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
    The polymerization of ethylene to linear polyethylene is represented by the reaction\[nC{{H}_{2}}=C{{H}_{2}}\xrightarrow{\,}\,{{(-C{{H}_{2}}-C{{H}_{2}}-)}_{n}}\] when 'n' has a large integral value. Given that average enthalpies of bond dissociation for \[C=C\] and \[C-C\] at 298K are + 590 and\[+331\,\,kJ\,\,mo{{l}^{-1}}\] respectively. Then the enthalpy of polymerization/mol of ethylene at298 K is

    A)  - 132 kJ                     

    B)  - 72 kJ

    C)  + 80 kJ                     

    D)  + 162 kJ

    Correct Answer: B

    Solution :

     \[n(C{{H}_{2}}=C{{H}_{2}})\xrightarrow{\ }\,{{(-C{{H}_{2}}-C{{H}_{2}}-)}_{n}},\]\[\Delta H=?\] Thus, ?n? double bonds are dissociated to form a molecule with 2n single bonds. \[\Delta H=\left( \frac{n}{n} \right)({{\Sigma }_{C=C}})-\left( \frac{2n}{n} \right)({{\Sigma }_{C=C}})\]\[=+590-2\times 331=-72\,kJ\,mo{{l}^{-1}}\]

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