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If the bond energies of H-H, Br-Br and H...

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

A

`-261KJ`

B

`+103KJ`

C

`+261KJ`

D

`-103KJ`

Text Solution

Verified by Experts

The correct Answer is:
d
`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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