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NEET Sample Paper NEET Sample Test Paper-93

  • question_answer
    AB, \[{{A}_{2}}\] and \[{{B}_{2}}\] are diatomic molecules. If the bond enthalpies of \[{{A}_{2}},\] AB and \[{{B}_{2}}\] are in the ratio 1:1:0.5 and enthalpy of formation of AB from \[{{A}_{2}}\]and \[{{B}_{2}}\] is \[-100kJ\,mo{{l}^{-1}}.\] What is the bond energy of \[{{A}_{2}}:\]

    A) \[200\,kJ\,mo{{l}^{-1}}\]         

    B)        \[100\,kJ\,mo{{l}^{-1}}\]

    C) \[300\,kJ\,mo{{l}^{-1}}\]         

    D)       

    Correct Answer: D

    Solution :

    [d] Let bond energy of \[{{A}_{2}}\] be x then bond energy of AB is also x and bond energy of\[{{B}_{2}}\]is\[{x}/{2.}\;\] Enthalpy of formation of AB is -\[100kJ/mol:\] \[{{A}_{2}}+{{B}_{2}}\to 2AB;\] \[\frac{1}{2}{{A}_{2}}+\frac{1}{2}{{B}_{2}}\to AB;\]\[{{\Delta }_{4}}=-100kJ\] or\[-100=\left( \frac{x}{2}+\frac{x}{4} \right)-x\]\[\therefore \]\[-100=\frac{2x+x-4x}{4}\]\[\therefore \]\[x=400kJ\]


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