The rate constant of a reaction is `1.5 xx 10^7 sec^(-1)` at `50^@`C and `4.5 xx 10^7 sec^(-1)` at `100^@ C`. Calculate energy of the activation for the reaction.

Explanation

`k_1=1.5xx10^7s^(-1),T_1=50^@=(50+273)K=323K`
`k_2=4.5xx10^7s^(-1)`,
`T_2=100^@C=(100+273)k=373k`
`E_a=?,R=8.314J" "k^(-1)mol^(-1)`
`log(k_2)/(k_1)=(Ea)/(2.303R)[(T_2-T_2)/(T_1T_2)]`
`log(4.5xx10^7s^(-1))/(1.5xx10^7s^(-1))=(Ea)/(2.303xx8.314JK^(-1)mol^(-1))`
`[(373-323)/(372xx323K)]`
`log3 =(Eaxx50)/(19.15JK^(-1)mol^(-1)xx120479K)`
`0.4771=(Eaxx50)/(19.15xx120479JKmol^(-1))`
`E_a=(0.4771xx19.15xx120479)/(50)Jmol^(-1)`
`=22015.042mol^(-1)`
`=22.015Jmol^(-1)`