Page 75 - Modeling of Chemical Kinetics and Reactor Design

Reaction Mechanisms and Rate Expressions 45

( +r H * ) = dC H * = kC CH 5 * − k C H * C CH 6 = 0 (1-198)
3
4
dt
2
2
From Equation 1-198 kC − k C C = 0. Therefore,
3 * 4 H * CH 6
2
CH 5
2
2
C = k 3 • C CH 5 *
H * k 4 C CH 6 (1-199)
2
From Equation 1-196, kC CH 6 − k C CH 3 * C CH 6 = 0. Therefore,
2
1
2
2
C * = k 1 (1-200)
CH 3 k 2
From Equation 1-197, kC * C + k C * C − k C 2 * = 0.
5
2
4
2
2
CH 3 CH 6 H CH 6 CH 5
2
Therefore,
kC 2 = k C C + k C C (1-201)
5 * 2 * CH 6 4 H * CH 6
2
2
CH 5 CH 3
2
Substituting Equations 1-199 and 1-200 into Equation 1-201 gives
2
kC 2 * = k • C • k 1 + kC • k 3 • C CH 5 *
5
2
4
2
2
2
CH 5 CH 6 k 2 CH 6 k 4 C CH 6
2
(1-202)
kC 2 * = k C + k C *
5
1
3
2
CH 5 CH 6 CH 5
2
2
Assuming k << k , therefore,
3 5
 k  05 .
C * =  1 • C CH 6  (1-203)
CH 5  k 5 2 
2
Substituting Equation 1-203 into Equation 1-195 gives
.
 k  05
+ ( r ) = k  1 C  (1-204)
CH 4 3 CH 6
2
 k 5 
2