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

If the end 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)to2HB r(g)` is

A

`+ 261` kJ

B

`- 103` kJ

C

`- 261` kJ

D

`+ 103` kJ

Text Solution

Verified by Experts

The correct Answer is:
B
`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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