m m 3 mol Pa = Pa or bar At this pressure the system must also be at approximately 1000 K.

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1 5. PHASES AND SOLUTIONS n Thermodynamics of Vapor Pressure 5.. At equilibrium, G(graphite) G(diamond); i.e., G 2 0. We are given G 2900 J mol. ( G/ P) T V V 2.0 g mol m 3 mol Holding T constant m 3 2 d G P 2 P VdP G 2 G V(P 2 P ) P 2 G 2 G V + P 5.2. V l J mol m 3 mol + 05 Pa Pa or bar At this pressure the system must also be at approximately 000 K. Molar mass density kg mol kg dm dm 3 mol V v kg mol kg dm dm3 mol From Eq. 5.8, dp dt P T S vap V v V l J K ( ) dm 3 mol 3.62 J dm 3 K P 3.62 J dm 3 K ( K) J m 3

2 96 n CHAPTER 5 Since J kg m 2 s 2 or N m and Pa kg m s 2 or N m 2 P Pa 5.3. ln P 2 P vap H m T 2 T R vap H m T 2 T T 2 T R ln vap H m /J mol vap H m /J mol (8.345) kj mol vap H m 42.7 kj mol The CRC Handbook gives kj mol Using the Clausius-Clapeyron equation, we get P ln Therefore, P atm e atm The molar volume of iodine, I 2, is V m Then from Eq. 5.23, kg mol 4.93 kg dm dm 3 mol m 3 mol ln Pg P g 2 V m (P P 2 ) RT ln Pg P g ( ) P g /Pg At 0.3 kpa the pressure is 33 Pa. Therefore, at kpa the pressure is Pa 984 Pa.

3 PHASES AND SOLUTIONS n The cubic expansion coefficient is a second-order transition since it can be expressed as α V T G P T P α T 5.7. From Eq. 5.6, vap H m T P 2 T R ln 2 P vap H m /J mol ( ) ln (200) (8.345) (2.7987) vap H m J mol 46.8 kj mol 5.8. ln P 2 P vap H m T 2 T R vap H m (T 2 T ) R T 2 T P 2 4 Torr/760 Torr atm Pa atm Pa At T K P Pa At T P Pa ln vap H m /J mol (48.5) (559.5) vap H m /J mol (8.345) ( ) kj mol The CRC Handbook value is 7.02 kj mol. The error is large, but considering the relative molecular mass of the compound, its high boiling point, and the wide range of T and P involved in the calculation, it is not surprising that the error is so large.

4 98 n CHAPTER Using Trouton s rule, vap H m 88 J K mol K kj mol % error % 5.0. Applying Trouton s rule, vap H T vap S (88 J K mol )( K) ln P 2 P 33.8 kj mol vap H T T 2 R ln P 2 atm 33.8 kj mol P J K mol atm Propanone is not particularly associated. Therefore, using T T b P, we have T (329.35) (2.825) K Thus T b at 98.5 kpa is K 5.2. At the triple point, the two vapor pressures must be equal since the liquid, solid, and vapor are all in equilibrium with each other. Therefore, we set 266 T trp T trp Solving for T trp, we obtain the triple point temperature: T trp K. Substituting the triple point temperature into one of the two equations given, we calculate the triple point pressure: ln P l P l Pa Since water is associated, the numerical value should be used in Eq T b K and P kpa, and thus T K The calculated boiling point at 0.3 kpa is thus K, an error of only 0.02 K Molar volume of water 8.05 g mol / g cm cm 3 mol

5 PHASES AND SOLUTIONS n 99 ln P/Torr cm 3 mol ( Torr Torr) cm atm K (760 Torr atm ) ( ) 8.08(732.46) ln P/Torr ln P Torr This is a rather small correction but may be necessary for very accurate work From Eq. 3.5, dg VdP SdT. Recognize that a change of state occurs at constant temperature. Therefore, differentiating Eq. 3.5 with respect to P at constant temperature, we obtain (see Eq. 3.9) G P T V. Now, for a change of state where G f G i G, there will be a corresponding change in volume, G P T V We start with the Clapeyron equation, Eq. 5.9, and substitute RT + M P rearrangement, Since dp P vap H m dt T(RT + M) T(RT + M) MT R M(RT + M) dp P vap H m dt M T R M Integration with vap H m constant gives vap H m dt (RT + M) for V; this gives, after ln P 2 P vap H m M ln T 2 ln P 2 P vap H m M ln T 2 T vap H m M T RT + M RT 2 + M 5.7. Molar mass of C 2 H 5 OH g mol ln RT 2 + M RT + M Molar volume of C 2 H 5 OH g mol g dm cm mol m 3 mol

6 00 n CHAPTER 5 Using Eq. (5.23), we get P/Pa ln Pa 00 Torr 760 Torr m 3 mol 00 Torr (0 325 Pa) J mol K K Pa 760 Torr ln P/Pa ln Pa Pa Pa 760 Torr 26 Torr 5.8. The volume change is V cm3 g m 3 kg. From the expression derived in Problem 5.5, we have d G VdP. Integrating both sides, we obtain 2 P2 P d G V dp. Let state be the standard state and state 2 be at equilibrium. Then, the expression is G G V(P P ). Now, at equilibrium, the Gibbs energy change G is zero. Also, using 0 5 Pa for the standard pressure P, we get 240 kj kg m 3 kg (P 0 5 ) Pa. Solving for P gives P Pa From Raoult s law (Eq. 5.26) we find, for toluene, P tol x tol P tol * bar 0. bar. and for benzene, P ben x ben P ben * bar bar. Therefore, the total pressure is P tot 0.bar bar 0.36 bar The molar heat of vaporization is vap H m 80 J g g mol J mol P2 vaph The Clausius-Clapeyron equation is ln. Substituting, we get P R T2 T

7 PHASES AND SOLUTIONS n ln, T which yields, upon simplification, ln T or ( /T ), from which, we get T 375 K or 02 C. n Raoult s Law, Equivalence of Units, and Partial Molar Quantities 5.2. P T 0.60(3.572) (9.657) kpa xvap toluene The molality m 2 is the amount of solute divided by the mass of solvent. If W is the mass of solvent, the solution contains m 2 W mol of solute and W /M mol of solvent. The mole fraction is thus m 2 W m 2 M x 2 W /M + m 2 W + m 2 M If we divide each term by its SI unit, (m 2 /mol kg ) (M /kg mol ) x 2 + (m 2 /mol kg ) (M /kg mol ) The customary unit for molar mass M is g mol, and we then obtain (m 2 /mol kg )(M /000 g mol ) x 2 + (m 2 /mol kg )(M /000 g mol ) (m 2 /mol kg )(M /g mol ) (m 2 /mol kg )(M /g mol ) If the solution is sufficiently dilute, the expression approximates to x 2 (m 2 /mol kg ) (M /g mol )/ Rearranging Eq. 5.22, and substituting gives

8 02 n CHAPTER 5 W 2 M 2 fus TW K f kg mol (20.0 K) (.00 kg).86 K kg mol kg. Converting the mass to a volume using density, and assuming that.00 kg of water has a volume of.00 dm 3, we get V kg/(.088 kg dm 3 ) dm 3. So, the volume ratio is dm 3 antifreeze to.0 dm 3 water (or approximately 3:5 volume ratio). The elevation of boiling point for this solution will be vap T K b m (0.667/ ) 5.48 K. This means that the solution will boil at K or 05 C If V is the volume of solution, concentration of solute c 2 ; amount of solute Vc 2 ; mass of solute Vc 2 M 2 where M 2 is the molar mass of solute. Mass of solution Vρ; mass of solvent Vρ Vc 2 M 2 ; amount of solvent (Vρ Vc 2 M 2 )/M. The mole fraction is therefore x 2 Vc 2 (Vρ Vc 2 M 2 )/M + Vc 2 c 2 M ρ + c 2 (M M 2 ) Dividing each term by its SI unit, (c 2 /mol m 3 ) (M /kg mol ) x 2 ρ/kg m 3 + (c 2 /mol m 3 ) [(M M 2 )/kg mol ] The more customary units are, for concentration, mol dm 3 ; for molar mass, g mol ; for density, kg dm 3 g cm 3. Then (000 c 2 /mol dm 3 ) (M /000 g mol ) x 2 (M 000 ρ/g cm 3 + (000 c 2 /mol dm 3 M 2 ) ) 000 g mol (c 2 /mol dm 3 ) (M /g mol ) 000 ρ/g cm 3 + (c 2 /mol dm 3 ) (M M 2 ) g mol If the solution is sufficiently dilute, the density of the solution is approximately that of the pure solvent, ρ, and the second term in the denominator can be neglected. x 2 c 2 M /ρ (c 2 /mol dm 3 ) (M /g mol ) 000 ρ /g cm It is difficult to work with the two laws in their normal form. Cast Henry s law into the form µ 2 µ RT ln x 2 using the arguments of Section 5.7 and Eq Then

9 PHASES AND SOLUTIONS n 03 dµ 2 /dx 2 RT/x 2 dµ 2 RT x 2 dx 2 Using the Gibbs-Duhem equation, x dµ + x 2 dµ 2 0 dµ x 2 dµ 2 /x RT/x dx 2 Since x + x 2, dx 2 dx dµ RT/x dx Integration gives µ RT ln x + constant Since x as µ goes to µ 0 µ RT ln x + µ Using Eq. 5.2, we get fus T.86 K kg mol.0 mol kg.86 K, which is not what is observed. NaCl completely ionizes in solution. Since colligative properties are, to a large extent, determined by the number of particles in solution rather than the actual identity of the species, we should realize that.0 mol of solid NaCl results in a solution containing 2.0 mol of particles. Therefore, we calculate that the temperature change is fus T.86 K kg mol 2.0 mol kg 3.72 K, which accounts for the observation From Problem 5.24, amount of solute Vc 2 ; mass of solvent Vρ Vc 2 M 2 Thus, the molality is m 2 c 2 ρ c 2 M 2 and the concentration in terms of m 2 is m 2 ρ c 2 + m 2 M 2 Dividing throughout by SI units, (m c 2 /mol m 3 2 /mol kg ) (ρ/kg m 3 ) + (m 2 /mol kg ) (M 2 /kg mol )

10 04 n CHAPTER 5 In terms of more usual units, 000 c 2 /mol dm 3 (m 2 /mol kg ) (000 ρ/g cm 3 ) g M 2 /000 mol + (m 2 /mol kg ) 000(m c 2 /mol dm 3 2 /mol kg ) (ρ/g cm 3 ) (m 2 /mol kg ) (M 2 /g mol ) In dilute solution, c 2 /mol dm 3 (m 2 /mol kg ) (ρ /g cm 3 ) where ρ is the density of the solvent. For aqueous solutions ρ g cm 3 and therefore the numerical values of the concentration and the molality, in the above units, are very similar Assuming that this is an ideal solution, we use Raoult s law with P Hg ( x M )P * Hg, where P * Hg Torr at the boiling point, and P Hg 754. Torr. Therefore, we calculate x M P Hg P * Hg Now, since /MW (.52/MW /200.59), we can solve for the molecular mass of the element, which gives MW 8.54 g mol. Comparison with the masses of elements shows that the metal is tin From Eq. 5.3, V NaCl /cm 3 mol V n V NaCl n H2 m O From Eq. 5.37, m m 2 n NaCl m dv NaCl dv H2 O n dv NaCl H2 O Integration from m 0 to m gives dv H 2 O m (55.508) m2 dm V H2 O V * H 2 O (55.508) m (55.508) m3 With V * H 2 O cm3, V H2 O /cm3 mol m m 3

11 PHASES AND SOLUTIONS n We first develop an expression for ( ρ/ n 2 )n. Since x 2 n 2 /(n + n 2 ), x 2 n 2 n and n (n + n 2 ) 2 ρ dρ n 2 n dx 2 x 2 n 2 n Substitution and division by (n + n 2 ) gives V 2 M 2 ρ (M x + M 2 x 2 ) x ρ 2 dρ dx 2 n (n + n 2 ) 2 dρ dx Substitution of x into the expression in Problem 5.23 gives ρ kg dm 3. Differentiating the ρ equation with respect to x 2 gives dρ dx x x x x kg dm 3 with M (H 2 O) kg mol and M 2 (methanol) kg mol, substitution with x gives V dm 3 mol Using Dalton s Law of partial pressures, P c y c P tot, where y c is the mole fraction of chloroform in the vapor phase. P c ( 0.88) Torr. Activity and activity coefficient: a c P c / P * c 40.3/ ; f c a c /x c 0.8/( 0.73) Using Eq. 5.80, 56.8 Torr Torr 56.8 Torr 2.5 g ( g mol ) M 2 (520.8 g) g mol /M 2 M g mol

12 06 n CHAPTER We prepare a graph of the data from which we see that the solution exhibits positive deviation from Raoult s law. P tot P iso P benz Pressure/Torr Mole fraction of Isopropanol From this graph, we construct the following table. a i P i /P * i, f i a i /x i. x I P I (Torr) a I P I 44.0 f I P I x I P B (Torr) a B P B 94.4 f B P B x B From Raoult s law, we have for propylene dibromide (p), P p x(l)p * p 0.600(28) Torr (33.33)Pa/Torr kpa For ethylene dibromide (e), P e x(l)p * e 0.40(72 Torr) (33.32)Pa/Torr 9.72 kpa The total pressure is P total kpa kpa 9.4 kpa. The mole fraction in the vapor phase is given by x P P total

13 PHASES AND SOLUTIONS n 07 For propylene dibromide, kpa x p (v) 9.4 kpa For ethylene dibromide, x e (v) 9.72 kpa 9.4 kpa First, calculate the moles of each present. For A, n g/89.5 g/mol mol. For B, n 000 g/85 g mol mol. Then mole fractions are calculated. For A, x / For B, x /( ) P total 520 Torr 430 Torr + P 2 P 2 90 Torr Then, from Henry s law, P 2 k x 2 k 90 Torr Torr a. Henry s law applies to the individual gas components, and we require the partial pressures of N 2 and O 2. Since the partial pressure is directly proportional to the mole fraction, for N 2 we have P N atm, and for O 2 we have P O atm. From Eq. 5.28, x(n 2 ) x(o 2 ) P N atm k N atm P O atm k O atm b. Since the mole fractions are so small, if we use mol of water as a reference, then there are (to a very good approximation) mol of N 2 and mol of O 2. The concentration calculation requires the volume of the solution, which is obtained from the density: c(n 2 ) c(o 2 ) mol (N 2 ) mol(h 2 O)[0.08 kg (H 2 O)/mol(H 2 O)] kg H 2 O dm M mol (O 2 ) mol(h 2 O)[0.08 kg (H 2 O)/mol(H 2 O)] kg H 2 O dm M

14 08 n CHAPTER Assume that the total vapor pressure of pure benzene is present in the total pressure of Torr. P P T P Torr P 2 k x 2 and k Torr Torr Torr x 2 x x Using Raoult s law P x P * Thus, the assumption is valid. Next, determine m 2. In 000 g of benzene, n x 2 n 2 n Solving for n 2 gives mol. Therefore, the molality is m. n Thermodynamics of Solutions From Raoult s law, x P /P * and x 2 x (P /P* ), from which we get x 2 P* P P * W 2 /M 2 (W /M ) + (W 2 /M 2 ) 8.04/M /M /M / /M M g mol. The correct value is 82.8 g mol.

15 PHASES AND SOLUTIONS n The vapor pressure has been reduced from to kpa, so that the mole fraction of the solvent is given by Raoult s law as x P kpa P * kpa 2 3 Let the amount of solute be n 2 mol; since n mol, the mole fraction of solvent is x + n Consequently, n 2 /2. Since 0.80 kg is half a mole, the molar mass is 0.60 kg mol 60 g mol From Eq. 5.78, P * P x 2 P* x 2 (3.3) x From Eq. 5.79, x 2 W 2 /M 2 W /M + W 2 /M 2 (M 2 W /M W 2 ) + M 2 W M W 2 x 2 8.0; M 2 M f T K f m 2.5 K m 2 /mol kg ; m mol kg mol % impurity / % 00.0% 2.7% 97.3 mol % pure a a γ γ Amount of NaCl.5 g/58.5 g mol 0.97 mol Amount of H 2 O 00.0 g/8.05 g mol 5.55 mol x n n + n , x a γ

16 0 n CHAPTER From Figure 5.3, the maximum value of the entropy is 5.76 J K mol. Then mix G id 300(5.76) 728 J mol. Therefore, the Gibbs energy of mixing ranges from 0 to.73 kj mol for a 50/50 mixture. Since this is a rather small driving force, in a nonideal solution where mix H 0, the value of mix H must be negative or only slightly positive for mixing to occur a H2 O P H2 O P * H 2 O kpa kpa Since a H2 O γ H 2 O x H 2 O γ H 2 O a H2 O x H2 O Find the value of µ A such that µ A + µ B ( G) is equal to the expression given in the problem. ln x A + n A n B a. µ A G n A n B,T,P µ* A + RT ln x A + RTnA RTn B Since and ln x B + Cn B (n A + n B ) n A n B (n A + n B ) 2 Cn A n B (n A + n B ) 2 ln x A n A ln x A n A n A n A + n B n B n A + n B n B n A n B n A n B n A n B n A (n A + n B ) (n A + n B ) µ A µ * A + RT ln x A + RT n A + n B n A + n + C B n B n A n B + n2 B n A n B (n A + n B ) 2 µ µ * A + RT ln x A + RT(0) + Cn 2 B (n A + n B ) 2 b. Write µ * A + RT ln x A + Cx2 B µ A µ * A + RT ln x A µ* A + RT ln γ A By comparison, RT ln γ A Cx 2 B or ln γ Cx2 B RT Thus γ e Cx2 B/RT when x B 0.

17 PHASES AND SOLUTIONS n This corresponds to a pure A. In a very dilute solution of A in B we also expect γ A. In that case µ * A lim x A 0 (µ A RT ln x A ) Substitution from above Therefore, µ * A lim x B (µ* A + Cx2 B ) µ* A + C µ µ * A + RT ln x A + C(x2 B ) µ* A + RT ln x A + RT ln γ A ln γ A C(x 2 B )/RT 0 when x B n Colligative Properties From Eq. 5.5, ln x ( ) x Rewriting Eq as P 2 (k ) c 2, the values for N 2 and O 2 may be substituted directly. We make the assumption that N 2 gives rise to 80% of the pressure and O 2 is responsible for the other 20%. Then, c(n 2 ) 0.80(0 325) Pa mol dm 3 Pa mol dm 3 c(o 2 ) 0.20(0 325) Pa mol dm 3 Pa mol dm 3 The total concentration is mol dm 3. This value of the concentration approaches the value of the molality. We may then use the molal freezing point depression expression. The result is fus T (.86) ( ) K Since the molecular weight is 60.06, c 0.00 mol dm 3 π crt n V RT 0.00 (8.345) (300.5) kpa kpa

18 2 n CHAPTER fus T K f m K From Eq. 5.5, substituting a for x as discussed in the paragraph after Eq. 5.00, we have ln a fus H m R T fus Hθ T * f R T * f ln a (3.255) 8.345(273.5) ; a x f a x The molality of the solute is calculated from vap T K b m 2 m fus T 0.9 K K b 6.26 K m 0.44 m 0.44 mol kg The mass of solute per kilogram of solvent is kg solute 0.50 kg bromobenzene Then for the solute, kg 0.44 mol and M kg 0.44 mol kg mol 39.4 g mol For the dissociation: A x B y xa z+ yb z Initial molalities: m 0 0 mol kg Molalities after dissociation: m αm xαm yαm mol kg Total molality m( α + xα + yα) mol kg Then i total molality initial molality α + xα + yα If ν x + y, i α + αν α ( ν) Then α i ν From our problem, fus T K i fus T/K f m K/[.86 (K m )0.00(m)].94

19 PHASES AND SOLUTIONS n 3 Since complete dissociation gives ν 2 for HCl, α i ν The electrolyte therefore appears to be 94% dissociated. This extent of dissociation is only apparent because of the nonideality of the solution From Eq. 5.34, π n 2 RT V Thus π m 2 /M 2 V RT where m 2 is the mass of solute dissolved in volume V and M 2 is the molar mass. Since the equations are good only for dilute solutions, we try to find a limiting value of M 2. From the preceding, M 2 (RT/π) (m 2 /V) and so lim m 2 V 0 and lim m 2 πv 0. A plot of (m 2 /πv)/g dm 3 atm against (m 2 /V)/g dm 3 gives a limiting value of m 2 /πv. The values of (m 2 /πv)/g dm 3 atm corresponding to the listed values are From the plot, the limiting value of m 2 /πv is about 2.9 g dm 3 atm. Thus M 2 RT m 2 V atm dm 3 K mol (K) (g dm 3 atm ) 30 g mol The molar mass of sucrose is 342 g mol, so that there is an error of about 9%. Ignoring the lowest concentration point and extrapolating leads to an error of about 4%. However, since the slope is expected to be fairly close to zero at infinite dilution, a better result cannot be obtained without more low-concentration data From Eq. 5.22, M 2.856(K kg mol ) 3.78 g 0.646(K) g kg mol 36.2 g mol m g/87.4 g mol mol m 250.0/ mol 3.88 x From Eq. 5.25,

20 4 n CHAPTER 5 ln x ln /T T K (/T /373.5) T K From Eq. 5.26, /T /373.5 vap T K b m K kg mol mol/ kg K The agreement is good. The difference between the two values indicates that the solution is sufficiently dilute for Eq to apply The Clapeyron equation (Eq. 5.9) may be used for this problem. Since the boiling point is given at a pressure for atm, it is appropriate to have the pressure of the gas given in atmospheres. Since mmhg Torr, log P/Torr log [P/atm atm (760 Torr) ] log P/atm + log atm (760 Torr) The derivative of this expression shows that it does not matter how pressure is expressed as long as we are considering only a change in pressure. Since d log P dt dp dt P log 0 e P dp dt P dp dt, d dt [ T T -2 ] We now can use the Clapeyron equation (Eq. 5.9). At the boiling point, P atm, T b K. dp dt ( atm) [427.3/(398.4) /((398.4) 3 ] atm K V m (liquid) kg mol 0.89 kg dm dm 3 mol V m (vapor) kg mol kg dm dm3 mol V V m (vapor) V m (liquid) dm 3 mol Rearranging and substituting into Eq. 5.9, we have vap H m T V m dp dt K( dm3 mol )( atm K ) 692 dm 3 atm mol 692 dm 3 atm mol 0 3 m 3 dm kj atm m kj mol

21 PHASES AND SOLUTIONS n Addition of the moles per liter of the salts listed in Table 5.5 gives a.0989 molar solution. Assuming that there is sufficient positive charge to cancel the negative charges present, no calculation need be made to determine van t Hoff s i factor. Although the temperature of the ocean is variable, we can assume an average value of 25 C near the surface. Then π crt.0989 mol dm 3 (8.345 J K mol )(298 K) J dm 3 Since J kg m 2 s 2 and Pa kg m s 2 π kg m 2 s 2 dm dm 3 /m kg m s MPa

Thermodynamic condition for equilibrium between two phases a and b is G a = G b, so that during an equilibrium phase change, G ab = G a G b = 0.

Thermodynamic condition for equilibrium between two phases a and b is G a = G b, so that during an equilibrium phase change, G ab = G a G b = 0. CHAPTER 5 LECTURE NOTES Phases and Solutions Phase diagrams for two one component systems, CO 2 and H 2 O, are shown below. The main items to note are the following: The lines represent equilibria between