Arrhenius Activation Energy Calculator
Find the activation energy from two rate constants measured at two temperatures using the two-point Arrhenius equation.
How it works
ln(k₂ ÷ k₁) = −Eₐ/R × (1/T₂ − 1/T₁) · Eₐ = R·ln(k₂/k₁) ÷ (1/T₁ − 1/T₂) · R = 8.314 J·mol⁻¹·K⁻¹
The two-point form of the Arrhenius equation extracts the activation energy from how much a reaction speeds up between two temperatures. A larger jump in the rate constant for the same temperature rise means a higher activation energy. Temperatures must be in kelvin, and k₁ and k₂ can be in any consistent units because only their ratio enters the formula.
Worked example
A reaction whose rate constant rises from k₁ = 1 at 300 K to k₂ = 10 at 350 K has Eₐ = 8.314 × ln(10) ÷ (1/300 − 1/350) = 8.314 × 2.3026 ÷ 0.000476 = 40,200 J/mol ≈ 40.2 kJ/mol.
Sources: Arrhenius equation — Wikipedia
Frequently asked questions
How is this different from solving for k?
This version uses two measured rate constants at two temperatures to find the activation energy Eₐ, rather than starting from a known Eₐ and pre-exponential factor to predict a single k.
Why must temperatures be in kelvin?
The Arrhenius exponent uses 1/T on an absolute scale. Celsius temperatures would give wrong (and even negative) reciprocals and break the calculation.
Do the units of k matter?
Only the ratio k₂/k₁ appears, so as long as both are in the same units the units cancel and the activation energy comes out correctly.
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