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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
      The standard enthalpy of formation of\[N{{H}_{3}}\]is\[-\,46.0\text{ }kJ\text{ }mo{{l}^{-1}}\]. If the enthalpy of formation of\[{{H}_{2}}\]from its atoms is\[-436\text{ }kJ\text{ }mo{{l}^{-1}}\]and that of\[{{N}_{2}}\]is\[-712\text{ }kJ\text{ }mo{{l}^{-1}}\], the average bond enthalpy of\[N-H\]bond in\[N{{H}_{3}}\]]       AIEEE  Solved  Paper-2010

    A) \[-1102\text{ }kJ\text{ }mo{{l}^{-1}}\]   

    B) \[-964\text{ }kJ\text{ }mo{{l}^{-1}}\]

    C)        \[+352\text{ }kJ\text{ }mo{{l}^{-1}}\]

    D)        \[+1056\text{ }kJ\text{ }mo{{l}^{-1}}\]

    Correct Answer: C

    Solution :

    \[\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}H(g)\xrightarrow{{}}N{{H}_{3}}\] \[\Delta {{H}_{f}}=\frac{1}{2}B-{{E}_{N-N}}+\frac{3}{2}B{{E}_{H-H}}-3.B.{{E}_{N-H}}\] \[-46=\frac{1}{2}\times (-712)+\frac{3}{2}\times (-436)-3\times x\] \[x=\frac{1056}{3}=352\,kJ/ml\]


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