• question_answer Direction: The bond dissociation energy of a diatomic molecule is also called bond energy. Bond energy is also called, the heat of formation of the bond from the gaseous atoms constituting the bond with reverse sign. Example$H(g)+Cl(g)\xrightarrow{\,}\,H-Cl(g),$$\Delta {{H}_{f}}=-431\,\,kJ\,\,mo{{l}^{-1}}$or bond energy of $H-Cl=-(\Delta {{H}_{f}})$$=-(431)=+431\,kJ\,mo{{l}^{-1}}$ When a compound shows resonance there occurs a fair agreement between the calculated values of heat of formation obtained from bond enthalpies and any other method. However deviation occur incase of compounds having alternate double bonds. Example                $\underset{(g)}{\mathop{{{C}_{6}}{{H}_{6}}}}\,\xrightarrow{\,}\underset{(g)}{\mathop{\,6C}}\,+\underset{(g)}{\mathop{6H}}\,$ Resonance energy = experimental heat of formation - calculated heat of formation If (i) $\Delta H_{f}^{o}$ (benzene) $=-358.5\,\,kJ\,\,mo{{l}^{-1}}$. (ii) Heat of atomization of graphite $=716.8\,\,kJ\,mo{{l}^{-1}}$. (iii) Bond energy of $C-H,\,\,C-C,\,\,C=C$ and $H-H$bonds are 490, 340, 620 and$436.9\,\,kJ\,\,mo{{l}^{-1}}$ respectively. The resonance energy (in $kJ\,\,mo{{l}^{-1}}$) of ${{C}_{6}}{{H}_{6}}$ using Kekule formula is A)  - 150                           B)  - 50 C)  - 250                           D)  +150

$6C(g)+3{{H}_{2}}(g)\xrightarrow{\,}\,{{C}_{6}}{{H}_{6}};$ $\Delta {{H}_{\exp }}=-358\,kJ$ $\Delta {{H}_{f}}$ can also be calculated as; $\Delta {{H}_{f}}({{C}_{6}}{{H}_{6}})=[6\times \Delta {{H}_{C}}(s)\xrightarrow{{}}$ ${{C}_{(g)}}+3\times \Delta {{H}_{H-H}}]$ $-[3BE\,of\,C=C+3BE\,of\,C=C\,+6BE\,of\,C-H]$  $=[6\times 716.8+3\times 436.9]$ $-[3\times 340+3\times 620+6\times 490]$ $\Delta {{H}_{resonance}}=\Delta {{H}_{f}}_{(\exp )}-\Delta {{H}_{f(calc)}}$ $=-358.5-(-208.5)=-150\,kJ\,mo{{l}^{-1}}$
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