7th edition 4E.5

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paytonm1H
Posts: 74
Joined: Fri Sep 28, 2018 12:18 am

7th edition 4E.5

Postby paytonm1H » Sat Feb 09, 2019 8:30 pm

Hi all. I am currently working on 4E.5 in the 7th edition textbook. The question asks to estimate the reaction enthalpy using bond enthalpies for the formation of benzene:
3C2H2(g) --> C6H6(g)
3 C-C triple bonds are broken and 6 C-C 1.5 bonds are formed. Why isn't the formation of 6 C-H bonds included?
Thanks!

Samantha Ito 2E
Posts: 64
Joined: Fri Sep 28, 2018 12:29 am

Re: 7th edition 4E.5

Postby Samantha Ito 2E » Sat Feb 09, 2019 8:35 pm

The C-H bonds of the 3C2H2 are never broken. Therefore, the C-H bonds do not need to be formed again to make C6H6.

Madeline Motamedi 4I
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

Re: 7th edition 4E.5

Postby Madeline Motamedi 4I » Sat Feb 09, 2019 9:16 pm

Keep in mind when using bond enthalpies you don't necessarily have to add every bond on the reactants side and subtract every bond on the products side. Instead you can just look at the lewis structures for the molecules and only add and subtract those that are going to be broken and formed.

Ruiting Jia 4D
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am

Re: 7th edition 4E.5

Postby Ruiting Jia 4D » Sun Feb 10, 2019 10:36 pm

paytonm1H wrote:Hi all. I am currently working on 4E.5 in the 7th edition textbook. The question asks to estimate the reaction enthalpy using bond enthalpies for the formation of benzene:
3C2H2(g) --> C6H6(g)
3 C-C triple bonds are broken and 6 C-C 1.5 bonds are formed. Why isn't the formation of 6 C-H bonds included?
Thanks!


how did you conclude that 6 C-C 1.5 bonds were formed?


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