Question
If the bond energies of $$H-H,$$ $$Br-Br$$ and $$H-Br$$ are $$433, 192$$ and $$364\,kJ\,mo{l^{ - 1}}$$ respectively, then $$\Delta {H^ \circ }$$ for the reaction $${H_2}\left( g \right) + B{r_2}\left( g \right) \to 2HBr\left( g \right)$$ is
A.
$$ - 261\,kJ$$
B.
$$ + 103\,kJ$$
C.
$$ + 261\,kJ$$
D.
$$ - 103\,kJ$$
Answer :
$$ - 103\,kJ$$
Solution :
$$\eqalign{
& {\text{For reaction,}} \cr
& {H_2}\left( g \right) + B{r_2}\left( g \right) \to 2HBr\left( g \right)\Delta {H^ \circ } = ? \cr} $$
$$\Delta {H^ \circ } = - \left[ {\left( {2 \times {\text{bond energy of }}HBr} \right)} \right.$$ $$\left. { - \left( {{\text{bond energy}}\,{\text{of}}\,{H_2} + {\text{bond energy of}}\,C{l_2}} \right)} \right]$$
$$\eqalign{
& \Delta {H^ \circ } = - \left[ {2 \times \left( {364} \right) - \left( {433 + 192} \right)} \right]kJ \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = - \left[ {728 - \left( {625} \right)} \right]kJ \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = - 103\,kJ \cr} $$