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The rate of a chemical reaction doubles ...

The rate of a chemical reaction doubles for an increases of 10 K in absolute temperature from 298 K. Calculate` E_e`.

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log `k_2/k_1=E_1/(2.303R)[(T_2-T_1)/(T_1T_2)]`
Substituting we get ,log 2=`(E_axx10)/(2.303xx8.314xx10^(-3)xx298xx308)`
`thereforeE_a=(0.3010xx2.303xx8.314xx10^(-3)xx298xx308)/10=52089KJmol^(-1)`
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