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T11-18 solution - 41 in No AStot< 0 so reaction does not...

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Unformatted text preview: 41 in No. AStot < 0 so reaction does not proceed to the right. e) ArU° = ArH° - Ar(PV) = ArH" — Ar(nR7) assuming ideal behavior for gases and ignoring the molar volume of solid graphite. Then Ar(nRT)— - Arn(RT)— - (+1) RT where Arn refers to change 1n gaseous moles. kJ ArU° = 133 ——1— (8 315 ——1J)Kmo (398 K) = 130 kJ/mol f) ArGo = ArHo— TArSO =133 5% —(398 K) (139 K1110 1) = 77 kJ/mol ArG" > 0 so reaction does not roceed to the ri ht. P g A O ) In K = _ r6 = 77,000 J/mol _ __23.3 RT (8 315 “WW3” K) U Q -. KP = 7.6><10~11 — P if Vdoes not change with P, which ls a AG = JP° VdP: WP _ P0) reasonable assumption for a solid or liquid. I!“ = 5 — 5° = u — 11° V(P—P°) is very small so a -— 6° and .0— may be replaced by 5° . ArG = Moao + #002 - Meac03 z M°Ca0 + M°c02 + RT 1n Pcoz - M°Cac03 31G = (M°Cao+ M°coz - M°Cac03) + RT 1n Pcoz = ArG° + RT ln PCO2 (PNOCI)2 = (1-2 atm)2 = 1'9 x 103 (PNO)2(PC12) (5.0 x 10-2 atm)2(3.0 x 10-1 atm) " 3 1 = l," Yes. Some NO and Cl would form. 42 0) Yes. Q = 1 Q < K p . In order to reach equilibrium, Q must increase. Thus more products will form; reaction proceeds to the right. 2 (1) Q __ (1.0 atm) _ —-—————-————~———————~ = 1.3x 105 (5.0 x 10-3 atm)2 (3.0 x 10~1 atm) Q > K p. Thus, more reactants must form to reach equilibrium. Reaction proceeds to the left. ‘ CA T12 Temperature Dependence of the Equilibrium Constant Exercises 52 _ _éfi [A J) 1 1n K1 — R T2 - T1 K1 = 870 T1 = 298K ArH°=—10.38 kJ/mol K2 = ? T2 = 328K 1n K2 = + ~——————10’380 ”gm” (3%?- — 559%) + In (870) = —0.383 + 6.768 = 6.385 K2 = 593 at 328K 3 o 2. 1n 14.5x10 ArH (1 1) 4.58x103=- R 3H—2—93 ArH° —- m —4 1'15 ‘ “8.3151/Kmole (‘1'744XI0 ) ArH° = + 54.9 kJ/mol ArG" = —RTln K = —(8.315 Elia) (295 K) In (4.58 x103) = —20.6 kJ/mol ArG° = ArH°— TArS" => ArS° = {AW} AS" = _ (—20.6 kJ/m20915—1248 kJ/mol) = + 255 Kim 3. dArH° = ArEpdT A1H¥ — M398 = A161» (T— 298) dan __ ArH" dT ' RT2 43 298 4GC = [@(CHn + (37» mm] — [mm + 3c? (112)] C—‘p(CH4) = 23.64 + 47.86><10‘3 T — 1.92x105/T2 C7(H20)= 30. 54 + 10 29 x 10-3 T @(CO) = 2841+ 4.10 ><10“3 T — 0.46 x105/T2 3><CTP(H2) = 81.84 + 9.78x10‘3T + 1.50x105/T2 ArCP =—56.07 + 44.27 ><10‘3 T — 0.88 ><105/T2 k] T ArHT = "206150F+ I [—56.07 + 44.27 x 10-3 T— . 1'5 01 088x 0]” 298 72 = —206,150 + T T T 0.8 15 g—56.07><TL+ +§4.427><103T2L +_8T><_L L] I 98 98 98 y/ H '\ 088x105 = —206,150 — 56.07 x T + 16,709 + 0.02214 T2 — 1,966 +*—T—— 295 0.88 x 105 = —191.7 311 —56.07 T+0.02214 T2+ +—T— 1 SOOAII0 298 5 _ _57372+ 1“ 191, __T__,2700 56*.T_07+ 02214 + 0.88;;10 ] W 500 500 500 5 500 4 1 191700 — 56.071n T + 0.02214 T — l w R T 98 2 T2 98 98 98 45 46 1 8.315 57.372 + {(—2263) — (29.0) + (4.5) — (—0.3)] 1 8.315 57.372 + [—2505] = 27.246 . _ 11 #3?) M aux/my) _ V _ —TAS—AA aT ‘ T2 ‘ T2 V _ :AL/ _ T2 A a U— TS dA = —PdV—Sd.T = —S 47 CA T13 Temperature and Pressure Dependence of Phase Equilibria for Pure Phases Exercises a 8T P = —S . Thus, for a given P, the value of G is increased when Tis decreased. Because S for liquids is less than S for gases, this effect is greater for gases. Consider Figure 1, but at a lower Tthan 373 K. gas at T< 373 K \ gas at 373 K / liquid at T< 373 K (Should be very close / to line at T: 373 K) liquid at 373 K P<1bar1 bar The intersection of the two dotted lines (T<37 3) will be at a lower P than 1 bar. Thus, the boiling point at pressures less than 1 bar correspond to T<373 K. G = H—TS, if the gas is ideal, H is not a function of pressure. At a fixed temperature, G will vary with P as S varies with P. S changes with P in a logarithmic fashion. (9G a __ The line in Figure 1 has a slope of 313) = V. But, for an ideal gas, V is not T _ RT _« constant as pressure changes: V — — . As PT, VJ, , so slope decreases. P 48 QI P at constant Tfor an ideal gas 35 ._ — _ 3. Recall that 55)]; = V. H20 is an unusual substance, in that V(solid) > V(liquid) at 273 K and pressures around 1 bar. (That's Why ice floats on water!) 20(s) H20(f ) P at the intersecting lines is the pressure at which 273 K is the melting point. 4-,——- block of ice \ masses 49 When P is increased on the ice (at its freezing point), the value of E for the solid is increased more than for the liquid (Vice > Vliquid at freezing point). Thus, 5(liquid) < 56%) and so the ice melts. The wire then slips down into the ice, and continues this process all the way through. ~— P dlanArHvap Izdlnpz dT R79 p1 Tz Aria ———P dT T1 RTQ ln & = — 9:19—an (L 1—) if Arfivap is independent of T. T2— T1 1 Thus, a plot of ln P vs T would have a slop pressure of the liquid at the temperature T. 6. A(s) :A(g) Ar}? = Arfisub e of — A—Jrljeva V(s)dP — —S_(s)dT = V(g)dP — ‘5"(g)dT at equilibrium [V<g)—V(s>]dP = [fie—T919] dT where P is the vapor — = ———— = —-—— at equilibrium (Ar—I; = TArE) dT Ar? TArTf (g) ~ Assume the gas is ideal and that 7(s) << 7 Then Ar? = V(g)—V(s) = 7(g) = R?T . 50 7 .' Analogous to Exercise 5, measure vapor pressure of the solid as a function of T. Then a plot: of In P vs. %, would have a slope of — 9%10 . 8. A(s) :zAm AYE = 111.?qu V(s)dP— ’s‘(s)dT= V(fl)dP~ "S“(mr at equilibrium [W0 — ”WgfldP = [T910 — 30)] d7 3—1;: = @3335- : Ar {1qu because (Arfi = T4»? at equilibrium) A r V TAr Vfus 9. i]: = ArH where A? = H(prod) — H(react) and A5; : V(prod) — V(react) . dT TAR? P _ T 2 2 —- '— [ dP = AYE I 9%,]: assuming ArH, ArV are independent of T. P1 ArV T1 Pz—Pl = ”Ar—I: (lnTz—ln T1) = Ar? ln % ArV A,.V 1 10. At 71 = 373 K, P1 = 1 bar. 72 = 353 K; AIH = 2255 g = 40,630 J/mol 111132 :_Ar_1E1m1_1 P] R T2 T1 ln P _ _ 40,630 J/mol l 1 2 ' 8.315 J/Kmol (353 ‘ 373) = —(4.886>< 103) (1.519 X104): —0.742 -. P2 = 0.476 bar 11. T1=353.4K P1=1bar T2 = 298.2 K P2 = 0.13 bar 12. 13. P2 ‘ J A}; _ Rlnfi _ (8315191101) In (0.13) r mp— L _1_ — _1_ *1_ (T1 _ T2) (353.4 ‘ 298.2) =32kJ/mol d In P Arm“; dT RTZ dlnP 2819.7 _3 _ Arma dT T2 +1.855><10 — ARTZ At 361°C Armap = 24.9 kJ/mol . Assume that the densities do not change with pressure. Solid H20 is 0.915 g/mL = 915 g/L = 50.8 Enrol 1 L V(s) = m = 0.0197m 751 . . . 1 — L L1qu1dH20181000% = 55.5 1%; wt) = 0.0180m—01 ArV = V(€)~V(s) = 4,0017% = —0.17>< 10-5m3/m01 Aims = 6024 J/mol at T1 = 273 K, P1 = 1 bar, P2 = 100 bar Ar}? A57 P2—P1 = Recall that 105i3 = 1 bar In 6024 J/rnol T2 100 bar — 1 bar = In —0.17 x 10-5 m3/mol 273 99 bar — (—3 544 x 104) In E ‘ ' 273 —2.79 x 10-3 + In 273 =1n 72 = 5.607 '. T2 = 272.2 K (lower than 273, as expected) 51 52 14. H20(s) 2 H20(g) at 273.15 °K. 15. AH for this process = Arfic’fus (melting) + Arman, (vaporization) (Hess' Law) . ~ _ ii JSJ. _ g . Angub — +44.8 mol + 6.024 mol — +50.8 mol d In P _ Ar—H'sub _ Ar—Hsub £1]: dT - RTZ =>dlnP_ R T2 T: ~15°C ’ T: ~15°C = 258 K J- dlnP = 91%12 % assuming ArHsub is T: 0°C V T: 0°C = 273 K independent of T. +50,800 J/mol l 1 1n PT=_15 °c — 1n PT=0 °C = -—_—J_~ (573“ —~ fig) 8.315 mol K In PT=2 58K = ln PT=273K — 1.301 P273K = w x 1.013255% : 6.1 x 10*3 bar ln PT=253K = “5.098— 1.301 = —6.399 PT=258K = 1.66 x 10-3 bar = 1.25 Torr 8G 97)]; 2 -§ ==> For solids, TS" is relatively constant as a function of T. d6 = Vdp — 's‘dT isle) > :S—(A) Cal A at high pressure A at 1 atm \ \ 4/ C at high pressure C at 1 atm 16. IfP T, thenET because 8i] = V- 310 T Line for C is raised more (at high pressure) than line for A because 7(C) > 7(A). Consider A i C o _ L1 o _ _J_ ArH _ 0.2 mol , AIS _ 4.2 Kmol Assuming ArH", ArS" are independent of T: At equilibrium: ArG" = O = ArHO — TArSO J J 0=2oom —T(4.2 mol K) => T— 48K ______________ 1 ~1 dT ._ _ _ _2.97 _1.4><10 ath P2 —P1 = 14 (298—48) = P2 -1 P -~ 3500 atm A37 and A7 are assumed to be independent of T and P. 53 54 CA T14 Vapor Pressure and One Component Phase Diagrams Exercises 1. At T Jz, vapor pressure i. Thus, some of the vapor condenses. Also, as T i, P l according to the ideal gas law. So, both effects lead to a dramatic decrease in gas pressure. Air has been expelled from the flask by vigorous heating and the water will boil at all temperatures because the vapor pressure of water is the external pressure. 2. Recall that g: = M: . TArV Arfifus < ArFI—vap For solid 3 liquid, Ar? > 0 but relatively small. For liquid : gas, Ar? > 0 but large. For solid :: gas, A? > 0 but large. Assume ArTT, Ar? (for solids, liquids) independent of T. (atm) 5.112 1.0 194.7 K 216.55 T (K) 3. If slope is negative, then freezing point of H20 decreases as P increases. 55 .i ngp— Eng = flap—Weir 'II Tis a constant. ngPg = 7mm 2%? = YT! The ratio 324 is a positive number. Vg Vg If the pressure on the liquid is inereased, dPfl, the vapor pressure, dPg, increases. A — Solid H20 at a very low temperature in equilibrium with a very low vapor pressure of H20(g) at a total pressure of 1 bar. B - As in A, but VP of H20(g) is slightly higher, and T is higher. C — Liquid H20 (above 0 0C) with some vapor in equilibrium at a total pressure of 1 bar. D - As in C, but VP of H20(g) is higher, and T is higher. E - T: 100 OC. Liquid is boiling. VP of gas = 1 bar. F - T> 100 °C. Vapor is present at total pressure of 1 bar. G — Solid is present at a low T. P is less than 1 bar. VP of gas present is less than P. H - At this T, solid is present, and VP of gas is equal to total pressure, P. I — Gas only is present at a total pressure of P. 56 CA T15 The Ideal Solution Exercises 1. This mixture is not expected to be ideal because the predominant intermolecular force in H20 is hydrogen bonding while that in dibutyl ether is dipole—dipole interaction. Because these types of interactions are different, the solution is not expected to be ideal. 2. Dalton's Law: P;- = Xi(vap) >< Pm Raoult's Law: P;- = X11501) >< Pi* Both laws involve the partial pressure of a component i in a mixture of gases. Dalton's Law relates this partial pressure to the total pressure of m mixture of ideal gases. However, Raoult‘s Law is used to describe the partial pressure of a component in equilibrium with a mixture of liquids. 3. Dalton's Law may be used: sz(vap) = -— moles of bz sz(vap) 1.00 1.00 0.00 0 0.200 0.263 0.400 0.445 0.800 0.818 0.100 0.263 0.800 0.928 4. P21); = 25 Torr Xtol(sol) = 0-50 Ptol = Xtol(sol) X Pfol = (0.50)(25 Torr) = 12.5 Torr Composition is generally described in terms of mole fractions. Using Dalton's Law for the vapor phase: Burma _1_2_~5_:9_r: _ _ _ 15 Torr _ 0.83 => Xxy_0.17 Xtol(vap) = Ptot ny = Xxflsol) X P33 = (0.50)(5 Torr) = 2.5 Torr 6. Assume that the mixture of two components, A and B, is ideal. Let XA( sol) be the mole fraction of A in solution. Then, from Raoult's Law, PA = XA(sol) >< PA* PB = X13001) >< PB* but XB(sol) = 1 — XA(sol) , SO PB = (1 - XA(sol))PB* and Pm: = XA(sol) >< PA* + (1 - XA(sol))PB* PA _ XA(solg >< PA* According to Dalton's Law, XA(vap) = Pm XA( ol)PA* +(1—XA(s 1))PB* S 0 Thus, PA* = PB* is the condition under which liquid and vapor compositions are identical. If PA* = 133* = P* then XA(s0l)P>I< XA(sol)P* - XA(sol) 13* + P* XA(vap) XA( sol 213* P * <3 H = XA(sol) 57 58 CA T16 Chemical Potential for a Component of a Solution Exercises 1. The chemical potential of a component in a mixture depends on the Gibbs energy of the pure liquid component and another term which is a function of the mole fraction of that component in the mixture. 2- MA(sol) = #342) + RT 111 XA(sol) Since X13450” < 1 for a mixture, the second term is negative, and the Gibbs energy of A in solution is less than that of pure A. The same holds true for B. 3. Gibbs energy of A decreases. 4. The mole fraction of the solvent decreases (from 1) when a solute is added. Thus, Gibbs energy decreases as seen from the equation in Exercise 1. 5. When pure water is boiling, the Gibbs energy of the liquid phase equals the Gibbs energy of the vapor phase. When salt is then added to the liquid, the Gibbs energy of water (liquid) is decreased, but the Gibbs energy of the vapor phase is unchanged because salt is not volatile (does not vaporize). Thus, boiling stops because the liquid phase has the lower Gibbs energy. Recall that 3%) = —S and that SM) < S(vap). If P T is raised, G is decreased for each phase. But G decreases more for vapor phase. Thus, at some higher T, G of both will become equal, as the G(mp) is decreasing faster. Thus, boiling occurs at a higher T. 59 CA T17 Partial Molar Quantities Exercises 1. Assume that the tennis balls are much smaller than the bowling balls. Then, the tennis balls are able (for the most part) to fill in the spaces present between the bowling balls. Thus, on average, adding a tennis ball will not increase the total volume very much at all: 32%) z 0 . Thus, total volume is not affected by ”BB adding tennis balls. The final volume could be the same as one of the components if the other component were small enough to fit into the spaces between (as in tennis balls and bowling balls). MAmz) = MM) + RT 111 XA MB(sol) = NEW) + RT ln XB — 3M: r a = all—Amlz _ A __ VA 3P )TflAflB — (9P + 3P [RTln XA] = 17:“) + 0 = 174%) molar volume of pure A Vim) = Em) molar volume of pure B AV—mix = Vinix — ”AVA" — ”BVB* but l7mix = ”Al—[A'F nBVB = nAl7A* + ”B VB* A Vnfix = O for ideal solutions Mia) ILA _ T — T _ + R In X A 3 (EA) T M _ * _ _ _T_ =__&=__3T___f=i1§kz;fl*a=_HA* 3T P,nA,nB T2 T2 T2 T2 EA = [if (~— molar entropy of pure "A" p Also :> 19—13 = Iii—13* T partial molar entropy of "A" in ideal solution 60 Thus, as for volumes, A Enix = 0. = —SA = —SA* +RlnXA and :9]; = 53* —RlnXB 4- A(—imix = Gmix “nAaA‘ ”B6B = mm + "BMB — nAuZZ — ”BIL; T T molar free energy of pure liquid But [Li = m* + RT 111 Xi, so mm = rum: + RT 1n XA) + mm; + RT 1n XB) my: — max»; = nARTln XA + n}; RTlnXB ASmix = Smix — 71A 32* - n3 3;; and from Exercise 3: —S3 = —SB*+R1nXB €35 : 89: «1qu ASmix 11A (5; — R In XA) + m; (53* — R 111 Kg) — nA SA* — nBSB* = —nAR1nXA — nBRlnXB 5. Afinfix = 0, AVmix = O as shown above in Exercise 3. WWW CA T18 Colligative Properties Exercises 1. Apr = 0.65 K Kbp = 5.03 K/molal To = 349.90 K AT], 1.30 K = 0.258 moles '. munknown = Kbp = m 1.29 x 10-2 moles dissolved in 50.00 g cc14 - _'_1-%_a_ — MW=96.1g/mole "1.29 ><10‘2 mol 2. This derivation is analogous to the development in CTQs- 9—1 1. XA(s0l) ‘- T A H dT I dln X1050» = ——er“8 T—z X=1 To —A H 1 1 1n XA(sol) = —rR—fQ§ (f - ‘T—o) “Arfifus T —T —AH ATE 1n XA(sol) = R ( 0 J = —"—r fus T-T0 R T02 when TO—T is a small number (ATfi, is small) Lt __ moles B e m— 1000 g solvent A 5 AT lnXA=1n(1—XB)=—XB = _Ar_Hf2.s. .42 / R T02 from Taylor series For a dilute solution of B in A i . ~ moles B _ Aerus 13ng "X3 ~ molesA _ R X T2 0 _ RT02 molesB _ kg A ATfp _ Aerus X moles A but molesA _ MWA ein kg/mol 1000 g solvent 61 62 U.) RTO2 moles B'MWA _ RTOZ-MWA molesB AT = ._ x m J?) Aerus kg A Aerus kg A RT Z-MWA AT =K ><m whereK = v—O————— fp fp B fp Aerus A<s> : Aw) AH°=AJifus Am :1: A; sol) AH° = 0 because Alzlmix for an ideal solution = O A(s) 3 A(sol) AHO = Arfifus for an ideal solution In Xl2(sol) = R —*—" Airfifus T‘To (T‘ To J T = 298 K Arfifus = 15,640 J/mol To = 386 K R = 8.315 J/mol K In X12(501) = —1.439 :> X12 = 0.237 for an ideal solution The solubility of "B" in "A" at a temperature Tis the mole fraction of B which gives rise to a freezing point depression of ATfp = TO — T. ...
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