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NEET AIPMT SOLVED PAPER SCREENING 2004

  • question_answer
    If the bond energies of \[\text{H}\text{H},\]\[\text{Br}\text{Br}\] and\[\text{H}\text{Br}\] are 433, 192 and 364 kJ \[mo{{l}^{-1}}\]respectively, then \[\Delta {{H}^{0}}\]for the reaction\[{{H}_{2}}(g)+B{{r}_{2}}(g)\xrightarrow[{}]{{}}2HBr(g)\]is:                                                                 

    A) \[-\text{261 kJ}\]    

    B) \[+\text{1}0\text{3 kJ}\]      

    C) \[+\text{261 kJ}\]   

    D) \[-\text{1}0\text{3 kJ}\]

    Correct Answer: D

    Solution :

    For reaction \[{{H}_{2}}(g)+B{{r}_{2}}(g)\xrightarrow[{}]{{}}2Hbr(g)\Delta {{H}^{o}}=?\] On the basis of bond energies of\[{{H}_{2}}\], \[Br,\]and \[HBr\], \[\Delta H\] of above is calculated as follows: \[\Delta H=-[2\times \]bond energy of \[HBr-\](bond energy of\[{{\text{H}}_{\text{2}}}+\] bond energy of\[C{{l}_{2}})]\] \[\Delta H=-[2\times (364)-(433)+192]kJ\] \[=-[728-(625)]\,kJ\]\[=-103\,kJ\]


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