JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • 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

    Correct Answer: A

    Solution :

     \[6C(g)+3{{H}_{2}}(g)\xrightarrow{\,}\,{{C}_{6}}{{H}_{6}};\] \[\Delta {{H}_{\exp }}=-358\,kJ\] \[\Delta {{H}_{f}}\] 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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