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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. |
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Answer» Correct Answer - `24.7kJ//"mol"` `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"` |
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