Rate contant k_2 of a reaction at 310 Kis two time of its rate constan – InterviewSolution

1.	Rate contant k_2 of a reaction at 310 Kis two time of its rate constant k_1at 300 K.Calculate activation energy of the reaction.(log 2=0.3010,log 1=0)
Answer» SOLUTION :LOG `k_2/k_1=(EA)/(2.303R)((T_2-T_1)/(T_1T_2))` LOG2=`(Ea)/(2.303xx8.314)[((310-300))/(310xx300)] Ea=(0.3010xx2.303xx8.314xx300)/10=53598J`

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