Homework Problem 8.73

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Rachel Lipman
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Joined: Fri Sep 25, 2015 3:00 am

Homework Problem 8.73

Postby Rachel Lipman » Thu Jan 14, 2016 3:22 pm

Starting with example (a), I understand the Carbon bonds are being broken, however, I am confused why the solutions manual explains the Carbon-Carbon bonds being broken as triple bonds. When I consult the Mean Bond Enthalpies Table (8.7), there are a few options for Carbon-Carbon bonds. Is there a certain explanation or tactic to determine what variation of the bonds are being broken, as well as being formed?

Reine Nakamura 1C
Posts: 38
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework Problem 8.73

Postby Reine Nakamura 1C » Thu Jan 14, 2016 8:03 pm

For these problems, it helps me to draw out the Lewis structures for both the products and reactants... If you draw out the C2H2 you see that there is indeed a triple bond between the 2 carbons and a single bond between each carbon and a hydrogen. Since there is a coefficient of 3 in front of the C2H2 there will be 3 of these triple bonds broken. As far as bonds formed, you can draw the C6H6 (benzene) which is a ring of carbons with a hydrogen bonded to each of the 6 carbons. Because benzene has resonance (alternates single and double bonds between carbons in the ring), each of the bonds are an average (about 1.5 bonds). This is denoted by the 2 C's with a dashed line on top of a solid line in Table 8.7. Benzene has 6 of these bonds, so you will multiply that value by 6. Finally, you will take the (sum of the mean bond enthalpies of the bonds broken) - (sum of the mean bond enthalpies of the bonds formed) to get the standard reaction enthalpy.

Maya Schnall 3J
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Joined: Fri Sep 25, 2015 3:00 am

Re: Homework Problem 8.73

Postby Maya Schnall 3J » Thu Jan 14, 2016 8:14 pm

It's extremely helpful to draw the Lewis Structures of these molecules before finding enthalpies because that will illustrate the specific bonds you have to break so that the product may be formed with its bonds. For the reactants, you'll see that the Lewis structure is H-C:::C-H meaning that you have to fully break that triple bond before creating any new bonds. (How you choose which enthalpy value to use)

Rachel Lipman
Posts: 45
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework Problem 8.73

Postby Rachel Lipman » Sun Jan 17, 2016 1:04 pm

Thank you Reine. An addition question I have is about calculating bond enthalpies in general. Is it most cases that bonds are broken and new ones formed?

Reine Nakamura 1C
Posts: 38
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework Problem 8.73

Postby Reine Nakamura 1C » Mon Jan 18, 2016 12:24 pm

Yes, by calculating the energy of the bonds broken and bonds formed, you can determine the enthalpy change for the reaction (calculate the value and see if it's exothermic or endothermic)

And as far as calculating bond enthalpies, Tables 8.6 and 8.7 on pages 300 and 301 have the values for various bonds. Just multiply those by the number of moles of bonds that you have for that particular type of bond. For example if you have 2H2 as a reactant and all the H-H bonds are being broken, you'd multiply the bond enthalpy for H2 (436 kJ/mol) by 2 mol.


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