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Topic: Are these binary mixtures going to react?  (Read 5192 times)

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Offline jcp

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Are these binary mixtures going to react?
« on: November 08, 2007, 06:26:42 AM »
Hi all;

I would like to know if anyone can assist me by telling me whether the pairs of chemicals below are going to react if they are heated to 100 degrees Celcius (373 Kelvin).

1.   1-Propanol + Butanoic Acid
2.   1-Propanol + Acetaldehyde(Ethanal)
3.   2-Propanone(Acetone) + 2-Butanol
4.   2-Propanone(Acetone) + Propanoic Acid.

I want to carry our binary vapour-liquid equilibrium measurements with the above mixtures and if they react with each other then it cant be done. I already know that the propanol and butanoic acid will react to form an ester - can anyone tell me about the other mixtures and what reactions I should look out for?

Any assistance will be greatly appreciated.

Thank you.

Jeremy
« Last Edit: November 09, 2007, 12:49:59 AM by jcp »

Offline ARGOS++

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Re: Are these binary mixtures going to react?
« Reply #1 on: November 08, 2007, 12:59:12 PM »
Dear JCP,

If you consult the following Kp-Table and I assume you meant "at 1.0 Atm.", how much would be left for any reaction?  Otherwise: Other conditions – other game.

Ethanal                  20.2° Celsius
Acetone                 56.7°     “
1-Propanol              97.2°     “
1-Butanoic  Acid  100 … 101° Celsius
1-Propanoic Acid     140.6°        “

 
Good Luck!
                    ARGOS++

P.S.: (Small Note: 100.0° Celsius should be 373 °Kelvin and a little bit.)

Offline jcp

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Re: Are these binary mixtures going to react?
« Reply #2 on: November 09, 2007, 01:00:55 AM »
Hi ARGOS++;

Thank you so much for replying to my message. My measurements are not all going to be at 1 atm (but higher), but for now I'd like to find out if they would react at 1 atm.

I'm sorry, but I dont know what information is given in your Kp table. What does this tell us? It seems that you've indicated the normal boiling points of the substances next to each of them - except for Butanoic Acid(for which i found the normal bp to be approx 164 Deg C).

If i was to perform the measurements at atmospheric pressure and 100 deg C, then only teh butanoic and propanoic acid will remain as liquids. The system is going to be closed to the atmosphere though at an elevated pressure so that vapour and liquid will exist in equilibrium with each other with both components of each binary mixture being present in both the vapour and liquid phases.

I hope I'm making sense...

JCP

Offline AWK

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Re: Are these binary mixtures going to react?
« Reply #3 on: November 09, 2007, 04:19:45 AM »
All these reaction are catalyzed by acids. in two cases acid is also a reagent. And it does not matter most of some reagents are in liquid state, because parts of reagent are also in gaseous state.
Od course, a degree of conversion will be depended od the equilibrium constant time, ratio of reagents and the presence of better catalyst
AWK

Offline ARGOS++

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Re: Are these binary mixtures going to react?
« Reply #4 on: November 09, 2007, 07:29:20 AM »

Dear JCP,

Sorry! – That I used the German Kp instead of Bp for Boiling point!

Initiated by “your” Bp of 164° Celsius I double checked it by NIST (163° +/- 2° Celsius) and by “CemExpert” (162° - 165° Celsius). So my primary Reference must be definitely wrong.
I have now corrected the citation (Thank you.).

The “Ok” tells that your first Reaction is possible as you told, but very “havy” and quite very slow, if not catalysed by a strong acid, as already AWK told too. So this should quite less influence your result as long as you can keep it clean to only two participants and you not attack it for too long at these conditions.

For all other it is really difficult to make any exact forecast, not only because the catalytic influence, also because the pressure that may exist. Additionally I think that there are very few - if, who have done the same experiment till jet.

Good Luck!
                    ARGOS++


Offline jcp

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Re: Are these binary mixtures going to react?
« Reply #5 on: November 12, 2007, 01:08:52 AM »
Ok - thank you very much for trying to help.

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