Chemical Reaction Engineering · Problem type

Second-order reaction in each reactor

Use this for “A second-order liquid reaction, k = 0.5 L/(mol·min), is fed at 2 mol/L. What space times reach 80% conversion in a CSTR and in a PFR?”

Chart

20 min4 minτCSTRCSTR space timeτPFRPFR space time
This is an example. Type your own values to start a new problem.
τCSTR (min)
×
k (L/(mol·min))
×
CA0 (mol/L)
=
(1
−
)
2
τPFR (min)
×
k (L/(mol·min))
×
CA0 (mol/L)
=
1
−

Formulas

τCSTR
=
X
÷
(
k
×
CA0
×
(1 −
X
)
2
)
20
=
0.8
÷
(0.5
×
2
×
(1 − 0.8)
2
)
τPFR
=
X
k
× CA0 × (1 −
X
)
4
=
0.8
0.5 × 2 × (1 − 0.8)

Assumptions

•
Second order in A, −rA = kCA², isothermal, liquid phase (constant density).
•
Both reactors are fed at CA0 and reach the same conversion X at steady state.
•
Space time τ = V ÷ v₀: multiply by the feed flow for each volume.

Step-by-step

WE KNOW
k = 0.5 L/(mol·min), CA0 = 2 mol/L, X = 0.8
FIND
CSTR space time (τCSTR), PFR space time (τPFR)
Step 1 · Find CSTR space time (τCSTR)
Use τCSTR = X ÷ (k × CA0 × (1 − X)²)
The
CSTR
works
at
the
exit
rate
kCA0²
(1 −
X
)
2
;
FA0X
=
(−rA)V,
divided
by
v₀,
gives
τ.
τCSTR
=
0.8
÷
(0.5
×
2
×
(1 − 0.8)
2
)
τCSTR
=
0.8
0.5 × 2 ×
0.2
2
τCSTR
=
0.8
0.5 × 2 × 0.04
τCSTR
=
0.8
0.04
τCSTR = 20 min
Step 2 · Find PFR space time (τPFR)
Use τPFR = X ÷ (k × CA0 × (1 − X))
Each
slice
works
at
its
own
rate:
τ
=
∫dX
÷
(kCA0
(1 −
X
)
2
)
from
0
to
X
,
which
is
X
kCA0(1 −
X
)
.
τPFR
=
0.8
0.5 × 2 × (1 − 0.8)
τPFR
=
0.8
0.5 × 2 × 0.2
τPFR
=
0.8
0.2
τPFR = 4 min
Check
Put the numbers back in. Both sides should match.
20
=
0.8
÷
(0.5
×
2
×
(1 − 0.8)
2
)
✓
4
=
0.8
0.5 × 2 × (1 − 0.8)
✓
Topic check
See what you know: 5 questions on this topic
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