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a) The rate of a particular reaction doubles when the temperature changes from 300 K to 310 K. Calculate the energy of activation of the reaction. [Given : R="8.314 JK"^(-1)" mol"^(-1)]. |
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Answer» Solution :a) `log""(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `log""(2k_(1))/(k_(1))=(E_(a))/(2.303xx8.314)[(300-310)/(300xx310)]` `log2=(E_(a))/(19.1471)[(10)/(93000)]` `E_(a)=53598J or 53.598kJ` |
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