# What is activation energy?

Finds the activation energy Ea of a reaction from rate constants at two temperatures with the Arrhenius equation, the rate constant at a new temperature from Ea, or Ea from one rate constant and the pre-exponential factor A.

- Page: https://www.acalculator.org/chemistry/activation-energy-calculator
- JSON spec: https://www.acalculator.org/chemistry/activation-energy-calculator.json
- Version: 1cbdea238fe8

## Default answer

Example with the default inputs (Find Activation energy, Temperature T₁ 555 K, Rate constant k₁ 0.000000352, Temperature T₂ 781 K, Rate constant k₂ 0.0395): Ea = 185.431 kJ/mol.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Find | Ea from two rate constants, k at a new temperature, or Ea from one rate constant and A. |
| t1 | Temperature T₁ | The first temperature. |
| k1 | Rate constant k₁ | The rate constant at T₁, in any unit (the same for k₂). |
| t2 | Temperature T₂ | The second temperature. |
| k2 | Rate constant k₂ | The rate constant at T₂, in the same unit as k₁. |
| ea | Activation energy Ea (kJ/mol) | The activation energy in kilojoules per mole. |
| a | Pre-exponential factor A | The frequency factor A, in the unit of k. |
| k | Rate constant k | The rate constant at temperature T, in the unit of A. |
| t | Temperature T | The temperature of the rate constant. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The activation energy in kJ/mol, or the rate constant at T₂. |
| eaKj | Activation energy (kJ/mol) | Ea in kilojoules per mole. |
| eaJ | Activation energy (J/mol) | Ea in joules per mole. |
| kNew | Rate constant k₂ | The rate constant at T₂, in the unit of k₁. |
| factor | k₂ ÷ k₁ | How many times faster the reaction is at T₂ than at T₁. |
| aValue | Pre-exponential factor A | k₁ × e^(Ea ÷ RT₁), in the unit of k; left out beyond 10^±300. |
| lnA | ln A | The natural logarithm of A. |

## Method

Ea = R × ln(k₂ ÷ k₁) ÷ (1 ÷ T₁ − 1 ÷ T₂); k₂ = k₁ × e^((Ea ÷ R)(1 ÷ T₁ − 1 ÷ T₂)); Ea = R × T × ln(A ÷ k); A = k₁ × e^(Ea ÷ (R × T₁)); R = 8.31446261815324 J/(mol·K).

## Assumptions

- Ea and A do not change with temperature over the range used.
- k₁, k₂, k and A are in the same unit; temperatures are absolute (the page converts °C and °F to kelvins).

## Worked examples

1. find = ea, t1 = 555, k1 = 0, t2 = 781, k2 = 0.0395 gives eaKj = 185.431, lnA = 25.3245, result = Ea = 185.431 kJ/mol. Source: OpenStax, Chemistry 2e, §12.5 Collision Theory (k = Ae^(−Ea/RT); ln(k₁/k₂) = (Ea/R)(1/T₂ − 1/T₁)), https://openstax.org/books/chemistry-2e/pages/12-5-collision-theory, Example 12.13 Determination of Ea: 1.8 × 10⁵ J/mol (180 kJ/mol); NIST, CODATA value: molar gas constant R = 8.314 462 618… J/(mol·K), exact, https://physics.nist.gov/cgi-bin/cuu/Value?r.
2. find = ea, t1 = 650, k1 = 1.66, t2 = 700, k2 = 7.39 gives eaKj = 112.986, result = Ea = 112.986 kJ/mol. Source: OpenStax, Chemistry 2e, §12.5 Collision Theory (k = Ae^(−Ea/RT); ln(k₁/k₂) = (Ea/R)(1/T₂ − 1/T₁)), https://openstax.org/books/chemistry-2e/pages/12-5-collision-theory, Example 12.13 Check Your Learning: 1.1 × 10⁵ J/mol.
3. find = k2, ea = 50, t1 = 298.15, k1 = 1, t2 = 308.15 gives kNew = 1.92426, result = k₂ = 1.92426. Source: OpenStax, Chemistry 2e, §12.5 Collision Theory (k = Ae^(−Ea/RT); ln(k₁/k₂) = (Ea/R)(1/T₂ − 1/T₁)), https://openstax.org/books/chemistry-2e/pages/12-5-collision-theory: k₂ = k₁ × e^((Ea/R)(1/T₁ − 1/T₂)).
4. find = ea1, a = 10,000,000,000,000, k = 0.001, t = 300 gives eaKj = 91.8948, result = Ea = 91.8948 kJ/mol. Source: OpenStax, Chemistry 2e, §12.5 Collision Theory (k = Ae^(−Ea/RT); ln(k₁/k₂) = (Ea/R)(1/T₂ − 1/T₁)), https://openstax.org/books/chemistry-2e/pages/12-5-collision-theory: Ea = RT ln(A/k) from k = Ae^(−Ea/RT).

## FAQ

### How do I calculate activation energy from two rate constants?

Use the two-point Arrhenius equation: Ea = R × ln(k₂ ÷ k₁) ÷ (1/T₁ − 1/T₂), with T in kelvins and R = 8.314 J/(mol·K). For N₂O₅: k = 1.66 at 650 K and 7.39 at 700 K gives Ea = 8.314 × ln(4.452) ÷ (1/650 − 1/700) = 1.13 × 10⁵ J/mol, or 113 kJ/mol.

### What is the activation energy formula?

From the Arrhenius equation k = A e^(−Ea/RT): Ea = RT ln(A ÷ k) for one temperature with A known, or Ea = R ln(k₂/k₁) ÷ (1/T₁ − 1/T₂) for two temperatures. A plot of ln k against 1/T is a line of slope −Ea/R.

### Why is my answer different from OpenStax’s 180 kJ/mol for HI?

OpenStax rounds 1/T to 1.80 × 10⁻³ and 1.28 × 10⁻³ K⁻¹, so the difference has only two figures (0.52 × 10⁻³), and it gives 1.8 × 10⁵ J/mol. With the unrounded 1/555 − 1/781 = 5.214 × 10⁻⁴ K⁻¹, the same data give 185.4 kJ/mol.

### How do I find the rate constant at another temperature?

Rearrange the two-point equation: k₂ = k₁ × e^((Ea/R)(1/T₁ − 1/T₂)). For Ea = 50 kJ/mol, warming from 25 °C to 35 °C multiplies k by e^(6013.62 × 1.08844 × 10⁻⁴) = 1.92, close to the rule of thumb that rates about double for each 10 °C.

### Do the rate constants need particular units?

No, only the same unit for both, because the formula uses their ratio. A has the unit of k. Temperatures must be absolute: the page converts °C and °F to kelvins exactly.

### Can activation energy be negative?

Elementary steps have Ea of 0 or more. If your rate constant falls as the temperature rises, the formula gives a negative Ea, which points to a more complex mechanism or a data error; the page shows it rather than hiding it.

## Sources

- OpenStax, Chemistry 2e, §12.5 Collision Theory: the Arrhenius equation k = Ae^(−Ea/RT), ln k = (−Ea/R)(1/T) + ln A, the two-point form ln(k₁/k₂) = (Ea/R)(1/T₂ − 1/T₁), Example 12.13 (HI: rate constants from 555 K to 781 K, Ea = 1.8 × 10⁵ J/mol) and its Check Your Learning (N₂O₅: 1.66 at 650 K, 7.39 at 700 K, 1.1 × 10⁵ J/mol). https://openstax.org/books/chemistry-2e/pages/12-5-collision-theory (retrieved 2026-10-19)
- NIST, CODATA value: molar gas constant R = 8.314 462 618… J/(mol·K), exact (N_A × k). https://physics.nist.gov/cgi-bin/cuu/Value?r (retrieved 2026-10-19)