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) is

Answer»

`+ 261` kJ
`- 103` kJ
`- 261` kJ
`+ 103` kJ

Solution :`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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