Relate a rate constant to temperature with k = A·e^(−Ea/RT): give any three of A, Ea (kJ/mol), T and k and solve for the fourth.
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From the article Arrhenius Equation — Finding Activation Energy from Rate Constants.
A reaction's rate constant doubles when the temperature rises from 300 K to 310 K. Find Ea.
k₂/k₁ = 2, so ln(k₂/k₁) = ln 2 = 0.6931.
1/T₁ − 1/T₂ = 1/300 − 1/310 = (310 − 300) ÷ (300 × 310) = 10 ÷ 93 000 = 1.0753 × 10⁻⁴ K⁻¹
Rearranging form 3: Ea = R × ln(k₂/k₁) ÷ (1/T₁ − 1/T₂)
Ea = 8.314 × 0.6931 ÷ (1.0753 × 10⁻⁴) = 5.7626 ÷ 0.00010753
Ea = 53 593 J mol⁻¹ ≈ 53.6 kJ mol⁻¹
So the familiar textbook statement "rate roughly doubles for every 10 °C rise" is a statement about reactions with Ea near 50–55 kJ mol⁻¹ around room temperature. It is a rule of thumb, not a law.
Worked in full in Arrhenius Equation — Finding Activation Energy from Rate Constants.
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