Moderators: Chem_Mod, Chem_Admin

Natalie LeRaybaud 1G
Posts: 54
Joined: Thu Jul 13, 2017 3:00 am


Postby Natalie LeRaybaud 1G » Tue Jan 16, 2018 5:45 pm

Can someone please explain how to go about doing this problem?

Benzene is more stable and less reactive than would be predicted from its Kekulé structures. Use the mean bond enthalpies in Table 8.7 to calculate the lowering in molar energy when resonance is allowed between the Kekulé structures of benzene.

Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.77

Postby ConnorThomas2E » Tue Jan 16, 2018 7:56 pm

In order to do this problem, you would first calculate the energies to break the bonds of 3 double bonded carbons and 3 single bonded carbons. You would then compare this value to 6 resonance carbon bonds, and see that the resonance structure requires more energy to break the bonds, and it is therefore more stable.

Ramya Lakkaraju 1B
Posts: 67
Joined: Fri Sep 29, 2017 7:03 am

Re: 8.77

Postby Ramya Lakkaraju 1B » Tue Jan 16, 2018 10:08 pm

In the table of enthalpies, there is a specific number for the bonds found in a benzene ring.

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

Who is online

Users browsing this forum: No registered users and 1 guest