The rate constant of a reaction is doubled when the temperature is raised from 298 K to 308 K. Calculate the activation energy.

The rate constant of a reaction is doubled when the temperature is rai

1.	The rate constant of a reaction is doubled when the temperature is raised from 298 K to 308 K. Calculate the activation energy.
Answer» `E_(a)=?k_(2)=2k_(1),T_(1)=298K,T_(2) =308K` `log ""(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` `log ""(k_(2))/(k_(1))=(E_(a))/(2.303xx8.314)((10)/(298xx308))` `0.3010 =(E_(a))/(2.303xx8.314)xx(10)/(298xx308)` `E_(a)=(0.3010xx2.303xx8.314xx298xx308)/(10)=52897.7 m ol ^(-1).` `E_(a)=52.8977kJ mol ^(-1).`

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The rate constant of a reaction is doubled when the temperature is raised from 298 K to 308 K. Calculate the activation energy.

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