What is the energy of activation of a reaction if its rate doubles when the temperature is raised from 290 K to 300 K?

Options

A. 12 kcal
B. 41 kcal
C. 13.8 kcal
D. 52 kcal

Answer

A

Explanation

`log(k_2)/(k_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
log 2=`(E_a)/(2.303xx2)[(300-290)/(290xx300)]` (`sincek_2=2k_1`)
`E_a`=12062 calories`~~`12 Kcal