Chemical Reaction Engineering

Nonisothermal reactors

Use this for “A = B with ΔHRx = −60 kJ/mol, ΣθCₚ = 250 J/(mol·K), fed at 300 K. What is the adiabatic temperature at 75% conversion?”

Chart

T(X) = 240X + 30000.20.40.60.810100200300400500600700800900Conversion XTemperature T (K)(0.75, 480)
T(0.75) = 480
This is an example. Type your own values to start a new problem.
ΔTad (K)
=
−(
ΔHRx (kJ/mol)
)
÷
ΣθᵢCₚᵢ (J/(mol·K))
T (K)
=
T₀ (K)
+
ΔTad (K)
×
X

Formulas

ΔTad
=
−
ΔHRx
ΣθᵢCₚᵢ
+240
=
−
−60,000
250
T
=
T₀
+
ΔTad
×
X
480 = 300 + 240 × 0.75
Formulas are worked in J/mol. The step-by-step shows the conversions.

Assumptions

•
Adiabatic and steady, with ΔCₚ = 0, so ΔHRx stays the same at every temperature.
•
ΣθᵢCₚᵢ counts A and every inert or product fed, per mole of A fed.
•
Exothermic (ΔHRx < 0): the reactor heats up as X rises; endothermic, it cools.

Step-by-step

WE KNOW
ΔHRx = −60 kJ/mol, ΣθᵢCₚᵢ = 250 J/(mol·K), T₀ = 300 K, X = 0.75
FIND
adiabatic temperature rise (ΔTad), reactor temperature (T)
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).
ΔHRx = −60 kJ/mol = −60,000 J/mol (1 kJ/mol = 1000 J/mol)
Step 2 · Find adiabatic temperature rise (ΔTad)
Use ΔTad = −ΔHRx ÷ ΣθᵢCₚᵢ
The heat one mole of A releases by reacting warms the feed that came with it: divide by its heat capacity.
ΔTad
=
−
−60,000
250
ΔTad = +240 K
Step 3 · Find reactor temperature (T)
Use T = T₀ + ΔTad × X
Only
the
share
X
of
A
has
reacted,
so
the
feed
warms
by
that
share
of
the
full
rise.
T
=
300
+
240
×
0.75
T
=
300
+
180
T
=
480
K
Check
Put the numbers back in. Both sides should match.
+240
=
−
−60,000
250
✓
480 = 300 + 240 × 0.75 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

More problem types

© 2026 SME. All rights reserved.