Saved Bookmarks
| 1. |
In general it is observed that the rate of a chemical reaction doubles with every 10 degree rise in temperature. If the generalization holds good for the reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for this reaction ? [R = 8.314 mol^(-1)JK^(-1)] |
|
Answer» Solution :`log.(K_(1))/(K_(2)) = (E)/(2.303 R)[(T_(2)-T_(1))/(T_(1) XX T_(2))]` `log.(2)/(1) = (E_(a))/(2.303 xx 8.314)[(305-295)/(295 xx 305)]` `0.3010 = (E_(a))/(19.14714)[(10)/(89975)]` `E_(a) = (0.3010 xx 19.147 xx 89975)/(10)` `E_(a) = 51906 J MOL^(-1)` `= 51.906 KJ//mol` |
|