Saved Bookmarks
| 1. |
The activation energy for a reaction when the temperature is raised from 300K to 310K is |
|
Answer» `50.6kJ mol^(-1)` `LOG (K_(2))/(K_(1)) = E_(a)/2.303[1/T_(1)-1/T_(2)]` `log(2K_(1))/(K_(1))=(E_(a))/(2.303R) [1/T_(1)-1/T_(2)]` `LOG2 = (E_(a))/(2.303 xx 8.314 Jk^(-1)mol^(-1))[1/(300K-1/310K)]` `0.30 = (E_(a))/(2.303 xx (8.314 J mol^(-1)) xx (9300)` `=53.6 kj mol^(-1)`
|
|