pchem6 - Chapter 8 Phase Diagrams and the Relative...

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Unformatted text preview: Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases fl)— : AS;apori.zatirm — AHJIaporization 62 = #58733]: + 21 965 dT vaporization A I/niaponzatwn 1 1 i T (ZilaporizationM ______—_ ‘ _ , K pgas pliquid f— 11378‘58733)? £15 _ 40.656x103 Jmol’l T . . d W” 373.15 le8.02xlO‘3 kg mol’l x 1 3 _ 1 3 0.58958 kg m' 958.66 kg m” = 3567 Pa K" 3 —l g1: z 40.656x10 J mol 2 3564 Pa K4 dT 1 Wm" 373.15 Kx18.02x10'3 kg mol'lx _~_—3_ 0.58958 kg m‘ -1 _ -1 thgivapor pressures of n-butane given in the following table to calculate the enthalpy of Relative Error = lOOXW = 0.061% 3567 Pa K’1 using a graphical method or a least squares fitting routine. P836) The densities of a given solid and liquid of molar mass 147.2 g mol'1 at its normal melting temperature of 372.54 K are 992.6 and 933.4 kg m‘3, respectively. If the pressur increased to 120. bar, the melting temperature increases to 375.88 K. Calculate AH?” and ASnfus’O” f this substance. £ gfi; AS $13M AT AV AT Asliusion zi‘f-M 1 __ 1 AT p liquid psolid 5 ASf‘ion :wxl472x10‘3kg mol’1 x ————-1————3——~———1~——§- 375.88 K~372.54 K 933 kg m‘ 992.6 kg m' = 33.5 JK'I mol‘1 AHf‘S’O" :2 TfmmnASfW = 372.54 Kx33.5 J K“ mor1 = 12.5x103 J moi" at 1 bar P8.3 7) The variation of the vapor pressure of the liquid and solid forms of anthracene near the triple point are given in the Appendix A data tables. Calculate the temperature and pressure at the triple point At the triple point, PsolidZPliquid. . ., . . ., . ., _ .. . . . . .. . .. $.QQ4 {llfifidirgfi {3.36545 Sfifiéwfi {3.8583 {ifi’flfigfi 3.98535 §.f’§(%<“} uares fit of In P versus 1/ T gives the result AH;;"”“”“”"" = 25.28 M mol‘l. 200 201 Chapter 8/ Phase Diagrams and the Relative Stability of Solids Liquids; and Gases ) 15 A f ( 203g) —" 228.(; kl n")! l and 30; (1420.1): ——237.1 M mol'l. Calculate the ng Equation (8.29), .3 —1 vapor pressure of water at this temperature. __ , . 27 2><26.43><108 Nm = 2.1l><106 Pa zit-Pr“: 2.5x10‘ m AG" = M}; (H20, g) —AG; (H20, 1) = ~228.6 kJ mol~1+237.l kJ mol‘1 -_ r + AP = 3.46X104 Pa + 211x106 Pa = 2.15x106 Pa = 8.5 k] l‘1 ’ fl _ 11 dro let, the vapor m0 ° “1 let AP ——> 0, the vapor pressure is 3.46x104Pa. For the sma p 111i : _é£ : _M\ : _3_429 or a very large dmp ’ P° RT 8.314 J mol‘1 K“ x298.15 K ‘ P =3.24><1o3 Pa 0‘ ressure is increased by the factor _ —] 153.82x103 kgmol X (2.15x106 —3.46x10‘*)Pa - ~ 0) 1594 kg m‘3 1 pressure shown schematically in 8.314 J mol'1 K' x318.15 K Figure 8 1 For this problem, 5,320,. = 48.0 J mol‘1 K4, 51320,, = 70.0 J mol“1 K“, and 52.0.5.1 = 188.8 J mol‘1 K“ a, By What amount does the chemical potential of water exceed that of ice at —3.50°C? b. By what amount does the chemical potential of water exceed that of steam at 103.50°C? 21) a 9 3 y - - P f th L P8 ’ . . . . t ‘ ' f the solid at the melting pom ' lculate the sublimation pressure 0 00°C 15 0.30844 Torr. Ca pressure at 25. A,“ : AG”) 2 * mAT at COHStam P (1 136°C) assumng 'ties do not change with temperature. All 2 m = —(SI:I,O,1 _ 52120,; )(AT) : “(70.0 JK‘lmOl'l — 48.0 JI<JHJOYI )(_3'500 C) I a. that the enthalpy Of sublimation and the heat cafe/1 be calCUIated from the equation AHosubliination (I) _ blimation at temperature can = 77.0 Jmol'l b, that the enthalpy of su :AHosublimation (TO) + ACP (fr—TO) b) A’“ : AG!” = a ($320,, —S;120,g)(AT) = —( 70.0 JK'lmol" — 188.8 JK-‘mol")x (350° C) = 416 .lmol‘1 a) If the enthalpy of sublimation is constant P AHsubllmanorl 1 1 ‘ In #T ff 45.0 C in equilibrium with its vapor. Use the tabulated value of the density and the surface t l ' 56.30xl03 J “1014 x 1 1 i f 1n P2 2111030844— tetrachloromethane at this temperature.) 8314Jm01“ K'1 m 298.15K I32 =56.1 Torr T A 2 3 ln ( )=A(l)~fi()~—=20.738—M=10.4527 Pa T 318.15—46.6667 ——+A(3) K P=3.46><104 Pa 202 203 Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases b) If the enthalpy of sublimation is given by AIf;blimarion (T) = AH:lblinialion (T0 ) + ACW1" (T — 7i) ) A P. P 7; RT2 7; RT2 1n 5:_AH’;;a170rlzali(m(Z)[i-i]+ACPY—l,[ifii]+£ln£ Pl R T2 7} R T2 71 71 3 -1 1n P2 =ln0.30844—56'30X10lJmOll x[ l _ 1 8.314JK' mol' 386.8 K 298.15 K +(36.9—54.4)JK" moi-1x29815 K [ 1 1 X _.——__* 8.314 J K" moi-I 386.8 K 298.15 K (36.9—54.4)J K" mor1 386.8 K + xln——— 8.314 J K" mol“ 298.15 K lnP2 =3.964 132 = 52.5 Torr phases are in equilibrium at 1 bar and 18°C. The densities for gray and White tin are 5750 and 7280 kg m , respectively, and the molar entropies for gray and White tin are 44.14 and 51.18 J K‘1 m0171, respectively. Calculate the temperature at which the two phases are in equilibrium at 200 bar. In going from latrn, 18°C to 200 atm, T AGgmy = VWAP — S gray AT AGwhile : I/mwhiIeAP _ SwhiteAT At equilibrium AGgray *AGWlme : 0 : (Kira); _ I/nlvhile)AP__(Sgray _ Swhue)AT 1 1 ( anmy - Vim” MP MS” [ pgmy _ pm. iAP AT : (Sgray *Swlzile) : transition ll8.7l><10‘3 kgmol'lx ;T—~;3 ><l99X105 Pa _ 5750 kg m" 7280 kg m‘ _ 12 3°C —7.04 JK“ morl _ ' T]. =5.7°c 204 . . . ' ' S Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and G356 given by pressure of krypton IS to the vapor reasonable approxrmation 8.44) A (P/Torr) = b — 0.05223(a/T) _ these .10810 a 2 10065 and b 2 7.1770 For liquid krypton, a = 9377.0 and b — 6.92387. Use For solid krypton, ii formulas to estimate 7 fusion, and sublimation of krypton. ' a orization, the triple point temperature and pressure and also the enthalpies of v p At the triple point the vapor pressures of the solid and liquid are equal. 9377.0 10065 = 6.92387 — 0.05223 x 1p 688 2!) 7.1770-0.05223 x ’l’ Tm=142K 10065 P'” = 7.1770 — 0.05223 x 142 = 3.4678 Torr Pip = 2.94><103 Torr 10$10 We next calculate the enthalpies of sublimation and vaporization. b) If the enthalpy of sublimation is given by AHsublimatitm (T) : AHSub/imalion (Ti) ) + ACP (T _ T6) In T2 AH\:ZP()rizalion (T1 ) + ACP (T __ Tl ) dT T7 va orization ; Pj‘I—d—g: - AHmp 2 dTZI I RT2 RT T P1 P T1 1 AC T P, AHIZapm-ization (T1) 3— 1 + AC P11 L_i]+ 1’ 111—1.?— 1” ?37 R T. R R T. R R 1 l 56.30x103 Jmor' X 1 _ 1 ] In P2 =1n0'30844_ 8.314 JK" mor1 386.8 K 298.15 K ' ' 1 (36.9—54.4)JK1 m61‘x298.15 KX 1 _ K) + 8 314 J K“ mol" 386.8 K 298.15 “ " 386.8 K + (36.9—54.4)JK n’llol Xln K 8.314 J K“ mol' 298.15 lnP2 =3.964 P2 = 52.5 Torr 205 From Equation 8.16 d In P _ AHizublimation dT RT2 dlnP _ dlnP dT 2 dlnP AHS"””’”"”"" 1 — W“? = *T “ _ ’" d *] d * dT R T T d1 0 gm P _ — AHZublimalion 0(1) 2.303 R T For this specific case AHSub/imalion R = 0.05223 x 2.303 x10065 AH;7“’"”“”‘“” = 10.1 x 103 J mol“ Following the same reasoning AHwb/imalian R = 0.05223 x 2.303 x 9377 Angb’i’W" = 9.38 x 103 J mol" AH’yy‘usmn : 1 Ergublimaflon __ a Ergaporization = 0.69 ><103 J mol’1 =10.07x103 J moi“ —9.38><103 J mol‘1 206 Chapter 9: Ideal and Real Solutions Problem numbers in italics indicate that the solution is included in the Student ’3 Solutions _Manual. Questions on Concepts (29,1) Why is the magnitude of the boiling point elevation less than that of the freezing point depression? The boiling point elevation is less than the freezing point depression because the chemical potential of the vapor is a much more steeply decreasing function of temperature than the solid, as seen in Figure 9.11a. This is due to the relation dlu = éSde at constant P and the fact that the molar entropy of a vapor is much larger than that of a solid. When the ‘uuquid curve is displaced down by the addition of the solute, (see Fig. 9.11a), the intersection of the ,uhquid curve with the lam” curve and the ,ugas curve determine the shift in the freezing and boiling temperatures. Because the magnitude of the slope of the ,ugas curve is greater than that of the ,usolid curve, Tb moves up less than the Tm moves down. Q9.2) Fractional distillation of a particular binary liquid mixture leaves behind a liquid consisting of both components in which the composition does not change as the liquid is boiled off. Is this behavior characteristic of a maximum or a minimum boiling point azeotrope? This behavior is characteristic of a maximum-boiling azeotrope. After initially giving off the more volatile component, the liquid remaining tends to the composition of the maximum boiling point at intermediate composition. After the more volatile component has boiled away, the azeotrope evaporates at constant composition. Q93) In the description of Figure 9.24, the following sentence appears: “At the point when the L2 phase disappears, the temperature increases beyond 94°C and the vapor composition changes along the i—j curve.” Why does the vapor composition change along the i—j curve? At the point when the L2 phase disappears, we are left with a dilute solution of butanol in water. The composition of the vapor is given by point i, and the composition of the solution is given by point g. As more vapor is produced, meno, decreases and the boiling temperature increases. The composition of the liquid and the vapor are given by the ends of a tie lie crossing the L1 + V region. We see that the liquid follows the composition on the curve g—k—O, and the vapor follows the composition along the curve i—j-O. (29.4) Explain why chemists doing quantitative work using liquid solutions prefer to express concentration in terms of molality rather than molarity. . Cha t 9 L p er / Ideal and Real SOlutio /1dea1 and Real Solutions What can you say about the composition of the solid below the eutectic temperature in Figure a microscopic scale? 68 per kilogram is t n'd silicon are not miscible in all proportions. Below the eutectic temperature, the solid is liter, wh' - . _ 1ch changes With the thermodynamic state. orrn and consists of small crystallites of gold contaminated with silicon and small crystallites of ficontaminated with gold. 1‘) Is a whale likely to get the bends when it dives deep into the ocean and resurfaces? Answer this ion by considering the likelihood of a diver getting the bends if he or she dives and resurfaces on ‘_§1ung full of air as opposed to breathing air for a long time at the deepest point of the dive. “a whale is unlikely to get the bends on one lungful of air. There is not enough N2 present in a single __gfu1 to yield a saturated solution of N2 in the blood. Breathing air for a long time at depth, however, _ 1‘ result in saturated blood for the diver. For this reason, the diver is more susceptible to getting the nds than the whale. {29.12) Why is the preferred standard state for the solvent in an ideal dilute solution the Raoult’s law standard state? Why is the preferred standard state for the solute in an ideal dilute solution the Henry’s mLxmg mixing ‘ 0 - G , . — R law standard state? Is there a preferred standard state for the solution in which xsolvenz = xwlme = 0.5? Imxmg " n T2 xi In x! Raoult‘s law defines the appropriate standard state of the solvent in a dilute solution because in this limit aAG . . . mixing 2 - ”21.ng = _ R - the solute has only a weak effect on the solvent. Henry’s law defines the appropriate standard state for . 6T 11 2x1. lnx, and ‘_ ‘ _ . P ’ the solute at a low solute concentration. However, neither standard state adequately describes the mixing = 6;“ng = 0 standard state for xsolvem = xsolute = 0.5. In this case, either standard state can be used, but neither has a 7"“? firm physical basis. [filming — AGmmg + TASmmmg : n R T Z x,- In x; _ T 11132 x. In x _ 0 (29.13) The entropy of two liquids is lowered if they mix. How can immiscibility be explained in terms of thermodynamic state functions? 1' Q93) What informa ‘ tron ca ‘ . . . . . . - - n be obtained from a t1e line in a P—Z phase diagram? A Single phase solution is more stable than two separate phases 1f AG = AH - TAS < 0. Even if AS IS Q9.9) You boil an ethanol~benzene mixture with x — ethanol Q9.14) Explain why colligative properties depend only on the concentration, and “Qt on the identity Of the solute molecules The origin of the colligative properties is the dilution of the solvent, which lowers its vapor pressure. This is the case because solute molecules at the surface of the liquid affect the relative rates of evaporation and condensation. This statistical likelihood of being near the interface doesn't depend on the identity of the solute—only the number in solution. Therefore, colligative properties (ideally) depend only on the number of species in solution. 208 209 f9/ Ideal and Real Solutions P 336 Torr a) for isopropanol P: = 1008 Torr for n—decane P; = 48.3 Torr Using the relationsPI. = yilzmp a. = P. /P.* and 7/ ~ 6” are shown below. J: = = 0.569 PA 591 Torr :51: 0.569 = 2.00 16,, 0.285 :5}: 496 Torr =0.986 PB 503 Torr 7/8 :1: 0.986 21.38 x3 0.715 you can from these observations. . 2) At a given temperature, a nonideal solution 211 x2 312 611 612 0.1312 0.0243 0.912 0.474 0.2040 0.0300 0.875 0.565 0.2714 0.0342 0.846 0.625 0.3360 0.0362 0.831 0.651 0.4425 0.0411 0.790 0.706 0.5578 0.0451 0.745 0.735 0.6036 0.0489 0.716 0.768 Yr 1.05 1.10 1.16 1.25 1.42 1.69 1.81 Y2 3.61 2.77 2.30 1.94 1.60 1.32 1.27 of the volatile components A and B has a vapor P9.3) Two liquids, A and B, are immiscible for T < 750°C and for T > 450°C and are completely miscible outside of this temperature range. Sketch the phase diagram, showing as much information as l t' ns Chapter 9/ Ideal and Real Solutio 19/ Ideal and Real SO u 10 T , 1 phase 52.17 0 cm3 m 01—1 and V15, = 57.0 cm3 mol‘1 95:: i ‘ ”H20 = 0.450 ”H20 + "E! 0 450, ng = 4 58 V = n ’1;qu “HM—VI! 45C H20 ~ 3 _ _ mol l d volume is given by = 3 75 molxl7 0 cm3 mol + 4 58 mol>< 57 0 cm t mixe g e to a = 325 cm3 8’ Kg 3 + ”E! ME! The figure IS drawn With a very narrow Single phase region near xA — 0 and xA — 00 Although the pHZO U 1cm3 —] problem states that the two liquids are immiscible, no two liquids are ever completely immiscible just as = 3.75 molxl8.02 g “101 X 0997 g for the reaction A 7—" B, the equilibrium constant can approach 0 or 00, but is never exactly equal to 14 lcm3 . . . . . 7 mo x————- these values because of the drivmg force of the entropy of mixmg. + 4-58 mOIX 46 0 g 0.7893 g P9.4) At 365 K, pure toluene and hexane have vapor pressures of 5.82 x 104 Pa and 1.99 x 105 Pa, respectively. =335 cm3 3 AVzV—V ...d=325 cm3—335 cm3=—10.cm ; 20. If the volume of the solution is a. Calculate the mole fraction of hexane in the liquid mixture that boils at 365 K at a pressure of 760. H P9.6) A solution is made up of 248.7 g of ethanol and 145.9 g of Torr. b. Calculate the mole fraction of hexane in the vapor that is in equilibrium with the liquid of part (a). Si 0 ethanol under these conditions? a) Prom] = xhcxPhex + (1 — xhex ) P * to] V : nhBOVHQO + methanol ethanol 5 1.01325x105 Pa:l.99x105 Paxhex +5.82x104 Pa(l~x,m) V = "H20 V1120 + "ethane/Vemmmz xhex : 0.306 _ 450.9 cm3 _ 1459 g _] x17.0 crn3mol'1 3 .1 . _ V _ ”H 01/“) _______l§_-9_2_____W__________ : 57.9 cm mol b) y. = . xi“ P“. Vi... : 2 ' = 248.7 g 1701 + (Pm _ P101 )xhax new” 46.04 ngl'l 0.306xl.99><105 Pa ’ ass if _ wn substance is measured at 298 K. Determine the molar m _ M 5.82x10“ Pa + 0.306x(1.99x105 pa -5.82x104 Pa) = 0.602 P9. 7) The osmotic pressure of an unkno - p . . h d 11 1 y g of the solution is 997 kg m‘3. P95) The partial molar volumes of water and ethanol in a solution with xHZO = 0.60 at 25°C are 17.0 RT and 57.0 cm3 mol‘I - ' c - 1 (2 nsoluleRT _ csolutepsolutionRT . M _____ psolulmn so u a 72' ~ " 9 solute V Msolute ‘17: ‘1 l 997 kg m’3x3l.2 kg m‘3x8.3l4Jmol K x298K 21.26X103 kg mol‘ _______________.______________ MAD/1118 _ 610 X 104 Pa , respectively. Calculate the volume change upon mixing sufficient ethanol with 2.00 mol of water to give this concentration. The densities of water and ethanol are 0.997 and 0.7893 g cm—3 respectively, at this temperature. 3 212 213 L R 1 Solutions Chapter 9/ Ideal and Real Solutions apter 9/ Ideal and ea e is 366 Torr. Calculate the ' 's 302.5 10°C the vapor pressure of ethyl bromide is 90.0 Torr and that of ethyl chloride 1 11) At "’ a = 0.33 assuming ideal behavior. a: . . . j 50 _ ASS total —xbenzene benzene hexane hexane ‘ ' n of ethyl chloride in the vapor is 0.82 and answer these questions: actlo . . . . .7 ' f ethyl chloride in the hquid. ' t l ressure and the mole fraction 0 = 0.33x244 Torr+(l ~0.33)x366 Torr = 3.3x102 Torr What 15 the to a p t er - is the overall composition of the system? # P1; _ 131:6 Lia) Bow} _ P1; +(PL:B “PI; )yEC,‘ 90.0 Torr—302.5 Torr = 302.5 Torr+(90.0 Torr—302.5 Torr)x0.820 = 212 Torr The first vapor is observed at a pressure of Bola] 2x141); +(1_XA)131: = va‘CP}; +(1_ xEC )PHB P ~13; 212.3 Torr—90.0 Torr __ total _ xfl?’ ‘m m = 0.575 b) We use the lever rule. P total R a PA -P8 172 Torr—57.6 Torr ”11:3(2153 —xEB) : n::;(y133 ‘ 253) 2158 _ x153 : (1“ 2156' ) — (1 _ x150" ) = xEC - ZEC yEB _Z}EB =(1_yl£(7)_(1— ZEC ) : Z1i‘('—_yEC therefore By Dalton's law, P02 = 0.20991” . At sea level, 72 3% _pH20VP02fsat 02 a : ”HZOxOZ 2 ”H20 E kgzM H20 we know that xEC = 0.575 and yEC = 0.820 -3 -3 3 no zw =2_23X104 mol 2 4.95x10 barxl8.02><10‘3 kg moi" mass = 72on = 2.23><10“1 molx32.0 g mol'I = 7.l4><10‘3 6.5 (0.820 —ZE(. ) = 55% —0.575) g Z EC 2 0.334 Z ‘ : (1— 215C): 0.666 At 18 000 feet “3 * _ d p* = 57.6 Torr. = a , _ . ' = 0,450 P — 98.5 Torr an 3 ideal solution at 298 K, Wlth ’01 ’ A - ‘ 822 _ szoV (Jr/5a! P112) A and B form an ~ h se ”02 ”H20x02 H10 kg” - kg: M H 0 a Calculate the partial pressures of A and B In the gas P a ' . 2 ' , ‘ ' 1 container. 998 kg m'3 xl0'3 m3 X0.2099><0.500 barx 0.71 b. A portion of the gas phase is remOVEd and condensed m a 56pm? 6, 8 K n0 2 _ = 8.34><10'5 mol - 'librium with this liquid sample at 29 . ‘ 4.95 bar x 18.02 ><10'3 kg mol 1 Calculate the partial pressures of A and B in equi mass = I102M = 8.34><10‘5 molx32.0 g mol'1 2 2.67><10‘3 g 215 214 Chapter 9/ Ideal and Real Solutions hapter 9/ Ideal and Real Solutions a) L 'larly the activity and activity coefficient for isooctane are given by "1““ , PA = xAP; = 0.450x98.5 Torr = 44.3 Torr ’ (1~ NEW, = WEE 2 1.411 4 __ = “’1'”- 43. orr PB =(1—xA)PB =0.550><57.6 Torr=31.7 Torr - P2 b) a2 1.411 =1420 72 = j; 21—09006 The composition of the initial gas is given by ‘ PA 44.3 Torr 7) If the solution were ideal, Raoult’s law would apply. 2 PA +PB : 44.3 Torr + 31.7 Torr yA = 0.583; y8 = 0.417 = 0.9006 x 130.4 Torr +(1— 0.9006) x 43.9 Torr 2121.8 Torr :- P9 16) Calculate the solubility of CO in 1 L of water if its pressure above the solution is 55.0 bar. The - —3 0 density of water at this temperature is 997 kg m . ”(to z "(:0 _ fig : _5_5'0_.1?.a_‘r__ = 942 ><lOT3 xco = ___—— _ H .84 x l 03 bar "co + ”H20 ”H20 km 5 For the portion removed, the new x A and x3 values are the previous y A PA = xAP; = 0.583x98.5 Torr = 57.4 Torr PB =(1—xA)1-§ = 0.417x57.6 Torr = 24.0 Torr and y B values. P9.l3) Describe what you would observe if you heated the liquid mixture at the composition corresponding to point i in Figure 9.241) from a temperature below T a to 118°C. _ PHZOV _ W ”H20 ‘ MH 0 _ 18.02x10'3kgmol‘] = 9.42x10—3 X553 2 0.521 mol 255.3 _ ”(:0 = x(,‘OnI-12() - P9 17) The binding of NADH to human liver mitochondrial isozyme was studied (Biochemistry 28 —1 - - - - - ' t 'thK=2.0x107M. the freezing point depression constant Kf. (1989) 5367) and it was deteirnined that only a Single binding Site is presen W1 2 K RMsolventTfirsioii 8-314 JITIOI—I I<-1 X18.02 X10.3 kg mOI-l X(273.15 K)2 f = ' = AHW, 6.008x103 Jmol“ Kf =1.861 K kg mol'1 What concentration of NADH is required to occupy 10% of the binding sites? Beginning with the expression for the fraction of occupied sites: — KCN...
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