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At 27^(@)C and 37^(@)C , the rates ofa reactionare given as 1.6 xx 10^(-2) mol L^(-1)s^(-1)and 3.2 xx 10^(-2)mol L^(-1)s^(-1) . Calculate the energy ofactivationfor thegivenreaction . |
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Answer» SOLUTION :The ration of specific rates at two DIFFERENT temperatures are gives as, `"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` Substituting the VALUES `k_(1),k_(2),T_(1),T_(2)` and molar GAS constant R , `0.301=(E_(a)xx1000)/(2.303xx8.314)xx(10)/(300xx310)` Energy of activation `=E_(a)=53" kJ mol"^(-1)`. |
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