Liquid Vapor Equilibrium

last update: October 12, 2000

file called ... ch407/LiqVap_remarks.htm

From my observations the experimental work seemed to be going fairly smoothly.

I has the early groups focus on systems that were far from ideal solutions. In this way we'd be able to explore azeotropes and activity coefficients for interesting systems. We chose combinations of a nonpolar hydrocarbon (cyclohexane) and a relatively polar alcohol (ethyl, isopropyl and n-propyl.) Methanol is too far from ideal-- it actually forms layers and complicates the system because we have many phases.

If we still have he opportunity, I'd like to have one group see if the concept of ideal solutions is borne out in practice. I'd suggest a mixture of cyclohexane and n-hexane. (Note that we really shouldn't use the standard hexane which is a mixture of isomers.) The normal boiling points are 69 C (n-hexane) and 81.4 (cyclohexane.) I assume they will separate reasonably well on the GC.

GC Response

• correction factor = [ mole fraction #1] / [(area#1) / (area#1+area#2)]
• ideally this is 1.00, meaning that the detector has equal molar response to both species.
• you should do 2-3 such mixtures and should find a consistent factor.
  • Later, when you analyze samples
  • Mole fraction of component #1 = (factor) x [(area#1) / (area#1+area#2)]
  • Mole fraction of component #2 = (1- X₁)

Experimental Data and Spreadsheet Form

• Assume we have a spreadsheet for the data (a sample is shown below)
• For each run you have a run identifying number (column A)
• You measure the reflux temperature (column B)
• You collect two samples, one representing the liquid phase and one representing the vapor phase.
• Chromatography analyzes each sample
• experimental values are the peak areas for each component:
  • for the liquid phase, Area#1(Column C) and Area#2 (Column D)
  • for the vapor phase, Area#1(Column E) and Area#2 (Column F)

  • 	A	B	C	D	E	F	G	H
    1	Liquid	Vapor	Spread	Sheet
    2			Liquid	Liquid	Vapor	Vapor	Liquid	Vapor
    3	sample	Temp	Area 1L	Area 2L	Area 1V	Area 2V	X1	Y1
    4	1a	65.2	5,421	1,202	6,315	1,495	**	**
    5	2a	64.7	4,903	1,604	5,502	2,167
    6	3a	63.8	5,235	2,423	4,993	2,047

• Now let's look at how we can use the data
  • ** the previous section shows how to convert areas into mole fraction
  • cell G4 would have a formula like =1.027 * C4/(C4+D4)
  • cell H4 is similar, but =1.027 * E4/(E4+E5)
  • the 1.027 is the experimentally found correction factor discussed above.

Plotting the Phase Diagram

• Everything you need is in the spreadsheet.
• Plot columns G and H vs. column B
  • Click and drag down column B
  • Press CTRL and click and drag down column G
  • Press CTRL and click and drag down column H
• Create a Chart or Graph
  • select graph type: scatter plot
  • select: markers but no line
  • The resulting graph is really 90^o from the usual phase diagram, so turn the paper so temperature is on the vertical axis and composition is on the horizontal axis.
    • (You can get a plot of B vs. G and B vs. H to appear on a graph without turning the page, but it is more complex to create. )
  • Don't try to force a trendline or other generate curve through the data-- there is no simple curve that will fit the data well.
  • With a pencil, sketch one line through all the liquid points
  • With a pencil (another color is helpful) sketch a line through all vapor points

Locating a Low Boiling Azeotrope on a Phase Diagram

• The lowest point lies at the azeotrope
• The x-axis gives the composition of the azeotrope
• at this point the vapor and liquid curves have joined
• The Y axis gives the temperature at the azeotrope
• Often, the boiling temperature changes very little in the region of the azeotrope.
• Small errors in the boiling point can make this curve uncertain
• one hint: You can often narrow down the azeotrope composition by a series of comparisons
  • when the liquid composition has X1 < azeotrope
    • the vapor has a higher mole fraction than the liquid
    • the vapor composition may be close, but will be less than that of the azeotrope
  • when the liquid composition has X1 > azeotrope
    • the vapor has a lower mole fraction than the liquid
    • the vapor composition may be close, but will be greater than that of the azeotrope

Comparing Data to Raoult's Law predictions

P1 (vapor) = X1(liquid) * P1^o(T)

You will need to compute the vapor pressure of the two pure liquids at the recorded boiling temperatures.

• Some data from the CRC and a calculation should be adequate
• A plot of ln (P) vs 1/T(K) has a straight line
• It's easy then to interpolate at any desired temperature
• Suggest you do this on a separate spreadsheet
  • Go to the CRC and use the index
  • look up vapor pressure of organic liquids
  • the table will list the boiling temperature of your liquid at 40 torr, 100 torr, 400 torr and 760 torr.
  • Convert data to two columns like C and D on the sample.
  • Plot C vs. D
  • Perform a linear fit (trendline, show equation)
  • If necessary, reformat the equation to give 4 significant figures
  • Then enter the temperatures of interest (boiling points you recorded)
  • convert column A to a formula = exp (intercept + slope * (1/B7+273.2))
  • repeat a similar pattern for the second liquid
  • finally, transfer these computed pressures to the first spreadsheet

Activity Coefficient Calculations

• Now, to the original spread sheet add two columns I and J
• these are the vapor pressure of pure 1 and pure 2 at the observed temperature
• Remember, column G has the mole fraction in the liquid
  • column K is ideal X1 (Raoult Law) = (I4*G4) /(I4*G4+(1-G4)*J4)
  • ratio of column H to column K is activity coefficient for species 1 (column L)
  • L4=H4/K4
• Need similar set of calculations t find the activity coefficient for liquid 2
  • Since the table does not compute X2 or Y2, we express them as (1-X1) or (1-Y1)
  • M4= X2 (Raoult's Law = (1-G4)*J4 /(I4*G4+(1-G4)*J4)
  • N4 is activity coefficient for liquid 2, N4= M4/(1-H4)

• Finally, plot column L and N vs. G (activity coefficient vs. composition)

1	A	B	C	D
2	Pressure	Temp (C)	1/T(K)	ln(P
3	100	47.2	=1/(B3+273.2)	=ln(A3)
4	400	58.2	.. fill	.. fill
5	760	66.3
6
7	***	65.2
8		63.3
9		61

What would you expect to see?

• When an alcohol is diluted with a hydrocarbon, the hydrogen bonding is destroyed
• this would make the liquid more volatile
• would make vapor pressure larger than predicted by Raoult's Law
• this would make the solution boil at a temperature well below the temperature predicted by Raoult's Law
• This would make the activity coefficient somewhere between 1.5-5.0 as the mole fraction of alcohol rises much above 0.5
• Of course, the activity coefficient goes to 1 as Mole fraction goes to 1.
• Diluting the hydrocarbon with the alcohol has a comparable effect.
• The hydrocarbon also has more tendency to vaporize and it's activity coefficients are greater than 1 also.

Compare your Results to Published Data

• Many such phase diagrams can be found in the Landorf-Bernstein Series (in the reserve collection in the science library)
• A copy of a few pages is kept in the lab near the apparatus for this experiment.