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InterviewSolution
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The rate constant for a first order reaction becomes six times when the temperature is raised from 350 to 400 K. Calculate the energy of activation for the reactions. |
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Answer» According to Arrhenius equation, `log k_(2)/k_(1)= E_(a)/(2.303 R) [1/T_(1)-1/T_(2)]` `k_(2)/k_(1) = 6, T_(1)=350K, T_(2)=400K , R= 8.314 JK^(-1)mol^(-1)` `log 6= log 6 = E_(a)/(2.303 xx (8.314 J mol^(-1)))[1/(350K)-1/(400K]]` `0.77815 = (E_(a))/(2.303 xx (8.314 J mol^(-1)))[(400-350)/(350 xx 400)]` `0.77815 = E_(a)/(2.303 xx (8.314 J mol^(-1)))[50/(350xx400)]` `E_(a) = ((0.77815 xx 2.303 xx (8.314 J mol^(-1)) xx 350 xx 400))/(50)` `41718 J mol^(-1) = 41.718 kJ mol^(-1)` |
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