Question

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

Answer 1

`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).`