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Topic: Molar enthalpy of vaporization  (Read 19211 times)

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Offline Kate

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Re: Molar enthalpy of vaporization
« Reply #15 on: March 16, 2008, 09:30:42 PM »
Well, I wasn't taking into consideration the atmospheric pressure. I got it all right now and the molar enthalpy of heat I got was more or less 43,2 kJ / mol (equal to 10.33 kcal/mol)

Equations I don't know. Handbook I have ;)

10.33 kcal/mol.

Are you sure ? Because I searched on the internet (wikipedia) and the value I found was 38.56 kJ/mol (equal to 9.21 kcal/mol)

Anyway, my question remains: does anybody know, besides the Clapeyron equation and Trouton's rule, other equations that can be used to predict the values of molar enthalpy of heat ?

Offline Borek

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Re: Molar enthalpy of vaporization
« Reply #16 on: March 17, 2008, 03:46:15 AM »
Equations I don't know. Handbook I have ;)

10.33 kcal/mol.

Are you sure ? Because I searched on the internet (wikipedia) and the value I found was 38.56 kJ/mol (equal to 9.21 kcal/mol)

10.33 at temp 20 deg C
9.41 at temp 78.3 deg C (boiling).
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Offline Kate

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Re: Molar enthalpy of vaporization
« Reply #17 on: March 17, 2008, 03:21:44 PM »
So, which of the two should I be interested the most ?

Offline Yggdrasil

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Re: Molar enthalpy of vaporization
« Reply #18 on: March 17, 2008, 03:53:06 PM »
Considering the temperatures used in your procedure, which one do you think will be more relevant?

Offline Kate

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Re: Molar enthalpy of vaporization
« Reply #19 on: March 17, 2008, 06:16:29 PM »
I'd say the second one, 9.41 at 78.3 ºC (the temperatures I used were 60.2 ºC and 52.8 ºC), but I hope not since the value I got is a bit different. Anyway, which one is it ?

EDIT: I have a physical chemistry book and it says that 43.5 kJ/mol is the standard enthalpy of heat at the transition temperature. I'm not sure what this transition temperature means but could they be referring to the temperature at which ethanol boils (78.35 ºC) ?
« Last Edit: March 17, 2008, 09:03:06 PM by Kate »

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