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JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    Given that :                 (i) \[{{\Delta }_{f}}{{H}^{0}}\]of\[{{N}_{2}}O\] is \[82\operatorname{k}\operatorname{j}\]\[\operatorname{mo}{{\operatorname{l}}^{-1}}\]                 (ii)  Bond energies of N\[\equiv \]N, N=N, O=O and N=O are 946,418,498 and 607 Kj\[{{\operatorname{mol}}^{-1}}\] respectively.                 The resonance energy of \[{{N}_{2}}\] O is     JEE Main Online Paper ( Held On 25  April 2013 )

    A)                 -88 Kj    

    B)                                        -66 Kj

    C)                                        -62 Kj                    

    D)                                        -44 Kj

    Correct Answer: A

    Solution :

                    \[{{N}_{2}}(g)+\frac{1}{2}{{O}_{2}}\to {{N}_{2}}O(g)\] \[N\equiv N(g)+\frac{1}{2}\left( O=O \right)\to N=\overset{+}{\mathop{N}}\,=O(g)\] \[\Delta H_{f}^{o}=\][Energy required for breaking of bonds] -[Energy released for forming of bonds] \[=(\Delta {{H}_{N\equiv N}}+\frac{1}{2}\Delta {{H}_{O=O}}-(\Delta {{H}_{N=N}}+\Delta {{H}_{N=O}})\] \[=(946+\frac{1}{2}\times 498)-(418+607)=170kJ\,mo{{l}^{-1}}\] Resonance energy \[=170-82=88kJ\,mo{{l}^{-1}}\]


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