Saved Bookmarks
| 1. |
At a certain temperature, the first order rate constant `k_(1)` is found to be smaller than the second order rate constant `k_(2)`. If `E_(a)(1)` of the first order reaction is greater than `E_(a)(2)` of the second order reaction, then as temperature is raised:A. `k_1` will increase faster than `k_2`, but always will remain less than `k_2`B. `k_2` will increase fasther than `k_1`C. `k_1` will increase than fasther than `k_2` and becomes equal to `k_2`D. `k_1` will increase faster `k_2` and becomes greater than `k_2` |
|
Answer» Correct Answer - A `(d(Ink))/(dT)=E_a/(RT^2)` As `E_a` increases rate of increase in k also increases so `k_f` will increase faster then `k_2` but always will remain less than `k_2` |
|