The activation energies of two reactions are `E_(1)` & `E_(2)` with `E_(1)gtE_(2)`. If temperature of reacting system is increased from `T_(1)` (rate constant are `k_(1)` and `k_(2)`) to `T_(2)` (rate constant are `k_(1)^(1)` and `k_(2)^(1)`) predict which of the following alternative is incorrect.

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)')`

Answer

B

Explanation

We know, `log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
i.e., if `E_(a)` is more, than `log"(k_(2))/(k_(1))` is more
If `E_(1) gt E_(2)`
`implies log"(k_(2)')/(k_(1)') gt log" (k_(2)')/(k_(2)')`
`implies (k_(1)')/(k_(2)') gt (k_(2)')/(k_(2)')`