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InterviewSolution
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The rate constant for a first order reaction increases from `4 xx 10^(-2)` to `8 xx 10^(-2)` when the temperature changes from `27^(@)`C to `37^(@)`C. Calculate energy of activation for the reaction. |
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Answer» According to Arrhenius equation, `(log)k_(2)/k_(1) = E_(a)/(2.303R)[1/T_(1)-1/T_(2)]` `k_(2)/k_(1) = (8xx 10^(-2))/(4 xx 10^(-2)) = 2, T_(1) = 273 + 27 = 300K, T_(2)= 273 + 37 = 310 K` `R= 8.314 JK^(-1)mol^(-1)` `log2 = (E_(a))/(2.303 xx (8.314 J K^(-1)mol^(-1)))[1/300K - 1/310K]` `E_(a) = (0.3010 xx 2.303 xx (8.314 J mol^(-1)) xx (300) xx (310))/10` `53598.59 J mol^(-1) = 53.6 kJ mol^(-1)` |
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