InterviewSolution
Saved Bookmarks
| 1. |
What is the activation energy for a reaction if the rate is doubled when the temperature is raised from `20^(@)` C to `35^(@)`C (R=`8.314 J mol^(-1)K^(-1))`A. 15.1 kJ `mol^(-1)`B. `342 kJ mol^(-1)`C. `269 kJmol^(-1)`D. `34.7 kjmol^(-1)` |
|
Answer» Correct Answer - D d) `k_(2)/k_(1) = 2, T_(1)=20+273=293 K`, `T_(2)=35 + 273=308`K `k=8.314 JK^(-1)mol^(-1)` `log k_(2)/k_(1) =E_(a)/(2.303 ) [1/T_(1)-1/T_(2)]` `log2= (E_(a))/(2.303 xx 8.314(JK^(-1)mol^(-1))[1/(293K)-1/(308K)]` `0.3010= (E_(a) xx 15)/(2.303 xx 8.314 (Jmol^(-1)) xx 293 xx 308)` `E_(a) = (0.3010 xx 2.303 xx 8.314 x 293 xx 308)/(15) (Jmol^(-1))` `=34.673 J mol^(-1) = 34.673 kJ mol^(-1)` |
|