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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

`+ 103 kJ`

B

261 kJ

C

`-103 kJ`

D

`-261 kJ`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta H = B.E(H_(2)) + B.E(Br_(2)) - 2B.E(HBr)`
`= 433 + 192 - 2 xx 364`
`= - 103 kJ`
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