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Enthalpy and stoichiometry

  • Thread starter Agent M27
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  • #1
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Homework Statement


The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is 5084.5 Kj. If the change in enthalpy is 5074.2 Kj, how much work is done during the combustion? Find work in Kj.



Homework Equations



[tex]\Delta[/tex]E=[tex]\Delta[/tex]H + (-P[tex]\Delta[/tex]V)

w=-P[tex]\Delta[/tex]V

The Attempt at a Solution



[tex]\Delta[/tex]E=5084.5Kj
[tex]\Delta[/tex]H=5074.2Kj

5084.5-5074.2=-1atm([tex]\Delta[/tex]V)
-10.3=[tex]\Delta[/tex]V

w=-(1)(-10.3)=10.3

10.3L*atm x 101.3J=1043.39J

[tex]\frac{1043.39}{1000}[/tex]= 1.04339 Kj

This is an online homework set, so when I input the answer it usually will give me a hint if I am close, but I am not getting anything. Can anyone spot my error? Thanks in advance.

Joe
 

Answers and Replies

  • #2
I don't know if you have already figured this out, but I thought I'd let you know I found your problem. The way you solved it, your change in volume ended up in kJ/atm when it should be in liters. I just did a track and took 10300J and divided it by 101.3 J to end up with about 101.68 L.

10.3kJ/atm x 1000J/1kJ x 1 Latm/101.3J= 101.68 L.

Hopefully that helped if you haven't already figured it out!
 

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