 2dn0O2 and dn0CO2 = ;2dn0O2 ,
respectively. Now we can ll in the rest of the table.
1 O2
*
reaction
CO
CO2
)
2
0
0
initial
nCO
nO2
n0CO2
change
2(x - n0O2 )
(x - n0O2 )
-2(x - n0O2 ))
equilibrium n0CO+ 2(x - n0O2 )
x
n0CO2 -2(x -n0O2 )
The equilibrium constant is written as:
n0CO
nT 2
K= 0
1
nCO n0O2 ! 2
nT nT
n0 n 2
K = 0 CO20 T 1
nCO(nO2 ) 2
1
The total number of moles of gaseous substances will change during a reaction. It is
important to use the total number of moles of the gaseous substances at equilibrium for
the equilibrium reaction constant, nT = n0CO2 + n0CO + x.
(n0CO2 ; 2(x ; n0O2 ))(n0CO2 + n0CO + x) 12
K=
(n0CO + 2(x ; n0O2 ))x 21
Example Problem 9.2
Given the molar enthalpies at 298K, H 298 , H 298 , and H 298, the molar entropies at
298K, S 298 , S 298 , and S 298, and the heat capacities, C , C , and C , calculate the
equilibrium reaction constant for the following reaction at an arbitrary temperature T.
a;
a;
b;
b;
c;
c;
p;a
p;b
p;c
a(g) + b(g) *
) 2c(g)
Solution 9.2 The reaction constant has the following relation to the molar gibbs free
energy of the reaction, G .
rxn
!
G
K = exp ;RT
rxn
2
= nnn
!
c
a
b
Now we can formulate G .
rxn
G
G = 2G ; G ; G
= 2G 298 ; G 298 ; G 298 + (2G ; G ; G )
rxn
rxn
c;
b;
c;
c
b
c
a
c
b
Now we can use C to calculate G .
p;i
i
= (2H 298 ; H 298 ; H 298)
;298(2S 298 ; S 298 ; S 298)
Z
Z
Z
+(2 C dT ; C dT ; C dT )
298
298
298
Z C
Z C
Z C
;T (2 298 T dT ; 298 T dT ; 298 T dT )
G
rxn
c;
b;
c;
p;c
T
b;
T
T
c;
c;
T
p;a
T
p;c
p;b
T
p;a
p;b
= (2H 298 ; H 298 ; H 298)298(2S 298 ; S 298 ; S 298)
Z 2C ; C ; C
+ (2C ; C ; C )dT ; T
dT
T
G
Z
rxn
c;
b;
c;
T
298
c;
T
p;c
p;a
The substitution is straightforward.
p;b
298
b;
p;c
p;a
c;
p;b