HW Question 8.73

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Sunia Khan 2E
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Joined: Fri Sep 25, 2015 3:00 am

HW Question 8.73

Postby Sunia Khan 2E » Sun Jan 17, 2016 9:55 pm

I understand for HW question 8.73 that you need to break the triple bond between the two carbons in order to form the 6 mol C-C bonds with resonance in benzene, but why don't you need to break the 3 mol of C-H bonds in order to form 6 mol of C-H bonds in benzene? Why are only the bonds broken and formed between two carbons accounted for when estimating the reaction enthalpy in this problem?

504650118
Posts: 11
Joined: Fri Sep 25, 2015 3:00 am

Re: HW Question 8.73

Postby 504650118 » Sun Jan 17, 2016 10:42 pm

You don't need to break the C-H bonds of C2H2 because these bonds reappear in the product. On the reactant side, there is a total of 6 C-H bonds. Each C2H2 has two C-H bonds, and there are 3 mols of C2H2 (2x3=6). The product, benzene, has a ring structure with 6 carbons, and each carbon has a H attached. Thus, there are 6 C-H bonds in benzene. You don't need to break any of them because the number of C-H bonds is the same on the reactant and product side. The bond enthalpy values for C-H bonds will cancel out.


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