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The rate of a reaction qudruples when the temperature changes from 293 K to 313 K.Calculation the energy of activation of the reaction assuming that it does not change with temperature. |
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Answer» SOLUTION :`T-(1)`=293 K rate=`r_(1)` means `k_(1)` and `E_(a)`=(?) `T_(2)`=313K rate =`r_(2)=4r_(1)` `k_(2)=4k_(1)` log `(K_(2))/(k_(1))`=log `(4k_(1))/(k_(1))` =log 4 log `(k_(2))/(k_(1))=(E_(a))/(2.303xx8.314)((313-293)/(313xx293))` `therefore 0.6021=(E_(a))/(2.303xx8.314)((20)/(313xx293))` `therefore E_(a)=(0.6021xx2.303xx8.314xx313xx293)/(20)` |
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