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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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Answer» Solution :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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