Reaction Rate & Arrhenius Calculator
Work with k = A·e^(−Ea/RT). Find the activation energy from rate constants at two temperatures, the rate constant at a new temperature, or fit an Arrhenius plot to your data; then use zero-, first- and second-order rate laws for concentrations, times and half-lives.
How to Use
- Choose what to work out: the activation energy from two points, k at a new temperature, the temperature for a target k, k from A and Ea, an Arrhenius plot of your own data, or a rate law and half-life.
- Enter rate constants in any unit (they only need to match each other) and temperatures in °C, K or °F; everything is converted to kelvin, and a temperature at or below absolute zero is rejected.
- Enter the activation energy in kJ/mol, J/mol or kcal/mol. The answer appears in the highlighted field; change its unit to convert it.
- For an Arrhenius plot, paste one temperature and rate constant per line, such as 300, 8.74e-4. The least-squares line through ln k against 1/T gives Ea from its slope and A from its intercept.
- For rate laws, pick zero, first or second order and either the time (to get the concentration left) or a target concentration (to get the time). The half-life is shown for every order.
Worked Example
Activation energy from two points. A rate constant doubles between 25 °C and 35 °C. In kelvin these are 298.15 K and 308.15 K, so 1/T₁ − 1/T₂ = 1.0884 × 10⁻⁴ K⁻¹. Ea = R ln(k₂/k₁) ÷ (1/T₁ − 1/T₂) = 8.314 × 0.6931 ÷ 1.0884 × 10⁻⁴ = 52.95 kJ/mol.
A first-order half-life. With k = 1.0 × 10⁻³ s⁻¹, t½ = ln 2 ÷ k = 693.1 s. After 1,000 s, [A] = 1.00 M × e^(−1.0) = 0.3679 M remains.
The common mistake: using °C instead of kelvin. Putting 25 and 35 straight into 1/T₁ − 1/T₂ gives 0.01143 and an activation energy of 504.3 J/mol, about 0.5 kJ/mol, a hundred times too small. The Arrhenius equation needs absolute temperature: with 298.15 K and 308.15 K the answer is 52.95 kJ/mol. A second trap is mixing kJ and J: R is 8.314 J/(mol·K), so Ea must be in J/mol inside the exponent.
Show Work
Formulas
Arrhenius and the Energy Barrier
Jacobus van ’t Hoff noticed in his 1884 book on chemical dynamics that the logarithm of a rate constant falls in a straight line against 1/T. In 1889 the Swedish chemist Svante Arrhenius explained it: only molecules with more than a threshold energy react, and the share of them grows as e−Ea/RT. Arrhenius won the 1903 Nobel Prize in Chemistry for his theory of electrolytic dissociation, and his rate equation is still the first thing measured for any new reaction.
Collision theory in the 1910s and transition-state theory, developed by Henry Eyring, Meredith Evans and Michael Polanyi in 1935, gave the pre-exponential factor A a physical meaning: how often molecules meet in the right orientation. Integrated rate laws were put on a firm footing earlier still: Ludwig Wilhelmy measured the first-order rate of sucrose inversion in 1850.
About This Tool
This calculator solves the Arrhenius equation for the activation energy, a rate constant or a temperature, works out k from A and Ea, and fits an Arrhenius plot to your own measurements by least squares, reporting Ea, A and R². Temperatures can be typed in °C, K or °F and are always converted to kelvin, so the classic °C mistake cannot happen. The rate-law mode covers zero-, first- and second-order reactions, giving the concentration after a time or the time to reach a concentration, with the half-life and a concentration–time graph.
The example rate constants and activation energies in the presets are illustrative numbers, not data for a particular reaction. Everything runs in your browser; nothing is sent anywhere.
Related tools: Gibbs Free Energy Calculator, Half-Life Calculator, and Chemical Equilibrium (ICE Table) Calculator.
Frequently Asked Questions
How do you calculate activation energy from two temperatures?
Use the two-point Arrhenius equation, ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), with T in kelvin. If a rate doubles between 25 °C (298.15 K) and 35 °C (308.15 K), Ea = 8.314 × ln 2 ÷ (1/298.15 − 1/308.15) = 52,949 J/mol, which is 52.95 kJ/mol.
Is it true that reaction rates double every 10 °C?
Only for activation energies near 53 kJ/mol at room temperature. From 25 °C to 35 °C the factor is 1.39 for Ea = 25 kJ/mol, 1.92 for 50, 2.67 for 75 and 3.70 for 100 kJ/mol. The rule is a rough guide for typical reactions, not a law.
How do I find a rate constant at a different temperature?
k₂ = k₁ × e^[(Ea/R)(1/T₁ − 1/T₂)]. With Ea = 50 kJ/mol and k₁ = 0.010 s⁻¹ at 25 °C, at 60 °C the exponent is 2.119, so k₂ = 0.010 × 8.323 = 0.08323 s⁻¹. To reach ten times the rate, 0.10 s⁻¹, the reaction needs 63.42 °C.
What is the half-life for zero, first and second order reactions?
Zero order: t½ = [A]₀ ÷ 2k. First order: t½ = ln 2 ÷ k, independent of concentration; k = 1.0 × 10⁻³ s⁻¹ gives 693.1 s. Second order: t½ = 1 ÷ (k[A]₀); 0.500 M with k = 0.20 M⁻¹s⁻¹ gives 10.0 s, and each later half-life is twice as long.
What does an Arrhenius plot show?
ln k plotted against 1/T is a straight line with slope −Ea/R and intercept ln A. For the example data from 300 K to 340 K the slope is −9,026 K, so Ea = 9,026 × 8.314 = 75.05 kJ/mol, and the intercept gives A = 1.02 × 10¹⁰ s⁻¹. A curved plot means the mechanism changes over the range.
How do I use the Reaction Rate & Arrhenius Calculator?
Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.
Does it cost anything or need an account?
No. The tool is completely free, there is no account to create, and it keeps working offline after the page first loads.
Is anything I type uploaded?
No. The tool works entirely on your device, so the values you enter never leave your browser.
Common Use Cases
Food storage
A spoilage reaction with an assumed Ea of 50 kJ/mol runs 4.61 times slower in a 4 °C fridge than at 25 °C.
Pressure cooking
With an assumed Ea of 100 kJ/mol, cooking at 120 °C instead of 100 °C speeds the reaction up 5.15 times.
Lab kinetics
Five rate constants from 300 K to 340 K fitted on an Arrhenius plot give Ea = 75.05 kJ/mol with R² = 0.999999.
Drug and reagent shelf life
A first-order decay with k = 1.0 × 10⁻³ s⁻¹ leaves 36.79% after 1,000 s and 12.5% after three half-lives, 2,079 s.
Second-order reactions
0.500 M of a reactant with k = 0.20 M⁻¹s⁻¹ falls to 0.100 M in 40.0 s.
Collision theory
At 500 K with Ea = 100 kJ/mol only 3.575 × 10⁻¹¹ of collisions have enough energy; with A = 1.0 × 10¹³ s⁻¹ that gives k = 357.5 s⁻¹.
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