Activation Energy Calculator from Two Rate Constants
By Santaji GadepatilinUpdated October 30, 2026
Activation energy is the energy barrier a reaction must overcome, and it controls how sharply the rate rises with temperature. This activation energy calculator uses the two-point form of the Arrhenius equation to find Ea from rate constants (or rates, or reaction times) measured at two temperatures — or the new rate constant once Ea is known.
How to use the activation energy calculator
Enter the rate constant measured at the first temperature and the rate constant at the second temperature. The calculator gives the ratio k₂/k₁ and the activation energy in kJ/mol. To predict a rate constant, enter k₁, both temperatures and Ea, and clear k₂ and the ratio.
Two-point Arrhenius equation
ln(k2 ÷ k1) = (Ea ÷ R) × (1/T1 − 1/T2)
Ea = R ln(k2 ÷ k1) ÷ (1/T1 − 1/T2)
R = 8.314 J/(mol·K); temperatures must be absolute (kelvin). The pre-exponential factor A cancels, so only the two measurements are needed — use the Arrhenius equation calculator to work with A directly.
Example
A rate constant rises from 0.0010 at 25 °C to 0.0040 at 45 °C:
- ln(0.0040 ÷ 0.0010) = ln 4 = 1.3863
- 1/298.15 − 1/318.15 = 2.1084 × 10⁻⁴ K⁻¹
- Ea = 8.314 × 1.3863 ÷ 2.1084 × 10⁻⁴ = 54.67 kJ/mol
How much faster a reaction runs when heated from 25 °C
| Activation energy | 25 → 35 °C | 25 → 75 °C |
|---|---|---|
| 20 kJ/mol | 1.30× | 3.2× |
| 40 kJ/mol | 1.69× | 10.1× |
| 50 kJ/mol | 1.92× | 18.1× |
| 60 kJ/mol | 2.19× | 32.3× |
| 80 kJ/mol | 2.85× | 103.0× |
| 100 kJ/mol | 3.70× | 328.1× |
| 150 kJ/mol | 7.13× | 5943.9× |
Notes
- Measure the two rates as far apart in temperature as practical — a small ΔT magnifies measurement error in Ea.
- A curved Arrhenius plot (ln k vs 1/T) means the mechanism changes with temperature; two points will then give only an average Ea.
- Typical values: enzyme reactions 30–60 kJ/mol, many organic reactions 50–100 kJ/mol, diffusion-controlled reactions below 20 kJ/mol.
Frequently asked questions
How do you calculate activation energy from two temperatures?
Ea = R × ln(k₂ ÷ k₁) ÷ (1/T₁ − 1/T₂), with R = 8.314 J/(mol·K) and temperatures in kelvin. A rate that quadruples from 25 °C to 45 °C gives Ea ≈ 54.7 kJ/mol.
Can I use reaction times instead of rate constants?
Yes, but invert them — the rate constant is inversely proportional to the time to reach the same conversion, so use k₁ = 1/t₁ and k₂ = 1/t₂.
Does the old rule "rate doubles every 10 °C" fit?
Only for activation energies around 50 kJ/mol near room temperature — that gives a factor of about 1.9 from 25 to 35 °C. Higher Ea means a steeper rise.
How do I predict the rate at another temperature?
Enter k₁, T₁, the new T₂ and the activation energy, and clear k₂ and the ratio — the calculator solves k₂ = k₁ × exp[(Ea ÷ R)(1/T₁ − 1/T₂)].