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The rate constant of a first order reaction at 300 K and 310 K are respectively 1.2 xx 10^(3) s^(-1) and 2.4 xx 10^(3) s^(-1). Calculate the energy of activation. (R = 8.314 J K^(-1) mol^(-1)) |
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Answer» SOLUTION :`LOG.(K_(2))/(K_(1)) = (E_(a))/(2.303 R) [(T_(2)-T_(1))/(T_(1)T_(2))]` `log.(2.4 XX 10^(-3))/(1.2 xx 10^(-3)) = (E_(a))/(2.303 xx 8.314) [(310 - 300)/(300 xx 310)]` `E_(a) = 53598 J` or `53.598 K J`. |
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