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The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_(a)) of the reaction assuming that it does not change with temperature.""[R=8.314JK^(-1)"mol"^(-1), log 4=0.6021] |
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Answer» Solution :`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` Substituting the values, we have `log 4=(E_(a))/(2.303xx8.314JK^(-1)"mol"^(-1))[(313K-293K)/(293K XX 313K)]` or `0.6021=(E_(a))/(19.147JK^(-1)"mol"^(-1))xx(20)/(91709K)` or `E_(a)=(0.6021xx19.147xx91709)/(20)"J mol"^(-1)=52863"J mol"^(-1)=52.86"kJ mol"^(-1)` |
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