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Manipal Medical Manipal Medical Solved Paper-2007

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

    A)  \[-261\text{ }kJ\]       

    B)  \[+103\text{ }kJ\]

    C)  \[+261\text{ }kJ\]       

    D)  \[-103kJ\]

    Correct Answer: D

    Solution :

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


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