Page 132 - Elements of Chemical Reaction Engineering 3rd Edition

104 Rate Laws and Stoichiometry Chap. 3

Rearranging to use POLYMATH yields

(E3-8.8)

For a pure N,O, feed, E = yAO 6 = l(2 - 1) = 1 .
We shall let xef represent the equilibrium conversion in a flow system. Equa-
tion (E3-8.8) written in the POLYMATH format becomes
f(xef) = xef -[kc*(l - xef)*(l + eps*xef)/4/cao] * * 0.5

This solution is also shown in Tables E3-8.2 and E3-8.3.
Note that the equilibrium conversion in a flow reactor (Le., Xet = 0.51), with
negligible pressure drop, is greater than the equilibrium conversion in a con-
stant-volume batch reactor (Xeb = 0.44 ). Recalling Le Chiitelier's principle, can you
suggest an explanation for this difference in X, ?
(c) Rate laws. Assuming that the reaction follows an elementary rate law, then

(E3-8.9)

1. For a flow system, C, = F,/v and C, = FB/v with u = v,(l +EX). Conse-
quently, we can substitute Equations (E3-8.5) and (E3-8.6) into Equation
(E3-8.9) to obtain

-rA = f(X) for a (E3-8.10)
flow reactor

Let's check to see if at equilibrium this equation reduces to the same equation
as that obtained from thermodynamics. At equilibrium -rA = 0:

0 =-[I ~ACAO -xt- Kc(1 4cA0x,2 +EX,) I
1 +EX,

Rearranging gives us

(E3-8.8)

It must agree with the value calculated from thermodynamic value and it does!
2. For a constant volume (V= V,) batch system, C, = N,/V, and
C, = NB/Vo. Substituting Equations (E3-8.2) and (E3-8.3) into the rate law,
we obtain the rate of disappearance of A as a function of conversion:

-rA = f(x) CAo(l-X)-- 1
for /a batch reactor (E3-8.11)
with V = V,