• # 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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