SME
Chemical Reaction Engineering · Problem type
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?”
Chart
Xₑ(350) ≈ 0.6167
This is an example. Type your own values to start a new problem.
ln(
K
÷
K₁
)
=
−(
ΔHRx (kJ/mol)
)
÷
R
×
(1
÷
T (K)
−
1
÷
T₁ (K)
)
Xₑ
=
K
÷
(1
+
K
)
Formulas
ln(
K
K₁
)
=
−
ΔHRx
8.314
×
(
1
T
−
1
T₁
)
ln(
1.609
50
)
=
−
−60,000
8.314
×
(
1
350
−
1
300
)
Xₑ
=
K
1 +
K
0.6167
=
1.609
1 + 1.609
Formulas are worked in J/mol. The step-by-step shows the conversions.
Assumptions
•
A ⇌ B, first order both ways, so Xₑ = K ÷ (1 + K); fed as pure A.
•
ΔHRx stays the same between T₁ and T (van’t Hoff); R = 8.314 J/(mol·K).
•
Exothermic (ΔHRx < 0): raising T lowers Xₑ. The adiabatic XEB line meets this curve at the highest reachable X.
Step-by-step
WE KNOW
ΔHRx = −60 kJ/mol, K₁ = 50, T₁ = 300 K, T = 350 K
FIND
equilibrium constant at T (K), equilibrium conversion (Xₑ)
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 equilibrium constant at T (K)
Use ln(K ÷ K₁) = −ΔHRx ÷ 8.314 × (1 ÷ T − 1 ÷ T₁)
Integrate
van’t
Hoff,
d
ln
K
dT
=
ΔHRx
RT²
,
from
T₁
to
T
,
then
undo
the
ln.
K
=
K₁
×
e^(−ΔHRx
÷
8.314
×
(
1
T
−
1
T₁
))
K
=
50
×
e^(−(−60,000)
÷
8.314
×
(
1
350
−
1
300
))
K
=
50
×
e
−(−60,000) ÷ 8.314 × (0.0028571 − 0.0033333)
K
=
50
×
e
−(−60,000) ÷ 8.314 × (−0.0004762)
K
=
50
×
e
60,000 ÷ 8.314 × (−0.0004762)
K
=
50
×
e
7,216.7 × (−0.0004762)
K
=
50
×
e
−3.4366
K
=
50
×
0.032174
K
=
1.609
Step 3 · Find equilibrium conversion (Xₑ)
Use Xₑ = K ÷ (1 + K)
At
equilibrium
K
=
CB
CA
=
X
1 − X
;
solve
for
X.
Xₑ
=
1.609
1 + 1.609
Xₑ
=
1.609
2.609
Xₑ = 0.6167
Check
Put the numbers back in. Both sides should match.
ln(
1.609
50
)
=
−
−60,000
8.314
×
(
1
350
−
1
300
)
✓
0.6167
=
1.609
1 + 1.609
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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