Enthalpy and stoichiometry


by Agent M27
Tags: enthalpy, stoichiometry
Agent M27
Agent M27 is offline
#1
Mar3-10, 07:49 PM
P: 171
1. The problem statement, all variables and given/known data
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.



2. Relevant equations

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

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

3. 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
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
nessiejo236
nessiejo236 is offline
#2
Mar9-10, 09:39 PM
P: 1
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!


Register to reply

Related Discussions
compound contains only carbon, hydrogen, nitrogen, and oxygen. Biology, Chemistry & Other Homework 2
Enthalpy of Formation vs. Bond Enthalpy Biology, Chemistry & Other Homework 7
Stoichiometry and reactions Biology, Chemistry & Other Homework 2
Enthalpy of Dissolution vs. Enthalpy of Solution Biology, Chemistry & Other Homework 4
Stoichiometry Introductory Physics Homework 18