8/13/2019 06 Atkins Chap06

1/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 117 #1

6 Phase diagrams

Answers to discussion questions

D6.1 Phase: a state of matter that is uniform throughout, not only in chemical composition but also in

physical state.

Constituent: any chemical species present in the system.

Component: a chemically independent constituent of the system. It is best understood in relation to the

phrase number of components which is the minimum number of independent species necessary to

define the composition of all the phases present in the system.

Degreeof freedom (orvariance): thenumber of intensive variables that canbe changed without disturbing

the number of phases in equilibrium.

D6.3 See Figs. 6.1(a) and (b).

Figure 6.1(a)

8/13/2019 06 Atkins Chap06

2/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 118 #2

118 SOLUTIONS MANUAL

Figure 6.1(b)

D6.3 See Fig. 6.2.

Figure 6.2

Solutions to exercises

E6.1(a) An expression for composition of the solution in terms of its vapor pressure is required. This is obtained

from Daltons law and Raoults law as follows

p = pA+ pB [Daltons law] =xApA+ (1xA)p

B.

Solving forxA,xA =ppBp

A pB

.

8/13/2019 06 Atkins Chap06

3/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 119 #3

PHASE DIAGRAMS 119

For boiling under 0.50 atm (50.7 kPa) pressure, the combined vapor pressure, p, must be 50.7 kPa;

hencexA =50.720.0

53.320.0= 0.920 ,xB = 0.080 .

The composition of the vapor is given by eqn 6.5,

yA =xAp

A

pB+ (pA p

B)xA

=0.92053.3

20.0+(53.320.0)0.920= 0.968 and

yB =1 0.968 = 0.032 .

E6.2(a) The vapor pressures of components A and B may be expressed in terms of both their composition in the

vapor and in the liquid. The pressures are the same whatever the expression; hence the expressions can

be set equal to each other and solved for the composition.

pA =yAp = 0.350p = xApA =xA (76.7 kPa),

pB =yBp = (1yA)p = 0.650p = xBpB =(1xA)52.0 kPa).

Therefore,yAp

yBp=

xApA

xBpB

.

Hence0.350

0.650=

76.7xA

52.0(1xA),

which solves toxA = 0.268 ,xB =1 xA = 0.732

and, since 0.350p = xAp

A,

p =xAp

A

0.350=

(0.268)(76.7 kPa)

0.350= 58.7 kPa .

E6.3(a) (a) Check to see if Raoults law holds; if it does the solution is ideal.

pA =xApA =(0.6589)(127.6 kPa) = 84.07 kPa,

pB = xApB =(0.3411)(50.60 kPa) = 17.26 kPa,

p = pA + pB = 101.3 kPa= 1 atm.

Since this is the pressure at which boiling occurs, Raoults law holds and the solution is ideal .

(b) yA =pA

p[6.4] =

84.07 kPa

101.3 kPa= 0.830 , yB =1 yA =1.0000.830 = 0.170 .

E6.4(a) (a) p(total) = pDE+ pDP[Daltons law] =xDEpDE+ xDPp

DP [Raoults law, 6.3].

xDE =zDE, xDP =1 zDE [system all liquid].

p(total) = (0.60)(22.9 kPa)+(0.40)(17.1 kPa) = 13.7+6.8 = 20.5 k Pa .

(b)yDE =

pDE

p[6.4] =

13.7 kPa

20.5 kPa= 0.67 , yDP =1 yDE = 0.33 .

8/13/2019 06 Atkins Chap06

4/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 120 #4

120 SOLUTIONS MANUAL

E6.5(a) The data are plotted in Fig. 6.3. From the graph, the vapor in equilibrium with a liquid of composition

(a) xM = 0.25 is determined from the tie line labeled a in the figure extending from xM = 0.25 toyM = 0.36 , (b) x0 = 0.25 is determined from the tie line labeled b in the figure extending from

xM =0.75 to yM = 0.82 .

Figure 6.3

E6.6(a) (a) Though there are three constituents, salt, water, and water vapor, there is an equilibrium condition

between liquid water and its vapor. Hence, C =2 .

(b) Disregarding the water vapor for the reasons in (a) there are seven species: Na+, H+, H2PO4

,

HPO24

PO34

, H2 O, OH. There are also three equilibria, namely,

H2PO4

H+ +HPO24 ,

HPO2

4 H+

+PO3

4 ,

H+OH H2O.

(These could all be written as Brnsted equilibria without changing the conclusions.) There are also two

conditions of electrical neutrality, namely,

[Na+] = [phosphates], [H+] = [OH] + [phosphates]

where [phosphates] = [H2PO4] + 2[HPO

24 ] + 3[PO

34 ]. Hence, the number of independent

components is

C =7 (3+2) = 2 .

E6.7(a) CuSO4 5H2O(s) CuSO4(s)+5H2O(g).

There are two solids, but one solid phase, as well as a gaseous phase; hence P = 2 . Assuming all the

water and CuSO4are formed by the dehydration, their amounts are then fixed by the equilibrium; hence

C =2 .

E6.8(a) (a) The two components are Na2SO4 and H2O (proton transfer equilibria to give HSO4, etc. do not

change the number of independent components) so C= 2 . There are three phases present (solid

salt, liquid solution, vapour), so P = 3 .

(b) The variance is F =C P +2 = 2 3 +2 = 1 .

8/13/2019 06 Atkins Chap06

5/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 121 #5

PHASE DIAGRAMS 121

Either pressure or temperature may be considered the independent variable, but not both as long as the

equilibrium is maintained. If the pressure is changed, the temperature must be changed to maintain theequilibrium.

E6.9(a) See Fig. 6.4.

Figure 6.4

E6.10(a) Refer to Fig. 6.26 of the text. Atb3 there are two phases with compositions xA =

0.18 and xA =

0.70;their abundances are in the ratio 0.13 (lever rule). Since C = 2 and P = 2 we have F = 2 (such as p

andx). On heating, the phases merge, and the single-phase region is encountered. Then F =3 (such as

p, T, andx). The liquid comes into equilibrium with its vapor when the isopleth cuts the phase line. At

this temperature, and for all points up tob1,C =2 andP = 2, implying that F =2 (for examplep,x).

The whole sample is a vapor above b1.

E6.11(a) The incongruent melting point (Section 6.6) is marked asT1 = 400C in Fig. 6.5(a). The composition

of the eutectic is marked as xe (0.30)in the figure. Its melting point isT2 (200C).

E6.12(a) The cooling curves are shown in Fig. 6.5(b). Note the breaks (abrupt change in slope) at temperatures

corresponding to pointsa1, a2, b1, b2. Also note the eutectic halt at b3.

E6.13(a) Refer to Fig. 6.6.

(a) The solubility of silver in tin at 800 C is determined by the pointc1(at higher proportions of silver,

the system separates into two phases). The point c1corresponds to 80 per cent silver by mass.

(b) See pointc2. The compound Ag3Sn decomposes at this temperature.

(c) The solubility of Ag3Sn in silver is given point c3 at 300C.

E6.14(a) (a) See Figs. 6.7(a) and (b).

(b) Follow line b in Fig. 6.7(a) down to the liquid line which intersects at point b1. The vapor pressure

atb1 is 620 Torr .

8/13/2019 06 Atkins Chap06

6/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 122 #6

122 SOLUTIONS MANUAL

Figure 6.5

Figure 6.6

(c) Follow linebin Fig. 6.7(a) down to the vapor line which intersects at point b2. The vapor pressure at

b2is 490 Torr . From points b1tob2, the system changes from essentially all liquid to essentially

all vapor.

(d) Consider tie lined; pointb1 gives the mole fractions of the liquid, which are

x(Hep)= 0.50 = 1 x(Hex), x(Hex)= 0.50 .

Pointd1 gives the mole fractions in the vapor which are

y(Hep) 0.28 = 1 y(Hex), y(Hex) 0.72 .

The initial vapor is richer in the more volatile component, hexane.

8/13/2019 06 Atkins Chap06

7/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 123 #7

PHASE DIAGRAMS 123

Figure 6.7(a)

Figure 6.7(b)

(e) Consider tie line e; pointb2 gives the mole fractions in the vapor, which are

y(Hep) = 0.50 = 1 y(Hex), y(Hex) = 0.50 .

Pointe1 gives the mole fractions in the liquid, which are

x(Hep) = 0.70 = 1 x(Hex), x(Hex) = 0.30 .

(f) Consider tie linef. The section,ll , from pointf1to the liquid line gives the relative amount of vapor;

the section,lv, from pointf1to the liquid line gives the relative amount of liquid. That is

nvlv =nlll [6.7] ornv

nl=

ll

lv

6

1.

Since the total amount is 2 mol, nv 1.7 andnl 0.3 mol .

8/13/2019 06 Atkins Chap06

8/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 124 #8

124 SOLUTIONS MANUAL

E6.15(a) The phase diagram is drawn in Fig. 6.8.

Figure 6.8

E6.16(a) The cooling curves are sketched in Fig. 6.9. Note the breaks and halts. The breaks correspond to changes

in the rate of cooling due to the freezing out of a solid, which releases its heat of fusion and thus slows

down the cooling process. The halts correspond to the existence of three phases and hence no variance

until one of the phases disappears.

Figure 6.9

E6.17(a) The phase diagram is sketched in Fig. 6.10.

Figure 6.10

8/13/2019 06 Atkins Chap06

9/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 125 #9

PHASE DIAGRAMS 125

(a) The mixture has a single liquid phase at all compositions.

(b) When the composition reaches x(C6F14) = 0.24 the mixture separates into two liquid phases of

compositionsx= 0.24 and0.48. Therelative amounts of thetwo phaseschange until thecomposition

reachesx= 0.48. At all mole fractions greater than 0.48 in C6F14, the mixture forms a single liquid

phase.

Solutions to problems

Solutions to numerical problems

P6.1 (a) The data, including that for pure chlorobenzene, are plotted in Fig. 6.11.

Figure 6.11

(b) The smooth curve through the x, Tdata crosses x = 0.300 at 391.0 K, the boiling point of the

mixture.

(c) We need not interpolate data, for 393.94 K is a temperature for which we have experimental data.

The mole fraction of 1-butanol in the liquid phase is 0.1700 and in the vapor phase 0.3691. According

to the lever rule, the proportions of the two phases are in an inverse ratio of the distances their mole

fractions are from the composition point in question. That is,

nliq

nvap=

v

l=

0.36910.300

0.3000.1700= 0.532 .

P6.3 pA =aApA =AxAp

A [5.45].

A =pA

xApA

=yAp

xApA

.

Sample calculation at 80 K:

O2(80 K)=0.11(100 kPa)

0.34(225 Torr)

760 Torr

101.325 kPa

,

O2(80 K)= 1.079.

8/13/2019 06 Atkins Chap06

10/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 126 #10

126 SOLUTIONS MANUAL

Summary:

T/K 77.3 78 80 82 84 86 88 90.2

O2 0.877 1.079 1.039 0.995 0.993 0.990 0.987

To within the experimental uncertainties the solution appears to be ideal( =1). The low value at 78 K

may be caused by nonideality; however, the larger relative uncertainty in y(O2) is probably the origin

of the low value.

A temperaturecomposition diagram is shown in Fig. 6.12(a). The near ideality of this solution is,

however, best shown in the pressurecomposition diagram of Fig. 6.12(b). The liquid line is essentially

a straight line as predicted for an ideal solution.

Figure 6.12(a)

Figure 6.12(b)

8/13/2019 06 Atkins Chap06

11/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 127 #11

PHASE DIAGRAMS 127

P6.5 A compound with probable formula A3B exists. It melts incongruently at 700C, undergoing the

peritectic reaction

A3B(s) A(s)+(A+B, l).

The proportions of A and B in the product are dependent upon the overall composition and the

temperature. A eutectic exists at 400C andxB 0.83. See Fig. 6.13.

Figure 6.13

P6.7 Theinformation hasbeen used to constructthe phase diagram in Fig. 6.14(a). In MgCu2the mass percent-

age ofMg is (100) 24.324.3+127

= 16 , and in Mg2Cu it is (100) 48.6

48.6+63.5= 43 . The initial point isa1,

corresponding to a liquid single-phase system. Ata2(at 720C) MgCu2 begins to come out of solution

and the liquid becomes richer in Mg, moving toward e2. At a3there is solid MgCu2+ liquid of compos-

ition e2 (33 per cent by mass of Mg). This solution freezes without further change. The cooling curve

will resemble that shown in Fig. 6.14(b).

1200

800

400

a

a1a2

a3e1

e2

e3

(a) (b)

Figure 6.14

8/13/2019 06 Atkins Chap06

12/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 128 #12

128 SOLUTIONS MANUAL

P6.9 (a) Eutectic: 40.2 at % Si at 1268C , Eutectic: 69.4 at % Si at 1030C [6.6].

Congruent melting compounds: Ca2Si mp=

1314CCaSi mp=1324C

[6.7].

Incongruent melting compound:CaSi2 mp=1040C meltsintoCaSi(s)andliquid(68at%Si).

(b) At 1000C the phases at equilibrium will be Ca(s) and liquid (13 at % Si) . The lever rulegives

the relative amounts:

nCa

nliq=

lliq

lCa=

0.20

0.20.13= 2.86 .

(c) When an 80 at% Si melt it cooled in a manner that maintains equilibrium, Si(s) begins to appear atabout 1250C. Further cooling causes more Si(s) to freeze out of the melt so that the melt becomes

more concentrated in Ca. There is a 69.4 at% Si eutectic at 1030 C. Just before the eutectic is

reached, the lever rule says that the relative amounts of the Si(s) and liquid (69.4% Si) phases are:

nSi

nliq=

lliq

lSi=

0.800.694

1.00.80= 0.53=relative amounts at T slightly higher than 1030C .

Just before 10.30C, the Si(s) is 34.6 mol% of the total heterogeneous mixture, the eutectic liquid is

65.4 mol%.

At the eutectic temperature a third phase appearsCaSi2(s). As the melt cools at this temperature, both

Si(s) and CaSi2(s) freeze out of the melt while the concentration of the melt remains constant. At a

temperature slightly below 1030C, all the melt will have frozen to Si(s) and CaSi2(s) with the relative

amounts:

nSi

nCaSi2=

lCaSi2

lSi=

0.800.667

1.00.80

= 0.665=relative amounts ofTslightly higher than 1030C .

Just under 1030C, the Si(s) is 39.9 mol% of the total heterogeneous mixture; the CaSi2 (s) is

60.1 mol%.

A graph of mol% Si(s) and mol% CaSi2(s) vs. mol% eutectic liquid is a convenient way to show relative

amounts of the three phases as the eutectic liquid freezes. See Fig. 6.15. Equations for the graph arederived with the law of conservation of mass. For the silicon mass,

nzSi = nliqwSi + nSixSi + nCaSi2ySi

wheren= total number of moles.

wSi =Si fraction in eutectic liquid = 0.694

xSi =Si fraction in Si(s) = 1.000

ySi =Si fraction in CaSi2(s) = 0.667

ZSi =Si fraction in melt = 0.800

8/13/2019 06 Atkins Chap06

13/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 129 #13

PHASE DIAGRAMS 129

This equation may be rewritten in mole fractions of each phase by dividing by n:

zSi = (mol fraction liq)wSi+ (mol fraction Si)xSi+ (mol fraction CaSi2)ySi.

Since,(mol fraction liq)+(mol fraction Si)+(mol fraction CaSi2) =1

or(mol fraction CaSi2) =1(mol fraction liq+mol fraction Si), we may write:

zSi = (mol fraction liq)wSi+ (mol fraction Si)xSi

+ [1(mol fraction liq+mol fraction Si)]ySi.

Solving for mol fraction Si:

mol fraction Si :=(zSi ySi)(wSi ySi)(mol fraction liq)

xSi

ySi

,

mol fraction CaSi2:=1(mol fraction liq+mol fraction Si).

These two eqns are used to prepare plots of the mol fraction of Si and mol fraction of CaSi2against the

mol fraction of the melt in the range 00.65.

0.6

0.5

0.4

0.3

0.2

0.1

00 0.1 0.2 0.3

mol fraction liq

Freezing proceeds toward left

Freezing of eutectic melt at 1030C

0.4 0.5 0.6 0.7

0.7

mol fraction CaSi2mol fraction Si

Figure 6.15

Solutions to theoretical problems

P6.11 The general condition of equilibrium in an isolated system is dS =0. Hence, if and constitute an

isolated system, which are in thermal contact with each other

dS= dS +dS =0. (a)

8/13/2019 06 Atkins Chap06

14/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 130 #14

130 SOLUTIONS MANUAL

Entropy is an additive property and may be expressed in terms ofUand V.

S= S(U,V).

The implication of this problem is that energy in the form of heat may be transferred from one phase to

another, but that the phases are mechanically rigid, and hence their volumes are constant. Thus, dV= 0,

and

dS=

S

U

V

dU +

S

U

V

dU =1

TdU +

1

TdU [3.45].

But, dU = dU ; therefore1

T=

1

Tor T =T .

Solutions to applications

P6.13 (i) Below a denaturant concentration of 0.1 only the native and unfolded forms are stable.

(ii) At denaturant concentration of 0.15 only the native form is stable below a temperature of about

0.70. At temperature 0.70 the native and molten-globule forms are at equilibrium. Heating above

0.70 causes all native forms to become molten-globules. At temperature 0.90, equilibrium between

molten-globule and unfolded protein is observed and above this temperature only the unfolded form

is stable.

P6.15 C= 1; hence, F =C P +2=3P.

Since the tube is sealed there will always be some gaseous compound in equilibrium with the condensed

phases. Thus when liquid begins to form upon melting, P =3 (s, l, and g) and F =0, corresponding to

a definite melting temperature. At the transition to a normal liquid, P = 3(l, l, and g) as well, so again

F =0.

P6.17 To examine the process of zone levelling with the phase diagram below, Fig. 6.16, consider a solid on

the isopleth through a1and heat the sample without coming to overall equilibrium. If the temperature

rises toa2, a liquid of compositionb2forms and the remaining solid is ata2. Heating that solid down an

isopleth passing through a2forms a liquid of compositionb3and leaves the solid at a3. This sequence

of heater passes shows that in a pass the impurities at the end of a sample are reduced while being

transferred to the liquid phase which moves with the heater down the length of the sample. With enough

passes the dopant, which is initially at the end of the sample, is distributed evenly throughout.

P6.19 The data are plotted in Fig. 6.17.

(a) As the solid compositionx(MgO) =0.3 is heated, liquid begins to form when the solid (lower) line

is reached at 2150C .

(b) From the tie line at 2200C, the liquid composition is y(MgO) = 0.18 and the solid x(MgO) =

0.35 .

8/13/2019 06 Atkins Chap06

15/16

8/13/2019 06 Atkins Chap06

16/16

TRAPP: CHAP06 2006/3/8 17:05 PAGE 132 #16

132 SOLUTIONS MANUAL

P6.21 (a) The data are plotted in Fig. 6.18.

(b) We need not interpolate data, for 6.02 MPa is a pressure for which we have experimental data. Themole fraction of CO2in the liquid phase is 0.4541 and in the vapor phase 0.9980. The proportions of

the two phases are in an inverse ratio of the distance their mole fractions are from the composition

point in question, according to the lever rule

nliq

nvap=

v

l=

0.99800.5000

0.50000.4541= 10.85 .