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### 8.73 (a)

Posted: Wed Jan 24, 2018 1:28 am
I am having some trouble with 8.73(a). The question is "Use the bond enthalpies in Table 8.6 and 8.7 to estimate the reaction enthalpy for a) 3C2H2 --> C6H6"
I looked up the molecular structure for both molecules. C2H2 has two carbons bonded with a triple bond and hydrogens attached to the other side. The bond enthalpy is 837 kJ. C6H6 is a hexagon of carbons with alternating double and single bonds between the ring, 3 of each, with a hydrogen on each. It seems like 3 triple bonds between two Cs were broken and 3 double bonds (enthalpy is 612kJ) and 3 single bonds (348kJ) are formed. I did 3(847kJ) - 3(612) -3(348) and I got -339kJ which is the wrong answer. the correct answer is -597 kJ/mol. Why is that? What am I doing wrong?

### Re: 8.73 (a)  [ENDORSED]

Posted: Wed Jan 24, 2018 7:43 am
Your procedure was correct. The issue is that C6H6 is benzene which has resonance. Therefore, its bonds are better described by the value 518 kJ/mol as written with an asterisk in Table 8.7 (pg. 300) since this specifically refers to mean bond enthalpy for $\textup{C-C} \leftrightarrow \textup{C=C}$ resonance such as found in benzene.