JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • 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 \[1:1\text{ }:0.5\] 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)  \[400\,kJ\,mo{{l}^{-1}}\]

    Correct Answer: D

    Solution :

     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\[~-100\text{ }KJ/mole\]: \[{{A}_{2}}+{{B}_{2}}\to 2AB;\]      \[\frac{1}{2}{{A}_{2}}+\frac{1}{2}{{B}_{2}}\to AB;\,\Delta =-100KJ\] 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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