Correct Answer - A
According to Arrhenius equation,
`"log"(k_(2))/(k_(1))E_(a)/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`k_(2)/(k_(1))=2,T_(1)=295k`
`T_(2)=305k`
`R=8.314Jk^(-1)mol^(-1)`
`:." "log2=(E_(a))/(2.303xx(8.314Jk^(-1)mol^(-1)))[(1)/(295k)-(1)/(305k)]`
`E_(a)=(0.3010xx2.303xx8.314xx295xx305)/(10)(Jmol^(-1))`
`=51855Jmol^(-1)=51.855kJmol^(-1)`