Page 128 - Elements of Chemical Reaction Engineering 3rd Edition

100 Rate Laws and Stoichiometry Chap. 3

mol
-
- 1485 kPa = 0.357 -
(8.3 14 Wa . dm3/mol. K)(500 K) dm3
We now evaluate E.
~=y~~S=(0.28)(1-1-$)= 0.14
-

@B -!x 0.1 (0.54 - OSX) mo,dm3
c, = c, (l+eX] 2 1-0.14X
=
cx
c -A=
o'lx mol/dm3
'- l+&X 1-0.14X

The concentrations of different species at various conversions are calculated
in Table E3-7.2 and plotted in Figure E3-7.1. Note that the concentration of N2 is
changing even though it is an inert species in this reaction.

TABLE E3-7.2. cOi.lqENTRATION AS A FUNCTION OF CONVERSION
C, (g moVdm3)
Species X = 0.0 X = 0.25 X = 0.5 X = 0.75 X = 1.0

so2 C, = 0.100 0.078 0.054 0.028 0.OOO
The concentration 02 C, = 0.054 0.043 0.031 0.018 0.005
of the inert is not so3 C, = O.OO0 0.026 0.054 0.084 0.116
constant
N2 c,= 0.203 0.210 0.218 0.227 0.236
c, = 0.357 0.357 0.357 0.357 0.357

We are now in a position to express -r, as a function of X. For example, ifthe rate
law for this reaction were first order in SO, (Le., A) and in 0, (Le., B), with
k = 200 dm3/mol.s, then the rate law becomes

Use Eq. (E3-7.2)
to obtain

Taking the reciprocal of -rA yields
- 0.5 [ 1 - 0.14-Y)'
I
I (E3-7.2)
-r, = (1 - X) (0.54 - 0.5X)
X