SME
Chemical Reaction Engineering · Problem type
Batch time for a conversion
Use this for “A first-order reaction with k = 0.2 per min runs in a batch reactor. How long until 90% of A has reacted?”
Chart
CA(11.51) = 0.2
This is an example. Type your own values to start a new problem.
t (min)
=
−ln(1
−
X
)
÷
k (min⁻¹)
CA (mol/L)
=
CA0 (mol/L)
×
(1
−
X
)
Formulas
t
=
−
ln(1 −
X
)
k
11.51
=
−
ln(1 − 0.9)
0.2
CA
=
CA0
×
(1
−
X
)
0.2 = 2 × (1 − 0.9)
Assumptions
•
First order, −rA = kCA, isothermal, constant volume (liquid phase).
•
Well mixed, so CA is the same everywhere; t is reaction time only, not filling or emptying.
•
k and t use the same time unit, minutes here.
Step-by-step
WE KNOW
k = 0.2 min⁻¹, X = 0.9, CA0 = 2 mol/L
FIND
batch time (t), final concentration (CA)
Step 1 · Find batch time (t)
Use t = −ln(1 − X) ÷ k
The
batch
balance
dX
dt
=
k
(1
−
X
)
gives
t
=
∫
dX
k
(1 −
X
)
from
0
to
X
,
which
is
−
ln(1 −
X
)
k
.
t
=
−
ln(1 − 0.9)
0.2
t
=
−
ln(0.1)
0.2
t
=
−
−2.303
0.2
t
=
2.303
0.2
t
=
11.51
min
Step 2 · Find final concentration (CA)
Use CA = CA0 × (1 − X)
At
constant
volume,
the
share
1
−
X
of
A
is
still
there.
CA = 2 × (1 − 0.9)
CA = 2 × 0.1
CA = 0.2 mol/L
Check
Put the numbers back in. Both sides should match.
11.51
=
−
ln(1 − 0.9)
0.2
✓
0.2 = 2 × (1 − 0.9) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Batch, CSTR and PFR design
Main lesson
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Use this for “Two CSTRs, each with 5 min space time, run a first-order reaction with k = 0.2 per min. What conversion leaves the second tank?”
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Use this for “20 mol/min of A at 2 mol/L reacts first order, k = 0.2 per min. Which needs less volume for 80% conversion? Read the CSTR as the rectangle and the PFR as the area under FA0 ÷ (−rA).”