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The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. |
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Answer» SOLUTION :Using the relation `"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`. In this CASE, `k_(2)=4k_(1)` `therefore "log"(4k_(1))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` Substituting the VALUES, we get `log 4=(E_(a))/(2.303x8.314)[(313-293)/(293xx313)] or 0.6021=(E_(a))/(19.147) xx (20)/(91709)` or `E_(a)=(0.6021xx19.147xx91709)/(20) or E_(a)=52862"J mol"^(-1)=52.862"kJ mol"^(-1)`. |
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