Help, how do you solve this?
2019-02-13 12:55 pm
A certain reaction has an activation energy of 65.25 kJ/mol. At what Kelvin temperature will the reaction proceed 7.00 times faster than it did at 303 K?
回答 (1)
2019-02-13 4:21 pm
✔ 最佳答案
Activation energy: Eₐ = 65.25 kJ/mol = 65250 J/molGas constant, R = 8.314 J/(mol K)
At T₁ = 303 K, k₁ = kₒ
At T₂ = ?, k₂ = 7kₒ
Arrhenius equation: ln(k₁/k₂) = (Eₐ/R) × [(1/T₂) - (1/T₁)]
ln[kₒ/(7kₒ )] = (65250/8.314) × [(1/T₂) - (1/303)]
(1/T₂) - (1/303) = (8.314/65250) ln(1/7)
(1/T₂) = (1/303) + (8.314/65250) ln(1/7)
The required temperature, T₂ = 1 / [(1/303) + (8.314/65250) ln(1/7)] = 328 K (to 3 sig. fig.)
收錄日期: 2021-05-01 22:28:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190213045545AAWXLmC