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CMC Medical CMC-Medical VELLORE Solved Paper-2012

  • question_answer
    If the bond energies of H?H, Br?Br and H?Br are 433 192 and 364 kJ \[mo{{l}^{-1}}\]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)  \[+\,261\,kJ\]                  

    D)  \[-103\,kJ\]

    Correct Answer: D

    Solution :

                    For the 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 \[HBr,\,\,\Delta H\]of above is calculated as follows \[\Delta H=-\][2 \[\times \]bond energy of \[HBr\]- (bond energy of \[{{H}_{2}}\]+ bond energy of\[C{{l}_{2}}\])] \[\Delta H=-[2\times (364)-(433)+192\,kJ]\]        \[=-\,[728-(625)]\,kJ=-103\,\,\text{k}J\]


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