Chapter 6 Phases Equilibria

1
Chapter 6 Phases Equilibria
6.1 Equilibrium Between Phases
Number of Components
The number of phases (p) is the number of phases
which are homogeneous and distinct regions
separated by definite boundaries from the rest of
the system.
ice cubes in water
two phases one solid and one liquid
a block of ice reduced to crushed ice still
consists of only one phase.
Example 6.1
How many phases are preset in the following
system at equilibrium?
Solution
Although there are two solid present, each has
its own structure and is separated by distinct
boundaries. Therefore, we have two solid phases
and one gas phase, a total of three phases.
2
The number of components (c) is the smallest
number of independent chemical constituents
needed to fix the composition of every phase in
the system.
H2O,NaCl,Na,Cl-
Water one component
WaterNaCl two component
The material balance and electroneutrality
conditions reduce the number of constituents to
two that is, c4-1-12.
Note that the number of components is unique.
If a chemical reaction can take place between
constituents of a solution, the number of
components is reduced by the number of
equilibrium conditions. Considering the system
We recognize three distinct chemical species.
This number of species is reduced by the
independent equilibrium condition and, therefore,
c3-12. Another reduction in the number of
components is possible if we start with pure
PCl5, in which case PCl3Cl2. In this case
the number of components is unity because of the
additional mathematical relation.
3
Example 6.2
How many components are present when ethanol and
acetic acid are mixed assuming the reaction to
occur to equilibrium?
Solution
At first sight we might predict two components
because there are two constituents, HOAc and
EtOH. However, these constituents react and
ethylacetate and water are also present at
equilibrium. This raises the number of components
from 2 to 4.
But now the equilibrium condition is applied and
reduces by 1 the number of components.
Furthermore, since EtOAc and HOH must be formed
in equal amounts, another mathematical condition
exists, reducing the number of components back
to 2 thus c4-22.
H2, O2, H2O three component room temperature
H2O2?H2O two component higher temperature
4
Problem 6.5
How many components are present in the system
CaCO3-CaO-CO2?
Solution
Starting with pure CaCO3, we have only one
component present. When two of the three species
are present, the third species is also present
but because of the equilibrium CaCO3?CaOCO2,
there are only two components.
Problem 6.6
How many components are present in the following
system?
Solution
There are four individual gases, and the
equilibrium equation reduces the number of
independent components to three.
5
Degrees of Freedom
The number of degrees of freedom (f) is the
number of intensive variables, such as
temperature, pressure, and concentration, that
can be independently varied without changing the
number of phases. Stated differently, the number
of degrees of freedom is the number of variables
that must be fixed in order for the condition of
a system at equilibrium to be completely
specified.
If the system has one degree of freedom, we say
it is univariant. If a system has two degrees of
freedom, it is a bivariant system. Thus pure
water is univariant since at any given
temperature, the pressure of vapor in equilibrium
with liquid is fixed (only the one variable,
temperature, may be varied independently).
Phase Rule
where the term 2 is for two variables,
temperature and pressure.
Considering c components and p phases
p-1 independent equations
concentration terms
temperature, pressure
6
Problem 6.3
Determine the number of degrees of freedom for
the following system a. A solution of potassium
chloride in water at the equilibrium pressure. b.
A solution of potassium chloride and sodium
chloride at 1 atm pressure. c. Ice in a solution
of water and alcohol.
Solution
a.
For KCl and H2O at the equilibrium pressure,
fc-p22-123 Since the equilibrium pressure is
specified, this reduces the number of degrees of
freedom to 2.
b.
Here, NaCl, KCl, and H2O are present. This is
actually a three-component system since the
solution contains Na, K, Cl-, and H2O. The
first three compositions are reduce to two
independent ones by the electroneutrality
condition. Therefore, fc-p23-124, but with
the restriction of constant pressure, the
variance is reduced by 1, and is therefore 3.
c.
Ice, water, and alcohol are only two components.
Consequently, fc-p22-222.
7
6.2 One-Component Systems
The stable form of sulfur at 1 atm pressure and
room temperature is a crystalline form called
(ortho)rhombic sulfur (???). As rhombic sulfur is
heated slowly at 1 atm, it transform to different
crystalline form called monoclinic sulfur (???)
at 368.55 K. The name allotrope (?????) refers to
each of the crystalline forms when this type of
transformation occurs in elements.
Figure 6.1
Monoclinic sulfur melts to the liquid along line
CE. However, if the rhombic form is rapidly
heated at 1 atm pressure, the transformation
temperature of 368.55 K is bypassed and the
rhombic form melts directly to liquid sulfur
at 387 K. When the rhombic sulfur is in
equilibrium with liquid, we have an example of
a metastable equilibrium since this
equilibrium position lies in the region of the
more thermodynamically stable monoclinic form
of sulfur. This is not a true equilibrium, but
it appears to be one because the process of
change into the more favored form is slow. In
this region both the rhombic and liquid forms
of sulfur can decrease their Gibbs energy by
converting to the favored monoclinic form.
The variation of vapor pressure with temperature
may be handled using the Clapeyron equation.
8
The intersection of lines BC and CE is brought
about by the difference in density of the two
crystalline forms. The density of rhombic sulfur
is greater than that of the monoclinic form,
which has a density greater than that of the
liquid. Therefore the lines BC and EC have
positive slopes.
triple point
Figure 6.1
The system is invariant. This means that the
equilibrium of the three phases automatically
fixes both the temperature and pressure of the
system, and no variable may be changed without
reducing the number of phases present.
curves AB, BC, BE, CE, and EF
The system is univariant. Thus under conditions
of two phases present either temperature or
pressure may be varied, but once one of them is
fixed along the respective equilibrium line, the
final state of the system is completed defined.
9
In the four regions where single phases exist
(namely rhombic, monoclinic, liquid, or vapor) we
have f1-122 and the system is bivariant. To
define the state of the system completely the two
variable temperature and pressure must both be
specified.
Figure 6.1
Point D is a metastable triple point with
two-phase equilibrium occurring along lines BD,
CD, and DE. The phase rule predicts the number of
degrees of freedom regardless of the fact that
the system is metastable, since the metastable
system, within the time scale of the
measurements, behaves as if it were at
equilibrium. Thus we find invariant and
univariant metastable systems as we did earlier
for the corresponding stable equilibria.
10
6.3 Binary System Involving Vapor
Liquid-Vapor Equilibria of Two-Component Systems
Figure 6.2
liquid curve
In the region marked Liquid, only one phase
exists. There are two components so that from
the phase rule, f2-1-23. Since figure 6.2 is
drawn at a specific temperature, one degree of
freedom is already determined. Only two more
variables are needed to specify the state of
the system completely. A choice of P and x2
then completes the description. In the region
marked Vapor, specific values of P and y2
will define the system.
vapor curve
11
The line, called a tie line, intersects that
liquid curve at l and the vapor curve at v. The
composition x2 and y2 corresponding to l and v
give the mole fractions of component 2 in the
liquid phase and vapor phase, respectively.
Figure 6.2
tie line
lever rule
In the region marked LiquidVapor, there are two
phases present and the phase rule requires
specification of only one variable since T is
fixed. This variable may be P, x2, or y2.
12
Example 6.3
Determine the mass percentage of carbon
tetrachloride CCl4 (P1114.5 Torr) in the vapor
phase at equilibrium in a 11 mole ideal solution
with trichloromethane CHCl3 (P2199.1 Torr) at
25 ?C.
Solution
Let the mole fraction of CCl4 in the vapor state
be y1 and that of CHCl3 be y2 above the solution
having mole fraction x10.5 CCl4. Then
The ratio is
So the mass percentage of carbon tetrachloride in
the vapor is
13
Figure 6.2
If the liquid of pure component 2 has a higher
vapor pressure than that of pure component 1, the
vapor will contain relatively more of component 2
than does the liquid that is in equilibrium with
it.
14
Problem 6.15
Calculate the composition of the vapor in
equilibrium at 323 K with a liquid solution of
0.600 mol fraction 2-methy-1-propanol (isobutyl
alcohol) and 0.400 mol fraction
3-methyl-1-butanol (isoamyl alcohol). The vapor
pressure of pure isobutyl alcohol is 7.46 kPa and
that of pure isoamyl alcohol is 2.33 kPa both at
323 K.
Solution
15
Liquid-Vapor Equilibrium in Systems Not Obeying
Raoults Law
Figure 6.3
When there are large, positive deviations from
Raoults law, there will be a maximum in the
total vapor pressure curve. The liquid-vapor
equilibrium will then be established, as shown in
Figure 6.3 for the chloroethene-ethanol system.
This curve shows a maximum at 38 mol ethanol at
a constant temperature of 313.1 K.
Figure 6.4
In a similar manner, there may be a minimum in
the total vapor pressure curve for negative
deviations from Raoults law. The liquid-vapor
equilibrium is established as shown in Figure 6.4
for the acetone-chloroform system. Here a minimum
occurs at approximately 58 mol chloroform at
308 K.
16
Temperature-Composition Diagrams Boiling Point
Curves
The right and left faces of the figure are the
respective vapor-pressure curves of the pure
liquids.
For a typical, completely miscible binary system,
the temperature and composition are represented
along the two horizontal axes, and pressure along
the vertical axis.
higher vapor pressure
Although the binary system may behave ideally,
the straight line from a P against x plot for the
total pressure of liquids is not a straight line
in the temperature-composition plot. (The vapor
pressure does not increase proportionally with T.)
17
Distillation
Figure 6.7
?p???,??l??,????????v,????2???????,???????,???????
?? 1???????????????1?,??????p?,?????v???D??????l?
??R,??? ???????level rule???
?p????p?,?????D(yD)??????R(xR),??????l?,?????xl
???,??????v,??????yv???,?????????,????D???,????
????l?,?????xl???,??????v,??????yv???,????
???????2???,???????????????????1????2,????????? (d
istillation)?
Vapor pressure of component 2 is higher than that
of component 1.
18
Problem 6.8
Answer the following equations, using the
accompanying figure. a.A liquid mixture consists
of 33 g of component A and 99 g of component B.
At what temperature would the mixture begin
to boil? b.Under the conditions in (a), what is
the composition of the vapor when boiling first
occurs? c.If the distillation is continued until
the boiling point is raised by 5.0 ?C, what would
be the composition of the liquid left in the
still?
60 ?C
53
88
Solution
a. wt of B99/(3399)10075, from the graph
at 75 B, the first vapor appears at 60 ?C. b.
The composition of the vapor is given by the
intersection of the tie line at the vapor curve.
In this case, the vapor has a composition of
88 B. c. The intersection of the liquid line at
65 ?C corresponds to 53 B in the liquid.
19
Figure 6.8
In this device preheated liquid mixture enters
the column about halfway up. The less volatile
components drop to the bottom boiler or still, A,
where they are reheated. As the vapor ascends the
column, the higher-boiling components begin to
condense while the lower-boiling material proceed
to higher stages. Thus a temperature gradient is
established, with the highest temperature at the
bottom of the column and the lowest at the top
from which the lowest-boiling solution may be
removed.
bubble-cap fractionating column
20
Figure 6.8
At each level in the column, such as B, vapor
from the level or plate below bubbles through a
thin film of liquid C, at a temperature slightly
lower than that of the vapor coming through the
bubble cap, D. Partial condensation of the vapor
occurs. The lower-boiling mixture remains as
vapor and moves to the next plate. Thus the vapor
leaving each plate is enriched in the more
volatile component compared to the entering vapor
from the plate below. The action of the vapor
condensing and then reevaporating is the same as
described for the behavior shown in Figure 6.7.
Excess liquid at each plate is returned to the
plates below via overflow tubes, E. The reflux
control allows part of the condensate to return
to the column, where it continues to undergo
further separation.
21
Azeotrope(???)
Mixtures corresponding to either a minimum or
maximum in the boiling point curves are called
azeotrope. Separation of azeotropic mixtures into
two components by direct distillation is
impossible. However, on pure component may be
separated as well as the azeotropic composition.
Figure 6.9
Figure 6.10
22
Figure 6.10
Considering Figure 6.10, if we heat a mixture
having the azeotrope composition p, the mixture
will boil unchanged no separation is
accomplished. If we now boil a mixture having
composition p, the first vapor to appear will
have v and is richer in cis-1,2-dichloroethene,
component 2. Continued fractionation, as
described before, will eventually produce pure
component 2 in the distillate with azeotrope as
the residual liquid in the pot. Boiling a
mixture at point p results in vapor having
composition v, which is much richer in
tetrehydrofuran, component 1, than before.
Fractionation can eventually yield pure component
1 at the head of the column and leave the
azeotropic mixture in the pot.
Similar results are obtained with a minimum
boiling azeotrope except that the azeotrope
composition is obtained at the head of the
column and pure component remains behind in the
still. Example of minimum boiling azeotrope are
far more numerous than those exhibiting a
maximum in their curves.
23
Azeotrope have sometimes been mistaken for pure
compounds because they boil at a constant
temperature. However, for an azeotrope a
variation in pressure changes not only the
boiling temperature but also the composition of
the mixture, and this easily distinguishes it
from a pure compound. This is demonstrated for
HCl in Table 6.3. One of the more useful
application of azeotrope in binary system is the
preparation of a constant composition mixture. In
the case of H2O and HCl, the constancy of the
composition of this azeotrope allows its use as a
standard solution of know composition.
24
Distillation of immiscible Liquids Steam
Distillation
Considering the behavior of two liquids whose
mutual solubility is so small that they may be
considered immiscible, in this case each liquid
exerts the same pressure as though it were the
only liquid present. Thus the total pressure
above the mixture at a particular temperature is
simply the sum of the vapor pressures of the two
components and remains so until one of the
components disappears.
If nA and nB are the amounts of each component
present in the vapor, the composition of the
vapor is
If PT is the total pressure, and PA and PB are
the vapor pressure of pure liquids A and B,
respectively, then
Since the ratio of partial pressure is a constant
at a particular temperature, the ratio nA/nB must
also be a constant. Thus the distillate is of
constant composition as long as both liquids are
present, and it boils at a constant temperature.
Water is often one component when this type of
distillation is used for purifying organic
compounds. This process, called steam
distillation, is frequently used for substances
that would decompose when boiled at atmospheric
pressure. What makes this process attractive is
the high yield of organic materials brought about
by the low molar mass of water and its convenient
boiling point, in contrast to the relatively high
molar masses of most organic substances.
25
Example 6.4
Toluene (methylbenzene) and water are immiscible.
If boiled together under atmospheric pressure of
755 Torr at 83 ?C, what is the ratio of toluene
to water in the distillate? The vapor pressure of
pure toluene and water at 83 ?C are 322 Torr and
400.6 Torr, respectively.
Solution
Problem 6.13
Under atmospheric pressure 1 kg of pure
naphthalene is to be prepared by steam
distillation at 372.4 K. What mass of steam is
required to perform this purification? The vapor
pressure of pure water at 372.4 K is 98.805 kPa.
Solution
26
Distillation of Partially Miscible Liquids
Figure 6.11
Generally partial miscibility at low temperature
is caused by large positive deviation from
Raoults law, in which case we expect to find
minimum in the boiling point-composition curve.
As the pressure on the system is reduced, the
boiling point curve generally intersects the
liquid-liquid equilibrium curve, resulting in the
curve for a typical system shown in Figure 6.11.
Any composition in the range from 0 to xa and
from xc to 1 will show the same behavior upon
boiling as already demonstrated for minimum
boiling azeotrope, with one exception. Two layers
are formed if liquid at point p is evaporated and
its vapor v condensed.
27
fc-p22-321-10
If solution of overall composition p is boiled
at Te, three phases will be in equilibrium
liquid phase L1 having composition xa, liquid
phase L2 having composition xc, and vapor having
composition yb. Thus, as vapor of composition yb
is removed, the composition of the two liquid
phases does not change, only the relative amount
of the two layers. In this particular case,
continued distillation will cause all the L2
layer to be consumed. When it is exhausted,
liquid L1 at a and vapor at b are left. The
temperature may be increased at this same
pressure then the liquid composition changes
along curve al and the vapor along bv. The last
drop of liquid disappears when l and v are
reached, and only vapor remains since this is the
original composition of liquid from which the
liquid p was produced.
Figure 6.11
f?
f?
f?
28
6.4 Condensed Binary Systems
Two-Liquid Components
Figure 6.12
The water-aniline system provides a simple
example of partial miscibility (see Figure 6.12).
If a small amount of aniline is added to pure
water at any temperature below 441 K, the
aniline dissolves in the water. If we work
at a constant temperature of 363 K, pure water
is present at point a and only one phase is
present as aniline is added. However, as more
aniline is added, point b on the solubility curve
is reached and then, in addition to phase L1
of composition b, a slight amount of a second
liquid phase L2 appears having composition e.
29
The composition of the L1 layer is a solution of
aniline in water and that of the L2 is a
solution of water in aniline. As more aniline
is added, the second liquid layer L2 becomes more
evident and continually increases with the
addition of aniline until the composition is
given by point e. Beyond point e, only one
phase is present. The same type behavior is
observed as water is added to pure aniline.
Figure 6.12
The composition of any point in the two-phase
region along the tie line between points b and e
is composed of varying proportions of
solution L1 and of solution L2. These
solutions are called conjugate solutions
(????). The composition of these layers
depends on the temperature. The amount of
the individual layers present may be
determined using the lever rule.
30
Constant composition lines, vertical on the
diagram in Figure 6.12, are known as isopleths.
As the temperature is increased from point c
along the isopleth cc or indeed from any
point left of the vertical dashed line joining
Tuc, the solubility of the aniline in water
layer, L1, grows as does the solubility of the
water in aniline layer, L2. As a result of
the change in solubility, the predominant
layer L1 increases at the expense of the L2
layer. Similar behavior is observed for the L2
layer when the temperature is increased from
point d.
Figure 6.12
A different behavior is observed with the
critical composition, which is the composition
corresponding to the highest temperature Tuc at
which two layer may coexist. (This temperature
is know as the upper consolute temperature or
critical solution temperature.) If the curve
is symmetrical, the relative size of the layers
remains constant as the temperature is raised
along the dotted line. Above Tuc, only one phase
exists.
31
Example 6.3
Calculate the ratio of the mass of the water-rich
layer to that of the aniline-rich layer, for a
20-wt water-aniline mixture at 363 K.
Solution
Using the lever rule, we have
32
Figure 6.13
Figure 6.13 shows an example of lower consolute
temperature in the water-triethylamine system.
The lower consolute temperature is 291.65 K,
and above this temperature two immiscible
layers exist. In this case the large positive
deviations from Raoults law responsible for the
immiscibility may be just balanced at the
lower temperature by large negative deviations
from Raoults law, which are normally
associated with compound formation.
Figure 6.14
Figure 6.14 shows the final type of
liquid-liquid equilibrium, called a
miscibility gap, is exhibited by the
water-nicotine system. In this case the
two-phase region is enclosed and has both an
upper and lower consolute temperature at
atmospheric pressure. This may be called a
closed miscibility loop. It has been shown for
this system that an increase of pressure can
cause total solubility.
33
An interesting example of liquid-liquid
solubility in two material normally solid,
and its application to a practical problem, is
afforded by the Pb-Zn system. Figure 6.15 gives
the phase diagram for this system to 1178 K,
above which boiling occurs. Ignoring the
details in the right- and left-hand extremes
of the diagram, we see that miscibility occurs
above the upper consolute temperature at
1071.1 K. The rather limited solubility of zinc
in lead may be used in the metallurgical
separation of dissolved silver in lead. The
zinc is added to the melted lead, which has an
economically recoverable amount of silver
dissolved in it. The melt is agitated to
effect thorough mixing. The zinc is then
allowed to rise to the surface of the lead and
is skimmed off. Because of the much higher
solubility of silver in Zn than Pb, most of the
silver will now be in the zinc. The zinc may
be boiled off from this liquid to give the
desired silver.
Figure 6.15
34
Solid-Liquid Equilibrium Simple Eutectic Phase
Diagrams
When a single-liquid melt formed from two
immiscible solids is cooled sufficiently, a solid
is formed. The temperature at which the solid is
first formed is the freezing point of the
solution, and this is dependent on the
composition. Such a case is provided by the
gold-silicon system shown in Figure 6.16.
A curve that represents the boundary between
liquid only and the liquid plus solid phase is
known as a liquidus curve. If we begin with pure
gold, the liquidus curve will start as xAu1
and drop toward the center of the figure. The
curve from the silicon side behaves in a
similar manner. The temperature of
intersection of the two curves, Te, is called
the eutectic temperature (????). The eutectic
composition xe has the lowest melting point
in the phase diagram. At the eutectic point,
three phase are in equilibrium solid Au, Solid
Si, and liquid. At fixed pressure the
eutectic point is invariant. This means that
the temperature is fixed until one of the
phases disappears.
Figure 6.16
35
The relationship between phases is easy to follow
if we isobarically cool the liquid represented
by point p in a single-liquid-phase region.
Figure 6.16
As the temperature is lowered, the first solid
appears at l at the same temperature as the
liquid. Since this is an almost completely
immiscible system, the solid is practically pure
gold. As the temperature is dropped further,
more crystals of pure gold form. The composition
of the liquid follows the line le, and the
overall composition between liquid and solid is
given by the lever rule. In this region of liquid
plus solid, either the temperature or the
composition may be varied.
fc-p22-321-10
fc-p22-222-11
When point s is reached, three phase are in
equilibrium solid Au, solid Si, and liquid of
composition xe. Since the system is invariant,
the liquid phase is entirely converted into the
two solids before the temperature may drop lower.
Finally, at point p, only two solid, Au and Si,
exist. The right-hand of the diagram may be
treated in the same way.
36
Figure 6.16
The mole fraction of silicon in the region near
pure silicon varies with temperature according
to the following equation
where T0 melting point of Si.
Problem 6.20
In Figure 6.16, a solution having composition p
is cooled to just above the eutectic temperature
(point s is about 0.18 xSi, and xe is 0.31 xSi)
calculate the composition of the solid that
separates and that of the liquid that remains.
Solution
42 solid 58 liquid
37
6.5 Thermal Analysis
The careful determination of phase boundaries,
particularly in complicated metallic systems, is
quite difficult and requires considerable effort.
One method that has proved useful for phase
determination is the technique of thermal
analysis. In this technique a series of mixtures
of known composition is prepared. Each sample is
heated above its melting point and, where
possible, is made homogeneous. Then the rate of
cooling of each sample is followed very closely.
Figure 6.17
Figure 6.17 shows a series of cooling curves in a
plot of T against time and shows how individual
points are used to form a phase diagram similar
to Figure 6.16.
38
For the two pure materials the rate of cooling of
the liquid melt is fairly rapid. When the melting
temperature is reached, there is generally a
little supercooling that is evidenced by a slight
jog in the curve. This is shown in curve 1. The
curve returns to the melting point and remains
there until all the liquid is converted to
solid. The temperature then drops more rapidly
for the solid than for liquid since the heat
capacity of a solid is generally lower than that
of a liquid. Thus it requires less heat removal
to cool the sample a fixed number of degrees.
Figure 6.17
Curve 2 represent a mixture of some B in A. The
mixture cools rapidly until point l is reached.
This point appears on the liquidus curve. Liquid
and solid are in equilibrium as the mixture cools
more slowly along line ls. This is because of the
heat that is released on solidification. At point
s, a horizontal region appears called the
eutectic halt. The liquid still present in the
system must completely solidify before the
temperature can drop farther. Once all the liquid
is converted to solid, the temperature will drop.
As with the pure materials, the cooling of two
solids is much more rapid than if liquid were
present.
39
Figure 6.17
The descriptions of curves 3 and 5 are the same
as for curve 2 except for the lengths of the
line ls and ls and for the time that the
system stays at the eutectic halt. This period
at the eutectic halt provides a means to
establish the eutectic temperature. In
general the eutectic halt will lengthen, and
lines like ls will shorten, as the
composition approaches the eutectic composition.
The reason for this is that for the eutectic
composition the cooling is rapid until the
eutectic temperature is reached. After all the
liquid is converted to solid, the mixture can
then cool further. The temperature for each
composition at which a change occurs in the
cooling curve is then used to establish a
point on the phase diagram, as is shown in the
right-hand portion of Figure 6.17.
40
6.6 More Complicated Binary Systems
Figure 6.18 shows the phase diagram of the
water- NaCl system. The most obvious
isothermal discontinuity occurs at -21.1 ?C,
which is the eutectic temperature at a
concentration of 23.3 wt. A second
discontinuity is a peritectic reaction at 0.1
?C and 61.7 wt, where there is exactly a 21
mole ratio of water to NaCl- ion pairs. This
combination is sometimes referred to as a
crystal dihydrate and more generally is termed
a phase compound. Such mixture have a precise
integral mole ratio defining the stoichiometry.
As a few crystal of NaCl are dissolved in liquid
water at room temperature, a liquid phase is
formed that contains both water and the ions Na
and Cl-. More added salt will dissolve up to a
limit called the solubility limit, shown along
EB in Figure 6.18. Up to this limit the salt
and water are completely miscible. At this limit
the solution is said to be saturated with
salt. As still more salt is added, the salt
concentration in the liquid phase does not
change, and any crystals added beyond the
solubility limit remain seemingly unchanged in
contact with the liquid.
41
The sufficient salt are mechanically mixed with
ice to form a 40 wt mixture of NaCl in
water maintained at -10 ?C. A liquid phase is
quickly formed, dissolving salt crystals up
to the solubility limit. This freezing point
curve for water is represented by the curve
AE. In this case, however, the crystals in
equilibrium with the saturated liquid are not
the same as the those formed above 0 ?C. These
latter crystals have the formula NaCl.2H2O.
They differ in composition and structure from
dry sodium chloride.
One use of eutectic systems such as just
described is to prepare a constant-temperature
bath at some temperature below that of melting
ice. If NaCl is added to ice, the ice melts.
Indeed, if this is done in an insulated
container, ice continues to melt with the
addition of NaCl until -21.1 ?C is reached. Then
the temperature of the system will remain
invariant until all the ice has been melted by
heat from an outside source.
Table 6.4 presents several eutectic compositions
involving different salts and water.
42
Solid Solutions
Only one type of situation is known in which the
mixture of two different substance may result
in an increase of melting point. This is the
case in which the two substances are
isomorphous. In terms of metallic alloys this
behavior is a result of the complete mutual
solubility of the binary components. This can
occur when the sizes of the two atoms of the
two components are about the same. Then atoms
of one type may replace the atoms of the other
type and form a substitutional alloy. An
example of this behavior is found in the Mo- V
system, the phase diagram for which is shown
in Figure 6.19. Addition of Mo to V will raise
the melting point.
Figure 6.19
The Cu-Ni system also forms a solid solution.
Copper melts at 1356 K, and addition of nickel
raises the temperature until for xNi1 the
temperature reaches 1726 K. An alloy known as
constantan, consisting of 60 wt Cu and 40 wt
Ni, has special interest since it is useful as
one component of a thermocouple for the
determination of temperature.
43
Partial Miscibility
Figure 6.20 shows an example of the limited
solubility for both components. Tin dissolves
lead to a maximum of approximately 2.5 mol or
1.45 wt . Tin is more soluble in lead,
dissolving to a maximum of 29 mol at 466 K.
The two-phase solid region is composed of these
two solid alloys in proportions dictated by
the lever rule. The situation is analogous to
that represented in Figure 6.11 concerning
liquid-vapor equilibrium where partial
miscibility occurs in the liquid phase.
Figure 6.20
Figure 6.11
2.5 mol
71 mol
44
Partial Miscibility
Figure 6.21
Figure 6.21 shows another type of system in
which partial miscibility occurs involves a
transition point. A transition temperature
exists along abc at which spinel, corundum,
and liquid of composition c coexist and the
system is invariant. At any temperature above the
transition temperature the spinel phase
disappears. Cooling of the liquid and corundum
phases in the a to b composition range results
in the formation of the spinel phase and
coexistence of two solid phase. Cooling in the
b to c range initially results in the
disappearance of corundum and formation of
spinel along with the liquid. Further,
cooling results in solid spinel only. However,
as the temperature falls further, corundum
makes an appearance again.
spinel??? corundum???
45
Figure 6.21
The region near 1300 K and 0.4 mol fraction Al2O3
appears to be similar to what has been described
as a eutectic point. However, where liquid would
be expected in a normal eutectic system, this
region is entirely solid. An invariant point such
as e surrounded solely by crystalline phases is
called a eutectoid (????). At the eutectoid,
phase reactions occur on change of heat resulting
in a change in proportion of the solid phases
exactly analogous to that at a eutectic point.
46
Problem 6.38
Describe what happens within the system
Mn2O3-Al2O3 in Figure 6.21 when a liquid of
xAl2O30.2 is cooled from 2100 K to 1200 K.
Solution
As liquid is cooled, solid spinel first appears
at about 1950 K in equilibrium with liquid. At
approximately 1875 K, all of the liquid converts
to solid spinel, the composition of which varies
according to the lever rule. As the temperature
falls to about 1400 K, a two-phase region appears
that is Mn3O4spinel. Below about 1285 K, the
spinel converts to corundum and Mn3O4corudum
coexist.
47
Compound Formation
Sometimes there are such strong interactions
between components that an actual compound is
formed. Two types of behavior can then be found.
In the first type, the compound formed melts into
liquid having the same composition as the
compound. This process is called congruent
melting (??). In the second type, when the
compound melts, the liquid does not contain melt
of the same composition as the compound. This
process is called incongruent melting (??).
Figure 6.22
In figure 6.22 for the system Tl2O-TlBO2 the
compound Tl3BO3 is formed and melts congruently
at 725 K. A eutectic occurs at 8.2 wt B2O3. Note
that the X axis is plotted as weight percent B2O3
over a very limited range. The easiest way to
interpret this figure is to mentally cut it in
half along the line representing pure Tl3BO3.
Then each half is treated as in Figure 6.16. The
left-hand portion introduces a new feature that
is often found out only in ceramic system such as
this but also in metallic systems. At 627 K a
reversible transformation occurs in the
crystalline structure of Tl2O. A form called ? is
stable below 627 K and the second form ? is
stable above the transition temperature all the
way to the melting point.
It appears that there are now two points in this
system that were previously described as eutectic
points. However, only the lowest one is referred
to as the eutectic point other are called
monotectic points.
48
In contrast to congruent melting, incongruent
melting occurs for each of the compounds in the
Au-K system shown in Figure 6.23. The composition
of each compound formed is given by the formula
alongside the line representing that compound.
If we examine the compound K2Au as it is heated,
we find that liquid of composition P is formed
at 543 K
Figure 6.23
Since the liquid is richer in potassium than is
solid KAu, some solid KAu will remain as
solid. Thus the reaction is known as phase
reaction or, more commonly, a peritectic reaction
(??? ?). The point P is know as the peritectic
point (???). This reaction is reversible if
liquid of the same total composition as K2Au
is cooled. Solid KAu begins to separate at l.
More solid KAu forms until the temperature of
543 K is reached. As heat is removed, the
reverse of the peritectic reaction just shown
occurs. From the lever rule, approximately 25
of the material initially exists as particle
of solid KAu surrounded by liquid of
composition P. Thus the KAu is consumed as the
reaction proceeds. As the last trace of liquid
and KAu disappears, the temperature is free to
drop and only K2Au is present.
If the starting melt has a composition that is
slightly rich in Au compared to K2Au (i.e., lies
to the left of K2Au), cooling into the two-solid
region will produce crystals of KAu surrounded
by the compound K2Au.
49
Problem 6.23
The following information is obtained from
cooling curve data on the partial system
Fe2O3-Y2O3
Sketch the simplest melting point diagram
consistent with these data. Label the phase
regions and give the composition of any
compounds formed.
Solution
50
Problem 6.26
The following data for the magnesium-copper
system is the result of analyzing cooling curve.
Pure copper melts at 1085 ?C while pure magnesium
melts at 659 ?C. Two compounds are formed, one at
16.05 wt Mg with a melting point of 800 ?C, and
the other at 43.44 wt Mg with a melting point of
583 ?C, respectively. Construct the phase diagram
from this information and identify the
compositions of the eutectics.
Solution
51
Problem 6.29
The aluminum-selenium system was determined from
thermal analysis. Al2Se3 melts congruently at
approximately 950 ?C and forms a eutectic both
with aluminum and with selenium at a very low
concentration of the alloying element and at a
temperature close to the melting point of the
base elements. Draw a diagram from this
information and give the composition of the
phases. Aluminum melts at 659.7 ?C and selenium
melts at approximately 217 ?C.
Solution
52
6.8 Ternary Systems
When we consider a three-component, one-phase
system, the phase rule allows for f c - p 2
3 - 1 2 4 degrees of freedom. These four
independent variables are generally taken as
pressure, temperature, and two composition
variables, since only two mole fractions are
necessary to define the composition. Thus the
composition of a three-component system can be
represented in two dimensions with T and P
constant. The composition is determined using the
fact that, from any point within an equilateral
triangle, the sum of the distances perpendicular
to each side is equal to the height of the
triangle. The height is set equal to 100 and is
divided into 10 equal parts. A network of small
equilateral triangles is formed by drawing lines
parallel to the three sides through the 10 equal
divisions.
53
Figure 6.26
Each apex of the equilateral triangle in Figure
6.26 represents one of the three pure
components, namely 100 A, 100 B, or 100 C.
The three sides of the triangle represent the
three possible binary systems and 0 of the
third component. Thus any point on the line
BC represents 0 A. A line parallel to BC
through P represents all possible
compositions of B and of C in combination
with 30 A. Here the percentage of A is read
from the value of the length of the line A'P.
Compositions along the other two sides are
read in a similar manner. Since the distance
perpendicular to a given side of the triangle
represents the percentage of the component in
the opposite apex, the compositions at P are
30 A, 50 B, and 20 C. Any composition of a
ternary system can thus be represented within
the equilateral triangle.
54
Liquid-Liquid Ternary Equilibrium
A simple example for demonstrating the behavior
of a three-component liquid system is the system
toluene-water-acetic acid. In this system,
toluene and acetic acid are completely miscible
in all proportions. The same is true for water
and acetic acid. However, toluene and water are
only slightly soluble in each other.
Figure 6.27
Their limited solubility causes two liquids to
form, as shown along the base of the triangle at
points p and q in Figure 6.27. Added acetic acid
will dissolve, distributing itself between the
two liquid layers. Therefore, two conjugate
ternary solutions are formed in equilibrium. With
temperature and pressure fixed in the two-phase
region, only one degree of freedom remains, and
that is given by the composition of one of the
conjugate solutions.
fc-p23-223-21
55
Because of the difference in solubility of acetic
acid in the two layers, however, the tie lines
connecting conjugate solutions are not parallel
to the toluene-water base. This is shown for the
tie lines p'q', p"q", etc. This type of curve is
called binodal. The relative amounts of the two
liquids are given by the lever rule.
Figure 6.27
As the two liquid solutions become more nearly
the same, the tie lines become shorter, finally
reducing in length to a point. This point
generally does not occur at the top of the
solubility curve and is called an isothermal
critical point or plait point, p in the
diagram. At p both layers are present in
approximately the same proportion, whereas at p"
only a trace of water remains in the toluene
layer. This curve becomes more complicated if the
other sets of components are only partially
miscible.
tie line
56
Problem 6.39
The isobaric solubility diagram for the system
acetic acid-toluene-water is shown in Figure
6.27. What phase(s) and their composition(s) will
be present if 0.2 mol of toluene is added to a
system consisting of 0.5 mol of water and 0.3 mol
of acetic acid? Give the relative amounts of
each phase.
Solution
The composition of the system is 30 mol acetic
acid, 50 mol water, and 20 mol toluene. The
system point is practically on the pq tie
line, and there are therefore two liquids
present. The ends of this line and thus the
concentrations of the two liquids are
approximately q 1 toluene, 37 acetic acid,
and 62 water p 95.5 toluene, 4 acetic acid,
and 0.5 water The relative amounts of the two
liquids are given by the lever rule
57
Solid-Liquid Equilibrium in Three-Component
Systems
The common-ion effect may be explained by use of
phase diagrams. Water and two salts with an ion
in common form a three-component system. A
typical phase diagram for such a system is shown
in Figure 6.28. Such systems as NaCl-KCl-H2O and
NH4Cl-(NH4)2SO4-H2O give this type of equilibrium
diagram. We now will see how each salt influences
the solubility of the other and how one salt may
actually be separated.
In Figure 6.28, A, B, and C represent
nonreactive pure components with C being the
liquid. Point a gives the maximum solubility of
A in C when B is not present. Point c gives
the maximum solubility of B in C in the
absence of A. Points along the line Aa
represent various amounts of solid A in
equilibrium with saturated solution a. Solutions
having composition between a and C are
unsaturated solutions of a in C. When B is
added to a mixture of A and C, the solubility of
A usually decreases along the line ab. In like
manner, addition of A to a solution B in C
usually decreases the solubility along the
line cb. This is the effect normally called
the common-ion effect. The meeting of these
curves at b represents a solution that is
saturated with respect to both salts.
Figure 6.28
liquid
solid
solid
58
In the region AbB, three phases coexist, the two
pure solids A and B and a saturated solution
of composition b. In the tie line regions, the
pure solid and saturated solution are in
equilibrium. Thus, if point d gives the
composition of the mixture, the amount of
solid phase present is given by the length of
the line de and the amount of saturated
solution is given by the length of the line dA.
Considering the regions of the three-component
systems with one solid phase present in
contact with liquid, we find a bivariant
system (no vapor phase present and the system
at constant pressure). Thus, at any given
temperature the concentration of the solution
may be changed. However, at the point b, two
solid phases are present and the system is
isothermally invariant and must have a definite
composition.
fc-p23-223-21
Figure 6.28
f?
fc-p23-322-20
Now consider what happens to an unsaturated
solution P as it is isothermally evaporated. The
overall or state composition moves along the line
Ch. Pure A begins to crystallize at point f with
the composition of the solution moving along fb.
At point g the solution composition is b and B
begins to crystallize. As evaporation and hence
removal of C continue, both solid A and solid B
are deposited until at h all the solution is
gone.
59
Figure 6.28
The process of recrystallization can also be
interpreted from Figure 6.28. Let A be the solid
to be purified from the only soluble impurity B.
If the original composition of the solid mixture
of A and B is h, water is added to achieve the
overall composition d. The mixture is heated,
thus changing the state to an unsaturated liquid
and effecting complete solubility. (Complete
solubility can usually be brought about by
increasing the temperature sufficiently.) When
the liquid is cooled, the impurity B stays in the
liquid phase as pure A crystallizes out along the
line equivalent to ab at the higher temperature.
Thus, when the solution returns to room
temperature, the crystals of pure A may be
filtered off.
60
Problem 6.36
The following questions refer to Figure 6.28 a.
If liquid C were added to the system, what
changes would occur if the system originally
contained 80 salt A and 20 salt B? b. What
changes would occur if the system originally
contained 50 salt A and 50 salt B upon the
addition of liquid. c. If liquid is added to an
unsaturated solution of salt A and salt B in
solution of composition lying at e, what
changes would occur?
Solution
a. As liquid C is added, the saturated liquid of
composition b would be in equilibrium with two
solids A and B. At approximately 25 C, when the
composition crosses the line Ab, the solid B
disappears and only solid A will be present in
equilibrium with liquid of composition b. b. The
two solid phases would not disappear until b is
passed at approximately 50 liquid. c. Added
liquid will cause dilution and solid salt will
cease to exist.
61
Another type of ternary system to be considered
involves one solid and two liquids. There are
many examples in organic chemistry in which the
addition of a salt to a mixture of an organic
liquid such as alcohol and water results in the
separation of two liquid layers, one rich in the
organic liquid, one rich in water. With the
addition of enough salt, a third layer, this time
a solid, also appears. By separating the organic
layer, a separation or salting out has been
achieved.
A typical salt-alcohol-water system is given in
Figure 6.29 for purposes of illustration. This
diagram is characterized by the formation of a
two-liquid region bfc. The solubility of salt in
water is given by point a. The nearness of
point d to 100 alcohol indicates a rather low
solubility of the salt in alcohol. As before,
the solubility of salt in water is changed by the
addition of alcohol along the line ab. In a
similar manner, water increases the solubility
of salt in alcohol along dc. These solutions, b
and c, are immiscible and form the ends of
the binodal curve bfe. This area enclosed by bfc
consists of two conjugate liquids having one
degree of freedom. The greatest solubility of
the two liquids occurs at the plait point f,
where the solutions are saturated with salt. Any
composition lying within the triangle salt, b,
c must yield salt and two liquid layers b and
c.
Figure 6.29
The "salting out" effect is easily demonstrated
by starting with solution g composed of only
water and alcohol. As salt is added point e is
reached, where two liquids x and y are formed.
Layer y containing the high proportion of alcohol
can now be separated.
62
Problem 6.41
In organic chemistry it is a common procedure to
separate a mixture of an organic liquid in water
by adding a salt to it. This is known as salting
out. The ternary system K2CO3-H2O-CH3OH is
typical. The system is distinguished by the
appearance of the two-liquid region abc. a.
Describe the phase(s) present in each region of
the diagram. b. What would occur as solid K2CO3
is added to a solution of H2O and CH3OH of
composition x? c. How can the organic-rich phase
in (b) be separated? d. How can K2CO3 be
precipitated from a solution having composition
y? e. Describe in detail the sequence of events
when a solution of composition F is evaoprated.
63
Solution
a.
Region System AEa
K2CO3water-rich saturated solution Aac
K2CO3conjugate liquids a and c abc
two conjugate liquids joined by tie
lines AcB K2CO3alcohol-rich
saturated solution
b. The state of the system will move along a line
joining x and A. Initially solution is formed as
more K2CO3 is added two layers a and c form, and
once beyond point z, K2CO3 ceases to dissolve so
that solid K2CO3 and the two liquids a and c
coexist. c. As long as two liquids exist, liquid
with composition in the region AcB is the
alcohol-rich layer and may be separated from the
water-rich layer by separatory funnel.
64
d. When water is added to unsaturated solution of
K2CO3 in alcohol, the state of the system moves
along the line joining y and D. Some K2CO3 will
precipitate as the state moves into the ABc
region and then redissolves as it moves into the
solution region again. e. On the evaporation
of F, the system composition follows a line drawn
from the water corner through F to the Ac line.
At the first composition line, two liquids form
and the compositions of the solutions move toward
a and c. When the system composition reaches the
ac line, K2CO3 begins to precipitate and is in
equilibrium with the conjugate liquids a and c.
Further reduction of water moves the ratio of
liquid a to liquid c in favor of c until the line
Ac is crossed, at which time solid K2CO3 is in
equilibrium with a single solution.