InterviewSolution
Saved Bookmarks
| 1. |
The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A.B.C.D. |
| Answer» Correct Answer - `53.59 kJ "mol"^(-1)` | |