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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Kate on March 12, 2008, 01:37:44 PM

Title: Molar enthalpy of vaporization
Post by: Kate on March 12, 2008, 01:37:44 PM: Hello. :)
The other day in class I did an experiment with the purpose of determining the molar enthalpy of vaporization (or heat) of ethanol based on the variation of its boiling point with the exterior pressure.
Basically, I need to calculate the molar enthalpy of heat of ethanol considering first the ideal vapor phase and then the real behaviour of the vapor phase. I'm not sure if anyone is familiar with this but, anyway, I was wondering if someone could shed me some light on this as I have no idea where to even start.
Thanks.
Title: Re: Molar enthalpy of vaporization
Post by: Yggdrasil on March 12, 2008, 01:53:48 PM: Sounds like reading about the Clausius-Clapeyron equation in your chemistry text or on wikipedia (http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation) would be a good starting point.
Title: Re: Molar enthalpy of vaporization
Post by: Kate on March 12, 2008, 02:35:01 PM: So, if I want to calculate the molar enthalpy of heat of ethanol considering the ideal vapor phase I have to use this equation...
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fb.imagehost.org%2F0244%2FCC1_8.jpg&hash=b1b44cb769408ef0e87a99c362661925fef7d4cf) (http://b.imagehost.org/download/0244/CC1_8.bmp)

...right ?

And in order to calculate the molar enthalpy of heat of ethanol considering the real behaviour of the vapor phase then I have to use this equation...

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fb.imagehost.org%2Ft%2F0244%2FCC.jpg&hash=e96dcfc861d67a774a4ddaf61c0644bf3f9d8968) (http://b.imagehost.org/view/0244/CC.png)

...right ? Or wrong ?
Title: Re: Molar enthalpy of vaporization
Post by: Yggdrasil on March 12, 2008, 03:20:43 PM: Both are equations that you posted are equivalent statements of the same equation. I'm not quite sure what they mean by ideal and real vapor pressure data.
Title: Re: Molar enthalpy of vaporization
Post by: Kate on March 12, 2008, 04:02:14 PM: Here are the values that I have:

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fb.imagehost.org%2F0245%2FT_8.jpg&hash=9042e2517d72e04814dd5689d200ee3ecd00e023) (http://b.imagehost.org/download/0245/T_8.bmp)

I used the second equation and the value I got for enthalpy was negative. Is this normal ? I'm assuming not. So what am I doing wrong ?
Title: Re: Molar enthalpy of vaporization
Post by: Kate on March 13, 2008, 03:36:14 PM: ANYONE ??
Title: Re: Molar enthalpy of vaporization
Post by: Yggdrasil on March 13, 2008, 05:28:06 PM: Shouldn't the vapor pressure increase with temperature?
Title: Re: Molar enthalpy of vaporization
Post by: Kate on March 13, 2008, 06:07:37 PM: According to my teacher (he was there when I did the experiment) no, it shouldn't. As you increase the pressure, the temperature (bp) starts lowering. What's your opinion ?
Title: Re: Molar enthalpy of vaporization
Post by: Yggdrasil on March 13, 2008, 09:33:13 PM: Does water boil at a lower temperature at sea level or at high altitude (where there is lower atmospheric pressure)?
Title: Re: Molar enthalpy of vaporization
Post by: Kate on March 14, 2008, 04:40:09 AM: The water boils at a lower temperature at higher altitudes. But the pressure values above don't correspond to the atmospheric pressure. The pressure values that I posted are the ones that I read using a manometer which was attached to some kind of vaccum machine (not sure how to say this).
Title: Re: Molar enthalpy of vaporization
Post by: Borek on March 14, 2008, 05:09:36 AM: Does it matter whether you decrease pressure in your system using pump or moving to high altitude?
Title: Re: Molar enthalpy of vaporization
Post by: Kate on March 14, 2008, 05:21:38 AM: Quote from: Borek on March 14, 2008, 05:09:36 AM
Does it matter whether you decrease pressure in your system using pump or moving to high altitude?
Oh, so the pump is used in order to decrease the pressure in the system. So, in order to calculate the pressure values to be used in the Clapeyron equation that correspond to the temperatures (bp) that I posted above, I must subtract to the atmospheric pressure the pressure values that I already know (that I read in the manometer). Is this it ?
Title: Re: Molar enthalpy of vaporization
Post by: Borek on March 14, 2008, 06:11:33 AM: I haven't seen your experimental setup, so I can't be sure what you have measured - but I doubt. I would expect bp to go down together with the pressure.
Title: Re: Molar enthalpy of vaporization
Post by: Kate on March 16, 2008, 07:46:45 PM: Thanks to both of you for helping me out. I have figured out what was wrong already. Anyway, apart from the Clapeyron equation and Trouton's rule, is there any other way to determine ethanol's molar enthalpy of heat ? Any other equations that can be used ? Thanks again.
Title: Re: Molar enthalpy of vaporization
Post by: Borek on March 16, 2008, 07:54:05 PM: Quote from: Kate on March 16, 2008, 07:46:45 PM
Thanks to both of you for helping me out. I have figured out what was wrong already.

Can you tell us? It can be usefull for someone one day.

Quote from: Kate on March 16, 2008, 07:46:45 PM
Any other equations that can be used ?

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

10.33 kcal/mol.
Title: Re: Molar enthalpy of vaporization
Post by: Kate 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)

Quote from: Borek on March 16, 2008, 07:54:05 PM
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 ?
Title: Re: Molar enthalpy of vaporization
Post by: Borek on March 17, 2008, 03:46:15 AM: Quote from: Kate on March 16, 2008, 09:30:42 PM
Quote from: Borek on March 16, 2008, 07:54:05 PM
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).
Title: Re: Molar enthalpy of vaporization
Post by: Kate on March 17, 2008, 03:21:44 PM: So, which of the two should I be interested the most ?
Title: Re: Molar enthalpy of vaporization
Post by: Yggdrasil on March 17, 2008, 03:53:06 PM: Considering the temperatures used in your procedure, which one do you think will be more relevant?
Title: Re: Molar enthalpy of vaporization
Post by: Kate 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) ?