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In terms of general stability, molecules are more stable when they have higher bond enthalpies. This is because the higher the bond enthalpy, the more energy is required to break that bond. With regard to this problem in particular, the Kekule structures, I think, are referring to the resonance structures of benzene. If you draw the lewis structure of benzene, you will notice that it has resonance structures and that they have 3 carbon double bonds and 3 carbon single bonds. In reality, the electrons are evenly shared between the six carbon bonds, creating 6 resonance stabilized bonds. If you find the sum of bond enthalpies of the resonance structures of benzene and compare it to benzene with the 6 resonance stabilized bonds, you will find that having 6 resonance stabilized bonds is more stable because it has a higher total bond enthalpy.
Yeah, in the question the lower bond energies = the lower molar energies. the 3 C-C bonds and C(double bond)C bonds has the value 2880kJ and the 6 resonance-stabilized bonds has 3108kJ so 3108kJ is more stable by 228kJ, as it takes that much more energy.
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