7 Simple mixtures. Solutions to exercises. Discussion questions. Numerical exercises

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1 7 Simple mixtures 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 E7.3(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 µ A µ A + RT ln x A or µ A µ A + RT ln a A, 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 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. 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 relation µ µ + RT ln a, where µ 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 volume V n A V A + n B V B n(x A V A + x B V B Total mass m n A M A + n B M B E7.5(b n(x A M A + (1 x A M B where n n A + n B m n x A M A + (1 x A M B kg(10 3 g/kg n ( (241.1g/mol + ( (198.2g/mol mol V n(x A V A + x B V B ( mol [( (188.2cm 3 mol 1 + ( ( cm 3 mol 1 ] 843.5cm 3 Let A denote water and B ethanol. The total volume of the solution is V n A V A + n B V B We know V B ; we need to determine n A and n B in order to solve for V A. Assume we have 100 cm 3 of solution; then the mass is m ρv ( g cm 3 (100 cm g of which (0.20 (96.87 g g is ethanol and (0.80 (96.87 g g is water. n A n B g g mol mol H 2O g mol ethanol g mol 1

2 98 INSTRUCTOR S MANUAL V n B V B V A 100 cm3 ( mol (52.2cm 3 mol cm 3 n A mol 18 cm 3 E7.6(b Check that p B /x B a constant (K B x B (p B /x B /kpa E7.7(b K B p/x, average value is kpa 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, n(a Therefore, x B corresponding to 1000 g 74.1 g mol mol n(b 0.25 mol 0.25 mol 0.25 mol mol using K B kpa [exercise 7.6(b] p kpa kpa E7.8(b K f RT 2 M fus H J K 1 mol 1 (354 K kg mol J mol K kg mol 1 K b RT 2 M vap H J K 1 mol 1 (490.9K kg mol J mol K kg mol 1 E7.9(b We assume that the solvent, 2-propanol, is ideal and obeys Raoult s law. x A (solvent p/p M A (C 3 H 8 O g mol g n A mol g mol 1 n A x A n A + n B n A n A + n B x A

3 SIMPLE MIXTURES 99 ( 1 n B n A 1 x A M B mol ( mol 8.69 g mol 27 3 g mol g mol 1 E7.10(b K f 6.94 for naphthalene mass of B M B n B n B mass of naphthalene b B b B T (mass of B K f so M B K f (mass of naphthalene T M B (5.00 g (6.94 K kg mol g mol 1 (0.250 kg (0.780 K n B E7.11(b T K f b B and b B mass of water n B Vρ ρ 10 3 kg m 3 (density of solution density of water n B V T K f K f 1.86 K mol 1 kg RT RTρ (1.86 K kg mol 1 ( Pa T (8.314 J K 1 mol 1 (288 K (10 3 kg m K T f C E7.12(b mix G nrt (x A ln x A + x B ln x B n Ar n Ne, x Ar x Ne 0.5, n n Ar + n Ne pv RT mix G pv ( 1 2 ln ln 2 1 pv ln 2 ( ( m 3 Pa (0.250 L 10 3 ln 2 L E7.13(b 17.3Pam J mix S mixg T mix G nrt x J ln x J [7.18] mix S nr J J n 1.00 mol mol 2.00 mol x(hex x(hep Therefore, 17.3J 273 K JK 1 x J ln x J [7.19] mixg T mix G (2.00 mol (8.314 J K 1 mol 1 (298 K (0.500 ln ln kj

4 100 INSTRUCTOR S MANUAL kj mix S 298 K +11.5JK 1 mix H for an ideal solution is zero as it is for a solution of perfect gases [7.20]. It can be demonstrated from mix H mix G + T mix S ( J + (298 K (11.5JK 1 0 E7.14(b Benzene and ethylbenzene form nearly ideal solutions, so mix S nr(x A ln x A + x B ln x B To find maximum mix S, differentiate with respect to x A and find value of x A at which the derivative is zero. Note that x B 1 x A so mix S nr(x A ln x A + (1 x A ln(1 x A use dlnx 1 dx x d dx ( x A mixs nr(ln x A + 1 ln(1 x A 1 nr ln 1 x A 0 when x A 1 2 Thus the maximum entropy of mixing is attained by mixing equal molar amounts of two components. n B n E 1 m B/M B m E /M E m B m E m E m B M E M B E7.15(b Assume Henry s law [7.26] applies; therefore, with K(N Torr and K(O Torr, as in Exercise 7.14, the amount of dissolved gas in 1 kg of water is ( 10 3 ( g p(n 2 n(n g mol ( mol (p/torr Torr For p(n 2 xp and p 760 Torr n(n 2 ( mol (x (760 x( mol and, with x 0.78 n(n 2 (0.78 ( mol mol 0.51 mmol The molality of the solution is therefore approximately 0.51 mmol kg 1 in N 2. Similarly, for oxygen, ( 10 3 ( g p(o 2 n(o g mol ( mol (p/torr Torr For p(o 2 xp and p 760 Torr n(o 2 ( mol (x (760 x(1.28 mmol and when x 0.21,n(O mmol. Hence the solution will be 0.27 mmol kg 1 in O 2.

5 SIMPLE MIXTURES 101 E7.16(b Use n(co 2 ( mol (p/torr, p 2.0(760 Torr 1520 Torr n(co 2 ( mol ( mol E7.17(b E7.18(b The molality will be about mol kg 1 and, since molalities and molar concentration for dilute aqueous solutions are approximately equal, the molar concentration is about mol L 1 M(glucose g mol 1 T K f b B K f 1.86 K kg mol 1 ( T (1.86 K kg mol 1 10 g/ g mol K kg Freezing point will be 0 C 0.52 C 0.52 C The procedure here is identical to Exercise 7.18(a. ln x B ( fush 1 R T 1 [7.39; B, the solute, is lead] T ( J mol 1 ( J K 1 mol K K x B , implying that x B 0.92 n(pb n(pb + n(bi, implying that n(pb x Bn(Bi 1 x B 1000 g For 1 kg of bismuth, n(bi mol g mol 1 Hence, the amount of lead that dissolves in 1 kg of bismuth is n(pb (0.92 (4.785 mol mol, or 11 kg E7.19(b 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. 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 cm/(gl m 4 kg Figure 7.1

6 102 INSTRUCTOR S MANUAL Therefore, M (8.314 J K 1 mol 1 ( K ( kg m 3 (9.81 m s 2 ( m 4 kg kg mol 1 E7.20(b Let A water and B solute. a A p A atm pa [42] atm γ A a A n A and x A x A n A + n B kg n A kg mol kg mol n B mol kg mol 1 x A γ A E7.21(b B Benzene µ B (l µ B (l + RT ln x B [7.50, ideal solution] RT ln x B (8.314 J K 1 mol 1 (353.3K (ln J mol 1 Thus, its chemical potential is lowered by this amount. p B a B pb [42] γ Bx B pb (0.93 (0.30 (760 Torr 212 Torr Question. What is the lowering of the chemical potential in the nonideal solution with γ 0.93? p A E7.22(b y A p A p A + p B 760 Torr p A (760 Torr ( Torr p B 760 Torr Torr Torr a A p A p A a B p B pb Torr ( ( Pa 1 atm Pa ( Pa γ A a A x A γ B a B x B Solutions to problems Torr ( 1 atm Pa ( Torr atm ( Torr atm Solutions to numerical problems ( V P7.3 V salt mol 1 [Problem 7.2] b H 2 O 69.38(b cm 3 mol 1 with b b/(mol kg 1 Therefore, at b mol kg 1,V salt 1.4cm 3 mol 1

7 SIMPLE MIXTURES 103 The total volume at this molality is V ( (34.69 ( cm cm 3 Hence, as in Problem 7.2, V(H 2 O ( cm3 (0.050 mol ( 1.4cm 3 mol mol Question. What meaning can be ascribed to a negative partial molar volume? P7.5 Let E denote ethanol and W denote water; then V n E V E + n W V W [7.3] For a 50 per cent mixture by mass, m E m W, implying that n E M E n W M W, or n W n EM E M W Hence, V n E V E + n EM E V W M W V M E V which solves to n E V E + M,n W EV W V M E M W + M E V W W n E 1 Furthermore, x E n E + n W 1 + M E M W cm 3 mol 1 Since M E g mol 1 and M W g mol 1, M E Therefore M W x E , x W 1 x E At this composition V E 56.0cm 3 mol 1 V W 17.5cm 3 mol 1 [Fig.7.1 of the text] 100 cm 3 Therefore, n E (56.0cm 3 mol 1 + (2.557 (17.5cm 3 mol mol n W (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 m E n E M E (0.993 mol (46.07 g mol g ethanol m W n W M W (2.54 mol (18.02 g mol g water At 20 C the densities of ethanol and water are, ρ E 0.789gcm 3,ρ W 0.997gcm 3. Hence, V E m E 45.7g 57.9cm3 ρ E 0.789gcm 3 of ethanol V W m W 45.7g 45.8cm3 ρ W 0.997gcm 3 of water

8 104 INSTRUCTOR S MANUAL The change in volume upon adding a small amount of ethanol can be approximated by V dv V E dn E V E n E where we have assumed that both V E and V W are constant over this small range of n E. Hence ( V (56.0cm 3 mol 1 (1.00 cm 3 (0.789gcm 3 (46.07 g mol cm 3 P7.7 m B T K f K 1.86 K/(mol kg mol kg 1 Since the solution molality is nominally mol kg 1 in Th(NO 3 4, each formula unit supplies ions. (More careful data, as described in the original reference gives ν 5to 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. 300 Extrapolate for K B 200 R B Henry p / Torr Raoult 100 Henry H A Raoult x A R H Figure 7.2 As in Problem 7.8, we need to form γ A p A x A pa and γ B p B x B pb for the Raoult s law activity coefficients and γ B p B for the activity coefficient of benzene on a Henry s law basis, with K x B K determined by extrapolation. We use pa 55 Torr, p B 264 Torr and K B 600 Torr to draw up

9 SIMPLE MIXTURES 105 the following table: x A p A /Torr p B /Torr a A (R [p A /p A ] a B (R [p B /p B ] γ A (R [p A /x A p A ] γ B (R [p B /x B p B ] a B (H [p B /K B ] γ B (H [p B /x B K B ] G E is defined as [Section 7.4]: G E mix G(actual mix G(ideal nrt (x A ln a A + x B ln a B nrt (x A ln x A + x B ln x B and with a γx G E nrt (x A ln γ A + x A ln γ B. For n 1, we can draw up the following table from the information above and RT 2.69 kj mol 1 : x A x A ln γ A x B ln γ B G E /(kj mol P7.11 (a The volume of an ideal mixture is V ideal n 1 V m,1 + n 2 V m,2 so the volume of a real mixture is V V ideal + V E 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 V E (n 1 + n 2 Vm E n ( 1n 2 a 0 + a 1(n 1 n 2 n 1 + n 2 n 1 + n 2 so V n 1 V m,1 + n 2 V m,2 + n ( 1n 2 a 0 + a 1(n 1 n 2 n 1 + n 2 n 1 + n 2 The partial molar volume of propionic acid is ( V a 0 n 2 2 V 1 V m,1 + n 1 p,t,n 2 (n 1 + n a 1(3n 1 n 2 n 2 2 (n 1 + n 2 3 V 1 V m,1 + a 0 x a 1(3x 1 x 2 x 2 2 That of oxane is V 2 V m,2 + a 0 x a 1(x 1 3x 2 x 2 1

10 106 INSTRUCTOR S MANUAL (b We need the molar volumes of the pure liquids V m,1 M 1 ρ g mol g cm cm3 mol g mol 1 and V m, g cm cm3 mol 1 In an equimolar mixture, the partial molar volume of propionic acid is V ( ( ( [3( ] (0.5 2 cm 3 mol cm 3 mol 1 and that of oxane is V ( ( ( [0.5 3(0.5] (0.5 2 cm 3 mol cm 3 mol 1 P7.13 Henry s law constant is the slope of a plot of p B versus x B in the limit of zero x B (Fig The partial pressures of CO 2 are almost but not quite equal to the total pressures reported above p CO2 py CO2 p(1 y cyc Linear regression of the low-pressure points gives K H 371 bar Figure 7.3 The activity of a solute is a B p B K H x B γ B so the activity coefficient is γ B p B y Bp x B K H x B K H

11 SIMPLE MIXTURES 107 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 p/bar y cyc x cyc γ CO P7.16 G E RT x(1 x{ (2x (2x 1 2 } with x 0.25 gives G E RT. Therefore, since mix G(actual mix G(ideal + ng E mix G nrt (x A ln x A + x B ln x B + ng E nrt (0.25 ln ln ng E 0.562nRT nRT 0.460nRT Since n 4 mol and RT (8.314 J K 1 mol 1 ( K 2.52 kj mol 1, mix G ( (4 mol (2.52 kj mol 1 4.6kJ Solutions to theoretical problems P7.18 x A dµ A + x B dµ B 0 [7.11, Gibbs Duhem equation] Therefore, after dividing through by dx A ( ( µa µb x A + x B 0 x A p,t x A p,t or, since dx B dx A,asx A + x B 1 ( ( µa µb x A x B 0 x A p,t x B p,t ( ( [ µa µb or, dlnx dx ] ln x A p,t ln x B p,t x Then, since µ µ + RT ln f ( ( ln fa ln fb p, ln x A p,t ln x B p,t ( ( ln pa ln pb On replacing f by p, ln x A p,t ln x B p,t If A satisfies Raoult s law, we can write p A x A pa, which implies that ( ln pa ln x A + ln p A ln x A p,t ln x A ln x A ( ln pb Therefore, 1 ln x B p,t which is satisfied if p B x B pb (by integration, or inspection. Hence, if A satisfies Raoult slaw,so does B.

12 108 INSTRUCTOR S MANUAL P7.20 ln x A fusg (Section 7.5 analogous to equation for ln x B used in derivation of eqn 7.39 RT dlnx A 1 dt R d ( fus G fush dt T RT 2 [Gibbs Helmholtz equation] xa T fus H dt dlnx A 1 T RT 2 T fush dt R T T 2 ln x A ( fush 1 R T 1 T The approximations ln x A x B and T 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 with a A in place of x A. At equilibrium µ A (p µ A(x A,p+ which implies that, with µ µ + RT ln a for a real solution, p+ µ A (p µ A (p + + RT ln a A µ A (p + V m dp + RT ln a A p p+ and hence that V m dp RT ln a A p For an incompressible solution, the integral evaluates to V m,so V m RT ln a A In terms of the osmotic coefficient φ (Problem 7.21 V m rφrt r x B x A n B n A For a dilute solution, n A V m V Hence, V n B φrt and therefore, with [B] n B φ[b]rt V Solutions to applications φ x A x B ln a A 1 r ln a A 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/gt M (a This expression may be written in the form Π cr T M which has the same form as the van t Hoff equation, but the unit of osmotic pressure (Π isnow force/area length/time 2 (mass length/(area time2 length/time 2 mass area

13 SIMPLE MIXTURES 109 This ratio can be specified in g cm 2. Likewise, the constant of proportionality (R would have the units of R/g energy K 1 mol 1 length/time 2 (mass length2 /time 2 K 1 mol 1 length/time 2 mass length K 1 mol 1 This result may be specified in gcmk 1 mol 1 R R g J K 1 mol m s kg m K 1 mol 1 ( 10 3 g kg R g cm K 1 mol 1 ( 10 2 cm m In the following we will drop the primes giving Π crt M and use the Π units of g cm 2 and the R units g cm K 1 mol 1. (b By extrapolating the low concentration plot of /c versus c (Fig. 7.4 (a to c 0wefind the intercept 230 g cm 2 /g cm 3. In this limit van t Hoff equation is valid so RT intercept or M n RT M n intercept M n ( g cm K 1 mol 1 ( K (230 g cm 2 /(gcm 3 M n g mol Figure 7.4(a

14 110 INSTRUCTOR S MANUAL (c The plot of Π/c versus c for the full concentration range (Fig. 7.4(b is very nonlinear. We may conclude that the solvent is good. This may be due to the nonpolar nature of both solvent and solute Figure 7.4(b (d Π/c (RT /M n (1 + B c + C c 2 Since RT/M n has been determined in part (b by extrapolation to c 0, it is best to determine the second and third virial coefficients with the linear regression fit (Π/c/(RT /M n 1 B + C c c R B 21.4cm 3 g 1, C 211 cm 6 g 2, standard deviation 2.4cm 3 g 1 standard deviation 15 cm 6 g 2 (e Using 1/4 for g and neglecting terms beyond the second power, we may write ( Π 1/2 ( RT 1/2 (1 + c 2 1 M B c n

15 SIMPLE MIXTURES 111 We can solve for B, then g(b 2 C. ( 1/2 Πc ( 1/ B c RT M n RT/M n has been determined above as 230 g cm 2 /g cm 3. We may analytically solve for B from one of the data points, say, /c 430 g cm 2 /gcm 3 at c 0.033gcm 3. ( 430 g cm 2 /gcm 3 1/2 230 g cm 2 /gcm B (0.033gcm 3 B 2 ( gcm cm 3 g 1 C g(b (22. 2cm 3 g cm 6 g 2 ( 1/2 /( Better values of B and C Π RT 1/2 can be obtained by plotting against c. This plot c M n is shown in Fig. 7.4(c. The slope is cm 3 g 1. B 2 slope 28. 0cm 3 g 1 C is then 19 6cm 6 g 2 The intercept of this plot should thereotically be 1.00, but it is in fact with a standard deviation of The overall consistency of the values of the parameters confirms that g is roughly 1/4 as assumed n Figure 7.4(c

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

m m 3 mol Pa = Pa or bar At this pressure the system must also be at approximately 1000 K. 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.95 0 6 m 3 mol Holding T constant