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 ?

Options

A. `(k_(1))/(k_(1))= (k_(2))/(k_(2))`
B. `(k_(1))/(k_(1)) gt (k_(2))/(k_(2))`
C. `(k_(1))/(k_(1)) lt (k_(2))/(k_(2))`
D. `(k_(1))/(k_(1))= (k_(2))/(k_(2))`=0

Answer

D

Explanation

[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))`