Question

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

• A. -261 kJ
• B. +103 kJ
• C. +261 kJ
• D. -103 kJ

Answer 1

Answer: (D)

For reaction
$H _{2}(g)+ Br _{2}(g) \longrightarrow 2 HBr (g) \Delta H^{\circ}=?$
On the basis of bond energies of $H _{2}, Br _{2}$ and $HBr , \Delta H$ of above is calculated as follows
$\Delta H=-[2 \times$ bond energy of $HBr -$ (bond
energy of $H _{2}+$ bond energy of $\left.\left.Cl _{2}\right)\right]$
$\Delta H =-[2 \times(364)-(433)+192 \,kJ$
$=-[728-(625)] \,kJ =-103\, kJ $