What is the activation energy for a reaction if its rate doubles when the temperature is raised from `20^@`C to `35^@`C?
(R=8.314 J `mol^(-1)K^(-1)`)

Options

A. 34.7 kJ `mol^(-1)`
B. 15.1 kJ `mol^(-1)`
C. 342 kJ `mol^(-1)`
D. 269 kJ `mol^(-1)`

Answer

A

Explanation

`log(k_2)/(k_1)=(E_a)/(2.303R)(1/T_1-1/T_2)`
`k_2=2k_1`,
`T_1=20+273=293 K` or `T_2=35+273=308` K
`R=8.314" "mol^(-1)K^(-1)`
log2=`(E_a)/(2.303xx8.314)((1)/(293)-(1)/(308))`
`0.3010=(E_a)/(19.147)xx(15)/(293xx308)`
`E_a=34673" "J " "mol^(-1)` or `34.7" "mol^(-1)`