1.

If the bond energies of H-H, Br-Br and H-Br are 433, 192 and 364 kJ "mol"^(-1) respectively, then DeltaH^@ for the reaction : H_(2(g)) + Br_(2(g)) to 2HBr_((g)) is

Answer»

`-261KJ`
`+103KJ`
`+261KJ`
`-103KJ`

Solution : `Delta_(R)H=sumB.E._("REACTANTS")-sumB.E._("PRODUCTS")`
`=[B.E_(H_(2))+B.E._(Br_(2))]-[2B.E._(HBr)]`
`=[433+192-2xx364]KJ=--103KJ`


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