Ch. 8 #67

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Ammarah 2H
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Joined: Fri Sep 29, 2017 7:06 am

Ch. 8 #67

Postby Ammarah 2H » Tue Jan 16, 2018 10:28 am

How do you do part C and D? Thank you!

Sohini Halder 1G
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Re: Ch. 8 #67

Postby Sohini Halder 1G » Tue Jan 16, 2018 12:27 pm


Benzene without resonance means that there are 3 single bonds and 3 double bonds in the Lewis structure you will use for reference.
It helps if you draw out the structure so you can visually see which bonds are being formed!

6 C(s) + 3 H2(g) --> C6H6(l)
Breaking: 3 H-H bonds, using energy
Forming: 6 C-H bonds, 3 C-C single bonds, and 3 C=C double bonds, releasing energy
Also using energy to convert C(s) to C(g), which is using enthalpy of sublimation given of C
Also releasing energy to convert C6H6(g) to final product form C6H6(l)

Using the tables provided, you get:
Breaking bonds: 1308 kJ
Forming bonds: -5352 kJ
C(s) to C(l): 4302 kJ
C6H6(g) to C6H6(l): -30.8 kJ

Therefore, the enthalpy of formation for liquid benzene without resonance is 227 kJ/mol


Essentially you are doing the same steps except now we are including resonance, so you are forming 6 C-C bonds that are intermediate (1.5) between single and double. Info for that is also given in the textbook table.

The enthalpy of formation for liquid benzene with resonance is -1 kJ/mol

Clara Rehmann 1K
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Joined: Fri Sep 29, 2017 7:03 am

Re: Ch. 8 #67

Postby Clara Rehmann 1K » Tue Jan 16, 2018 12:30 pm

You are calculating the enthalpy of formation, which means you need to calculate the enthalpies for each bond broken and formed. Approach the calculation for C6H6 the same way you did parts (a) and (b) - in part (c), take into account the bonds in a standard structure, and in part (d), take into account the bonds of a resonance structure.

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

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