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

Explanation

Given, `T_(1) = 10^(@)C + 273 = 283 k`
`k_(T_(1)) = 4.5 xx 10^(3) sec^(-1)`
`E_(a) = 60 kJ mol^(-1) = 6 xx 10^(3) J mol^(-1)`
`T_(2) = ?`
`k_(T_(2)) = 1.5 xx 10^(4) sec^(-1)`
We know that
`log.(k_(T_(1))/(k_(T_(1)))) = (E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1) xx T_(2))]`
On putting values
`log.(1.5 xx 10^(4))/(4.5 xx 10^(3)) = (60 xx 10^(3) J mol^(-1))/(2.303 xx 8.314)((T_(2)- 283)/(283 T_(2)))`
`0.523 = 3133.63((T_(2)-283)/(283 T_(2)))`
`(T_(2) - 283)/(283 T_(2)) = 1.67 xx 10^(-4)`
`T_(2) - 283 = 0.0472 T_(2)`
`0.953 T_(2) = 283`
`T_(2) = 297 K`
`T_(2) = 24^(@) C`