8.73 Bond Enthalpy of Benzene

[Melissa Kulon 2D]

For problem 8.73, I'm looking at both the tables, 8.6 and 8.7, and neither have included the bond enthalpy for benzene and, looking at the notes from my discussion and at the solution manual, they both use an already calculated total value for benzene. How did they get this value?

[Christine Van 2E]

Hi!

Benzene has a ring structure of 6 carbons and one hydrogen attached to each carbon. Each C-C bond in the ring structure has resonance, and its bond enthalpy can be found in the tables. The bond enthalpy for C-C with a dashed line on top (indicating resonance) is 518 kJ/mol. For this problem in part (a), you will need to break 3 mols of C-C triple bonds with a bond enthalpy of 837 kJ/mol for each triple bond and you will form 6 mols of C-C resonance bonds.

Hope that helps!

[Aya Ghoneum 1H]

For Problem 8.73 and any other problem involving calculating reaction enthalpy using bond enthalpies, you need to look up tables (Table 8.7 in this case) for the bond enthalpies of the bonds being broken and formed. The total reaction enthalpy will be equal to all of the bonds being broken subtracted by all of the bonds being formed. For this problem, and for all problems involving bond enthalpies, it is most helpful to draw the Lewis structures of all the reactants and products (that makes it easier to see and count the number of bonds being broken and being formed). You will see that in ethyne, there is one carbon-carbon triple bond (837 kJ/mol) and two carbon-hydrogen (412 kJ/mol) single bonds. For benzene, there are six carbon-hydrogen single bonds (412 kJ/mol) and 6 carbon-carbon partial double bonds (518 kJ/mol). Taking the stoichiometric coefficients into account, the calculation should look like this:

[3(837 kJ/mol)+6(412 kJ/mol)]- [6(518 kJ/mol)+6(412 kJ/mol)]= -597 kJ/mol

I hope this helps!

[Stevie Wisz]

Regarding this problem, I just need clarification.... the more energy/enthalpy the more stable the molecule? For instance, the benzene structure that is higher in kj is the most stable, correct? :)

[Chem_Mod]

The higher the enthalpy, the more energy is required to break the bond. So yes, more stable.