A) \[~-14\text{ kcal}\]
B) \[~-28\text{ kcal}\]
C) \[~-\,42\text{ kcal}\]
D) \[~-\,56\text{ kcal}\]
Correct Answer: B
Solution :
Key Idea: \[\Delta H=\sum{\Delta \Eta =}\sum{{{H}_{R}}}-\sum{{{H}_{P}}}\] Given bond energy of C -C = 80 kcal, C =C = 145 kcal C-H= 98 kcal, H-H= 103 kcal \[H-\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,=\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,-H+H-H\xrightarrow{{}}H-\underset{H}{\overset{H}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-\underset{H}{\overset{H}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-H\] \[\Delta H=[(BE\,of\,4C-H\,bond)\] \[+\,(Be\,of\,C=C)+BE\,of\,H-H]\] \[-[BE\,of\,6C-H\,bond\,+\,BE\,of\,C-C]\] \[=[(98\times 4)+(145)+(103)-[(6\times 98)+80]\] \[=640-668\] \[=-28\,kcal\]
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