## 8.77

sofiakavanaugh
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
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### 8.77

“Benzene is more stable and less reactive than would be predicted from its Kekulé structures. Use mean bond enthalpies to calculate the lowering in molar energy when resonance is allowed between the Kekulé structures of benzene.”

I don’t really know what they are talking about in this problem (molar energy, kekule structures, etc), and would appreciate any help on this one.

Thanks!

Tasnia Haider 1E
Posts: 55
Joined: Sat Jul 22, 2017 3:01 am

### Re: 8.77

This problem is just allowing you to see how benzene'S resonance allows it to have a lower molar energy than the different resonance structures that we can draw it of the c-c and c=c bonds. The kekule structure is the ones that we've drawn out with the 3 single and 3 double carbon bonds, while resonance structure is talking about the delocalized electrons that are moving around.

Ozhen Atoyan 1F
Posts: 50
Joined: Thu Jul 27, 2017 3:01 am
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### Re: 8.77

If you do not let the wording confuse, you you have to do the same as 8.67 (c) and (d). This means you have to calculate the enthalpy of benzene with and without resonance. You do this with the mean bond enthalpies that can be found in a chart. It is one of the methods that our professor explained to calculate the enthalpy with the bonds for a given reaction.

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

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