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

Explanation

We have ,
`log. (k_(2))/(k_(1)) =E/(2.303R) ((T_(2)-T_(1))/(T_(1)T_(2)))" " (Eqn .12)`
`log 2 = E/(2.303 xx2) ((300-290)/(290xx300))=" " (R = 2 "cal /K/Mole")`
E = 12062 calories
`approx 12 kcal ` .