Activation energy (Ea) is the energy barrier that reacting particles must get over before old bonds break and new ones form. Only collisions with at least this much energy can turn into products, so Ea controls how fast a reaction runs and how sensitive the rate is to temperature. A match does not burst into flame on its own because the combustion barrier is high; the friction of striking it supplies the energy for the first molecules, and the heat released then keeps the reaction going.
The Arrhenius equation k = A·e^(−Ea/RT) links the rate constant k to Ea, the absolute temperature T and the frequency factor A (the rate constant you would get if every collision had enough energy). Rearranged, it gives Ea = −RT·ln(k/A). In the lab A is often unknown, so the usual method is to measure k at two temperatures and use the two-point form, which cancels A.
This calculator handles both routes, converts the result between kJ/mol, J/mol, kcal/mol and eV per molecule, and also reports the Boltzmann factor (the fraction of collisions energetic enough to react) and how much faster the reaction would run 10 K warmer.
How to use the Activation Energy Calculator
- Calculate from
- Temperature T
- Rate constant k — For other reaction orders use any unit – it only has to match A.
- Frequency factor A (same unit as k)
- First temperature T₁
- Rate constant k₁ at T₁
- Second temperature T₂
- Rate constant k₂ at T₂
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Choose whether you know k and A at one temperature or k at two temperatures. The calculator converts temperatures to kelvin, applies Ea = −RT·ln(k/A) or Ea = R·ln(k₂/k₁)/(1/T₁ − 1/T₂), and in the two-point mode back-calculates A from the first point. k and A must use the same unit; for reactions that are not first order the concentration part of the unit cancels in the ratio.
Worked example
One temperature: a decomposition has k = 4.5 × 10⁻⁴ s⁻¹ at 600 K and A = 1.0 × 10¹³ s⁻¹. Then k/A = 4.5 × 10⁻¹⁷, ln(k/A) = −37.640, and Ea = −8.3145 × 600 × (−37.640) = 187 773 J/mol ≈ 187.8 kJ/mol. A 10 K rise would speed it up about 1.85 times.
Two temperatures: k₁ = 1.0 × 10⁻³ s⁻¹ at 300 K and k₂ = 8.0 × 10⁻³ s⁻¹ at 330 K. ln(k₂/k₁) = ln 8 = 2.0794 and 1/300 − 1/330 = 3.0303 × 10⁻⁴ K⁻¹, so Ea = 8.3145 × 2.0794 ÷ 3.0303 × 10⁻⁴ ≈ 57.06 kJ/mol, and A = k₁·e^(Ea/RT₁) ≈ 8.6 × 10⁶ s⁻¹.
How temperature sensitivity depends on Ea
| Ea (kJ/mol) | Rate increase 25 → 35 °C | Fraction of collisions with E ≥ Ea at 25 °C |
|---|---|---|
| 25 | 1.39× | 4.2 × 10⁻⁵ |
| 50 | 1.92× | 1.7 × 10⁻⁹ |
| 75 | 2.67× | 7.3 × 10⁻¹⁴ |
| 100 | 3.70× | 3.0 × 10⁻¹⁸ |
Inputs explained
| Field | What to enter |
|---|---|
| Calculate from | Choose one: k and frequency factor A at one temperature, Rate constants at two temperatures (two-point method) |
| Temperature T | Units: K, °C, °F |
| Rate constant k | Units: s⁻¹, min⁻¹, h⁻¹, ms⁻¹ (per millisecond). For other reaction orders use any unit – it only has to match A. |
| Frequency factor A (same unit as k) | Units: s⁻¹, min⁻¹, h⁻¹, ms⁻¹ (per millisecond) |
| First temperature T₁ | Units: K, °C, °F |
| Rate constant k₁ at T₁ | Units: s⁻¹, min⁻¹, h⁻¹, ms⁻¹ (per millisecond) |
| Second temperature T₂ | Units: K, °C, °F |
| Rate constant k₂ at T₂ | Units: s⁻¹, min⁻¹, h⁻¹, ms⁻¹ (per millisecond) |
What each result means
- Frequency factor A (from the first point) = [Rate constant k₁ at T₁]×exp([Activation energy Ea]×1000÷(8.314462618×[First temperature T₁]))
- Fraction of collisions energetic enough, e^(−Ea/RT) = exp(-[Activation energy Ea]×1000÷(8.314462618×[Reference temperature]))
- Rate increase for a 10 K rise = exp([Activation energy Ea]×1000÷8.314462618×(1÷[Reference temperature] – 1÷([Reference temperature] + 10)))
Formula
Ea = −R·T·ln(k ÷ A) · two-point form: Ea = R·ln(k₂ ÷ k₁) ÷ (1/T₁ − 1/T₂) · R = 8.314 J/(mol·K)
Worked example
Inputs — Calculate from: k and frequency factor A at one temperature; Temperature T: 600 K; Rate constant k: 0.00045 s⁻¹; Frequency factor A (same unit as k): 10,000,000,000,000 s⁻¹.
Results — Activation energy Ea = 187.7732 kJ/mol; Fraction of collisions energetic enough, e^(−Ea/RT) = 4.5000e-17; Rate increase for a 10 K rise = 1.853 ×; Barrier height = High – needs heating or a catalyst.
Key terms
- Activation energy (Ea)
- The minimum energy per mole needed for reactants to reach the transition state and turn into products.
- Frequency (pre-exponential) factor A
- The rate constant in the limit of every collision having enough energy; it reflects collision frequency and orientation.
- Boltzmann factor
- e^(−Ea/RT), the fraction of molecular collisions with energy at least Ea.
- Transition state
- The highest-energy arrangement of atoms along the reaction path.
Frequently asked questions
What is activation energy?
It is the minimum energy (per mole) that colliding particles need for a reaction to happen – the height of the energy barrier between reactants and the transition state.
How do I calculate activation energy from two temperatures?
Use Ea = R·ln(k₂/k₁) ÷ (1/T₁ − 1/T₂) with temperatures in kelvin. For k rising from 1.0 × 10⁻³ to 8.0 × 10⁻³ s⁻¹ between 300 K and 330 K, Ea ≈ 57.1 kJ/mol.
What activation energy doubles the rate for a 10 °C rise?
Around room temperature (25 → 35 °C) a rate doubling corresponds to Ea ≈ 53 kJ/mol – the origin of the old rule of thumb.
Can activation energy be negative?
For elementary steps no. A negative apparent Ea appears in some multistep reactions (for example when a pre-equilibrium shifts back on heating) or simply from swapped data.
What units is activation energy given in?
Usually kJ/mol or J/mol; older texts use kcal/mol (1 kcal = 4.184 kJ). Per molecule it is often quoted in eV (1 eV per particle = 96.485 kJ/mol).
How does a catalyst affect activation energy?
A catalyst opens another reaction path with a lower barrier. It lowers Ea for both the forward and reverse reactions, so equilibrium is reached faster but the equilibrium position is unchanged.
How do I get Ea from an Arrhenius plot?
Plot ln k against 1/T. The points fall on a straight line with slope −Ea/R, so Ea = −slope × 8.314 J/(mol·K).