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

From Arrhenius equation , we obtain
log `(k_(2))/(k_(1)) = (E_(a))/(2.303 R) ((T_(2) - T_(1))/(T_(1)T_(2)))`
Also , `k_(1) = 4.5 xx 10^(3) s^(-1)`
`T_(1) = 273 + 10 = 283` K
`k_(2) = 1.5 xx 10^(4) s^(-1)`
`E_(a) = 60 kJ mol^(-1) = 6.0 xx 10^(4) J mol^(-1)`
Then ,
log `(1.5 xx 10^(4))/(4.5 xx 10^(3)) = (6.0 xx 10^(4) J mol^(-1))/( 2.303 xx 8.314 J K^(-1) mol^(-1)) ((T_(2) - 283)/(283 T_(2)))`
`implies 0.5229 = 3133. 627 ((T_(2) - 283)/(283 T_(2)))`
`implies (0.5229 xx 283 T_(2))/(3133 . 627) = T_(2) - 283`
`implies 0.0472 T_(2) = T_(2) - 283`
`implies 0.9528 T_(2) = 283`
`implies T_(2) = 297 . 019 K ` (approximately)
= 297 K
= `24^(@)C`
Hence , k would be `1.5 xx 10^(4) s^(-1)` at `24^(@)C`
Note : There is a slight variation in this answer and the one given in the NCERT textbook .