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Topic: Water/Ethanol Azeotrope  (Read 30021 times)

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Offline farlsbarkley

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Water/Ethanol Azeotrope
« on: December 15, 2009, 08:46:22 PM »
This was a question I got wrong on a test.  I didn't understand my professor's explanation, and I was hoping for some assistance.

The question is below.  I got part (a) correct (the diagram), but apparently my answer for (b) was totally wrong.

Question:
At 1 atm, water boils at 100 C and ethanol boils at 78.5 C.  There is an azeotrope that boils at 78.2 C that is 0.96 ethanol by mole fraction.

a)  Draw a boiling point vs. mole fraction (of ethanol) diagram for ethanol/water, showing the boiling points of the pure phases and the azeotrope, as well as (approximate) lines defining the liquid and vapor compositions and the liquid/vapor coexistence region.

b)  What would the composition of the first and last drops distilled be for a perfect fractional distillation if the mixture of the liquid was (in mole fraction ethanol): 0.99 and 0.1.  (4 answers needed)

My Answer:
This is the diagram I drew for the first part:
http://img8.imageshack.us/img8/9361/azeotrope.jpg

For part (b), I answered that the last drops distilled from either mixture (0.99 or 0.10) would have the azeotropic composition (96% ethanol).  However, I could not say what the composition of the first drops would be, except that the first drop distilled from 0.99 ethanol would have a composition less than 0.99 but more than 0.96, and the first distilled from the 0.10 mixture would have a composition more than 0.10 but less than 0.96.

I thought my reasoning directly followed from this diagram in Atkins:

http://img694.imageshack.us/img694/5229/azeotropeatkins.jpg

but apparently I'm wrong.  My professor said the first drop distilled will have the azeotropic composition in both cases, and the last drop distilled will be the specified compositions (ie. 0.99 or 0.10), but I have no clue why that is.

Can someone assist me?

Offline Grundalizer

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Re: Water/Ethanol Azeotrope
« Reply #1 on: December 16, 2009, 12:22:00 AM »
Wow, what class is this from?  What college?  What year?  I'm a Jr. and haven't done anything like this.  Sorry I can't help you, but now I'd like to learn about this stuff. 

Offline farlsbarkley

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Re: Water/Ethanol Azeotrope
« Reply #2 on: December 16, 2009, 05:47:38 AM »
This was from my physical chem (really classical thermodynamics) undergraduate class at CUNY.  I'm not sure what year it's normally taken, because I'm not really a chem major.  I'm taking it to prepare to go back to grad school for nanotech in the fall.  If you're interested, take a look at the thermo material at MIT's OpenCourseWare site:

http://ocw.mit.edu/OcwWeb/Chemistry/5-60Spring-2008/CourseHome/index.htm

My class didn't get into the statistical mechanics, so we only covered the equivalent of the first 23 lectures, but those notes and videos helped me tremendously.

I still can't wrap my head around this azeotrope problem.  I watched the MIT lecture at http://www.youtube.com/watch?v=wCSl5eeMSDY&feature=SeriesPlayList&p=A62087102CC93765 and he discusses them around 28:30.  He seems to be saying that if you take a sample of liquid with a composition that lies to the left of the B.P. curve's minimum and heat it, it will begin to boil at the point where the vertical line meets the B.P. curve.  At that temp, the liquid will have the original composition and if you draw a horizontal tie line to the dew point curve it will tell you the composition of the vapor.  You would then ignore the original liquid sample and condense the vapor (not changing its composition) to a new liquid.  That newly condensed liquid is what I assume my prof meant by the "first drop."  Since that first drop has a composition farther to the right, it is richer in whichever component is indicated by the horizontal axis.  If you repeat the process, you eventually reach the minimum (which is the azeotrope composition), and the vapor composition will be the same as the liquid composition, preventing further distillation (the "last drop").

I'm beginning to think that either I was right, or I'm misinterpreting something.  I really hope somebody can clarify.

Offline savy2020

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Re: Water/Ethanol Azeotrope
« Reply #3 on: December 16, 2009, 09:10:21 AM »
For part (b), I answered that the last drops distilled from either mixture (0.99 or 0.10) would have the azeotropic composition (96% ethanol).  However, I could not say what the composition of the first drops would be, except that the first drop distilled from 0.99 ethanol would have a composition less than 0.99 but more than 0.96, and the first distilled from the 0.10 mixture would have a composition more than 0.10 but less than 0.96.

My professor said the first drop distilled will have the azeotropic composition in both cases, and the last drop distilled will be the specified compositions (ie. 0.99 or 0.10), but I have no clue why that is.
As a high school senior I've learnt about colligative properties but I'm not familiar with the azeotropes{I know what they mean but quantitatively I've no idea how to proceed}
But for the past hour or so I've searched the internet and learnt something interesting-Thanks to you :)

First of all check this site http://www.chemguide.co.uk/physical/phaseeqia/nonideal.html {you might be knowing all that}
After reading that I feel that what you did is indeed correct.
So cheers ;D

Concerning part A, I would like to tell you that the azeotropic mixture is 96% by mass. The mole fraction comes out to be 0.895 as pointed out in the site I mentioned above. So in your diagram as the horizontal axis is XEtOh the point of azeotropic boiling should be at XEtOH=0.895

« Last Edit: December 16, 2009, 09:37:55 AM by savy2020 »
:-) SKS

Offline savy2020

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Re: Water/Ethanol Azeotrope
« Reply #4 on: December 16, 2009, 02:40:24 PM »
After reading that I feel that what you did is indeed correct.
So cheers ;D
OHh!! After about 3hrs I again thought of that..
Now I feel that what I thought to be correct is indeed incorrect. :-[  :(
I'm sorry for that.
Your professor is correct.

Let me explain how ...                           {I've attached the part of Atkins explaining the diagram.}
Now you take some mixture of composition a1 (as in the figure and text of the book). Then it is heated it to it's boiling point which corresponds to the point a2. When this liquid at a2 vaporises it forms vapor at a2' having different composition compared to the liquid at a2. This vapor at a2' comes to equillibrium with it's condensed form. i.e, it is condensed to liquid at a3. {This liquid doesn't mix with the liquid in the distillation flask because it is formed in the fractionating column}.
Now this liquid again comes to equillibrium with it's vapor at a3' and so on...Thus finally the drop distill out as vapor at b whose composition is azeotropic composition.
So the first drop in both the cases has azeotropic composition as said by your professor.
[I feel that it's correct. If not please point out my mistake I'll be glad to rectify ]

Now about the last drop.
First you think of it and put forth your ideas. I'm exhausted.

BUT I thank you Very very much. I've learnt these things today because of you!!! ;D
:-) SKS

Offline savy2020

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Re: Water/Ethanol Azeotrope
« Reply #5 on: December 16, 2009, 04:07:35 PM »
This was from my physical chem (really classical thermodynamics) undergraduate class at CUNY.  I'm not sure what year it's normally taken, because I'm not really a chem major.  I'm taking it to prepare to go back to grad school for nanotech in the fall.  If you're interested, take a look at the thermo material at MIT's OpenCourseWare site:

http://ocw.mit.edu/OcwWeb/Chemistry/5-60Spring-2008/CourseHome/index.htm

My class didn't get into the statistical mechanics, so we only covered the equivalent of the first 23 lectures, but those notes and videos helped me tremendously.

I still can't wrap my head around this azeotrope problem.  I watched the MIT lecture at http://www.youtube.com/watch?v=wCSl5eeMSDY&feature=SeriesPlayList&p=A62087102CC93765 and he discusses them around 28:30.  He seems to be saying that if you take a sample of liquid with a composition that lies to the left of the B.P. curve's minimum and heat it, it will begin to boil at the point where the vertical line meets the B.P. curve.  At that temp, the liquid will have the original composition and if you draw a horizontal tie line to the dew point curve it will tell you the composition of the vapor.  You would then ignore the original liquid sample and condense the vapor (not changing its composition) to a new liquid.  That newly condensed liquid is what I assume my prof meant by the "first drop."  Since that first drop has a composition farther to the right, it is richer in whichever component is indicated by the horizontal axis.  If you repeat the process, you eventually reach the minimum (which is the azeotrope composition), and the vapor composition will be the same as the liquid composition, preventing further distillation (the "last drop").

I'm beginning to think that either I was right, or I'm misinterpreting something.  I really hope somebody can clarify.
Hey!! I've somehow missed this post. Sorry for that. I've seen it just now.
OK. The actual problem is that you're misinterpreting the first and last drop.
First drop of the liquid which evaporates and distills out is what's meant by 'the first drop'
and the last drop is last drop of the solution which distills.
Now I think it'll be clear
:-) SKS

Offline farlsbarkley

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Re: Water/Ethanol Azeotrope
« Reply #6 on: December 16, 2009, 05:04:06 PM »
Savy - I really appreciate your interest in this.  First of all, as is mentioned in the excerpt from Atkins you supplied, you were absolutely right about the azeotrope forming at 96% ethanol by mass not mole fraction.  I double checked, and my professor actually made the mistake in the problem wording - so my diagram was only right based on the info given.

That doesn't affect the rest of the problem though, which I am starting to understand with your help.  My problem was essentially that I have not actually performed this experiment, and so, as you pointed out, I misinterpreted the first and last drop.  I looked more into the procedure and it is clearer now why the first drop is the azeotrope composition (because as the substance condenses and vaporizes on each segment of the fractionating column, it gets closer to the minimum of the graph).  I had originally thought that the "first drop" referred to the very first drop of vapor condensed at the bottom of the fractionating column, though I had no clue as to the experimental setup.

So that's all well and good, but I'm still unsure as to how to determine the last drop composition.  Again, not having done the experiment, I'm not sure how to think about it.

Thanks again for your efforts.  I'm glad at least one of us is learning something!

Offline savy2020

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Re: Water/Ethanol Azeotrope
« Reply #7 on: December 17, 2009, 09:22:58 AM »
So that's all well and good, but I'm still unsure as to how to determine the last drop composition.  Again, not having done the experiment, I'm not sure how to think about it.
You don't need to do the experiment to know something. You can, most of the time,  predict the results using theory.
The attached diagram is a good one representing the apparatus I've found in google search.

Regarding the last drop, first try to answer these questions ;)
As the distillation occurs, do you think that the concentration of the solution remains the same?
If not how will it change?{i.e, whether concentration increases/decreases} and why?
Finally what will you be left with in the distillation flask?

Did you go through the site I've mentioned above? It is one of the web links provided at the atkins orc
:-) SKS

Offline clriley

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Re: Water/Ethanol Azeotrope
« Reply #8 on: February 28, 2010, 08:28:52 PM »
We just did this in lab with Ethyl acetone and Ethanol. What these guys have said is indeed correct and it really helps when you look at the graph.

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