The decomposition of `A` into product has value of `k` as `4.5xx10^(3)s^(-1)` at `10^(@)C` and energy of activation of `60kJmol^(-1)`. At what temperature would `k` be `1.5xx10^(4)s^(-1)?`

Explanation

According to Arrhenius equation,
`log k_(2)/k_(1) = E_(a)/(2.303 R) (T_(2)-T_(1))/(T_(1)T_(2))`
`k_(1) = 4.5 xx 10^(3)s^(-1)`, `k_(2) = 1.5 xx 10^(4)s^(-1)`, `T_(1) = 10^(@)`C = 283 K.
`(log)(1.5 xx 10^(4))/(4.5 xx 10^(3)) = (60000 J mol^(-1))/(2.303 xx (8.314 J mol^(-1)) (T_(2)-283)/(283 T_(2))`
`log 3.333 = 3133.62(T_(2)-283)/(283 T_(2))`
`(0.5288)/(3133.62) = (T_(2)-283)/(283 T_(2))` or `(0.5288 xx 283)/(3133.62) = (T_(2) - 283)/(283 T_(2))`
`0.5288/3133.62 = (T_(2) - 283)/(283 T_(2))` or `(0.5288 xx 283)/(3133.62) = (283)/(0.9524)`
`T_(2) = 297.19 K= (297.19 - 273.0) = 24.19^(@)`C.