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)?`

Options

A. 273.15 k
B. `24.01^@C`
C. 280.39 K
D. `45.29^@C`

Answer

B

Explanation

According to Arrhenius equation,
`"log"(k_(2))/(k_(1))=E_(a)/(2.303R)xx(T_(2)-T_(1))/(T_(1)T_(2))` `"log"(1.5xx10^(4))/(4.5xx10^(3))=(60000J"mol"^(-1))/(2.303xx(8.314J"mol"^(-1)))((T_(2)-283)/(283T_(2)))`
`log3.333=3.1333.62((T_(2)-283)/(283T_(2)))`
`T_(2)=297.01k=(297.01-273.0)^@C=24.01^@C`
Temperature =`24.01^@C`