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`.


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