InterviewSolution
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InterviewSolution
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For a decomposition, the values of rate constants at two different temperature are given below: `k_(1) = 2.15 xx 10^(-8)L mol^(-1)s^(-1)` at 650 K `k_(2) = 2.39 xx 10^(-7) L mol^(-1)s^(-1)` at 700 K Calculate the value of activation energy for the reaction (R=`8.314 JK^(-1) mol^(-1))` |
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Answer» According of Arrhenius equation: `(log)k_(2)/k_(1) = E_(a)/(2.303R)[1/T_(1)-1/T_(2)]` `k_(1) = 2.25 xx 10^(-8) L mol^(-1)s^(-1), k_(2)=2.39xx10^(-7)Lmol^(-1)s^(-1)` `T_(1) = 650K,T_(2)= 700K, R=8.314 JK^(-1)mol^(-1)` `log(2.39 xx 10^(-7))/(2.15 xx 10^(-8))= E_(a)/(2.303 xx (8.314 JK^(-1)mol^(-1)))[(320K-300K)/((320K)xx(300K))]` `0.6020 = (E_(a) xx 20)/((2.303) xx (8.314 J mol^(-1)) xx (320) xx (300))` `E_(a) = (0.6020 xx (2.303) xx (8.314J mol^(-1)) xx 320 xx 300)/(20)` `=55327.6 J mol^(-1) = 55.328 kJmol^(-1)` |
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