ln(k₂ ÷ k₁) = −(Eₐ ÷ R) × (1/T₂ − 1/T₁) — fill in every field except the one you want; the empty field is solved for.
Activation energy (Eₐ) = 65.145 kJ/mol
- Convert T₁: 25 °C =
298.15 K - Convert T₂: 45 °C =
318.15 K - Rearrange:
Eₐ = −8.3145 × ln(k₂ ÷ k₁) ÷ (1/T₂ − 1/T₁) ÷ 1000 - Substitute:
−8.3145 × ln((0.012) ÷ (0.0023)) ÷ (1/(318.15 K) − 1/(298.15 K)) ÷ 1000=65.145 kJ/mol
How it works
The two-point form, ln(k₂ ÷ k₁) = −(Eₐ ÷ R)(1/T₂ − 1/T₁), comes from the Arrhenius equation k = A e^(−Eₐ/RT) written for two temperatures, so the pre-exponential factor A cancels. Temperatures are converted to kelvin; Eₐ is worked in kJ/mol with R = 8.3145 J/(mol·K).
k₁ and k₂ can be in any unit as long as it is the same for both — you can even use rates or reciprocal times measured under otherwise identical conditions. The equation assumes Eₐ does not change with temperature.
Frequently asked questions
- How do I calculate activation energy from two rate constants?
- Measure k at two temperatures and solve Eₐ = −R ln(k₂/k₁) ÷ (1/T₂ − 1/T₁). The calculator does this when you leave Eₐ empty.
- Does a 10 °C rise always double the rate?
- Only roughly, for reactions with activation energies around 50 kJ/mol near room temperature. Reactions with higher Eₐ are more sensitive to temperature.
- How do catalysts fit in?
- A catalyst provides a pathway with lower Eₐ, so k is larger at every temperature.
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