A) -14 kcal
B) -28 kcal
C) -42 kcal
D) -56 kcal
Correct Answer: B
Solution :
Key Idea: \[\Delta H=\Sigma \,{{H}_{R}}-\Sigma {{H}_{P}}\] Given bond energy of \[C-C=80\,kcal,\,\,C=C=145\,\,kcal\] \[C-H=98\,\,kcal,\,\,H-H=103\,\,cal\] \[H-\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}\,=\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}\,-H+H-H\xrightarrow{{}}H-\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{C}}\,}}\,-\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{C}}\,}}\,-H\] \[\Delta H=\] [(BE of \[4\,C-H\]bond) + (BE of \[C=C\]) + BE of \[H-H\]] -[BE of 6 \[C-H\] bond + BE of \[C-C\]] \[=[(98\times 4)+(145)+(103)]-[(6\times 98)+80)]\] = 640 - 668 = - 28 kcal
You need to login to perform this action.
You will be redirected in
3 sec
