Acids, equilibrium & kinetics · HL
Arrhenius Eₐ (HL)
Estimate activation energy from two rate constants.
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Number formats & units
Use decimal dots in datasets. Single values also accept decimal commas and scientific notation such as 1.2e-3 or 1.2 × 10⁻³.
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Result & working
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Formula & example
The relationship
What the variables mean
| Symbol | Meaning |
|---|---|
| Activation energy. | |
| Pre-exponential factor in the same units as k. |
Worked example
- Temperature T₁ / K
- 298.15
- Rate constant k₁
- 0.0100
- Temperature T₂ / K
- 318.15
- Rate constant k₂
- 0.0300
43.32269 kJ mol⁻¹
Follow the example steps
- ln(k₂/k₁) = ln(0.03/0.01) = 1.098612
- 1/T₁ − 1/T₂ = 0.000210845
- Eₐ = R ln(k₂/k₁)/(1/T₁ − 1/T₂) = 43322.69 J mol⁻¹ = 43.32269 kJ mol⁻¹
- A = k₁e^(Eₐ/RT₁) = 388871.7
Common mistakes
Convert both temperatures to kelvin.
Incorrect approach
20 °C, 30 °C: 1/20 − 1/30
Correct approach
293.15 K, 303.15 K: 1/293.15 − 1/303.15
More checks before you submit
- Temperature must be in K.
- The units of A follow the units of k.
Method notes
This two-point form assumes an Arrhenius model over the temperature interval. A multi-point plot of ln k against 1/T is usually stronger evidence.