Information regarding phase
equilibria can be predicted by a simple rule(“Gibbs phase rule”)
f =
c – p + 2
Where c is the number of
components and p is the number of phases present in the system. The degrees of
freedom, f or variance, gives the number of variables (e.g., pressure,
temperature, composition, etc.) that must be given to completely describe the
system, or to locate the state of the system on the phase diagram.
For
ternary systems (i.e., consisting of three components), we have c = 3 and f = 5
– p. if the system consists of only one phase, f = 4. The required four
variables for describing such system are : two for describing the relative
composition (mass fractions) and one of the pairs (P,V), (P,T) or (T,V). note
that if only two mass fractions x1 and x2, are given, the
third can be obtained by x3 = 1 – x1 – x2.
Also in practice, (P,T) pair is chosen. If the system separates into two
different phases, only f = 5 - 2 = 3 variables are needed ( one mass fraction
and (P,T)).
Figure 1 : A
triangular phase diagram showing the representation of the mass fractions for
ternary systems. The colours indicate how concentrations for different species
should be read from the diagram. The point marked in the diagram (•) represents
30% 1-butanol, 10% water and 60% acetic acid. The one-phase and two-phase
regions have been separated by a black line. The line drawn is only
demonstration and does not correspond to experimental observation. Pressure and
temperature are assumed to be fixed.
Phase
diagrams for ternary systems are usually represented using a triangle shown in
Figure 1. This graph accounts for the fact that only two variables are
required. Along the phase boundary only one variable is required.
Figure 2 : Drawing of
a tie line in a triangular phase diagram. The diagram may contain one or many
tie lines. The point ’A’ denotes the composition of phase 1, ’B’ denotes the
initial composition of the system, and ’C’ the composition of phase 2.
Regions
where one or two phases appear have also been indicated in Figure 2. Note that
the line drawn is hypothetical, the real curve will be determined in this
experiment. When the solution is stirred, the transition from one region to another
can be observed by appearance (or disappearance) of cloudiness or turbidity in
the solution. The turbidity results from scattering of light by the large
number of very small “oily” droplets of the second phase that are produced when
the system is stirred. Sometimes it is easier to see this when stopping the
stirring briefly.
If the
three components are mixed to give an overall system composition that falls in
the 2-phase region, the system will separate into two phases: a phase rich in
water and another rich in 1-butanol. The compositions of the phases that form
are given by the intersections of a tie line with the phase boundary. The tie line
must also contain the point describing the overall system composition. A
graphical demonstration and an interpretation of a tie line are given in figure
2.
Note
that in the case of figure 2 only the mass fraction of 1-butanol must be given
when the system remains on the phase boundary line. This determines the mass
fractions for water and acetic acid. Hence the phase rule holds with f = 5 − p
= 3 (i.e., mass fraction for 1-butanol, temperature and pressure). If the
system was initially in the two-phase region, the tie line uniquely connects
the points along the phase separating line. Given the point ’A’ in figure 2
(depending on the 1-butanol mass fraction in phase 1), the points ’B’ and ’C’
are uniquely determined. Thus, only the 1-butanol mass fraction in phase 1,
temperature and pressure are required for complete description of the system,
which had separated into two phases. This is again in accordance with the phase
rule.
Apparatus :
Retort stand and
clamp
Burette
Pipette
Conical flask
Conical flask stopper
Chemicals :
Ethanol
Toluene
Distilled water
Experimental
procedure :
1.
Mixtures
of ethanol and toluene were prepared in sealed containers measuring 100 cm3
containing the following percentages of ethanol (in percent) : 10, 25,
35, 50, 65, 75, 90 and 95.
2.
20
mL of each mixture were prepared by filling a certain volume using a burette.
3.
Each
mixture were titrated with water until cloudiness was observed due to the
existence of a second phase.
4.
A
little water was added and shaked well after each addition.
5.
The
room temperature was measured.
6.
The
percentage based on the volume of each component were calculated when the
second phase started to appeared or separated.
7.
The
points were plotted onto a triangular paper to give a triple phase diagram at
the recorded temperature.
8.
The
experiment was repeated once again.
RESULT AND GRAPH
Ethanol
|
Toluene
|
Water
|
||||
Percentage (%)
|
Volume (mL)
|
Percentage (%)
|
Volume (mL)
|
Trial 1 (mL)
|
Trial 2 (mL)
|
Average, x (mL)
|
10
|
2
|
90
|
18
|
1.3
|
5.5
|
3.40
|
25
|
5
|
75
|
15
|
0.9
|
1.4
|
1.15
|
35
|
7
|
65
|
13
|
1.4
|
1.6
|
1.50
|
50
|
10
|
50
|
10
|
2.0
|
1.9
|
1.95
|
65
|
13
|
35
|
7
|
3.7
|
3.0
|
3.35
|
75
|
15
|
25
|
5
|
4.3
|
4.5
|
4.40
|
90
|
18
|
10
|
2
|
10.9
|
11.2
|
11.05
|
95
|
19
|
5
|
1
|
16.1
|
16.0
|
16.05
|
Total volume of mixture (mL)
|
Ethanol
|
Toluene
|
Water
|
|||
Volume (mL)
|
Percentage (%)
|
Volume (mL)
|
Percentage (%)
|
Volume (mL)
|
Percentage (%)
|
|
23.40
|
2
|
8.5
|
18
|
76.9
|
3.40
|
14.5
|
21.15
|
5
|
23.6
|
15
|
70.9
|
1.15
|
5.4
|
21.50
|
7
|
32.6
|
13
|
60.5
|
1.50
|
7.0
|
21.95
|
10
|
45.6
|
10
|
45.6
|
1.95
|
8.9
|
23.35
|
13
|
55.7
|
7
|
30.0
|
3.35
|
14.3
|
24.40
|
15
|
61.5
|
5
|
20.5
|
4.40
|
18.0
|
31.05
|
18
|
58.0
|
2
|
6.4
|
11.05
|
35.6
|
36.05
|
19
|
53.0
|
1
|
2.8
|
16.05
|
44.5
|
Questions
1. Does
the mixture containing 70% ethanol, 20%water and 10% toluene (volume) appear
clear or does it form two layers?
The mixture appears clear.
2. What
will happen if you dilute 1 part of the mixture with 4 parts of
(a)
water
Two phases are formed. This is
because the point of the mixture lies within the boundary of the curve.
(b)
toluene
One phase is formed. This is
because the point of the mixture lies outside the boundary of the curve.
(c)
ethanol
One phase is formed. This is
because the point of the mixture lies outside the boundary of the curve.
Discussion
Each apex (corner) of the triangle represents 100% composition
of the component respectively (ethanol, water or toluene) while the other two
components are 0%. The area within the triangle represents all the possible
combinations of ethanol, water and toluene to give three-component-system. In
this experiment, the system consists of three components which are ethanol
toluene and water but only one phase. Based
on Gibbs’ Phase Rule, F=C-P+2, it can be calculated that F=3-1+2=4. Four
degrees of freedom include the temperature, pressure and the concentrations of any
two out the three components are required. Only two concentrations are required
because the concentration of the third component can be calculated. The
experiment is conducted under constant temperature and pressure.
Toluene is soluble in ethanol but in the case of water and
toluene, they form a two phase system because they are only slightly miscible. However,
ethanol is completely miscible in both toluene and water. Thus, the addition of
sufficient amount of ethanol to the toluene-water system would result in a
single phase. In this experiment, ethanol and toluene are mixed in certain
compositions which they will form a colourless solution because they are
completely miscible. Then, water is added where at first it results in two
phase, and further addition of water until a certain amount will result in one
phase. This is because the presence of ethanol increases the solubility of the liquid pair (toluene-water) until at the point where it results in a
homogenous mixture (one phase). The shape of the graph plotted is binodal. The
region bounded by the curve represents two phase (heterogeneous) while the remaining
region outside the curve represents single phase (homogenous).
The results obtained are not very accurate because the graph
plotted should have a perfect binodal curve and looked like this.
 |
| Example of a perfect binodal curve |
The inaccuracy of the data is due to some errors that occurred
during the experiment. One of the errors is parallax error. Parallax error may
occur when the eye level is not perpendicular to the reading of the measuring
cylinder or burette. This causes an inaccurate reading which affects the
measurement. Another error may be the over-titration of distilled water from
the burette into the toluene-ethanol mixture. The end point may be overshot
because the cloudiness was hard to determine initially so excess water was
titrated. This causes the volume of distilled water required to turn the colourless
mixture to cloudy being higher than required, which in turn affects the
calculations. Besides, prolonged exposure of the ethanol-toluene mixture to air
before sealing causes some of the chemicals to evaporate due to their
volatility nature. This also affects the data because there will be lesser ethanol-toluene
mixture.
Several precaution steps have to be taken in order to reduce the
number of errors and to ensure a more accurate experiment. Firstly, the eye of
the observer has to be directly perpendicular to the reading of the apparatus
(measuring cylinder and burette). This is important to avoid parallax error. Furthermore,
the distilled water has to be titrated drop by drop carefully so that the cloudiness
can be observed without overshooting the end point. The conical flask has to swirled
consistently throughout the titration process in order to mix the substances
evenly. Moreover, the ethanol-toluene mixture has to be sealed right away with
the aluminium foil to reduce the evaporation of the chemicals.
CONCLUSION
The
phase diagram for ethanol, toluene and water system was determined. The
experiment was done by mixing different properties of ethanol and toluene. Then
water was titrated into the mixture. As the mixture turns cloudy, it indicates
that two phase system were established. The reaction of water, ethanol, and
toluene will appears as two phase’s system due to the decreasing in solubility
of the mixture. From the experiment, as the volume of toluene decrease more
water needed to break the homogeneity.
REFERENCES
1. Oxtoby, D. W., et al. 2008. Principles of Modern Chemistry. Thomson Brooks Cole.
2. Florence A. T. & Attwood D. 2006. Physicochemical Properties of Pharmacy. 3rd Edition. Great Britain: Pharmaceutical Press.
3.http://chemhail.wordpress.com/2009/07/05/application-of-phase-rule-to-three-component-systems/
4. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/phase.php
5. https://www.csun.edu/~jeloranta/CHEM355L/experiment5.pdf
Comment not appropriate!
ReplyDeleteHowever this was a very good explanation of three phase diagrams.