The rate constant for the decomposition of `N_2O_6`,
`N_2O_6 rarr N_2O_4+1/2O_2`
is `3.46 xx 10^(-5)` at `25^@`C and `4.87 xx 10^(-3)` at `65^@`C. Calculate the energy of activation of the reaction.

Explanation

`k_1=3.46xx10^(-5),T_1=25^@`C
`=(25+273)K=298K`
`k_2=4.87xx10^(-3)`
`T_2=65^@C=(65+273)K=338K`
Ea=? , R=8.314 J`K^(-1)"mol"^(-1)`
`log(k_2)/(k_1)=(Ea)/(2.303R)=[(T_2-T_1)/(T_1T_2)]`
`log(4.87xx10^(-3))/(3.46xx10^(-5))=(Ea)/(2.303xx8.314JK^(-1)mol^(-1))`
`[(338-298)/(338xx298K)]`
`log(4.87)/(3.46)=(Eaxx40)/(19.15JK^(-1)mol^(-1)xx100724K)`
`log140.75=(Eaxx40)/(1928864.6 " "mol^(-1))`
`2.1483=(E_axx40)/(1928864.6J" "mol^(-1))`
`E_a=(2.1483xx1928864.6)/(40)J" "mol^(-1)`
`=103594.49" J mol"^(-1)`
`103.6 Jmol^(-1)`