Calculating Bond Enthalpies - #7.115

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Mara Lockhart 3J
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Calculating Bond Enthalpies - #7.115

Postby Mara Lockhart 3J » Mon Jan 19, 2015 11:08 am

On part (b) of problem 7.115, we are asked to use bond enthalpies to calculate the enthalpy of combustion of each fuel, assuming they burn to produce gaseous CO2 and gaseous H2O. For CH4, I know we have to find all of the enthalpies and add them together, but why do the products, CO2 and H20, have negative bond enthalpies? Is this because the bonds are formed?

Kayla Denton 1A
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Re: Calculating Bond Enthalpies - #7.115

Postby Kayla Denton 1A » Mon Jan 19, 2015 1:15 pm

Yes! Bond forming is an exothermic process because it is the reverse of bond breaking, which is endothermic (as heat is required to break bonds).

Denise 3L
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Re: Calculating Bond Enthalpies - #7.115

Postby Denise 3L » Mon Jan 19, 2015 4:48 pm

For part e of this question, how did they find the heat per mole of CO2 released for each gas?

Kayla Denton 1A
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Re: Calculating Bond Enthalpies - #7.115

Postby Kayla Denton 1A » Mon Jan 19, 2015 4:50 pm

It's the delta H of combustion divided by the number of moles of CO2 gas produced for each substance (which can be found by the balanced equation of combustion for each substance—methane, ethanol, and octane).

Denise 3L
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Re: Calculating Bond Enthalpies - #7.115

Postby Denise 3L » Mon Jan 19, 2015 5:05 pm

Ahh, because the answer for octane threw me off in the solutions manual. If the delta h of combustion for octane is -5471kj/mol and there are 2 mols of C02, shouldnt it be -2735.5 k/mol CO2 instead of -684kj/mol?

Neil DSilva 1L
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Re: Calculating Bond Enthalpies - #7.115

Postby Neil DSilva 1L » Mon Jan 19, 2015 5:20 pm

is the combustion reaction for octane, but since enthalpies are per mole, the reaction for which enthalpy is calculated is: . So there's 8 CO2 produced and you would divide the enthalpy of combustion by 8 to get -684 kJ/mol CO2 produced.

Denise 3L
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Re: Calculating Bond Enthalpies - #7.115

Postby Denise 3L » Mon Jan 19, 2015 8:29 pm

OH I forgot to do the balanced reaction, thank you so much!


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