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easy chem,easy point

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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... 顯示更多 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

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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

其他解答:

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)7638E748CCC65837