I think the answers is d, but not sure is it right or not. please correct me if i,m wrong.
1.)The rate constant for a reaction at 40.0 Celsius is exactly three times the rate constant at 20.0 Celsius. Calculate the activation energy for this reaction.
The Arrhenius equation is:
k = Ae-Ea/RT
which can be re-written as:
ln(k2/k1) = (Ea/R) ((1/T1) - (1/T2))
or
ln(k) = -(Ea/R)(1/T) + ln A
The activation energy, Ea =
a.)3.00 kJ/mol
b.)0.366 kJ/mol
c.)3.20 kJ/mol
d.)41.9 kJ/mol
e.)366 kJ/mol
easy chem,easy point
2010-04-03 6:48 am
回答 (2)
2010-04-03 8:59 am
✔ 最佳答案
The answer is (d).T1 = 273 + 40 = 313 K
T2 = 273 + 20 = 293 K
R = 8.314 J mol^-1 K^-1
k1/k2 = 3
ln(k1/k2) = (-Ea/R)[(1/T1) - (1/T2)]
ln(3) = (-Ea/8.314)[(1/313) - (1/293)]
Ea
= -(8.314)ln(3) ÷ [(1/313) - (1/293)]
= 41900 J mol^-1
= 41.9 kJ mol^-1
參考: 老爺子
2010-04-03 7:19 am
T1=20Celsius
T2=40Celsius
k2=3k1
so
ln(3) = Ea/R(1/20-1/40)
Ea= ln3 X 40 X 8.314 = ~366J/mol --> = 0.366kJ/mol = (B)
T2=40Celsius
k2=3k1
so
ln(3) = Ea/R(1/20-1/40)
Ea= ln3 X 40 X 8.314 = ~366J/mol --> = 0.366kJ/mol = (B)
收錄日期: 2021-05-01 01:01:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100402000016KK08528