Which equation is true to calculate the energy of ativation, if the rate of reaction is doubled by increasing temperature from `T_(1) K "to" T_(2) K`?

Options

A. `log_(10)((k_(1))/(k_(2)))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
B. `log_(10)((k_(2))/(k_(1)))=(E_(a))/(2.303R)[(1)/(T_(2))-(1)/(T_(1))]`
C. `log_(10).(1)/(2)=(E_(a))/(2.303R)[(1)/(T_(2))-(1)/(T_(1))]`
D. `log_(10).2=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`

Explanation

We know that, `log_(10)((k_(2))/(k_(1)))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
When reaction rate becomes double then `(k_(2))/(k_(1))` will be equal to 2
Then, `log_(10).2=(E_(a))/(2.3030R)[(1)/(T_(1))-(1)/(T_(2))]`