Question
Given that bond energies of $$H-H$$ and $$Cl-Cl$$ are $$430\,kJ\,mo{l^{ - 1}}$$ and $$240\,kJ\,mo{l^{ - 1}}$$ respectively and $$\Delta {H_f}$$ for $$HCl$$ is $$ - 90\,kJ\,mo{l^{ - 1}}.$$ Bond enthalpy of $$HCl$$ is
A.
$$290\,kJ\,mo{l^{ - 1}}$$
B.
$$380\,kJ\,mo{l^{ - 1}}$$
C.
$$425\,kJ\,mo{l^{ - 1}}$$
D.
$$245\,kJ\,mo{l^{ - 1}}$$
Answer :
$$380\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& \Delta {H_{{\text{reaction}}}} = {\Delta _{H - H}} + \Delta {H_{Cl - Cl}} - 2\Delta {H_{HCl}} \cr
& {\text{or}}\,\,\Delta {H_{H - Cl}} = \frac{{430 + 240 - \left( { - 90} \right)}}{2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{760}}{2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 380\,kJ\,mo{l^{ - 1}} \cr} $$