The activation energies of two reactions are E_(1) and E_(2) (E_(1) gt E_(2)) . If the temperature of the system is increased from T_(1) to T_(2) , the rate constant of the reaction changes from k_(1) to k_(1) in the first reaction and k_(2) to k_(2) in second reaction predict which of the following expression is correct ?

The activation energies of two reactions are E_(1) and E_(2) (E_(1) gt

1.	The activation energies of two reactions are E_(1) and E_(2) (E_(1) gt E_(2)) . If the temperature of the system is increased from T_(1) to T_(2) , the rate constant of the reaction changes from k_(1) to k_(1) in the first reaction and k_(2) to k_(2) in second reaction predict which of the following expression is correct ?
Answer» `(k_(1))/(k_(1))= (k_(2))/(k_(2))` `(k_(1))/(k_(1)) GT (k_(2))/(k_(2))` `(k_(1))/(k_(1)) lt (k_(2))/(k_(2))` `(k_(1))/(k_(1))= (k_(2))/(k_(2))`=0 Solution :[B ] log `(k_(1))/(k_(1)) = (E_(1))/(2.303R) [(T_(2)-T_(1))/(T_(1)T_(2))]` log`(k_(2))/(k_(2))=(E_(2))/(2.303R)[(T_(2)-T_(1))/(T_(1)-T_(2))]` SINCE `E_(1) gt E_(2)` `("log"(k_(1))/(k_(1)))/("log"k_(2)/(k_(2)))gt1` or ` (k_(1))/(k_(1))gt (k_(2))/(k_(2))`