1

LAB 3: BINARY SOLID-LIQUID EQUILIBRIUM IN TWO-COMPONENT SYSTEM

Solid–Liquid Phase Equilibrium (Ref: Gallus et. al. J.Chem. Ed., 78, 961-964, 2001.)

The theoretical background for solid–liquid binary phase diagrams with no solid

solubilities is quite simple. A system with one or more compounds Ai in two phases (s) and ( )

that are in mutual contact is in thermodynamic equilibrium if the chemical potentials μ

l

i are the

same in both phases:

(1) ))(,,)(( ),)(( ll iii xTpμTpsμ =

In the following, we consider the compounds to be completely insoluble in the solid phase

(2) ),)(( ),)(( TpsμTpsμ ii∗=

where is the chemical potential of the pure solid compound A),)(( Tpsμi

∗i at pressure p and

temperature T. Complete miscibility in the liquid phase leads to

(3) ))(),(ln(),)(( ))(,,)(( lllll iiiii xfRTTpμxTpμ += ∗

where represents the chemical potential of the pure liquid compound A),)(( Tpμi l∗

i. While the activity coefficients in real systems depend on the composition of the liquid phase, in systems exhibiting nearly ideal behavior ≈ 1, which, as a result of the experimental findings, can be considered fulfilled in the following. The temperature T at which the two coexisting phases are in equilibrium at constant pressure is then related to the mole fraction of the liquid melt by

))(,,)(( ll ii xTpf)(lif

)(lix

1

m

)(ln1 −

∗∗ ⎟⎟⎠

⎞⎜⎜⎝

⎛Δ

−= liii

xH

RT

T (4)

where is the molar heat of fusion and is the melting point of the pure substance A

)()(m shhH iii −=Δ ∗ l ∗iT

i. In deriving eq (4), is considered to be constant in the temperature range investigated. The coexistence curve between the liquid mixture and the pure solid A

∗Δ iHm

i in the diagram extends from the pure substance to the eutectic composition; that is, 1 ≤

≤ for i = {1,2}. ))(,( lixT )(lix)(E, lixThe eutectic point in an ideal binary system (where = 1) with

given quantities and can be obtained by numerical analysis (e.g. using Maple or Mathematica) or by a simple iterative approximation (Fig. 1).

)),(( EE, Txi l )()( 21 ll xx +∗

iT ∗Δ iHm

The two liquidus curves described by eq (4) cross at the eutectic composition when = . Table 2 lists eutectic compositions and temperatures that have been

calculated for our binary systems using the data of the pure components as given in Table 1. The cryoscopic constant is the slope of the liquidus curve at = 1.

))(( 1 lxT ))(( 2 lxT

∗∗ Δ ii HTR m2 /)( )) )l(( lixT (ix

2

Figure 1. Calculated liquidus curves (bold lines) in an ideal binary solid–liquid phase diagram assuming no solid solubility. The two curves (dotted lines) described by eq. (4) intersect at the eutectic point at

. The light solid straight lines show the asymptotic freezing point depression near the pure components. This diagram represents the binary system p-dichlorobenzene (1)–phenanthrene (2).

)(E,1 lx

Table 1. Properties of some compounds

Table 2. Compositions and temperatures of all 15 eutectic binary systems

The Phase Rule

3

The simplest possible two-component phase diagram is one in which no solid solutions or

compounds occur (Fig. 2).

R

QS

c3

Temperature

R

QS

c3

R

QS

c3

Temperature

Figure 2 Two solids and one liquid, no solid solutions.

According to the phase rule, a single phase in a two-component system may possess three degrees

of freedom:

F = C – P + 2 = 2 – 1 + 2 = 3 (5)

when F is the degree of freedom, C is the number of component, and P is the number of phase.

These are temperature, pressure, and the composition of one of the two components.

Since studies of solid-liquid equilibrium are nearly always carried out at constant

pressure, one degree of freedom is lost and the phase rule becomes

F = C – P + 1 = 2 – P + 1 = 3 – P (6)

Thus, temperature and the composition are the variables, and these may be plotted on ordinary

graph paper.

4

THERMODYNAMIC ANALYSIS

Let us continue with this point of view, and think of the system at point a2 as a saturated

solution of solid B in the solvent A. At point a2, then the following equilibrium is established

between solid B and a solution of mole fraction xB : B

B (solid) = B (solution) (7)

Since the system is at equilibrium, ΔG = 0, so that we can write

ΔG = 0 = GB (solution) – GB BB (solid) (8)

If we choose as the standard state a solution of unit mole fraction (xB = 1), then the

relationship between the free energy per mole at any x

B

BB

and the free energy per mole in the

standard state is given by

GB (solution) = GB BB

o (solution) + RT ln xB (9) B

when Equation (6) is substituted in eq (8), the result is

GB (solution) – GB Bo (solution) = RT ln xBB (10)

This may be rearranged slightly to give

RT

Gx

)solution( ln B

BΔ−

= (11)

Recalling that 2P

)/( TH

TTG

−=⎟⎠⎞

⎜⎝⎛

∂∂ at constant pressure, we differentiate eq (11) to obtain

2B

p

B )solution(

lnRT

HTx Δ

=⎟⎠

⎞⎜⎝

⎛∂

∂ (12)

Equation (12) may also be written in a slightly different from:

R

HTx )solution(

)(1/

ln B

p

B Δ=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂ (13)

At this point let us write Equation (4) in two steps to indicate first the solution process

involving the fusion of the solute, and then the ideal mixing of the liquid solute with the already

liquid solvent, for which ΔH (mix) = 0:

B (solid) = B (liquid) = B (solution) (14)

5

← ΔH (fus) →← ΔH (mix) →

← ΔH (soln) →

Heat of Solution

When the solution process is ideal, ΔH (mix) = 0, so that ΔH (solution) = ΔH (fusion),

and eqn (13) can be written as

R

HTx )fusion(

)(1/

ln B

p

B Δ=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂ (15)

Consequently, data from phase diagrams such as Fig. 1 can be used to determine heats of fusion

of both components,since a plot of ln xA against (1/T) gives a straight line of slope equal to the

heat of fusion of A, and a plot of ln xB against (1/T) gives a straight line of slope equal to the

heat of fusion of B (both divided by –R). In this manner heats of fusion for many nonelectrolytes

and metals can be determined, if the two components form an ideal solution ⎯ which is not

uncommon.

B

Ideal Solubility

Integration of Equation (13) leads to another interesting interpretation of ideal

solubilities:

R

H

TT

x fusion)(

11

ln B

oB

B Δ=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

(16)

This may be written explicitly in terms of ln xB, where xB BB is the solubility of B in A at

temperature T, ΔHB (fus) is the heat of fusion of B, and is the melting point of pure B: B

oBT

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

TTRH

x 11 fusion)(

lno

B

BB (17)

What is remarkable about this equation is that it states that the solubility of B expressed in mole

fraction is the same in all solvents, and that the solubility depends only on two properties of the

solute B: the heat of fusion of B and the melting point of pure B. Whether A or B is considered

the solvent or solute is immaterial; it is conventional to look upon the substance present in the

larger amount as the solvent. Consequently, we can rewrite eq (17) for substance A:

6

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

TTRH

x 11 fusion)(

ln oA

AA (18)

Equations (17) and (18) are valid for ideal solutions. To check the validity of these equations, let

us examine the solubility of p-dibromobenzene in a number of solvents in which it forms nearly

ideal solutions, as shown in Table 3. The solubility in mole fraction agrees quite well with the

mole fraction calculated with eq (17). Not surprisingly, the agreement is best with bromobenzene

as solvent.

Table 3 Solubility of p-dibromobenzene at 25°C [ΔH (fusion) = 20.29 kJ/mol, T = 359.1 K ]

Measured* Solvent g/100g xA

xA calculated with eq (17)

CS2 47.4 0.225 0.2497 C6H6 45.6 0.217 0.2497

C6H5Br 32.5 0.243 0.2497 * Data from Stephen et al., 1963.

Calculation of the Liquids Curve

Equations (17) and (18) as written might be described as giving the solubility ( xA or xB )

of a solute at temperature T. Alternatively, we can look at T as the temperature at which the solid

and liquid phases are in equilibrium (eq. (14)). The phase diagram is a plot of this temperature

against composition (mole fraction). Let us rearrange eqs (17) and (18) so that the solid-liquid

equilibrium temperature T is expressed explicitly in terms of composition and two properties of

the solvent-heat of fusion and melting point:

B

1

A

Ao

A )fusion(ln 1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛

Δ−=

HxR

TT (19)

( ) 1

B

Ao

B )fusion(1ln 1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛

Δ−

−=H

xRT

T (20)

7

CHEMICALS AND APPARATUS

1. Component A: Naphthalene

2. Component B: Diphenylamine

3. Beaker (600 mL)

4. Test tubes (∼ 25×200 mm)

5. Thermometer (0 - 110°C)

6. Clamp and ring stand

7. Aluminum foil

Fig. 2 Some ideals of how to set-up the freezing-point apparatus

8

PROCEDURES

1. Make two series of runs: A-rich series, beginning with pure A, followed by systems made

by successive additions of B to the previous run beginning with pure A; and B-rich series

similarly prepared. Weight to ± 0.01 g. Some approximate weights are given in Table 4.

Table 4 Approximate Range of Compositions

Start: 3 g Pure A Start: 3 g pure B Run Add B (g) wt.% A Run Add A (g) wt.% B 1A 0.00 100 1B 0.00 100 2A 0.33 90 2B 0.33 90 3A 0.42 80 3B 0.42 80 4A 0.54 70 4B 0.54 70 5A 0.71 60 5B 0.71 60 6A 1.00 50 6B 1.00 50

2. Heat the sample tube of mixture in a water bath until it is melted. Then remove the sample tube

from the water bath and measure the temperature periodically (e.g., every minute) until the

system is essentially solid. The relationship between the cooling curves and the phase

diagram is shown in Fig. 4.

3. When a pure substance cools, the temperature falls smoothly until the freezing point is reached;

then the temperature halts, as shown for curve 1 (Fig. 4a), when the solid A begins to

solidify.

Fig. 4 (a) Cooling curves, and (b) Phase diagram (two solids and one liquid, no solid solutions).

Cooling curves

Temperature Liquid cooling

Liquid freezing

Solid cooling

curve1 curve 2

(a) (b) Phase diagram

9

For a one-component system, at constant pressure, the phase rule is

F = C − P + 1 = 2 − P

When two phases are in equilibrium, the system is invariant and the temperature is fixed, that is,

at the melting point. Above and below the melting point, there is only one phase, so F = 1, and

the temperature is no longer fixed; it varies and must be specified to fix the state of the system.

Runs 1A and 1B in Table 4 are of this type.

For a two-component system, at constant pressure, the phase rule is

F = C − P + 2 = 3 − P

When a homogeneous melt cools (curve 2 in Fig. 4a), a temperature is reached at which the

solution is saturated with respect to one of the components, A in this case. At this temperature, A

begins to precipitate out of the solution and the rate of cooling decreases. It decreases because

heat is given up to the system by the exothermic process:

A (solution) → A (solid) ΔH is negative (21)

Above the two-phase line, there is one phase, so that F = 3 − 1 = 2 .

Below the two-phase line, there are two phase in equilibrium, so that F = 3 − 2 = 1 .

However, at the eutectic temperature, the solution is not only saturated with respect to

A, but it is also saturated with respect to B, which now begins to precipitate out of solution

simultaneously with A. Since three phases are in equilibrium, F = 3 − 3 = 0, and the system is

invariant. The temperature is fixed and the cooling halts at the eutectic temperature.

The cooling curves for all the systems in Table 4 are of this type except those for runs

1A and 1B, which are one-component systems. Measuring the cooling curve for a two-component

system of known composition reveals the eutectic temperature as well as the temperature on the

two-phase curve. The cooling curves of all the two-component systems show a halt at the same

(eutectic) temperature. They also show an inflection point at a higher temperature, which depends

upon the composition.

10

TREATMENT OF RESULTS

1. Graph the cooling curves and phase diagram (the same as Fig. 4). Tabulate the compositions,

observed eutectic temperatures, and two-phase temperatures for the various runs.

2. Construct a phase diagram from your observations. Tabulate and plot ln x versus 1/T for both

A and B.

3. Calculate the heat of fusion.

4. Compare your heats of fusion with the literature values. With the literature values for the heat

of fusion and melting points of pure A and B, construct phase diagram based upon ideal

solution behavior (eqs 19 and 20).

5. Compare your observed phase diagram, eutectic temperature, and composition with the

calculated values and calculated phase diagram.

SAFETY CONSIDERATIONS

Some of the above compounds are known carcinogens and should not be allowed to

contact skin. Others smell bad and it may be advantageous to carry out the measurements in a

hood. Take care not to touch a hot test tube.

REFERENCES

1. J. Gallus, Q. Lin, A. Zumbühl, S.D. Friess, R. Hartmann, and E.C. Meister, “Binary Solid–Liquid Phase Diagrams of Selected Organic Compounds”, J.Chem. Ed., 78, 961-964, 2001.

2. D.P. Shoemaker, C.W. Garland, and J.W. Nibler, Experiments in Physical Chemistry, 5th

ed., McGraw-Hill, New York, 1994, pp. 195-197, 238-246; and 6th ed., pp. 179-182, 215-

222.

3. G.P. Matthews, Experimental Physical Chemistry, Clarendon Press, Oxford, 1985, pp. 46-

48, 52-53.

4. P.P. Blanchette, J. Chem. Ed., 64, 267-269, 1987.

11

Appendix

ตัวอยางการวิเคราะหขอมูลการทดลอง

(0.62, 30.95)

0102030405060708090

0 0.2 0.4 0.6 0.8 1

mole fraction of diphenylamine, xB

Free

zing

poi

nt [o C

]

Data taken from: http://www.williams.edu/chemistry/epeacock/EPL_CHEM_366/366_LAB_WEB/

Liquid-solid phase diagram for naphthalene-diphenylamine system

A(s) + l

A(s) + B(s)

A(l) + B(l)

Runwt.%

A xB Runwt.%

B xB

1A 100 0.00 1B 100 1.002A 90 0.08 2B 90 0.873A 80 0.16 3B 80 0.754A 70 0.25 4B 70 0.645A 60 0.34 5B 60 0.536A 50 0.43 6B 50 0.43

Near the eutectic pointB(s) + l

eutectic point

A = naphthaleneB = diphenylamine

การทดลองกอนหนา

12

(0.62, 30.95)

0102030405060708090

0 0.2 0.4 0.6 0.8 1

mole fraction of diphenylamine, xB

Free

zing

poi

nt [o C

] ใหเลือกจุดซีกขวาของ eutectic point (ฝง B rich) มาพลอต กราฟระหวาง 1/Tf กับ ln xB

⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ+⎟

⎠⎞

⎜⎝⎛ Δ−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

oB

BBB

oB

BB

fus)(1fus)( ln

11 fus)( ln

RTH

TRHxor

TTRHx

Slope =R

H fus)(BΔ−

Y = Slope ⋅ X + Y-intercept

( ) kJ/mol 4.17K 2093.9K mol

J 8.314fus)(B +=⎟⎠⎞

⎜⎝⎛=Δ∴ --H

Slope = RH fus)(BΔ

−

y = -2093.9x + 6.4117R2 = 0.999

-0.5

-0.4

-0.3

-0.2

-0.1

00.003 0.0031 0.0032 0.0033

1/T [K-1]

ln x

B

คํานวณ

13

เชนเดียวกัน เลือกขอมูลซีกซายของ eutectic point

(0.62, 30.95)

0102030405060708090

0 0.2 0.4 0.6 0.8 1

mole fraction of diphenylamine, xB

Free

zing

poi

nt [o C

]

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

TTRHx 11 fus)( ln o

A

AAY-axis X-axis

ก็จะไดกราฟเสนตรง

y = -2127.6x + 6.0237R2 = 0.9989

-1

-0.8

-0.6

-0.4

-0.2

00.0028 0.0029 0.003 0.0031 0.0032 0.0033

ln x

A

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

TTRHx 11 fus)( ln o

A

AA

( ) kJ/mol 7.17K 2127.6K mol

J 8.314fus)(A +=⎟⎠⎞

⎜⎝⎛=Δ --H∴

14

ถาเกิดจุดท่ีเราเลือก ครอม eutectic point กราฟจะเกิดจุดหกัใกลๆ eutectic composition ทําใหไดกราฟไมเปนเสนตรง

y = -1971.6x + 5.9249R2 = 0.3443

-0.8

-0.6

-0.4

-0.2

00.003 0.0031 0.0032 0.0033 0.0034

1/T [K-1]

ln x

B