The activation energy of an uncatalyzed reaction is 92kJ/mol . The addition of a catalyst lowers the activation energy to 65kJ/mol .
Assuming that the collision factor remains the same, by what factor will the catalyst
increase the rate of the reaction at 16
answer: kc/k = 7.6×104
Assuming that the collision factor remains the same, by what factor will the catalyst
increase the rate of the reaction at 132
answer: kc/k = 3000
請問怎樣得出這答案,,,求詳細過程,THANKS
化學(reaction+catalyst)
2014-12-13 10:00 pm
回答 (1)
2014-12-15 9:25 pm
✔ 最佳答案
Rate constant k = Ae^(Ea/RT) where A is the collision factor, Ea the activation energy, R is gas constant (8.314) and T in kelvin.At T=(273+16)=289
(kc/k) = Ae^(-65000/RT) / Ae^(-92000/RT)
ln(kc/k) = (-65000/RT) - (-92000/RT) = 27000/(8.314*289) = 11.2371428357
(kc/k) = 75897.79 => 7.6 x 10^4
At T=(273+132)=405
(kc/k) = Ae^(-65000/RT) / Ae^(-92000/RT)
ln(kc/k) = (-65000/RT) - (-92000/RT) =27000/(8.314*405) = 8.0186031593
(kc/k) = 3036.93 => 3 x 10^3
收錄日期: 2021-05-01 01:18:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141213000051KK00042