For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. `1.0xx10^(6)s^(-1) and 9.2 kJ mol^(-1)`B. `6.0 s^(-1) and 16.6kJ mol^(-1)`C. `1.0xx10^(6)s^(-1) and 16.6 kJ mol^(-1)`D. `1.0xx10^(6) s^(-1) and 38.3 kJ mol^(-1)`

For a first order reaction `A rarr P`, the temperature `(T)` dependent

1.	For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. `1.0xx10^(6)s^(-1) and 9.2 kJ mol^(-1)`B. `6.0 s^(-1) and 16.6kJ mol^(-1)`C. `1.0xx10^(6)s^(-1) and 16.6 kJ mol^(-1)`D. `1.0xx10^(6) s^(-1) and 38.3 kJ mol^(-1)`
Answer» Correct Answer - D According to Arrhenius equation `K=Ae^(-E_(a)//RT)` Taking logarithm of both the sides, we get log `k=logA-(E_(a))/(2.303RT)` We are given log `k=6.0-(2000)(1)/(T)` comparing the two, we obtain log `A=6.0` or `A=10^(6)sec^(-1)` `(E_(a))/(2.303R)=2000` or `E_(a)=(2000)(2.303)(8.14JK^(-1)mol^(-1))` `=38.3KJmol^(-1)`

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