Page 484 - Elements of Chemical Reaction Engineering Ebook

Sec. 8.3 Nonisotherrnal Continuous-Flow Reactors 455

Energy balance. Recalling Equation (8-30), we have
Q - Ws - FAO ,x(@, ep,(T - To)) - FAoXIAHi, (TR) + A?,,(T - TR)] = 0 (8-30)

From the problem statement
Adiabatic: Q = 0
No work: W = 0
Atp = ?p, - Cp, = 141 - 141 = 0
Applying the conditions above to Equation (8-30) and rearranging gives

(E8-6.9)

c@,C,,, = cpA+@l?pI
=

= 159.5 J/mol.K

- - 6900)x
(
T = 330+
159.5
I T = 330 + 43.3X 1 (E8-61.10)
Substituting for the activation energy, T, , and k, in Equation (E8-6.3), we obtain
[& - $11

k = 31.1 exp[-
1
1 k = 31.1 erp[7906 (-]I I
I
(E8-6.11)

Substituting for AH,, , T2, and K,(T,) In Equation (E8-6.4) yields

(E8-6.12)

Recalling the rate law gives us
[ ( 41
-r, =kC,, I - 1 + - x (E8-6.7)

At equilibrium
-rA = 0