What is activation energy?
Enter a reaction’s rate constant at two temperatures, and the calculator gives its activation energy from the two-point Arrhenius equation, with the pre-exponential factor A. It also predicts the rate constant at a new temperature from Ea, and finds Ea from one rate constant and A.
- Answer
- Ea = 185.431 kJ/mol
Ea = 185.431 kJ/mol.
- Activation energy (kJ/mol)
- 185.431
- Activation energy (J/mol)
- 185,431
- k₂ ÷ k₁
- 112,216
- Pre-exponential factor A
- 99,604,600,000
- ln A
- 25.3245
Answer: Ea = 185.431 kJ/mol. Ea = 185.431 kJ/mol.
How to calculate
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.
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.
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).
- 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
Each example is checked against the calculator on every build.
- Find Activation energy, Temperature T₁ 555 K, Rate constant k₁ 0, Temperature T₂ 781 K, Rate constant k₂ 0.0395 gives Activation energy (kJ/mol) 185.431, ln A 25.3245, Answer 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
- Find Activation energy, Temperature T₁ 650 K, Rate constant k₁ 1.66, Temperature T₂ 700 K, Rate constant k₂ 7.39 gives Activation energy (kJ/mol) 112.986, Answer 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
- Find k at a new T, Activation energy Ea (kJ/mol) 50, Temperature T₁ 298.2 K, Rate constant k₁ 1, Temperature T₂ 308.2 K gives Rate constant k₂ 1.92426, Answer 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))
- Find Ea from A, Pre-exponential factor A 10,000,000,000,000, Rate constant k 0.001, Temperature T 300 K gives Activation energy (kJ/mol) 91.8948, Answer 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)
How it works
Arrhenius equation (OpenStax §12.5). k = A e^(−Ea ÷ RT), with T in kelvins and R = 8.31446261815324 J/(mol·K) exactly (NIST). OpenStax rounds R to 8.314.
- Activation energy from two points: Ea = R × ln(k₂ ÷ k₁) ÷ (1 ÷ T₁ − 1 ÷ T₂). The page also gives the pre-exponential factor, A = k₁ × e^(Ea ÷ (R × T₁)), with ln A = ln k₁ + ln(k₂ ÷ k₁) × T₂ ÷ (T₂ − T₁), and k₂ ÷ k₁.
- Rate constant at a new temperature: k₂ = k₁ × e^((Ea ÷ R)(1 ÷ T₁ − 1 ÷ T₂)), with Ea typed in kJ/mol and turned into J/mol (× 1,000). The page also gives k₂ ÷ k₁.
- Activation energy from A: Ea = R × T × ln(A ÷ k).
In the two activation energy modes, Ea is shown in kJ/mol and J/mol.
Exact values. Temperatures are read back as the exact decimals typed (°C + 273.15, °F converted exactly), and 1 ÷ T₁ − 1 ÷ T₂ = (T₂ − T₁) ÷ (T₁ × T₂) is worked exactly. The logarithm and powers of e are double-precision numbers.
Rules the page checks.
- Temperatures from 1 K to 100,000 K; rate constants and A from 10⁻¹⁰⁰ to 10¹⁰⁰; Ea from −10,000 to 10,000 kJ/mol.
- The two temperatures must differ, or there is no answer.
- A k₂ above 10³⁰⁰ or below 10⁻³⁰⁰ has no answer. A is left out when it is above 10³⁰⁰ or below 10⁻³⁰⁰ (|ln A| above 300 × ln 10); ln A is always shown. k₂ ÷ k₁ is left out in the new-temperature mode when |(Ea ÷ R)(1 ÷ T₁ − 1 ÷ T₂)| is 700 or more.
How results are written. The sentence reads “Ea = 185.431 kJ/mol” or “k₂ = 1.92426”: 6 significant figures, trailing zeros dropped, thousands commas, the minus sign −. Ea is written m × 10ⁿ below 0.000001 or from 10²¹; k₂ is written m × 10ⁿ below 0.001 or from 1,000,000 (judged before rounding) and plainly between. The rows write a number below 0.000001 or from 10¹⁵ as m × 10ⁿ (ln A = 2 × 10⁻⁸ for k₁ = 1 at 300 K and k₂ = 1.00000001 at 600 K).
Assumptions. Ea and A stay the same over the temperature range; k₁, k₂, k and A share one unit.
Worked examples by hand
HI (OpenStax Example 12.13, first and last rows). ln(0.0395 ÷ 3.52 × 10⁻⁷) = 11.6282; 1/555 − 1/781 = 5.21392 × 10⁻⁴ K⁻¹. Ea = 8.31446261815324 × 11.6282 ÷ 5.21392 × 10⁻⁴ = 185.431 kJ/mol (OpenStax, with 1/T rounded: 1.8 × 10⁵ J/mol). ln A = ln(3.52 × 10⁻⁷) + 11.6282 × 781 ÷ 226 = 25.3245, so A = 9.96046 × 10¹⁰ L/(mol·s).
N₂O₅ (Check Your Learning). ln(7.39 ÷ 1.66) = 1.49331; 1/650 − 1/700 = 1.0989 × 10⁻⁴ K⁻¹. Ea = 8.31446261815324 × 1.49331 ÷ 1.0989 × 10⁻⁴ = 112.986 kJ/mol (OpenStax: 1.1 × 10⁵ J/mol).
k at 35 °C for Ea = 50 kJ/mol. T₁ = 298.15 K, T₂ = 308.15 K: (50,000 ÷ 8.31446261815324) × (1/298.15 − 1/308.15) = 6,013.62 × 1.08844 × 10⁻⁴ = 0.654544; k₂ = 1 × e^0.654544 = 1.92426.
Ea from A. A = 10¹³, k = 0.001 at 300 K: Ea = 8.31446261815324 × 300 × ln(10¹⁶) = 91.8948 kJ/mol.
Other questions people ask
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.