SME
Chemical Reaction Engineering · Problem type
Adiabatic CSTR steady states
Use this for “E = 80 kJ/mol, k = 5 × 10⁻⁵ per min at 300 K, τ = 10 min, fed at 300 K with ΔTad = 240 K. Compare XMB and XEB at 360 K.”
Chart
XMB(360) ≈ 0.0949
XMB(T) = XEB(T)
The curves cross at T ≈ 300.1, T ≈ 383.3 and T ≈ 539.7.
This is an example. Type your own values to start a new problem.
ln(
k (min⁻¹)
÷
k₁ (min⁻¹)
)
=
E (kJ/mol)
÷
R
×
(1
÷
T₁ (K)
−
1
÷
T (K)
)
XMB
=
τ (min)
×
k (min⁻¹)
÷
(1
+
τ (min)
×
k (min⁻¹)
)
XEB
=
(
T (K)
−
T₀ (K)
)
÷
ΔTad (K)
T goes
Where T goes
worked out (not typed)
Formulas
ln(
k
k₁
)
=
E
8.314
×
(
1
T₁
−
1
T
)
ln(
0.01049
5 ×
10
-5
)
=
80,000
8.314
×
(
1
300
−
1
360
)
XMB
=
τ
×
k
1 +
τ
×
k
0.0949
=
10
×
0.01049
1 + 10 × 0.01049
XEB
=
T
−
T₀
ΔTad
0.25
=
360 − 300
240
T
goes
=
T
rises
if
XMB
>
XEB,
steady
state
if
XMB
=
XEB,
else
T
falls
0.0949
<
0.25,
so
T
falls
Formulas are worked in J/mol. The step-by-step shows the conversions.
Assumptions
•
First order in A, liquid, well mixed, adiabatic; the steady states are where XMB = XEB.
•
XMB > XEB: the reaction makes more heat than holding T needs, so T rises; below, T falls.
•
Of three crossings, the middle one is unstable: a small upset sends the reactor to another.
Step-by-step
WE KNOW
E = 80 kJ/mol, k₁ = 5 × 10⁻⁵ min⁻¹, T₁ = 300 K, τ = 10 min, T₀ = 300 K, ΔTad = 240 K, T = 360 K
FIND
rate constant at T (k), mole-balance conversion (XMB), energy-balance conversion (XEB), where T goes (T goes)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to J/mol).
E = 80 kJ/mol = 80,000 J/mol (1 kJ/mol = 1000 J/mol)
Step 2 · Find rate constant at T (k)
Use ln(k ÷ k₁) = E ÷ 8.314 × (1 ÷ T₁ − 1 ÷ T)
Arrhenius
from
the
known
temperature:
k₁
times
e
raised
to
(
E
R
)(
1
T₁
−
1
T
).
k
=
k₁
×
e^(
E
÷
8.314
×
(
1
T₁
−
1
T
))
k
=
5 ×
10
-5
×
e^(80,000
÷
8.314
×
(
1
300
−
1
360
))
k
=
(5
×
10
-5
)
×
e
80,000 ÷ 8.314 × (0.00333333 − 0.00277778)
k
=
(5
×
10
-5
)
×
e
80,000 ÷ 8.314 × 0.00055555
k
=
(5
×
10
-5
)
×
e
9,622.32 × 0.00055555
k
=
(5
×
10
-5
)
×
e
5.34568
k
=
(5
×
10
-5
)
×
209.7
k
=
0.01049
min⁻¹
Step 3 · Find mole-balance conversion (XMB)
Use XMB = τ × k ÷ (1 + τ × k)
The CSTR mole balance FA0X = kCA0(1 − X)V, divided by FA0 and solved for X.
XMB
=
10
×
0.01049
1 + 10 × 0.01049
XMB
=
10
×
0.01049
1 + 0.1049
XMB
=
10
×
0.01049
1.1049
XMB
=
0.1049
1.1049
XMB = 0.0949
Step 4 · Find energy-balance conversion (XEB)
Use XEB = (T − T₀) ÷ ΔTad
The
adiabatic
energy
balance:
to
sit
at
T
,
the
share
reacted
must
have
heated
the
feed
from
T₀
.
XEB
=
360 − 300
240
XEB
=
60
240
XEB = 0.25
Step 5 · Find where T goes (T goes)
Use T goes = T rises if XMB > XEB, steady state if XMB = XEB, else T falls
The
reaction
makes
too
little
heat
to
hold
T
,
so
the
reactor
cools
toward
a
lower
steady
state.
0.0949
<
0.25:
T
falls
T
goes
=
T
rises
if
0.0949
>
0.25,
steady
state
if
0.0949
=
0.25,
else
T
falls
T
goes
=
T
falls
Check
Put the numbers back in. Both sides should match.
ln(
0.01049
5 ×
10
-5
)
=
80,000
8.314
×
(
1
300
−
1
360
)
✓
0.0949
=
10
×
0.01049
1 + 10 × 0.01049
✓
0.25
=
360 − 300
240
✓
0.0949
<
0.25,
so
T
falls
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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Main lesson
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Equilibrium conversion and temperature
Use this for “A ⇌ B has ΔHRx = −60 kJ/mol and K = 50 at 300 K. What is the equilibrium conversion at 350 K?”