SOLUTIONS OF EXERCISES 301 central role of Gibbs energy 301:1 characteristic Numerical values in SI units: m.p = 2131.3; ΔS = 7.95; ΔH = 16943; ΔCP = −11.47. 301:2 a coexistence problem 1) f = M[T,P] – N[2 conditions] = 0 ; this excludes the first option; 2) by eliminating lnP from the two equilibrium equations in the ideal-gas approximation, the invariant point should follow from [60000 + 16 (T/K) = 0]; this excludes the second option. 301:3 trigonometric excess Gibbs energy – spinodal and binodal . -1 2 RTspin(X) = (4000 J mol )π X(1−X)sin2πX; limits at zero K are X = 0 (obviously) and X = 0.75; at 0.75 Tc the binodal mole fractions are about 0.11 and 0.52 301:4 from experimental equal-G curve to phase diagram ΔΩ = 1342 J.mol-1; Xsol = 0.635; Xliq = 0.435; see also van der Linde et al. 2002. 301:5 an elliptical stability field * The property ΔG A , which invariably is negative, passes through a maximum, when plotted against temperature. The maximum is reached for a temperature which is about ‘in the middle of the ellipse’. The property ΔΩ is positive. 301:6 a metastable form stabilized by mixing The stable diagram has three three-phase equilibria: αI + liquid + β at left; eutectic; liquid + β + αII at right; peritectic; αI + β + αII (below the metastable eutectic αI + liquid + αII); eutectoid. 301:7 the equal-G curve and the region of demixing For the left-hand case, where the EGC has a minimum, the stable phase diagram has a eutectic three-phase equilibrium. The eutectic three-phase line is just above the EGC minimum and below the intersection of EGC and BIN. For the right-hand case, the EGC having a maximum and a minimum, the stable phase diagram has a peritectic three-phase equilibrium. The three-phase line is below the left-hand intersection of EGC and BIN. M.H.G. Jacobs and H.A.J. Oonk, Equilibrium Between Phases of Matter: 219 Supplemental Text for Materials Science and High-Pressure Geophysics, DOI 10.1007/978-94-007-1948-4, © Springer Science+Business Media B.V. 201 2 220 301:8 crossed isodimorphism 301:9 activities and Gibbs-Duhem For this case, the Gibbs-Duhem equation comes down to XA d(lnaA) + XB d(lnaB) = 0, from which it follows that, when one of the activities decreases, the other has to increase; in other words, if one of the activities has a minimum for given X, the other must have a maximum for the same X. The G curve is not convex over the whole X range – typically of a miscibility gap. See also Oonk (1981) pp 94-95. 301:10 statements about solubilities Yes for the first statement, provided that the solubility is expressed in mole fraction. A metastable form has a higher solubility than the stable form; the solubility is dependent of the choice of solvent; the ratio, indeed, is independent of that choice. Solutions 221 301:11 a different analysis of the region of demixing in AgCl + NaCl From the special condition two equations are obtained, which are (1) RT’ ln[X’/(1−X’)] + g1(T=T’)(1−2X’) = 0 (2) RT’ ln[Y’/(1−Y’)] + g1(T=T’)(1−2Y’) = 0 Equation (A) is (1) minus (2). The nine function values, Equation (A), yield the result . -1 g1 = [9100 – 2.8(T/K)] J mol . The 18 function values generated by (1) and (2) yield virtually the same result. 301:12 partial pressure versus concentration . [A] = 100 XA P/ RT = 100 PA / RT Using the symbol Kp for the equilibrium constant in partial pressures, the equations become o ln Kc = ln Kp + 2 ln(RT/100) = −ΔG / RT + 2 ln(RT/100) o 2 d ln Kc / dT = ΔH / (RT) + 2/T 301:13 reminiscence of James Boswell In the state where there are equal amounts of the three substances and the pressure indicated by the manometer is 3 bar, the partial pressures of all three substances are 1 bar. Accordingly, the equilibrium constant in terms of partial pressures (Kp) has the value of 1 bar-1. The equilibrium temperature, as a result, is the solution of ΔGo = ΔHo – TΔSo = − 195274 + 185.48T = 0. The solution is T = 1053 K. The numerical value of the equilibrium constant in terms of 3. -1 concentration is Kc = 87.55 dm mol . The equilibrium state at T = 1103 K has mole fractions XSO3 = 0.25; XSO2 = 0.39; XO2 = 0.36; and pressure P = 3.25 bar. The volume of the vessel is V = 87.55 dm3. -1 The values of the equilibrium constants at 1000 K are Kp = 3.2477 bar and Kc = 270.03 3. -1 dm mol . After the addition of one mole of O2, the equilibrium state has P = 3.61 bar, nSO3 = 1.404; nSO2 = 0.596; nO2 = 1.798. 301:14 vapour pressures over liquid mixtures of methanol and methyl acetate The maximum is at about X = 0.67 and P = 73.22 kPa; these values involve the g1 value of 2842 J.mol-1 shown in the table below. The quadruplets that follow give (1) the experimental liquid composition; (2) component A’s partial excess Gibbs energy; (3) component B’s partial excess Gibbs energy; (4) the integral excess Gibbs energy: [0.0392; 6; 2594; 107] [0.0805; 34; 2342; 220] [0.1184; 60; 2152; 308] [0.1715; 107; 1883; 411] [0.2330; 180; 1612; 514] [0.3810; 438; 1054; 673] [0.4830; 686; 746; 715] [0.6200; 1096; 410; 671] [0.7060; 1414; 247; 590] [0.9240; 2392; 24; 204]. 222 The ensemble of quadruplets involves the g1 and g2 values shown in the table below. The quadruplets that follow give (1) the experimental liquid composition; (2) the value of Q; (3) XEGC; and (4) the value of the excess Gibbs energy: [0.0392; 2.902; 0.0795;211] [0.085; 2.048; 0.1486; 372] [0.1184; 1.572; 0.2039; 476] [0.1715; 1.145; 0.2704; 570] [0.2330; 0.847; 0.3369; 642] [0.3810; 0.520; 0.4662; 710] [0.4830; 0.451; 0.5411; 711] [0.6200; 0.477; 0.6369; 659] [0.7070; 0.566; 0.7000; 597] [0.9240; 1.278; 0.9029; 254]. The g1 and g2 are shown in the table below. -1 . -1 g1 / J mol g2 / J mol ΔP / kPa ΔX maximum 2842 0.106 0.0040 partial props 2879 25 0.300 0.0055 EGC fraction 2877 34 0.299 0.0055 One can observe that the results of the second and third rounds of calculation, although they resemble one another, do not show an improvement with respect of the outcome of the simplistic calculation in the first round. In more detail, and concentrating on pressure: in all three cases there is a systematic deviation between calculated and experimental pressures. For the last two cases, nine of the ten calculated pressures are higher than the experimental ones. The origin of the lack of agreement between experimental and calculated equilibrium states is the subject of the two following exercises. Is it a question of methodology, or rather a lack of compatibility between thermodynamic description and the nature of the experimental data? See also Figurski et al. (1997) 301:15 a consistency test The rule is not fully satisfied: the area at the negative side is about 1.06 times the area at the positive side. 301:16 assessment of methodology . -1 . -1 g1 / J mol g2 / J mol ΔP / hPa ΔX maximum 2728 2.00 0.0028 partial props 2750 100 0.00 0.0000 EGC fraction 2779 150 1.55 0.0016 In the first task the calculated pressures at the A side of the system are somewhat too low, and at the B side somewhat too high. This is an expression of the need to introduce Solutions 223 a positive g2. The partial-properties method gives a one-to-one reproduction of the data set. The result of the EGC-fraction method corresponds to an overestimation of the asymmetry of the excess property. Clearly, it is difficult to locate the exact position of the EGC. Notwithstanding these remarks, the GE values provided by the three approaches are the same within a few percent. In order to fully comply with the thermodynamic description, the experimental two- phase region, Exc 14, should be somewhat broader. As a matter of fact, the phase diagram for the same system at the same temperature measured by V.Bekarek (1968) Collect. Czech Chem. Commun. 33:2608 (figures 546; 547; 549 in Ohe 1989) has a broader two-phase region than the one in Exc 14. 301:17 dimolybdenum nitride The Gibbs energy change of the formation reaction changes from a negative (spontaneous formation at 298.15 K; 1 bar)) to a positive value (spontaneous decomposition at 1400 K; 1 bar). If the property were to change with temperature in a linear manner, it would become zero at 1001 K. o . -1. -1 . -1. -1 The values of Δf S : − 89.81 J K mol at 298.15 K and – 66.43 J K mol at 1400 K. The heat capacity property is o . -1. -1 Δf Cp = a + bT = (0.548 + 0.020438 T / K) J K mol . The Gibbs energy of formation changes sign between 961 K and 962 K. Ignoring deviation from ideal-gas behaviour, and assuming that the vessel does not expand, the pressure at 1400 K should be at least 211 bar, which comes down to 45 bar at 298.15 K.