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JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

  • question_answer
    Bond enthalpies of \[AB,\,\,{{A}_{2}}\] and \[{{B}_{2}}\] respectively are in the ratio 1:1:0.5. If \[\Delta H\] for the formation of AB is -200 kJ, bond enthalpy of \[{{A}_{2}}\] will be (in kJ/mol)

    A)  100                            

    B)  300             

    C)  800                            

    D)  400

    Correct Answer: C

    Solution :

     Let the bond dissociation energies of AB, \[{{A}_{2}}\] and \[{{B}_{2}}\], be\[x,\,\,x\] and \[0.5\,\,x\] respectively. \[\frac{1}{2}{{A}_{2}}+\frac{1}{2}{{B}_{2}}\to AB;\] \[{{\Delta }_{f}}H=-200\,\,kJ/mol\] \[\Delta H=\frac{1}{2}\,\Delta H({{A}_{2}})+\frac{1}{2}\,\Delta H({{B}_{2}})-\Delta H(AB)\] \[-200=\frac{1}{2}x+\frac{0.5x}{2}-x\]\[x=800\,\,kJ/mol\]


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