Chapter 6   Phases Equilibria
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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.

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Chapter 6 Phases Equilibria

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Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

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

Water+NaCl

two component

The material balance and electroneutrality conditions reduce the number of constituents

to two; that is, c=4-1-1=2.

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, c=3-1=2.

~Another reduction in the number of components is possible if we start with pure PCl5, in

which case [PCl3]=[Cl2]. In this case the number of components is unity because of the

additional mathematical relation.


Chapter 6 phases equilibria

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 c=4-2=2.

H2, O2, H2O

three component

room temperature

H2+O2 H2O

two component

higher temperature


Chapter 6 phases equilibria

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CaO+CO2, 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.


Chapter 6 phases equilibria

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 bivariantsystem. 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


Chapter 6 phases equilibria

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, f=c-p+2=2-1+2=3

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, f=c-p+2=3-1+2=4, 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, f=c-p+2=2-2+2=2.


Chapter 6 phases equilibria

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 (斜方硫). As rhombic sulfur is heated slowly at 1 atm, it transform to different crystalline form called monoclinic sulfur (單斜硫) at 368.55 K. The name allotrope (同素異性體) 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.


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

In the four regions where single phases exist (namely rhombic, monoclinic, liquid, or vapor) we have f=1-1+2=2 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.


Chapter 6 phases equilibria

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, f=2-1-2=3.

~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


Chapter 6 phases equilibria

This 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

lever rule

In the region marked Liquid+Vapor, 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.


Chapter 6 phases equilibria

Example 6.3

Determine the mass percentage of carbon tetrachloride CCl4 (P1*=114.5 Torr) in the vapor phase at equilibrium in a 1:1 mole ideal solution with trichloromethane CHCl3 (P2*=199.1 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 x1=0.5 CCl4. Then

The ratio is

So the mass percentage of carbon tetrachloride in the vapor is


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

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


Chapter 6 phases equilibria

Liquid-Vapor Equilibrium in Systems ~Not Obeying Raoult’s Law

Figure 6.3

When there are large, positive deviations from Raoult’s 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 Raoult’s 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.


Chapter 6 phases equilibria

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.)


Chapter 6 phases equilibria

Distillation

Figure 6.7

將p點加熱,達到l點時,此時蒸氣之組成為v,其化合物2所佔有比例較多,若將此蒸氣移除,則剩餘液體內化合物 1所佔有比例較多。當溫度上升至1時,則系統移動至p’點,蒸氣組成由v移動至D而液體組成由l移動至R,此時蒸

氣與液體含量為level rule決定,

將p點加熱至p’點,蒸氣組成為D(yd)而液體組成為R(xR),將蒸氣冷凝至l’點,得到組成為xl’的液體,再將其汽化至v’,可得到組成為yv’的蒸氣,再將其冷凝至l’’點,得到組成為xl’’的液體,再將其汽化至v’’,可得到組成為yv’’的蒸氣,如此可得到接近純化合物2的蒸氣,依此方式最後可以將混合物分離成純化合物1與化合物2,上述的過程稱為

蒸餾(distillation)。

Vapor pressure of component 2 is higher than that of component 1.


Chapter 6 phases equilibria

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 B=99/(33+99)*100=75%, 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.


Chapter 6 phases equilibria

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


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

Azeotrope(共沸物)

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


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

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 nAand 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.


Chapter 6 phases equilibria

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


Chapter 6 phases equilibria

Distillation of Partially Miscible Liquids

Figure 6.11

Generally partial miscibility at low temperature is caused by large positive deviation from Raoult’s 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.


Chapter 6 phases equilibria

f=c-p+2=2-3+2=1-1=0

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 which the liquid p’

was produced.

Figure 6.11

f=?

f=?

f=?


Chapter 6 phases equilibria

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 of a second liquid phase L2

appears having composition e.


Chapter 6 phases equilibria

~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

conjugated solutions (共軛溶液).

~The composition of these layers depends on the

temperature. The amount of the individual layers

present may be determined using the lever rule.


Chapter 6 phases equilibria

~Constant composition lines, vertical on the

diagram in Figure 6.12, are known asisopleths.

~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.


Chapter 6 phases equilibria

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


Chapter 6 phases equilibria

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 Raoult’s law responsible for the

immiscibility may be just balanced at the lower temperature

by large negative deviations from Raoult’s 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.


Chapter 6 phases equilibria

~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


Chapter 6 phases equilibria

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 xAu=1 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 (共熔溫度).

~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


Chapter 6 phases equilibria

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 my be varied.

f=c-p+2=2-3+2=1-1=0

f=c-p+2=2-2+2=2-1=1

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.


Chapter 6 phases equilibria

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


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

Figure 6.17

~The descriptions of curves 3 and 5 are the same as for curve 2 except for the lengths

of the line l’s’ and l”s” 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 l’s’ 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.


Chapter 6 phases equilibria

Figure 6.18

B

A

E

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 2:1 mole

ratio of water to Na+Cl- 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 more 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.


Chapter 6 phases equilibria

Figure 6.18

B

A

E

~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 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 bath at some temperature below that of meltingice. 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.


Chapter 6 phases equilibria

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 xNi=1 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.


Chapter 6 phases equilibria

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%


Chapter 6 phases equilibria

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:尖晶石

corundum:金剛砂


Chapter 6 phases equilibria

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(共熔合金). 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.


Chapter 6 phases equilibria

Problem 6.38

Describe what happens within the system Mn2O3-Al2O3 in Figure 6.21 when a liquid of xAl2O3=0.2 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 Mn3O4+spinel. Below about 1285 K, the spinel converts to

corundum and Mn3O4+corudum coexist.


Chapter 6 phases equilibria

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 (合熔). 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 (分熔).

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.


Chapter 6 phases equilibria

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 (過渡反

應). The point P is know as the peritectic point (分熔點).

~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.


Chapter 6 phases equilibria

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


Chapter 6 phases equilibria

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


Chapter 6 phases equilibria

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


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

CH3OCOOH

C7H8

H2O

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.


Chapter 6 phases equilibria

CH3OCOOH

C7H8

H2O

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.


Chapter 6 phases equilibria

CH3OCOOH

F

C7H8

H2O

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 p’’q’’ 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:


Chapter 6 phases equilibria

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 Aarepresent 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


Chapter 6 phases equilibria

~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.

Figure 6.28

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.


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

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. ~Thesesolutions, 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.


Chapter 6 phases equilibria

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.


Chapter 6 phases equilibria

Solution

a.

Region System

AEa K2CO3+water-rich saturated solution

AaC K2CO3+conjugate liquids a and c

abc two conjugate liquids joined by tie lines

AcB K2CO3+alcohol-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.


Chapter 6 phases equilibria

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.


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