Arrhenius Equation Calculator
Solve the Arrhenius equation for the rate constant k or the activation energy Ea from temperature.
Exponential factor e^(−Ea ÷ RT) = 7.132 × 10⁻¹⁴. k carries the same units as A.
The Arrhenius equation is k = A·e^(−Ea ÷ RT), linking a reaction’s rate constant to temperature. With A = 1×10¹³, Ea = 75 kJ/mol (75 000 J/mol), R = 8.314 and T = 298 K, the exponent is −30.27, so e^(−30.27) = 7.13×10⁻¹⁴ and k = 1×10¹³ × 7.13×10⁻¹⁴ ≈ 0.71 s⁻¹.
What the Arrhenius equation describes
Reactions speed up when you heat them, and the Arrhenius equation quantifies that dependence. It says the rate constant k equals a pre-exponential factor A (how often reactant molecules collide with the right orientation) multiplied by an exponential term, e^(−Ea ÷ RT). That exponential is the fraction of collisions energetic enough to clear the activation barrier Ea. Raising the temperature T or lowering Ea makes the exponent less negative, so a larger fraction of collisions succeed and k grows.
Ea in J/mol · R = 8.314 J·mol⁻¹·K⁻¹ · T in kelvin · k shares the units of A
Worked example
A first-order reaction has A = 1×10¹³ s⁻¹ and an activation energy of 75 kJ/mol. Find k at 298 K (25 °C).
- 1 Convert Ea to J/mol. Multiply by 1000: 75 kJ/mol × 1000 = 75 000 J/mol, so the units match R (J·mol⁻¹·K⁻¹).
- 2 Use kelvin for temperature. T = 25 °C + 273.15 ≈ 298 K. The equation only works with absolute temperature.
- 3 Compute the exponent −Ea ÷ (R·T). −75 000 ÷ (8.314 × 298) = −75 000 ÷ 2477.6 = −30.27.
- 4 Take the exponential. e^(−30.27) = 7.13×10⁻¹⁴ — the fraction of collisions with enough energy.
- 5 Multiply by A. k = 1×10¹³ × 7.13×10⁻¹⁴ ≈ 0.71 s⁻¹.
How k rises with temperature (A = 1×10¹³, Ea = 75 kJ/mol)
Same reaction, warmed step by step. A rise of a few tens of degrees can multiply k several-fold — the origin of the “≈ doubling every 10 °C” rule of thumb.
| Temperature | −Ea ÷ (R·T) | e^(−Ea ÷ RT) | k (s⁻¹) |
|---|---|---|---|
| 298 K (25 °C) | −30.27 | 7.13×10⁻¹⁴ | 0.71 |
| 308 K (35 °C) | −29.29 | 1.91×10⁻¹³ | 1.91 |
| 318 K (45 °C) | −28.37 | 4.79×10⁻¹³ | 4.79 |
| 350 K (77 °C) | −25.77 | 6.40×10⁻¹² | 64.0 |
Higher T, lower Ea — and the two-point form
Higher temperature and lower activation energy both speed a reaction up. Both make the exponent −Ea ÷ RT less negative, so e^(−Ea ÷ RT) — the fraction of sufficiently energetic collisions — grows and k with it. A catalyst works by lowering Ea, which is why it can accelerate a reaction without adding heat.
The gas constant R. Use R = 8.314 J·mol⁻¹·K⁻¹ whenever Ea is in J/mol, which is why the calculator converts your kJ/mol entry by ×1000. Mixing kJ with the J-based R is the most common mistake.
Two-point form. If you know k at two temperatures you can find Ea without A: ln(k₂ ÷ k₁) = −(Ea ÷ R)(1 ÷ T₂ − 1 ÷ T₁). This tool uses the single-point form, solving for k from A, Ea and T, or for Ea from a measured k, A and T.