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

Options

A. `-261 KJ`
B. `+103 KJ`
C. `+261 KJ`
D. `-103 KJ`

Answer

D

Explanation

`Delta_r H = sum B.E._("reactants")- sumB.E._("Products")`
`=[B.E._((H_2))-sumB.E._((Br_2))]-[2B.E._((H.Br))]`
`=[433+192 -2 xx 364] KJ=-103KJ`