Page 293 - Advanced thermodynamics for engineers

12.11 DISSOCIATION PROBLEMS 281

Evaluating the energy contained in the products at 2772 K gives
U P ðT P Þ U P ðT s Þ¼ 5195240 kJ

Constituent CO 2 H 2 O CO H 2 O 2 N 2
125654.9 99196.2 70702.8 65594.3 74678.0 69948.5
u 2772
7041.6 7223.3 6011.8 6043.1 6032.0 6025.8
u 298
Difference 118613.3 91972.9 64690.8 59551.2 68646.0 63922.7
n 7.2240 8.8596 0.776 0.1404 1.8472 52.222

These values are within 0.05% and hence satisfy the energy equation.

ALTERNATIVE METHOD FOR CALCULATING THE CHEMICAL EQUATION
The approach used above to calculate the coefficients in the chemical equation was based on the
degrees of dissociation of the two reactions occurring in this example. It was shown previously that this
approach is often the best for manual solution, but that a more general approach, in which a system of
simultaneous equations is developed, is better for computer solution. This second method will be
outlined below for one of the steps in the previous example.
First, Eqn (12.109) can be replaced by

C 8 H 18 þ 13:889ðO 2 þ 3:76N 2 Þ / aCO 2 þ bH 2 O þ cCO þ dH 2 þ eO 2 þ fN 2 (12.131)
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
n R ¼67:111 n P ¼aþbþcþdþeþf
If the first iteration of the previous approach is used, then T P ¼ 2800 K and, as given in Eqn
(12.110),

p rCO 2
K p r1 ¼ 1=2 ¼ 6:58152
p rCO p
rO 2
p rCO p rH 2 O
Kp r2 ¼ ¼ 6:8295
p
p rCO 2 rH 2
Now, by definition,
1=2
a n
p rCO 2 P 1=2
¼ ¼ p (12.132)
K p r1 1=2 1=2 1=2 0
p rCO p ce p
rO 2 P
which gives
2
a n P
K 2 ¼ p 0 (12.133)
p r1 2
c e p P
For the water gas reaction
p rCO p rH 2 O cb
¼ ¼ 6:8295 ¼ (12.134)
K p r2
p ad
p rCO 2 rH 2