Page 119 - Modeling of Chemical Kinetics and Reactor Design

Thermodynamics of Chemical Reactions 89

Component Heat of formation ∆H (cal/gmol)
f
A –26,416
B 0.0
C –48,080
D 0.0

Component a b × 10 –2 c × 10 –5 d × 10 –9
A 6.726 0.04001 0.1283 –0.5307
B 6.952 –0.04576 0.09563 –0.2079
C 4.55 2.186 –0.291 –1.92
D 0 0 0 0
∆ –16.08 2.23751 –0.61056 –0.9735

Solution

The heat of reaction at the standard state 298 K is:

∆H ( K)=α o +α o − α o −α o
rxn 298 3 ∆H FC 4 ∆H FD 1 ∆H FA 2 ∆H FB

1 48 080)+ (
1 26 416) − (
∆H ( K)=− ( 0 0) −− ( 2 0)
rxn 298 , ,
,
− 21 664 cal mol
623
∆H ( K)= ∆H ( K)+ ∫
p
rxn 623 rxn 298 ∆C dT
298

623
∫
= ∆H o (298 K ) + (∆a + ∆b ×10 −2 T + ∆c ×10 −5 T 2 + ∆d ×10 −9 T 3 ) dT
rxn
298
Substituting ∆H (
o
rxn 298 K) and the parameters ∆a, ∆b, and ∆c in
Equation 2-110 gives

∆a T T ) + (
2
2
T
∆H T ( )= ∆H o T ( )+ ( − ∆b T − )
rxnT rxnT R R R
2
∆c
∆d
4
3
3
4
T
+ ( T − ) + ( T − )
T
R
R
3 4