Arrhenius Equation Calculator

Solve the Arrhenius equation for rate constant, activation energy, frequency factor, or temperature.

Supports exponential and logarithmic forms with per-mole and per-molecule bases, plus eight energy units.

Updated October 1, 2026

CreatorFrank Zhao

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What does this calculator do?

The Arrhenius equation links a chemical reaction's rate constant to temperature. This calculator solves it in every direction: give it any three of the rate constant $k$ , the frequency factor $A$ , the activation energy $E_a$ , and the absolute temperature $T$ , and it returns the fourth — then re-solves instantly if you change your mind about any input.

It handles both forms of the equation: the per-mole form that uses the gas constant $R$ , and the per-molecule form that swaps in the Boltzmann constant $k_B$ . It also offers the natural-logarithm version of each, which is what you want when you are reading a slope off an Arrhenius plot rather than plugging in a single temperature.

Typical users are chemistry students checking homework, lab researchers turning measured rates into activation energies, and anyone sizing how much faster a reaction runs after heating it. If you need just the energy barrier itself, there is also a dedicated Activation Energy Calculator.

How to use this calculator

1Pick an equation form at the top: per mole or per molecule, and exponential or ln. The fields below rearrange to match your choice.
2Fill in three of the four fields. Whatever you leave empty is the quantity the calculator solves for — it turns blue to show it is being derived.
3Check the unit selectors: energy defaults to kJ/mol (eV/molecule in the per-molecule form), temperature to °C, and $A$ and $k$ to the ×10⁰ scale, where you can pick any power of ten from ×10⁻³⁴ to ×10³⁰.
4Read the answer in the blue field. Change any input and everything re-solves.

Worked example — find k

A first-order reaction has $A = 1\times10^{10}\ \mathrm{s^{-1}}$ and $E_a = 50\ \mathrm{kJ/mol}$ . What is the rate constant at 26.85 °C?

T = 26.85 + 273.15

=

300\ \mathrm{K}

k = A\,e^{-E_a/(R\,T)}

=

10^{10}\,e^{-50000/(8.314\times300)}

\approx

19.70\ \mathrm{s^{-1}}

How to read the result

In the example above, a rate constant near 19.7 s⁻¹ means the reaction has a half-life of roughly $\ln 2 / k \approx 0.035\ \mathrm{s}$ — it is essentially complete in a fraction of a second. Larger $k$ always means a faster reaction. The units of $k$ follow the reaction order (s⁻¹ for first order, M⁻¹·s⁻¹ for second order), and $A$ always shares those units — the yellow note in the calculator spells out the pattern.

Formulas and the four equation forms

Svante Arrhenius proposed in 1889 that reaction rates rise with temperature because more collisions carry enough energy to clear the barrier. The rate constant is:

k = A\,e^{-E_a/(R\,T)}

where $k$ is the rate constant, $A$ is the frequency factor (how often collisions happen with the right geometry), $E_a$ is the activation energy, $T$ is the absolute temperature in kelvins, and $R$ is the molar gas constant, 8.314 462 618 J·mol⁻¹·K⁻¹ (exact).

On a per-molecule basis you replace $R$ with the Boltzmann constant, 1.380 649 × 10⁻²³ J·K⁻¹ (exact) and quote $E_a$ in joules per molecule:

k = A\,e^{-E_a/(k_B\,T)}

The two energies are related by Avogadro's number: $E_a^{\mathrm{molecule}} = E_a^{\mathrm{mol}} / N_A$ . So 50 kJ/mol is about $8.3\times10^{-20}$ J/molecule. The calculator keeps the two fields separate so you never mix the bases by accident.

The ln form and the Arrhenius plot

Taking the natural log turns the exponential into a straight line:

\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A

y = \ln k

,

m = -E_a/R

,

x = 1/T

,

c = \ln A

This is why the ln modes show $-E_a/R$ and $1/T$ as their own fields: the first is the plot's slope, the second its x-axis. Slope has units of kelvins here (the K suffix on $-E_a/R$ is not a temperature — it is J/mol ÷ J·mol⁻¹·K⁻¹). Enter $-E_a/R$ directly and the calculator back-solves $E_a$ for you; enter $E_a$ and the slope updates.

Every direction it solves

Exponential forms: solve for $k$ , $A$ , $E_a$ , or $T$ — four directions each.
ln forms: solve for $\ln k$ , $\ln A$ , the slope $-E_a/R$ , or $T$ .
Cross-conversions run in the background: $A \leftrightarrow \ln A$ , $k \leftrightarrow \ln k$ , $T \leftrightarrow 1/T$ , and $E_a \leftrightarrow -E_a/R$ .

Worked examples

Example 1 — recover the activation energy from a measured rate

A kinetic experiment gives $k = 0.5$ and you know $A = 1\times10^{10}$ at $T = 26.85\ ^\circ\mathrm{C}$ . In per-mole (exponential) form, fill those three fields and read $E_a$ :

E_a = -R\,T\,\ln\!\left(\frac{k}{A}\right)

=

-8.314\times300\times\ln(5\times10^{-11})

\approx

59.16\ \mathrm{kJ/mol}

That is a plausible barrier for a modest organic reaction. If the result comes out negative, don't panic — see the limitations section below.

Example 2 — build the Arrhenius plot

With $E_a = 50\ \mathrm{kJ/mol}$ and $A = 1\times10^{10}$ , switch Show Arrhenius plot? to Yes. The line has slope and intercept:

m = -E_a/R = -50000/8.314 \approx -6014\ \mathrm{K}

c = \ln A = \ln 10^{10} \approx 23.03

The Change graph options checkbox lets you set the x-range in 1/T units (K⁻¹). A default span of 0.01 to 0.1 K⁻¹ covers roughly 10 K to 100 K — widen or narrow it to zoom into the temperature window you care about.

Example 3 — per-molecule form

In per-molecule (exponential) form, with $A = 1000$ , $E_a = 1\times10^{-20}\ \mathrm{J/molecule}$ , and $T = 300\ \mathrm{K}$ :

k = 1000\,e^{-10^{-20}/(1.381\times10^{-23}\cdot300)}

\approx

89.4

Notice how tiny $E_a$ looks on a per-molecule basis — that same barrier is about 6.0 kJ/mol once you multiply back by Avogadro's number, just divided across individual molecules. The energy unit dropdown defaults to eV in this mode, where 1 eV ≈ 1.602 × 10⁻¹⁹ J.

Common mistakes to avoid

Temperature must be absolute. The formula needs kelvins. The calculator converts °C and °F for you, but when you check the math by hand, 26.85 °C is 300 K — feeding 26.85 into $e^{-E_a/(RT)}$ directly gives a wildly wrong answer.
Energy units default to kJ/mol, not J/mol. The internal math uses joules per mole, so switching the dropdown between J, kJ, MJ, Wh, kWh, ft-lb, kcal, or eV only changes how the number is displayed — the value itself never jumps when you change units.
Use the ×10ⁿ selector for A and k. Frequency factors are huge and rate constants can be tiny. Instead of typing 24 zeros, switch the scale to ×10⁹ and type 6.71. Both $A$ and $k$ share the same scale selector style.
k > A is a valid state. If the rate constant exceeds the frequency factor, the back-solved $E_a$ comes out negative. That is arithmetic, not an error — check the limitations section for when it is also physically real.
The plot range is in 1/T, not T. Start and End are reciprocal kelvins. If you set Start after End, the calculator tells you so under the field instead of drawing a backwards graph.

Frequently asked questions

QCan I enter temperature in Celsius or Fahrenheit?

Yes. The temperature field offers °C, °F, and K. Internally everything is computed in kelvins, so switching the unit only changes the display — your inputs and results keep their meaning.

QWhat are the units of A and k?

They are always the same as each other, and they follow the reaction order $n$ : M¹⁻ⁿ·s⁻¹. Zero order gives M·s⁻¹, first order gives s⁻¹, second order gives M⁻¹·s⁻¹. The calculator does not force a unit — it leaves that convention to you and states it in the note under the inputs.

QWhen should I use per mole vs per molecule?

Use per mole whenever your energy is quoted in kJ/mol or cal/mol — the everyday convention in chemistry. Switch to per molecule when you are working with molecular-scale energies (J/molecule, eV), for example when comparing with computational chemistry output. The equation is identical apart from $R$ versus $k_B$ .

QWhat does the ln mode give me that the exponential mode doesn't?

The ln form exposes the plot directly. If you have a slope from fitting $\ln k$ against $1/T$ in lab, type that slope into $-E_a/R$ and the calculator recovers $E_a$ — no manual multiplication by $-R$ needed.

QWhy does my activation energy come out negative?

Mathematically, it happens whenever $k > A$ . Physically, a negative apparent activation energy does occur — for example in enzyme-catalyzed reactions near their optimum temperature, or in some diffusion- controlled and multi-step mechanisms. The calculator accepts it; whether your specific measurement is trustworthy is a separate question (see limitations).

QCan the calculator plot my measured data points?

The plot draws the theoretical line implied by your $A$ and $E_a$ (or $\ln A$ and slope). To overlay experimental points, fit their $\ln k$ vs. $1/T$ trend first, then enter the resulting slope and intercept here — the line will pass through your data by construction.

Limitations

Single-barrier assumption. The equation assumes one elementary step with a temperature-independent activation energy. Reactions with parallel pathways, equilibria before the rate-determining step, or temperature-varying $E_a$ will show curvature on an Arrhenius plot, and a straight-line fit across the whole range will bias your result.
Deviations from Arrhenius behavior are real. Quantum tunneling, diffusion-limited reactions, and enzyme kinetics near the optimum temperature can produce negative apparent activation energies or curved plots. A negative $E_a$ from this calculator is arithmetically valid — confirm it matches your system's chemistry before relying on it.
The frequency factor is a catch-all. $A$ lumps together collision frequency and molecular orientation. Fitted values of $A$ and $E_a$ are strongly correlated — small errors in one inflate the other, so treat both as a pair rather than as independent truths.
Educational and estimation use. Results here support coursework, back-of-envelope checks, and engineering estimates. For safety- critical or publication-grade kinetic parameters, validate against primary literature and your own measurements under controlled conditions.

Curious how the same equation shows up outside the lab? The Crickets Chirping Thermometer uses the Arrhenius relationship to estimate outdoor temperature from chirp rates — a live demonstration of temperature-dependent kinetics.

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