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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