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Topic: Heat of formation and heat of vaporization  (Read 4935 times)

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Offline Enantiomer

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Heat of formation and heat of vaporization
« on: October 11, 2010, 01:41:25 AM »
Hi guys,
I'm working on a lab right now to figures out the resonance energy in benzene and one part of the lab asks

"Using standard heats of formation for benzene and cyclohexane found in the tables, and your ∆fH of your hydrocarbon (1,5,9-trans-trans-cis-cyclododecatriene) values you previously calculated (this value was 7291 kJ/mol) calculate standard enthalpy change and energy change for the following gas phase reaction at (STP).  Report your answers in kJ/mol.
 (C12H18)g  ==> (C6H6)g + (C6H12)g
Show that this requires the standard enthalpy of vaporization of cyclododecatriene    "
That last part is where I'm lost, I realize that my original value calculated was for the formation of cyclododecatriene based off of it being a liquid but I'm not sure if I should be using the vapor value to show a conversion of the liquid to the vapor or what? 
Using trouton's rule I calculated my Heat of vaporization to be 45.76 kJ/mol
I think it might be asking me to compare the enthalpic value of cyclododecatriene as a liquid to the gas giving me:
-124.6kJ/mol+82.93kJ/mol-7291kJ/mol = -7332.6
vs.
-124.6kJ/mol+82.93kJ/mol-45.76 kJ/mol = -87.43
to give us a much smaller entropic value

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