SME
General Chemistry II · Problem type
Activation energy from two temperatures
Use this for “A rate constant doubles, from 1.00 × 10⁻³ s⁻¹ at 25 °C to 2.00 × 10⁻³ s⁻¹ at 35 °C. Find the activation energy and the frequency factor A.”
Diagram
ln
k
against
1
T
is
straight:
ln
k
=
ln
A
−
Eₐ/(RT).
Its
slope
through
the
two
readings
is
ln(k₂ ÷ k₁)
1
T₂
−
1
T₁
=
−6,370
K.
Eₐ = −R × slope = 8.314 J/(mol·K) × 6,370 K = 52.9 kJ/mol.
This is an example. Type your own values to start a new problem.
ln(
k₂ (s⁻¹)
÷
k₁ (s⁻¹)
)
=
(
Eₐ (kJ/mol)
÷
R)
×
1
T₁ (K)
−
1
T₂ (K)
A
Frequency factor
calculated · s⁻¹
Formulas
ln(
k₂
)
−
ln(
k₁
)
=
Eₐ
×
1,000
8.314
×
(
1
T₁
−
1
T₂
)
ln(0.002)
−
ln(0.001)
=
52.9
×
1,000
8.314
×
(
1
298.15
−
1
308.15
)
k₁
=
A
×
e
−Eₐ × 1,000 ÷ (8.314 ×
T₁
)
0.001
=
1.89 ×
10
6
×
e
−52.9 × 1,000 ÷ (8.314 × 298.15)
Assumptions
•
Eₐ and A stay the same between the two temperatures, so ln k against 1/T is a straight line with slope −Eₐ ÷ R.
•
R = 8.314 J/(mol·K) and Eₐ is in kJ/mol, so Eₐ is multiplied by 1000 to put it in J/mol.
•
Temperatures are in kelvins (°C + 273.15): 25 °C is 298.15 K.
•
k₁, k₂ and A share one unit, s⁻¹ here; only the ratio k₂ ÷ k₁ sets Eₐ.
Step-by-step
WE KNOW
k₁ = 0.001 s⁻¹, T₁ = 298.15 K, k₂ = 0.002 s⁻¹, T₂ = 308.15 K
FIND
activation energy (Eₐ), frequency factor (A)
Step 1 · Find activation energy (Eₐ)
Use ln(k₂) − ln(k₁) = Eₐ × 1,000 ÷ 8.314 × (1 ÷ T₁ − 1 ÷ T₂)
The
rise
in
ln
k
over
the
change
in
1
T
is
the
slope,
−
Eₐ
R
.
Multiply
by
R,
then
divide
by
1000
for
kJ.
Eₐ
=
8.314
×
ln(
k₂
) − ln(
k₁
)
1,000 × (1 ÷
T₁
− 1 ÷
T₂
)
Eₐ
=
8.314
×
ln(0.002) − ln(0.001)
1,000 × (1 ÷ 298.15 − 1 ÷ 308.15)
Eₐ
=
8.314
×
0.6931
1,000 × 1.0884 ×
10
-4
Eₐ = 52.95 kJ/mol
Step 2 · Find frequency factor (A)
Use k₁ = A × e−Eₐ × 1,000 ÷ (8.314 × T₁)
Only
the
share
e
−Eₐ/RT
of
collisions
has
energy
Eₐ.
Divide
k₁
by
that
share.
A
=
k₁
×
e
Eₐ × 1,000 ÷ (8.314 ×
T₁
)
A
=
0.001
×
e
52.95 × 1,000 ÷ (8.314 × 298.15)
A
=
0.001
×
e
52.95 × 1,000 ÷ 2,478.8191
A
=
0.001
×
e
52,950 ÷ 2,478.8191
A
=
0.001
×
e
21.361
A
=
0.001
×
(1.8922
×
10
9
)
A
=
1.89 ×
10
6
s
-1
Check
Put the numbers back in. Both sides should match.
ln(0.002)
−
ln(0.001)
=
52.95
×
1,000
8.314
×
(
1
298.15
−
1
308.15
)
✓
0.001
=
1.89 ×
10
6
×
e
−52.95 × 1,000 ÷ (8.314 × 298.15)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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