Activation energy is the concept at the heart of chemical kinetics. It explains why reactions speed up with temperature, why enzymes make life possible, and why you can store petrol in a tank without it exploding: the reactant molecules need enough energy to climb over the barrier before products can form.
This calculator covers four interconnected tools:
- Arrhenius equation — solve for k, Eₐ, A, or T in any direction
- Two-point Eₐ extraction — find activation energy from rate constants measured at two temperatures
- First-order half-life — convert between k and t₁/₂ for radioactive decay, drug clearance, and unimolecular reactions
- Eyring–Polanyi — transition-state theory rate constant from ΔH‡ and ΔS‡
Every mode shows the full substitution so you can follow the arithmetic step by step.
How it works
The Arrhenius equation
The Arrhenius equation (1889) quantifies how the rate constant k depends on temperature:
k = A · exp(−Eₐ / (R·T))
where A is the pre-exponential factor (same units as k, often s⁻¹ for unimolecular reactions), Eₐ is the activation energy in J/mol, R = 8.314 J/(mol·K), and T is in Kelvin. Taking the natural log linearises the equation:
ln k = ln A − Eₐ/(R·T)
A plot of ln k versus 1/T is a straight line with slope −Eₐ/R — the classic Arrhenius plot.
Two-point method
Writing the Arrhenius equation at two temperatures and subtracting eliminates ln A:
ln(k₂/k₁) = −(Eₐ/R) · (1/T₂ − 1/T₁)
This is the standard undergraduate method: measure k at a low and a high temperature, plug both pairs in, and recover Eₐ without needing to know A at all.
First-order half-life
For any first-order reaction (or first-order process such as radioactive decay):
t₁/₂ = ln(2) / k ≈ 0.6931 / k
The half-life is independent of concentration — only k matters. Given t₁/₂ the reverse gives k = ln(2) / t₁/₂.
Eyring–Polanyi (transition-state theory)
Going deeper than the empirical Arrhenius equation, transition-state theory expresses k in terms of the activation enthalpy ΔH‡ and activation entropy ΔS‡:
k = (kB·T / h) · exp(−ΔG‡ / (R·T))
where ΔG‡ = ΔH‡ − T·ΔS‡ is the Gibbs energy of activation, kB = 1.381×10⁻²³ J/K, and h = 6.626×10⁻³⁴ J·s. The prefactor kB·T/h ≈ 6.25×10¹² s⁻¹ at 298 K sets the universal speed limit for any elementary step.
Worked example
Problem: The rate constant for the thermal decomposition of N₂O₅ is k = 3.46×10⁻⁵ s⁻¹ at 25 °C (298 K) and k = 1.35×10⁻³ s⁻¹ at 45 °C (318 K). Find Eₐ.
Using the two-point formula:
- ln(k₂/k₁) = ln(1.35×10⁻³ / 3.46×10⁻⁵) = ln(39.02) ≈ 3.664
- 1/T₂ − 1/T₁ = 1/318 − 1/298 = 3.145×10⁻³ − 3.356×10⁻³ = −2.11×10⁻⁴ K⁻¹
- Eₐ = −R · 3.664 / (−2.11×10⁻⁴) = 8.314 × 17 360 ≈ 144.3 kJ/mol
Enter k₁ = 3.46e-5, T₁ = 298, k₂ = 1.35e-3, T₂ = 318 in the Two-Point tab and the calculator gives the same answer with full working shown.
Formula reference
| Equation | Formula | Constants used |
|---|---|---|
| Arrhenius | k = A·exp(−Eₐ/(R·T)) | R = 8.314 J/mol·K |
| Two-point Eₐ | Eₐ = −R·ln(k₂/k₁)/(1/T₂ − 1/T₁) | R = 8.314 J/mol·K |
| First-order half-life | t₁/₂ = ln(2)/k | — |
| Eyring–Polanyi | k = (kB·T/h)·exp(−ΔG‡/(RT)) | kB = 1.381×10⁻²³ J/K, h = 6.626×10⁻³⁴ J·s |
| Gibbs of activation | ΔG‡ = ΔH‡ − T·ΔS‡ | — |
All calculations run entirely in your browser — no data is transmitted or stored.