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Offline farzin007

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Using standard enthalpies of formation
« 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


Offline Ligander

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Re: Using standard enthalpies of formation
« Reply #1 on: January 26, 2010, 03:36:03 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



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...

Offline farzin007

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Re: Using standard enthalpies of formation
« Reply #2 on: February 01, 2010, 02:44:21 PM »
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

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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?

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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


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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


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