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InterviewSolution
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The rate constant for the first order decomposition of a certain reaction is described by the equation `log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)` (a) What is the energy of activation for the reaction? (b) At what temperature will its half-life period be `256 min`? |
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Answer» Arrhenius equation is given by , `k = Ae^(-E_(a) //RT)` `implies ` ln k = ln A `- (E_(a))/(RT)` `implies` ln k = log A `-(E_(a))/(RT)` `implies ` log k = log A `- (E_(a))/(2.303 RT) " " (i)` The given equation is log k = `14. 34 - 1.25 xx 10^(4) K//T " " (ii)` From equation (i) and (ii) , we obtain `(E_(a))/(2.303RT) = (1.25 xx 10^(4)K)/(T)` `implies E_(a) = 1.25 xx 10^(4) K xx 2.303 xx R` =` 1.25 xx 10^(4) K xx 2.303 xx 8.314 J K^(-1) mol^(-1)` = `239339.3 J mol^(-1)` (approximately) = `239. 34 kJ mol^(-1)` Also , when `t_(1//2)` = 256 minutes `k = (0.693)/(t_((1)/(2)))` = `(0.693)/(256)` = `2.707 xx 10^(-3) "min"^(-1)` =`4.51 xx 10^(-5) s^(-1)` It is also given that log k = `14.34 - 1.25 xx 10^(4) K//T` `implies` log `(4.51 xx 10^(-5)) = 14.34 - (1.25 xx 10^(4)K)/(T)` `implies "log" (0.654 - 05) = 14.34 - (1.25 xx 10^(4)K)/(T)` `implies (1.25 xx 10^(4)K)/(T) = 18. 686` `implies T = (1.25 xx 10^(4) K)/(18.686)` = 668. 85 K = 669 K (approximately) |
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