easy chem,easy point

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/RTwhich can be re-written as:ln(k2/k1) = (Ea/R) ((1/T1) - (1/T2))orln(k) = -(Ea/R)(1/T) ln AThe activation energy

Ea =a.)3.00 kJ/molb.)0.366 kJ/molc.)3.20 kJ/mold.)41.9 kJ/mole.)366 kJ/mol
The answer is (d).T1 = 273 40 = 313 KT2 = 273 20 = 293 KR = 8.314 J mol^-1 K^-1k1/k2 = 3ln(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 參考資料 老爺子
T1=20CelsiusT2=40Celsiusk2=3k1soln(3) = Ea/R(1/20-1/40)Ea= ln3 X 40 X 8.314 = ~366J/mol --