| Direction: The bond dissociation energy of a diatomic molecule is also called bond energy. Bond energy is also called, the heat of formation of the bond from the gaseous atoms constituting the bond with reverse sign. |
| Example\[H(g)+Cl(g)\xrightarrow{\,}\,H-Cl(g),\]\[\Delta {{H}_{f}}=-431\,\,kJ\,\,mo{{l}^{-1}}\]or bond energy of \[H-Cl=-(\Delta {{H}_{f}})\]\[=-(431)=+431\,kJ\,mo{{l}^{-1}}\] When a compound shows resonance there occurs a fair agreement between the calculated values of heat of formation obtained from bond enthalpies and any other method. However deviation occur incase of compounds having alternate double bonds. |
| Example \[\underset{(g)}{\mathop{{{C}_{6}}{{H}_{6}}}}\,\xrightarrow{\,}\underset{(g)}{\mathop{\,6C}}\,+\underset{(g)}{\mathop{6H}}\,\] |
| Resonance energy = experimental heat of formation - calculated heat of formation |
A) 309.17 kJ
B) 206 kJ
C) 109.2 kJ
D) 275.8 kJ
Correct Answer: A
Solution :
Given \[S(s)+3{{F}_{2}}(g)\xrightarrow{\,}S{{F}_{6}};\] \[\Delta H=-1100\,kJ\] ?(i) \[\frac{1}{2}{{F}_{2}}(g)\xrightarrow{\,}\,F(g);\] \[\Delta H=80\,\,kJ\] ?(ii) \[S(s)\xrightarrow{\,}\,S(g);\,\,\Delta H=+275\,kJ\] ?(iii) Subtracting Eq. (iii) from Eq. (ii) with 6 gives \[S(g)+3{{F}_{2}}(g)\,\beta \,\xrightarrow{\,}\,6{{F}_{6}};\]\[\Delta H=-1100-275\] \[=-1375\,\,kJ\] ?(iv) Subtracting Eq. (ii) with 6 gives \[3{{F}_{2}}(g)\xrightarrow{\,}\,6F(g);\,\,\Delta H=6\times 80\,kJ\] \[=480\,kJ\] ?(v) Subtracting Eq. (v) from Eq. (iv). \[S(g)+6F(g)\xrightarrow{\,}\,S{{F}_{6}};\] \[\Delta H=-1375-480=-1855\,kJ\] or \[S{{F}_{6}}\xrightarrow{\,}\,S(g)+6F(g);\] \[\Delta H=+1855\,kJ\] Average \[BE=\frac{1855}{6}=309.17\,kJ\]
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