Rate constant of a reaction changes by `2%` by `0.1^(@)C` rise in temperataure at `25^(@)C`. The standard heat of reaction is `12.1kJ"mol"^(-1)` . Calculate `E_(a)` or reverse reaction.

Answer

`24.7kJ//"mol"`

Explanation

`log.((k_(2))/(k_(1)))=(E_(a))/(2.303RT)((1)/(T_(1))-(1)/(T_(2)))`
`log.(102)/(100)=(E)/(2.303xx8.314)((1)/(298)-(1)/(298.1))`
`E=1.463xx10^(5)J//"mol"=146.kJ//"mol"`
`Delta=E_(f)-E_(b)`
`131.6=146.3-E_(b)`
`E_(b)=24.7 kJ//"mol"`