Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: farzin007 on January 26, 2010, 02:23:15 PM
-
I'm using standard enthalpies of formation to determine the enthalpy change for the the combustion of octane. Here is my work:
2 C8H18 + 25 O2 → 16 CO2 + H2O
∆Hºf = 250.1 (C8H18)
∆Hºf = 0 (O2)
∆Hºf = -393.5 (CO2)
∆Hºf = -241.8 (H2O(g))
∆Htarget = [ ( -393.5 kJ / mol×K × 16 mol ) + ( -241.8 kJ / mol×K × 1mol ) ] - [ ( 250.1 kJ / mol×K × 16 mol ) + ( 0 kJ / mol×K × 2mol ) ]
= - 10539.4 kJ
Am I doing this right? Was I correct to use the gas instead of liquid molar enthalpy for H2O?
Thanks in advanced
-
I'm using standard enthalpies of formation to determine the enthalpy change for the the combustion of octane. Here is my work:
2 C8H18 + 25 O2 → 16 CO2 + H2O
∆Hºf = 250.1 (C8H18)
∆Hºf = 0 (O2)
∆Hºf = -393.5 (CO2)
∆Hºf = -241.8 (H2O(g))
∆Htarget = [ ( -393.5 kJ / mol×K × 16 mol ) + ( -241.8 kJ / mol×K × 1mol ) ] - [ ( 250.1 kJ / mol×K × 16 mol ) + ( 0 kJ / mol×K × 2mol ) ]
= - 10539.4 kJ
Am I doing this right? Was I correct to use the gas instead of liquid molar enthalpy for H2O?
Thanks in advanced
a) Wikipedia says that standard enthalpies of formation of different isooctanes lie between -50 and -55, and it is very far from +250
b) In your equation you have 36 hydrogenes in the left side and only two at the right side.
c) Standard enthalpy of formation is measured in kJ / mol , but not in kJ / mol×K
d). and so on.... too many mistakes...
-
OK, I think I've corrected my errors:
chemical equation
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
Standard molar entropies ( kJ / mol ):
products
∆Hºf = -393.5 (CO2)
∆Hºf = -241.8 (H2O(g))
reactants
∆Hºf = 0 (O2)
∆Hºf = -49.8 (C8H18)
formula
∆Htarget = ∑∆Hknown
∆Hº = [ ( -393.5 kJ / mol × 16mol ) + ( -241.8 kJ / mol × 18mol ) ] -
[ ( -49.8 kJ / mol × 16 mol ) + ( 0 kJ / mol × 25mol ) ]
= - 9843.6
--------
Did I do it all correctly?
This is for the combustion for Octane in fuel for a car engine. so my question is, should I be using the enthalpy of H2O(g) or H2O(l) for water?
--------
for the second part of my assignment, i need to determine Q for octane. Here is the data I will be using:
Octane:
m - mass (grams): 1
T1 - initial temperature (ºC): 22.75
T2 - final temperature (ºC): 31.9
c - constant (kj / kg °C): 4.84
here is how i derived Q:
Q = m × c × ∆T
Q = 0.001 kg × 4.84 kJ/kg °C × 9.15 °C
Q = 44.286 kJ
--------
Am I doing this part correctly? is "c" suppose to be represented in kj / kg °C or KJ/°C? how do i compare these Q values to the ∆Hº to determine the percentage error? Thanks again