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The Arrhenius equation shows how a reaction's rate constant grows with temperature: k = A e^(−Ea/RT), where Ea is the activation energy. Comparing two temperatures removes the unknown factor A.
Enter the rate constant at one temperature, the activation energy and a second temperature to get the new rate constant. The default Ea of 52.9 kJ/mol is the value for which the rate roughly doubles between 25 °C and 35 °C — the famous 'rate doubles for every 10 °C' rule.
| Rate constant at T₁ (k₁) | 0.02 |
| Temperature T₁ | 25 °C |
| Temperature T₂ | 35 °C |
| Activation energy (Ea) | 52.9 kJ/mol |
| Rate constant at T₂ (k₂) | 0.0399744 |
| k₂ ÷ k₁ (rate increases by ×) | 1.99872 |
| T₁ (K) | 298.15 |
| T₂ (K) | 308.15 |
The rate constant rises to 0.0399744, about 1.99872 times faster.
Try this example →| Rate constant at T₁ (k₁) | 0.02 |
| Temperature T₁ | 25 °C |
| Temperature T₂ | 35 °C |
| Activation energy (Ea) | 100 kJ/mol |
| Rate constant at T₂ (k₂) | 0.0740559 |
| k₂ ÷ k₁ (rate increases by ×) | 3.7028 |
| T₁ (K) | 298.15 |
| T₂ (K) | 308.15 |
With Ea = 100 kJ/mol, the same 10 °C rise makes the reaction 3.7028 times faster.
Try this example →Rearranged, Ea = 2.303R × log(k₂/k₁) × T₁T₂ ÷ (T₂ − T₁). Measure k at two temperatures and you can find the activation energy. Plotting ln k against 1/T gives a straight line of slope −Ea/R.
k = A e^(−Ea/RT). It relates the rate constant to temperature and activation energy.
The minimum energy that colliding molecules need for a reaction to happen.
For typical activation energies around 50 kJ/mol, a 10 °C rise near room temperature about doubles the fraction of molecules with enough energy.
Yes. A catalyst provides a pathway with lower activation energy, so k increases.