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The rate of a reaction quadruples when the temperature changes from `293K` to `313K`. Calculate the energy of activation of the reaction assuming that it does not change with temperature. |
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Answer» Correct Answer - `E_(a)=52.8 mol^(-1)` Let `k` at `293K=k_(1),k at 313K=k_(2)` `:.k_(2)=4k_(1) or (k_(2))/(k_(1))=4` Using Arrhenius equation, `log``(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` `log4=(E_(a))/(2.303xx8.314J K^(-1)mol^(-1))xx((313-293))/(313xx293)` `0.606=(E_(a))/(2.303xx8.314)xx(10)/(313xx293)` Solvie for `E_(a):` `:. E_(a)=52860J mol^(-1)=52.86mol^(-1)` |
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