In a Arrhenius equation for a certain reaction, the values of `A and E_(a)` ( energy of activation)are `4xx10^(13)s^(-1)` and `98.6KJ mol^(-1)`, respectively. If the reaction of first order at, what temperature will its life periof be `10 min`.

Answer

`T = 311.35 K`

Explanation

Half life `= (0.693)/(k)`
or `10 xx 60 = (0.693)/(k)`
or `k = (0.693)/(600)`
Since,
`k = Ae^(-E_(a)//RT)`
`logk = logA - (E_(a))/(2.303 RT)`
or `log.(0.693)/(600) = log 4 xx 10^(13) - (98.6 xx 10^(3))/(2.303 xx 8.314 xx T)`
or `T = 311.35 K`