Arrhenius Equation Calculator

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How to Use the Arrhenius Equation Calculator

1. 1. Choose the mode. Use the first to apply k = Ae^(−Ea/RT) directly, the second when you have rate constants at two temperatures, the third for "how much faster at 10 °C warmer?", and the fourth to fit several measurements.
2. 2. Pick the unknown and enter the rest. Activation energy can be in kJ/mol, J/mol, kcal/mol or cal/mol, and temperature in K, °C or °F. Rate constants take any unit, and A uses the same one; type it in the label box if you want it shown.
3. 3. Check the temperature. It is converted to kelvin first, and a value at or below absolute zero is rejected.
4. 4. Read the working. Each line is shown with your numbers and R = 8.314 462 618 J/(mol·K), so you can compare it with your own solution.

The Arrhenius Equation

k = A × e^(−Ea / RT)

ln k = ln A − (Ea / R)(1 / T)

ln(k₂ / k₁) = (Ea / R)(1 / T₁ − 1 / T₂)

k is the rate constant, A the pre-exponential (frequency) factor, Ea the activation energy in J/mol, R the gas constant and T the absolute temperature in kelvin. The term e^(−Ea/RT) is the fraction of molecular collisions that carry at least the activation energy, so it is always between 0 and 1. That is why k can never exceed A, and why the calculator reports an error if you ask for an Ea from a k larger than A.

The linear form is what makes the equation practical. Plot ln k against 1/T for measurements at several temperatures and you get a straight line with slope −Ea/R and intercept ln A (OpenStax Chemistry 2e, chapter 12). The gas constant is exact since the 2019 SI redefinition: R = 8.314 462 618… J/(mol·K) (NIST), and the calculator uses it in full rather than 8.314 or 8.31.

Worked Examples

k from A and Ea (the default inputs). With A = 1.00 × 10¹³ s⁻¹, Ea = 75.0 kJ/mol and T = 25.0 °C (298.15 K), Ea ÷ RT = 75 000 ÷ (8.314 462 618 × 298.15) = 30.3, so k = 1.00 × 10¹³ × e^(−30.25) = 0.725 s⁻¹. Only 7.25 × 10⁻¹⁴ of collisions are energetic enough.

Ea from two temperatures. Decomposing N₂O₅, k = 1.66 L mol⁻¹ s⁻¹ at 650 K and 7.39 L mol⁻¹ s⁻¹ at 700 K (the data in an OpenStax check-your-learning problem). Ea = R ln(7.39 ÷ 1.66) ÷ (1/650 − 1/700) = 8.314 462 618 × 1.49 ÷ 1.099 × 10⁻⁴ = 113 kJ/mol, and the pre-exponential factor comes out at 1.99 × 10⁹ L mol⁻¹ s⁻¹.

Rate change per 10 °C. For Ea = 50.0 kJ/mol starting at 25.0 °C, k₂ ÷ k₁ = e^[(50 000 ÷ 8.314 462 618)(1/298.15 − 1/308.15)] = 1.92. The widely quoted "rate doubles every 10 °C" corresponds to an Ea of about 53 kJ/mol near room temperature. A reaction with 25 kJ/mol speeds up only 1.39-fold, and one with 100 kJ/mol 3.70-fold.

An Arrhenius plot. Fitting the five HI-decomposition measurements from the same OpenStax chapter (555–781 K) gives a line with slope −22 287 K, so Ea = 185 kJ/mol, A = 8.83 × 10¹⁰ L mol⁻¹ s⁻¹ and R² = 0.9996. The textbook's rounded graphical estimate is 180 kJ/mol; a least-squares fit uses every point and is less sensitive to rounding.

Common Mistakes

• Using Celsius for T. The equation needs kelvin. A 10 °C step is 10 K, but 25 °C is 298.15 K, not 25.
• Mixing kJ and J. Ea in kJ/mol must be divided into R in J/(mol·K) only after converting it to joules. The calculator converts from the unit menu.
• Swapping the pairs. Keep each rate constant with its own temperature. Reversing the order of both pairs gives the same Ea, but mixing them gives the wrong sign.
• Rounding R or 1/T early. 1/T₁ − 1/T₂ is a small difference of large numbers; rounding either value to three figures can shift Ea by several kJ/mol.
• Trusting two points. Two measurements always fit a line exactly. Three or more temperatures let you see whether the data are truly Arrhenius.

Frequently Asked Questions

What is the Arrhenius equation?

k = A·e^(−Ea/RT). It relates a reaction's rate constant k to the absolute temperature T through the activation energy Ea and the pre-exponential factor A. Raising the temperature or lowering Ea increases k.

How do you calculate activation energy from two temperatures?

Use Ea = R × ln(k₂/k₁) ÷ (1/T₁ − 1/T₂), with T in kelvin and R = 8.314 J/(mol·K). The answer comes out in J/mol; divide by 1000 for kJ/mol. For rate constants of 1.66 and 7.39 at 650 K and 700 K, Ea is about 113 kJ/mol.

How much does the rate increase for every 10 °C rise?

It depends on Ea and on the starting temperature. A common rule of thumb is a doubling or tripling, which matches Ea of roughly 50–80 kJ/mol near 25 °C. At higher temperatures the same Ea gives a smaller factor, since 1/T changes less per degree.

What are the units of the pre-exponential factor?

The same as the rate constant. For a first-order reaction that is s⁻¹; for a second-order reaction L mol⁻¹ s⁻¹. The exponential part is dimensionless, so A carries all the units.

Can the activation energy be negative?

Rarely. A negative apparent Ea means k decreases with temperature, which happens in some barrierless radical reactions and multi-step mechanisms. In an introductory problem it almost always signals that the data were entered in the wrong order.

Why plot ln k against 1/T?

Taking the natural log of the Arrhenius equation turns it into a straight line, ln k = ln A − (Ea/R)(1/T). The slope gives Ea, the intercept gives A, and the scatter shows whether the reaction is following one mechanism over the temperature range.