HW 8.73 (Bond Enthalpies

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Palmquist_Sierra_2N
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HW 8.73 (Bond Enthalpies

Postby Palmquist_Sierra_2N » Tue Jan 17, 2017 10:51 am

I'm kind of confused with the concept of bond enthalpies.
Estimate the reaction enthalpy for
a) 3 C2H2(g)-->C6H6(g)
b) CH4(g)+4 Cl2(g)-->CCL4(g)+4 HCL(g)

Ishita_Dubey_2D
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Re: HW 8.73 (Bond Enthalpies

Postby Ishita_Dubey_2D » Tue Jan 17, 2017 2:30 pm

To calculate the enthalpy of a reaction using bond enthalpies you would have to add together all of the bond enthalpies of new bonds formed (which will be a negative value because formation of bonds releases energy) and add this value to the total of the bond enthalpies of all of the bonds broken (which will be a positive value because breaking a bond requires an input of energy). Personally, it helps me a little more to draw out the structures of the reactants and products to see exactly which bonds are broken and which ones are formed. For example, for part a, 3 carbon-carbon triple bonds (reactants) are broken to form 3 carbon-carbon single bonds and 3 carbon-carbon double bonds in the resulting hydrocarbon ring structure.

Ann Zhang_1M
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Re: HW 8.73 (Bond Enthalpies

Postby Ann Zhang_1M » Thu Jan 19, 2017 1:21 am

use the sum of bond enthalpies of all products minus the sum of bond enthalpies of all reactants.
and do not forget to multiply each bond with their moles

Zoe Robertson 2H
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Re: HW 8.73 (Bond Enthalpies

Postby Zoe Robertson 2H » Tue Mar 14, 2017 12:12 pm

Why don't we include the C-H bonds? Is it because there are 3 moles of 2 C-H bonds broken and 6 C-H formed in the benzene ring, so they cancel out?

104781135
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Re: HW 8.73 (Bond Enthalpies

Postby 104781135 » Sat Mar 18, 2017 3:37 pm

I'm not getting the correct answer for part A. I get a -369kJ/mol but the answer is -597kJ/mol? Can anyone help me with this please?

104781135
Posts: 13
Joined: Wed Sep 21, 2016 2:56 pm

Re: HW 8.73 (Bond Enthalpies

Postby 104781135 » Sat Mar 18, 2017 3:49 pm

104781135 wrote:I'm not getting the correct answer for part A. I get a -369kJ/mol but the answer is -597kJ/mol? Can anyone help me with this please?



Never mind, I figured it out. C6H6 is benzene, and all of benzene's carbon bonds are resonance bonds, which have a different bond enthalpy than a single carbon bond or triple carbon bond.


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