InterviewSolution
Saved Bookmarks
| 1. |
If the bond energies of H - H, Br - Br and HBr are 433, 192 and 364 kJ `mol^(-1)` respectively, the `DeltaH^(@)` for the reaction, `H_(2)(g)+Br_(2)(g)rarr2HBr(g)` isA. `+ 261` kJB. `- 103` kJC. `- 261` kJD. `+ 103` kJ |
|
Answer» Correct Answer - B `H-H+Br-Brrarr2H-Br` `{:(433+192,2xx364),(625,728):}` Energy absorbed = Energy released Not energy released = 728-625=103 kJ i.e., = `DeltaH=-103 kJ`. |
|