The rate constant of a first order reaction can be determined by the equation: `logk=12.6-(4267)/(T)K.` Calculate the energy of activation `(E_(a))` and frequency factor (A) of the reaction.

Explanation

We know , `logk=logA-(E_(a))/(2.303R)xx(1)/(T)" ".... [1]`
Comparing the given equation with equation [1] gives, `logA=12.6`
or, `A=10^(12.6)s^(-1)(because` the units of k and A are the same and for a first order reaction the unit of k is `s^(-1)` ) and `(E_(a))/(2.303R)=4267K`
` thereforeE_(a)=4267xx2.303xx8.314"J.mol"^(-1)=81.7"kJ.mol"^(-1)`
`therefore` The activation energy = `81.7"kJ.mol"^(-1)` and frequency factor `=10^(12.6)s^(-1)=3.98xx10^(12)s^(-1)`