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The rates of most reaction double when their temperature is raised from `298K` to `308K`. Calculate their activation energy. |
Answer» `2.303 "log" K_(2)/K_(1)=E_(a)/R[(T_(2)-T_(1))/(T_(1)T_(2))]` `K_(2)/K_(1)=2, T_(2) =308 K, T_(1)=298K` `:. 2.303 log 2=E_(a)/8.314xx10/(308xx298)` `E_(a)=52.903xx10^(3) J` or `E_(a)=52.903 kJ ` |
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