The rate constant for a reaction at 230 ⁰C is found to be exactly twice the value at 220 ⁰C. Calculate the activation energy.?
回答 (1)
2020-10-12 6:11 pm
Gas constant, R = 0.008314 kJ / (mol K)
When T₁ = (273 + 220) K = 493 K, k₁ = k
When T₂ = (273 + 230) K = 503 K, k₂ = 2k
Arrhenius equation: ln(k₁/k₂) = (Eₐ/R) [(1/T₂) - (1/T₁)]
ln(1/2) = (Eₐ/0.008314) [(1/503) - (1/493)]
Eₐ = ln(1/2) × 0.008314 / [(1/503) - (1/493)] kJ/mol
Activation energy, Eₐ = 143 kJ/mol
When T₁ = (273 + 220) K = 493 K, k₁ = k
When T₂ = (273 + 230) K = 503 K, k₂ = 2k
Arrhenius equation: ln(k₁/k₂) = (Eₐ/R) [(1/T₂) - (1/T₁)]
ln(1/2) = (Eₐ/0.008314) [(1/503) - (1/493)]
Eₐ = ln(1/2) × 0.008314 / [(1/503) - (1/493)] kJ/mol
Activation energy, Eₐ = 143 kJ/mol
收錄日期: 2021-04-24 08:03:24
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