Question

If the bond energies of $H - H , Br - B r$ and $H - Br$ are $433,192$ and $364\, kJ\, mol ^{-1}$ respectively, then $\Delta H^{\circ}$ for the reaction,
$H_2(g) +Br_2 (g) \to 2HBr (g) $ is

• A. $-261 \,kJ$
• B. $+103\, kJ$
• C. $+621\, kJ$
• D. $-103\, kJ$

Answer 1

Answer: (D)

Required reaction
$H _{2}(g)+ Br _{2}(g) \longrightarrow 2 HBr (g)$
Given, $ BE ( H - H )=433 kJ\, mol ^{-1}$
$BE ( Br - Br )=192 \,kJ\, mol ^{-1}$
$BE ( H - Br )=364\, kJ\, mol ^{-1}$
$ \because \,\Delta H_{R}^{\circ} =\Sigma BE$ (Reactants) -\Sigma BE (Products)
$\Delta H_{R}^{\circ} =(433+\text { 192) }-(2 \times 364)$
$ \Delta H_{R}^{\circ} =625-728 $
$ \Delta H_{R}^{\circ} =-103\, kJ \,mol ^{-1} $