solutions 153Short Answer: 10.Suppose for an uncatalyzed reaction, it is observed that the rate constant, k, at T= 300 K and T = 1000 K are 3.88×10−5s−1and 5.98×107s−1, respectively. a)Assuming the Arrhenius law holds for the rates, what are the activation energy and pre-exponential factor for the reaction? 11lnlnRTEAKa−=and 22lnlnRTEAKa−=Hence, ⎟⎟⎠⎞⎜⎜⎝⎛−−=121211lnTTREKKaEa= ()()00233333.06.28314.81000130011088.31098.5ln314.811ln572112=⎟⎠⎞⎜⎝⎛−××=⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛−TTKKR= 100 kJ. 11/1/1RTERTEaaeKAAeK=⇒=−= (3.88×10−5) ()()300314.8000,100e= 1.00×1013sec−1. −b)Suppose that the reaction is catalyzed at T = 500 K and it is found that kcat(500) = 4.40×107s−1. Assuming the catalytic mechanism is entirely due to lowering of activation energy, how much has the transition state energy been lowered? Solution: ln K= ln A −Ea/RTln 4.40×107= ln 1.00×1013- )500)(314.8('EaSolve this equation to get Ea’ = 51.3 kJ Therefore ∆Ea = Ea – Ea’ = 48.7 kJ c)At what temperature does the uncatalyzed reaction proceed at the same rate as the catalyzed reaction at T = 500 K? Solution: Kcat= ARXERTEacaAeKe//−−==caaacaEETXRXERTE=⇒=. Ea= 100 kJ and caE= 51.3kJ ()⎟⎠⎞⎜⎝⎛=kJ51.3kJ100K500oX= 975K. .9 K.