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The rate constant for the first order decompoistion 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 periof be `256 min`? |
Answer» Step I. Calculation of activation energy `(E_(a))` According to Arrhenius equation , k=`Ae^(-Ea//RT)`. log k `= log A - (E_(a))/(2.303 RT)` The given equ. Is log k `= 14.34 - (1.25 xx 10^(4)(K))/(T)` On comparison: `E_(a)/(2.303 RT) = (1.25 xx 10^(4)(K))/(T)` `E_(a) = 1.25 xx 10^(4) xx 2.303 xx 8.314 (J mol^(-1))` `=239,339 J mol^(-1) = 239.339 kJ mol^(-1)` Step II. Calculation of desired temperature For the first order reaction , k `= 0.693/t_(1//2) = 0.693/(256 min) = 0.693/(256 xx 60s) = 4.51 xx 10^(-5) s^(-1)` According to Arrheneius theory, log k `= 14.34 - (1.25 xx 10^(4))/T` , `-4.35 = 14.34 -(1.25 xx 10^(4))/(18.69) = 669 K` |
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