Saved Bookmarks
| 1. |
Assuming that energy of activation for most of the reactions is 52 kJ, what conclusion you draw about the effect of temperature on the rate of a reaction ? (Based on Arrhenius equation) |
| Answer» Solution :Substituting `T_(1)=300" K", T_(2)=310" K",E_(a)=52000" J mol"^(1)` in the Arrhenius equation, viz., `log""(k_(2))/(k_(1))=(E_(a))/(2.303R)((1)/(T_(1))-(1)/(T_(2)))`, we get `k_(2)~~2k_(1)`. This SHOWS that for `10^(@)` rise of TEMPERATURE, the rate of REACTION is nearly doubled. | |