Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap07

E7.2b All the colligative properties are a result of the lowering of the chemical potential of the solvent due to the presence of the solute.. The lowering of the chemical potential resu

Solutions to exercises

Discussion questions

E7.1(b) For a component in an ideal solution, Raoult’s law is:p = xp∗ For real solutions, the activity,a,

replaces the mole fraction,x, and Raoult’s law becomes p = ap∗

E7.2(b) All the colligative properties are a result of the lowering of the chemical potential of the solvent

due to the presence of the solute This reduction takes the form µA = µA ∗+ RT ln xAorµA =

µA ∗+ RT ln aA, depending on whether or not the solution can be considered ideal The lowering of the chemical potential results in a freezing point depression and a boiling point elevation as illustrated

in Fig 7.20 of the text Both of these effects can be explained by the lowering of the vapour pressure

of the solvent in solution due to the presence of the solute The solute molecules get in the way of the solvent molecules, reducing their escaping tendency

E7.3(b) The activity of a solute is that property which determines how the chemical potential of the solute

varies from its value in a specified reference state This is seen from the relationµ = µ−−+ RT ln a,

whereµ−−is the value of the chemical potential in the reference state The reference state is either the hypothetical state where the pure solute obeys Henry’s law (if the solute is volatile) or the hypothetical state where the solute at unit molality obeys Henry’s law (if the solute is involatile) The activity of the solute can then be defined as that physical property which makes the above relation true It can

be interpreted as an effective concentration

Numerical exercises

E7.4(b) Total volumeV = nAVA+ nBVB= n(xAVA+ xBVB)

Total massm = nAMA+ nBMB

= n(xAMA+ (1 − xA)MB) where n = nA+ nB

m

xAMA+ (1 − xA)MB = n

n = (0.3713) × (241.1 g/mol) + (1 − 0.3713) × (198.2 g/mol)1.000 kg(103g/kg) = 4.670¯1 mol

V = n(xAVA+ xBVB)

= (4.670¯1 mol) × [(0.3713) × (188.2 cm3mol−1) + (1 − 0.3713) × (176.14 cm3mol−1)]

= 843.5 cm3

E7.5(b) Let A denote water and B ethanol The total volume of the solution isV = nAVA+ nBVB

We knowVB; we need to determinenAandnBin order to solve forVA

Assume we have 100 cm3of solution; then the mass is

m = ρV = (0.9687 g cm−3) × (100 cm3) = 96.87 g

of which(0.20) × (96.87 g) = 19.374 g is ethanol and (0.80) × (96.87 g) = 77.496 g is water.

nA= 77.496 g

18.02 g mol−1 = 4.30 mol H2O

nB= 19.374 g

46.07 g mol−1 = 0.4205 mol ethanol

V − nBVB

nA = VA= 100 cm3− (0.4205 mol) × (52.2 cm3mol−1)

3

= 18 cm3

E7.6(b) Check thatpB/xB= a constant (KB)

(pB/xB)/kPa 8.2 × 103 8.1 × 103 8.3 × 103

KB= p/x, average value is 8.2 × 103kPa

E7.7(b) In exercise 7.6(b), the Henry’s law constant was determined for concentrations expressed in mole

fractions Thus the concentration in molality must be converted to mole fraction

m(A) = 1000 g, corresponding to

n(A) = 1000 g

74.1 g mol−1 = 13.5¯0 mol n(B) = 0.25 mol

Therefore,

xB= 0.25 mol

0.25 mol + 13.5¯0 mol = 0.018¯2

usingKB= 8.2 × 103kPa [exercise 7.6(b)]

p = 0.018¯2 × 8.2 × 103

kPa= 1.5 × 102

kPa

E7.8(b) Kf= RT  ∗2M

fusH =

8.314 J K−1mol−1× (354 K)2× 0.12818 kg mol−1

18.80 × 103J mol−1

= 7.1 K kg mol−1

Kb= RT  ∗2M

vapH =

8.314 J K−1mol−1× (490.9 K)2× 0.12818 kg mol−1

51.51 × 103J mol−1

= 4.99 K kg mol−1

E7.9(b) We assume that the solvent, 2-propanol, is ideal and obeys Raoult’s law

xA(solvent) = p/p∗=49.62

50.00 = 0.9924

MA(C3H8O) = 60.096 g mol−1

nA= 250 g

60.096 g mol−1 = 4.1600 mol

xA= nA

nA+ nB nA+ nB= nA

xA

nB= nA

1

xA− 1



= 4.1600 mol

1

0.9924 − 1



= 3.186 × 10−2mol

3.186 × 10−2mol = 27¯3 g mol−1= 270 g mol−1

E7.10(b) Kf = 6.94 for naphthalene

MB=mass of Bn

B

nB= mass of naphthalene · bB

bB=T

Kf

so MB= (mass of B)× Kf

(mass of naphthalene) × T

MB=(5.00 g) × (6.94 K kg mol (0.250 kg) × (0.780 K)−1)= 178 g mol−1

E7.11(b) T = KfbB and bB= nB

mass of water = nB

Vρ

ρ = 103kg m−3 (density of solution ≈ density of water)

nB= V

 RTρ Kf = 1.86 K mol−1kg

T = (1.86 K kg mol−1) × (99 × 103Pa)

(8.314 J K−1mol−1) × (288 K) × (103kg m−3) = 7.7 × 10−2K

Tf = −0.077◦C

E7.12(b) mixG = nRT (xAlnxA+ xBlnxB)

nAr= nNe, xAr= xNe= 0.5, n = nAr+ nNe=RT pV

mixG = pV1

2ln12+1

2ln12

= −pV ln 2

= −(100 × 103Pa) × (0.250 L)



1 m3

103L



ln 2

= −17.3 Pa m3= −17.3 J mixS = −mixG

17.3 J

273 K = 6.34 × 10−2J K−1

E7.13(b) mixG = nRT

J

xJlnxJ[7.18] mixS = −nR

J

xJlnxJ[7.19] = −mixG

T

n = 1.00 mol + 1.00 mol = 2.00 mol

x(Hex) = x(Hep) = 0.500

Therefore,

mixG = (2.00 mol) × (8.314 J K−1mol−1) × (298 K) × (0.500 ln 0.500 + 0.500 ln 0.500)

= −3.43 kJ

mixS = +3.43 kJ

298 K = +11.5 J K−1

mixH for an ideal solution is zero as it is for a solution of perfect gases [7.20] It can be demonstrated

from

mixH = mixG + T mixS = (−3.43 × 103J) + (298 K) × (11.5 J K−1) = 0

E7.14(b) Benzene and ethylbenzene form nearly ideal solutions, so

mixS = −nR(xAlnxA+ xBlnxB)

To find maximummixS, differentiate with respect to xAand find value ofxAat which the derivative

is zero

Note thatxB= 1 − xAso

mixS = −nR(xAlnxA+ (1 − xA) ln(1 − xA))

used lnx

dx =

1

x

d

dx (mixS) = −nR(ln xA+ 1 − ln(1 − xA) − 1) = −nR ln xA

1− xA

= 0 when xA=1

2

Thus the maximum entropy of mixing is attained by mixing equal molar amounts of two components

nB

nE = 1 = mB/MB

mE/ME

mE

mB = ME

MB =106.169

78.115 = 1.3591

mB

mE = 0.7358

E7.15(b) Assume Henry’s law [7.26] applies; therefore, with K(N2) = 6.51 × 107Torr andK(O2) = 3.30 ×

107Torr, as in Exercise 7.14, the amount of dissolved gas in 1 kg of water is

n(N2) =



103g

18.02 g mol−1



×

2)

6.51 × 107Torr



= (8.52 × 10−7mol) × (p/Torr)

Forp(N2) = xp and p = 760 Torr

n(N2) = (8.52 × 10−7mol) × (x) × (760) = x(6.48 × 10−4mol)

and, withx = 0.78

n(N2) = (0.78) × (6.48 × 10−4mol) = 5.1 × 10−4mol= 0.51 mmol

The molality of the solution is therefore approximately 0.51 mmol kg−1 in N2 Similarly, for oxygen,

n(O2) =



103g

18.02 g mol−1



×

p(O2)

3.30 × 107Torr



= (1.68 × 10−6mol) × (p/Torr)

Forp(O2) = xp and p = 760 Torr

n(O2) = (1.68 × 10−6mol) × (x) × (760) = x(1.28 mmol)

and whenx = 0.21, n(O2) ≈ 0.27 mmol Hence the solution will be 0.27 mmol kg−1 in O2

E7.16(b) Use n(CO2) = (4.4 × 10−5mol) × (p/Torr), p = 2.0(760 Torr) = 1520 Torr

n(CO2) = (4.4 × 10−5mol) × (1520) = 0.067 mol

The molality will be about 0.067 mol kg−1and, since molalities and molar concentration for dilute aqueous solutions are approximately equal, the molar concentration is about 0.067 mol L−1

E7.17(b) M(glucose) = 180.16 g mol−1

T = KfbB Kf = 1.86 K kg mol−1

T = (1.86 K kg mol−1) ×



10 g/180.16 g mol−1

0.200 kg



= 0.52 K

Freezing point will be 0◦C− 0.52◦C= −0.52◦C

E7.18(b) The procedure here is identical to Exercise 7.18(a).

lnxB= fusH

1

T∗ −T1

 [7.39; B, the solute, is lead]

=



5.2 × 103J mol−1

8.314 J K−1mol−1



×

1

600 K − 1

553 K



= −0.088¯6, implying that xB= 0.92

n(Pb) + n(Bi) , implying that n(Pb) =

xBn(Bi)

1− xB

For 1 kg of bismuth,n(Bi) = 1000 g

208.98 g mol−1 = 4.785 mol

Hence, the amount of lead that dissolves in 1 kg of bismuth is

n(Pb) = (0.92) × (4.785 mol)

1− 0.92 = 55 mol, or 11 kg

Comment It is highly unlikely that a solution of 11 kg of lead and 1 kg of bismuth could in any

sense be considered ideal The assumptions upon which eqn 7.39 is based are not likely to apply The answer above must then be considered an order of magnitude result only

E7.19(b) Proceed as in Exercise 7.19(a) The data are plotted in Fig 7.1, and the slope of the line is 1.78 cm/

(mg cm−3) = 1.78 cm/(g L−1) = 1.78 × 10−2m4kg−1.

6 8 10

12

Figure 7.1

(1.000 × 103kg m−3) × (9.81 m s−2) × (1.78 × 10−2m4kg−1) = 14.0 kg mol−1

E7.20(b) Let A = water and B = solute.

aA= pA

p∗ A

[42]=0.02239 atm

0.02308 atm = 0.9701

γA= aA

xA

and xA= nA

nA+ nB

nA= 0.920 kg

0.01802 kg mol−1 = 51.0¯5 mol nB= 0.122 kg

0.241 kg mol−1 = 0.506 mol

xA= 51.0¯5

51.05 + 0.506 = 0.990 γA= 0.9701

0.990 = 0.980

E7.21(b) B= Benzene µB(l) = µ∗B(l) + RT ln xB[7.50, ideal solution]

RT ln xB= (8.314 J K−1mol−1) × (353.3 K) × (ln 0.30) = −353¯6 J mol−1

Thus, its chemical potential is lowered by this amount

pB= aBp∗B[42]= γBxBpB∗ = (0.93) × (0.30) × (760 Torr) = 212 Torr

Question What is the lowering of the chemical potential in the nonideal solution withγ = 0.93?

E7.22(b) yA= pA

pA+ pB = pA

760 Torr = 0.314

pA= (760 Torr) × (0.314) = 238.64 Torr

pB= 760 Torr − 238.64 Torr = 521.36 Torr

aA= pA

p∗ A

(73.0 × 103Pa) × 1 atm

101 325 Pa

×760 Torr atm

= 0.436

aB= p pB∗

B

(92.1 × 103Pa) × 1 atm

101 325 Pa

×760 Torr atm

= 0.755

γA= aA

xA =0.436

0.220= 1.98

γB= aB

xB = 0.755

0.780 = 0.968

Solutions to problems

Solutions to numerical problems

P7.3 Vsalt=

∂V

∂b



H 2 O

mol−1[Problem 7.2]

= 69.38(b − 0.070) cm3mol−1 withb ≡ b/(mol kg−1)

Therefore, atb = 0.050 mol kg−1, Vsalt= −1.4 cm3mol−1

The total volume at this molality is

V = (1001.21) + (34.69) × (0.02)2cm3= 1001.22 cm3

Hence, as in Problem 7.2,

V (H2O) = (1001.22 cm3) − (0.050 mol) × (−1.4 cm3mol−1)

Question What meaning can be ascribed to a negative partial molar volume?

P7.5 Let E denote ethanol and W denote water; then

V = nEVE+ nWVW[7.3]

For a 50 per cent mixture by mass,mE= mW, implying that

nEME = nWMW, or nW= nEME

MW

Hence,V = nEVE+nEMEVW

MW

which solves tonE= V

VE+MEVW

MW

VEMW+ MEVW

Furthermore,xE= nE

nE+ nW = 1

1+ ME

MW

SinceME= 46.07 g mol−1andMW = 18.02 g mol−1, ME

MW = 2.557 Therefore

xE= 0.2811, xW = 1 − xE= 0.7189

At this composition

VE = 56.0 cm3mol−1 VW= 17.5 cm3mol−1[Fig.7.1 of the text]

(56.0 cm3mol−1) + (2.557) × (17.5 cm3mol−1) = 0.993 mol

nW = (2.557) × (0.993 mol) = 2.54 mol

The fact that these amounts correspond to a mixture containing 50 per cent by mass of both components

is easily checked as follows

mE= nEME= (0.993 mol) × (46.07 g mol−1) = 45.7 g ethanol

mW= nWMW= (2.54 mol) × (18.02 g mol−1) = 45.7 g water

At 20◦C the densities of ethanol and water are,ρE = 0.789 g cm−3, ρW = 0.997 g cm−3 Hence,

VE =mE

ρE = 45.7 g

0.789 g cm−3 = 57.9 cm3

of ethanol

VW =mW

ρW = 45.7 g

0.997 g cm−3 = 45.8 cm3 of water

The change in volume upon adding a small amount of ethanol can be approximated by

V = dV ≈ VEdnE≈ VEnE

where we have assumed that bothVEandVWare constant over this small range ofnE Hence

V ≈ (56.0 cm3mol−1) ×



(1.00 cm3) × (0.789 g cm−3) (46.07 g mol−1)



= +0.96 cm3

Kf = 0.0703 K

1.86 K/(mol kg−1) = 0.0378 mol kg−1

Since the solution molality is nominally 0.0096 mol kg−1in Th(NO

3)4, each formula unit supplies

0.0378

0.0096 ≈ 4 ions (More careful data, as described in the original reference gives ν ≈ 5 to 6.)

P7.9 The data are plotted in Figure 7.2 The regions where the vapor pressure curves show approximate

straight lines are denoted R for Raoult and H for Henry A and B denote acetic acid and benzene respectively

R

B

300

200

100

0

xA

0.8 1.0 Henry

Henry

Raoult

Raoult

Extrapolate

for KB

Figure 7.2

As in Problem 7.8, we need to formγA = pA

xAp∗A andγB =

pB

xBpB∗ for the Raoult’s law activity

coefficients andγB = pB

xBK for the activity coefficient of benzene on a Henry’s law basis, withK

determined by extrapolation We usep A∗ = 55 Torr, p B∗ = 264 Torr and K B∗ = 600 Torr to draw up

the following table:

A]

B]

A]

B]

GEis defined as [Section 7.4]:

GE = mixG(actual) − mixG(ideal) = nRT (xAlnaA+ xBlnaB) − nRT (xAlnxA+ xBlnxB)

and witha = γ x

G E = nRT (xAlnγA+ xAlnγB).

Forn = 1, we can draw up the following table from the information above and RT = 2.69 kJ mol−1:

P7.11 (a) The volume of an ideal mixture is

Videal= n1Vm,1 + n2Vm,2

so the volume of a real mixture is

V = Videal+ VE

We have an expression for excess molar volume in terms of mole fractions To compute partial molar volumes, we need an expression for the actual excess volume as a function of moles

VE= (n1+ n2)VE

m = n n1n2

1+ n2

a0+a1n (n1− n2)

1+ n2



soV = n1Vm,1 + n2Vm,2+ n1n2

n1+ n2

a0+a1(n1− n2)

n1+ n2

 The partial molar volume of propionic acid is

V1=

∂V

∂n1



p,T ,n2

= Vm,1+ a0n22

(n1+ n2)2 +a1(3n1− n2)n22

(n1+ n2)3

V1= Vm,1 + a0x2

2+ a1(3x1− x2)x2

2

That of oxane is

V2= Vm,2 + a0x2

1+ a1(x1− 3x2)x2

1

(b) We need the molar volumes of the pure liquids

Vm,1= M1

ρ1 = 74.08 g mol−1

0.97174 g cm−3 = 76.23 cm3mol−1

andVm,2= 86.13 g mol−1

0.86398 g cm−3 = 99.69 cm3mol−1

In an equimolar mixture, the partial molar volume of propionic acid is

V1= 76.23 + (−2.4697) × (0.500)2+ (0.0608) × [3(0.5) − 0.5] × (0.5)2cm3mol−1

= 75.63 cm3mol−1

and that of oxane is

V2= 99.69 + (−2.4697) × (0.500)2+ (0.0608) × [0.5 − 3(0.5)] × (0.5)2cm3mol−1

= 99.06 cm3

mol−1

P7.13 Henry’s law constant is the slope of a plot ofpB versusxBin the limit of zeroxB (Fig 7.3) The

partial pressures of CO2are almost but not quite equal to the total pressures reported above

pCO 2 = pyCO 2 = p(1 − ycyc)

Linear regression of the low-pressure points givesKH= 371 bar

0 20

40

60

80

Figure 7.3

The activity of a solute is

aB= pB

KH = xBγB

so the activity coefficient is

γB= pB

xBKH = yBp

xBKH

where the last equality applies Dalton’s law of partial pressures to the vapour phase A spreadsheet applied this equation to the above data to yield

P7.16 GE = RT x(1 − x){0.4857 − 0.1077(2x − 1) + 0.0191(2x − 1)2}

withx = 0.25 gives GE= 0.1021RT Therefore, since

mixG(actual) = mixG(ideal) + nGE

mixG = nRT (xAlnxA+ xBlnxB) + nGE= nRT (0.25 ln 0.25 + 0.75 ln 0.75) + nGE

= −0.562nRT + 0.1021nRT = −0.460nRT

Sincen = 4 mol and RT = (8.314 J K−1mol−1) × (303.15 K) = 2.52 kJ mol−1,

mixG = (−0.460) × (4 mol) × (2.52 kJ mol−1) = −4.6 kJ

Solutions to theoretical problems

P7.18 xAdµA+ xBdµB= 0 [7.11, Gibbs–Duhem equation]

Therefore, after dividing through by dxA

xA

∂µ

A

∂xA



p,T + xB

∂µ

B

∂xA



p,T = 0

or, since dxB= −dxA, asxA+ xB= 1

xA

∂µ

A

∂xA



p,T − xB

∂µ

B

∂xB



p,T = 0 or,

∂µ

A

∂ ln xA



p,T =

∂µ

B

∂ ln xB



p,T

d lnx = dx

x

Then, sinceµ = µ−−+ RT ln f

p−  −,

∂ ln fA

∂ ln xA



p,T =

∂ ln fB

∂ ln xB



p,T

On replacingf by p,

∂ ln pA

∂ ln xA



p,T =

∂ ln pB

∂ ln xB



p,T

If A satisfies Raoult’s law, we can writepA= xApA∗, which implies that

∂ ln pA

∂ ln xA



p,T = ∂ ln xA

∂ ln xA +∂ ln pA∗

∂ ln xA = 1 + 0 Therefore,

∂ ln p

B

∂ ln xB



p,T = 1 which is satisfied ifpB= xBp∗B(by integration, or inspection) Hence, if A satisfies Raoult’s law, so does B

P7.20 lnxA= −fusG

RT (Section 7.5 analogous to equation for lnxBused in derivation of eqn 7.39)

d lnxA

dT = −

1

d

dT

fusG T



= fusH

RT2 [Gibbs–Helmholtz equation]

xA

1

d lnxA= T

T∗

fusH dT

RT2 ≈fusH

R

T

T∗

dT

T2

lnxA= −fusH

1

1

T∗



The approximations lnxA≈ −xBandT ≈ T∗then lead to eqns 33 and 36, as in the text

P7.22 Retrace the argument leading to eqn 7.40 of the text Exactly the same process applies withaAin

place ofxA At equilibrium

µ∗A(p) = µA(xA, p + )

which implies that, withµ = µ∗+ RT ln a for a real solution,

µ∗A(p) = µ∗A(p + ) + RT ln aA= µ∗A(p) + p+

p Vmdp + RT ln aA

and hence that p+

p Vmdp = −RT ln aA

For an incompressible solution, the integral evaluates toVm, soVm= −RT ln aA

In terms of the osmotic coefficientφ (Problem 7.21)

Vm = rφRT r = xB

xA = nB

nA φ = − xA

xB

lnaA= −1

rlnaA

For a dilute solution,nAVm≈ V

Hence,V = nBφRT

and therefore, with [B]= nB

Solutions to applications

P7.24 By the van’t Hoff equation [7.40]

Π = [B]RT = cRT

M

Division by the standard acceleration of free fall,g, gives

Π

8 =c(R/g)T

M

(a) This expression may be written in the form

Π=cRT

M

which has the same form as the van’t Hoff equation, but the unit of osmotic pressure (Π) is now force/area

length/time2 = (mass length)/(area time2)

length/time2 = mass

area

Solutions to applications

P7.24 By the van’t Hoff equation [7.40]

Π = [B]RT = cRT

M

Division by the standard... −fusG

RT (Section 7.5 analogous to equation for lnxBused in derivation of eqn 7.39)

d lnxA... −xBandT ≈ T∗then lead to eqns 33 and 36, as in the text

P7.22 Retrace the argument leading to eqn 7.40 of the text Exactly the same process applies