Chemistry and laboratory
Arrhenius equation calculator
Activation energy from two rate constants, a rate constant at a new temperature, or k from A and Ea, with Q10, using k = A e^(−Ea/RT).
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Result
- Activation energy Ea
- 52.9 kJ/mol
- Pre-exponential factor A (unit of k)
- 1.89 × 10⁹
- Q₁₀ (rate change per 10 K)
- 2.00
Working
- ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)
- Ea = R ln(k₂/k₁) ÷ (1/T₁ − 1/T₂) = 8.314462618 × ln(2/1) ÷ (1/298.15 − 1/308.15) = 52948.9 J/mol
- A = k₁ e^(Ea/RT₁) = 1.89 × 10⁹
- Assumes Ea and A do not change over the temperature range (a straight Arrhenius plot). Rate constants can be in any unit, as long as both use the same one; A has the unit of k.
- Rounded to 3 significant figures.
Method and limits
Choose what to find. Temperatures may be in °C, K or °F; they are converted to kelvin. Rate constants can be in any unit as long as both use the same one.
k = A e^(−Ea ÷ RT) ln(k₂ ÷ k₁) = (Ea ÷ R)(1/T₁ − 1/T₂)
Ea = R ln(k₂ ÷ k₁) ÷ (1/T₁ − 1/T₂) Q₁₀ = (k₂ ÷ k₁)^(10 K ÷ (T₂ − T₁))
R = 8.314 462 618 J mol⁻¹ K⁻¹ is exact. The two-point form needs no value of A; once Ea is known, A = k₁ e^(Ea/RT₁). Q₁₀ is the factor by which the rate changes for a 10 K rise, a common way to describe biological temperature sensitivity.
Limits
The Arrhenius equation treats Ea and A as constant. Over wide temperature ranges, or for enzymes that denature, the plot of ln k against 1/T curves and two points can mislead; fit several points instead. Answers keep the significant figures of the least precise number you enter (at least 3, at most 6).
Worked example
A reaction twice as fast at 35 °C as at 25 °C: Ea = 8.314 × ln 2 ÷ (1/298.15 K − 1/308.15 K) = 52 900 J/mol = 52.9 kJ/mol, the origin of the rule of thumb that rates double every 10 °C near room temperature.
References
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