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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

`+261kJ`

B

`-103kJ`

C

`-261kJ`

D

`+103 kJ`

Text Solution

Verified by Experts

The correct Answer is:
B

`H- H+ Br- Br rarr 2H- Br`
`433+192 " " 2 xx 364`
`625 " " 728`
Energy absorbed= Energy released
Net energy released = 728-625= 103kJ
i.e, `Delta H= -103kJ`
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