Punjab Medical Punjab - MET Solved Paper-2005

  • question_answer \[C{{H}_{2}}=C{{H}_{2(g)}}+{{H}_{2(g)}}\to C{{H}_{3}}-C{{H}_{3(g)}}\] The heat of reaction is [bond energy of\[C-C=80\,\,Kcal,\]\[C=C=145\,\,kcal,\]\[C-H=98\,\,kcal,\,\,H-H=103\,\,kcal]\]

    A) \[-14\,\,kcal\]

    B) \[-28\,\,kcal\]

    C) \[-42\,\,kcal\]                   

    D) \[-56\,\,kcal\]

    Correct Answer: B

    Solution :

    The required thermochemical equation is\[H-\overset{\begin{smallmatrix}  H \\  | \end{smallmatrix}}{\mathop{C}}\,=\overset{\begin{smallmatrix}  H \\  | \end{smallmatrix}}{\mathop{C}}\,-H(g)+H-H(g)\xrightarrow[{}]{{}}H-\underset{\begin{smallmatrix}  | \\  H \end{smallmatrix}}{\overset{\begin{smallmatrix}  H \\  | \end{smallmatrix}}{\mathop{C}}}\,-\underset{\begin{smallmatrix}  | \\  H \end{smallmatrix}}{\overset{\begin{smallmatrix}  H \\  | \end{smallmatrix}}{\mathop{C}}}\,-H(g)\] \[=\Delta H\]{Bond energies of reactants} - {Bond energies of products} \[=[B.E.\,\,of\,\,4(C-H)+Be\,\,of\,\,(C=C)+B.E.\,\,of\]\[(H-H)\}-\{B.E.\,\,of\,\,6(C-H)+B.E.\,\,of\,\,(C-C)\}\]\[=B.E.\,\,of\,\,(C=C)+B.E.\,\,of\,\,(H-H)-B.E.\,\,of\]\[2(C-H)-B.E.\,\,of\,\,(C-C)\] \[=145+103-2\times 98-80\] \[=248-276=-28\,\,kcal\].

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