| If (i) \[\Delta H_{f}^{o}\] (benzene)\[=-358.5\,\,kJ\,\,mo{{l}^{-1}}\]. |
| (ii) Heat of atomization of graphite\[=716.8\,\,kJ\,mo{{l}^{-1}}\]. |
| (iii) Bond energy of \[C-H,\,\,C-C,\,\,C=C\] and \[H-H\] bonds are 490, 340, 620 and\[436.9\,\,kJ\,\,mo{{l}^{-1}}\] respectively. The resonance energy (in \[kJ\,\,mo{{l}^{-1}}\]) of \[{{C}_{6}}{{H}_{6}}\] using Kekule formula is |
A) - 150
B) - 50
C) - 250
D) +150
Correct Answer: A
Solution :
| [a] \[6C(g)+3{{H}_{2}}(g)\xrightarrow{\,}\,{{C}_{6}}{{H}_{6}};\]\[\Delta {{H}_{\exp }}=-358\,kJ\] |
| \[\Delta {{H}_{f}}\] can also be calculated as; |
| \[\Delta {{H}_{f}}({{C}_{6}}{{H}_{6}})=[6\times \Delta {{H}_{C}}(s)\xrightarrow{{}}\]\[{{C}_{(g)}}+3\times \Delta {{H}_{H-H}}]\] |
| \[-[3BE\] of \[C=C+3BE\] of \[C=C\,+6BE\] of \[C-H]\] |
| \[=[6\times 716.8+3\times 436.9]\] |
| \[-[3\times 340+3\times 620+6\times 490]\] |
| \[\Delta {{H}_{resonance}}=\Delta {{H}_{f}}_{(\exp )}-\Delta {{H}_{f(calc)}}\] |
| \[=-358.5-(-208.5)=-150\,kJ\,mo{{l}^{-1}}\] |
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