Thus:
CAo = CAo
Fm = Vo
2.02 160.9
= 0.0126
Kmol 3
m From (3)
τ
=
∫XAf 1/RT + k2 CA dXA
CAo
k C
0
Put
1 RT
k1
τ
1
=
1 8.314x519.3
=
0.00189
=
0.8492
k2
=
=
CAO
∫0.874 0
Since CA
A
=
CAO
12394
& XAF = 0.874
0.00189+12394 CA 0.8492 CA
(1-XA)
dXA
So (4) becomes:
(1+0.33XA)
τ
= CAO
∫0.874 0
Put CAO =
τ
(1+0.33XA)
CAO (1-XA)
(0.00189(1+0.33XA)+12394CAO(1-XA))
dXA
(1+0.33XA)
0.0126 & simplify
= 1.178
∫0.874 0.00189(1+0.33XA) +0.0126x12394(1-XA) dXA 0
(1-XA)
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