Page 113 - Modeling of Chemical Kinetics and Reactor Design

Thermodynamics of Chemical Reactions 83
(
∆H ( K)=− 22 040 − 15 455 428 298)
−
rxn 428 , .
×
.
2
2
+ 17 124 10 − 3 ( 428 − 298 )
2
×
.
3
3
− 4 887 10 − 6 ( 428 − 298 )
3
∆H ( 428 K)=− 22 040 − 2 009 15 808 08 84 61
+
−
,
.
,
.
.
rxn
.
− 23 325 68 kcal
,
kmolN
2
1 kcal = 4.184 kJ
kJ
.
,
=− 97 594 65
kmol N
2
∆H ( 428 K)= 1 kgmolN   − , . kJ  
2
97 594 65
rxn
3 kgmolH  kmolN 
2 2
kJ
,
=−32 531 55 at 428K
.
kg mol H 2
When the reaction proceeds at a constant temperature T, and the
reactants and products remain at the standard state (represented by
the superscript o), we can use Equation 2-44 to calculate the true
equilibrium constant:
∆G = –RT ln K (2-44)

o
= ∆H – T∆S o
r
where

T
rs ∫
∆H = ∆H + ∆C dT
o
o
o
r
p
Ts