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Rate contant k2 of a reaction at 310 K ...

Rate contant `k_2` of a reaction at 310 K is two time of its rate constant `k_1`at 300 K.Calculate activation energy of the reaction.(log 2=0.3010,log 1=0)

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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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