SOLUTIONS OF EXERCISES central role of Gibbs energy

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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. g1 9100 2 8 T K J mol 1, 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. ln Kc ln Kp 2 ln RT 100 Go RT 2 ln RT 100,d ln Kc dT Ho RT 2 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. Go Ho T So 195274 185 48T 0, The solution is T 1053 K The numerical value of the equilibrium constant in terms of. concentration is Kc 87 55 dm3 mol 1, 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. The values of the equilibrium constants at 1000 K are Kp 3 2477 bar 1 and Kc 270 03. dm3 mol 1 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. 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. g1 J mol 1 g2 J mol 1 P kPa X,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. g1 J mol 1 g2 J mol 1 P hPa X,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, The values of f So 89 81 J K 1 mol 1 at 298 15 K and 66 43 J K 1 mol 1 at 1400 K. The heat capacity property is, f Cp o a bT 0 548 0 020438 T K J K 1 mol 1 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, 301 18 the Helmholtz energy in its capacity of equilibrium arbiter. 0 20 3991 8504 12495,0 25 4157 8498 12655,0 30 4324 8413 12737. 0 35 4490 8261 12751,0 40 4656 8049 12705,0 45 4822 7781 12603. 0 50 4989 7460 12499, This time i e at constant T and V the Helmholtz energy is the equilibrium arbiter. Exc 108 11,301 19 the Wilson equation,EA RT ln XA AB XB XB AB XA AB XB BA XB BA XA.
EB RT ln XB BA XA XA AB XA AB XB BA XB BA XA, For X 0 the expression of the derivative of EA RT with respect to X takes the form. AB 1 AB BA2 BA2 0, and it means that Raoult s law read the condition EA XB XB 0 0 is satisfied The. condition is not satisfied if BA is put equal to zero the Wilson parameters should not be. put equal to zero, Observe that if both of the two parameters were put equal to zero the excess Gibbs. energy would be equal to minus the ideal Gibbs energy of mixing and the complete. Gibbs energy of mixing would be zero for each value of X. Unlike e g the expression,G E RT X 1 X g1 RT g2 RT 1 2X. the Wilson equation is not applicable to systems with limited miscibility The second. derivative of the complete Gibbs energy of mixing is positive for every X no matter the. values of AB and BA,AB 0 01 0 2 0 4 0 6 0 8 1 0 1 2 1 4 1 6 1 8 2 0.
0 01 0 683 0 15 0 27 0 38 0 49 0 59 0 70 0 81 0 95 1 11 1 34. 0 2 0 597 0 511 0 11 0 20 0 29 0 37 0 45 0 55 0 66 0 85 1 26. 0 4 0 520 0 434 0 357 0 08 0 16 0 22 0 30 0 39 0 54 1 90 0 01. 0 6 0 453 0 367 0 290 0 223 0 07 0 13 0 19 0 32 0 13 0 24 0 28. 0 8 0 394 0 308 0 231 0 164 0 105 0 06 0 14 0 14 0 18 0 22 0 25. 1 0 0 342 0 255 0 178 0 112 0 053 0 000 0 05 0 09 0 13 0 19 0 22. 1 2 0 294 0 208 0 131 0 064 0 005 048 095 0 04 0 08 0 11 0 15. 1 4 0 250 0 164 0 087 0 020 038 091 139 182 0 04 0 07 0 10. 1 6 0 210 0 124 0 047 020 079 131 179 222 262 0 03 0 06. 1 8 0 173 0 087 0 010 057 116 168 216 259 299 336 0 03. 2 0 0 139 0 053 024 091 150 203 250 294 334 371 405. Solutions 225, For given values of its parameters the two most important characteristics of the Wilson. equation are its numerical value at the equimolar composition and its degree of. asymmetry with respect to that composition These characteristics are displayed in the. table on p 224 for a number of parameter pairs Along the diagonal of the table with the. bold numbers the values of the two parameters are the same and it means that the. Wilson function is symmetrical Below the table s diagonal function values are given. for the equimolar composition these values do not change if the values of the two. parameters are interchanged As regards the asymmetry of the function the numbers in. italics above the diagonal represent function value for X 0 25 function value for X. 0 75 function value for X 0 75 The same numbers are obtained when the Wilson. parameters are interchanged and at the same time component A B is made to change. into component B A, As regards the excess Gibbs energy of liquid mixtures of water A and methanol B. almost quantitative agreement is obtained for AB 1 406 and BA 0 257. 301 20 Marius Ramirez s view from the arc, The arc has b 0 000449 K 1 h 0 0366 and Tmax 317 K Difference in heat capacity. amounts to 120 J K 1 mol 1 Heat of vaporization at 298 15 K is 80 6 kJ mol 1. 301 21 the Rackett equation,2 7 ln A1 1 T Tc 5 7 Tc. Obviously and also in view of the validity range the information is on liquids under. their own equilibrium vapour pressure The influence of pressure therefore is hidden. in the equation Strictly speaking,1 V dV dT dP dT, where the symbol is for isothermal compressibility and dP dT for the slope of the.
liquid vapour equilibrium curve, V 89 416 cm3 mol 1 1 14 x 10 3 K 1 Pc RTc A2 4 90 x 106 Pa 49 bar. 301 22 the Simon equation,A bar b P bar T K,Babb 4266 4 44 34 0 26. footsteps 4048 4 60 38 0 31, NB In the original paper Simon and Glatzel 1929 the relation between T and P was. given as log a p c log T b and three data pairs were used to evaluate the. values of the system dependent constants a b and c. For sodium molybdate the change in molar volume on melting is 6 1 cm3 mol 1. 301 23 an exploratory calculation,e g for kyanite, GIII T P 2443881 84 467 T K 298 15 44090 P GPa J mol 1. e g for I II,T K 1085 882 P GPa, Triple point coordinates P 0 405 GPa T 728 K the coordinates given by.
Tonkov 1992 are 0 55 GPa and 893 K taken from Richardson et al 1969 NB One. can read in Tonkov that the experimental triple point reported in seven independent. investigations varies from about 0 2 GPa 693 K to 0 9 GPa 663 K. In the phase diagram pressure axis horizontal and temperature axis vertical the. sillimanite field is at the upper side the andalusite field is at the left hand side and. the kyanite field at the right hand side,301 24 fluid carbon dioxide. G 21 5 kJ mol 1 RT ln 176 fugacity coefficient f P 0 88 Angus gives f P. 301 25 compounds in the role of component, 1 1 X mole of A2BC X mole of AB2C A s mole fraction in the liquid. mixture x A 2 1 X X 4 2 X 4,x B 1 X 4 x C 1 4,For left hand liquidus A2BC 2A B C. K x A 2 x B x C 2 X 2 1 X 64 Ko 4 64,R ln K Ko R ln 2 X 2 1 X 4 HA2BC 1 T 1 To. For reasons of symmetry the eutectic point is at X 0 5 Apparently it is assumed that. heat capacities may be ignored the calculated eutectic temperature is 1479 1 K At X. 0 and X 1 the two liquidi have a horizontal start,2 For the second task x A 3 2X 4 2X x B 1 4 2X.
Left hand liquidus,R ln 256 3 2X 3 27 4 2X 4 HA3B 1 T 1 ToA3B. right hand liquidus,R ln 4 3 2X 4 2X 2 HA3B 1 T 1 ToAB. Eutectic point at X 0 812 T 1468 9 K,Again horizontal start of liquidi. 3 By the change in melting point to 1400K AB has become an incongruently. melting compound And because it is not so wise to take an incongruently melting. end member it is recommendable to change the definition of the system e g to. Solutions 227,The liquidus for the compound A3B is given by. R ln 256 27 1 X 3X HA3B 1 T 1 ToA3B,and the liquidus for AB by.
R ln 4 1 X X HAB 1 T 1 ToAB, Crystallization starts at 1437 5 K A3B is formed At the three phase equilibrium. temperature about 1394 K the last part of the liquid reacts with the last part of the. precipitated A3B to yield AB that is to say if during crystallization the precipitated. solid is not removed,301 26 Professor Geus s nickel coin. The calculated pressure for 0 is 548 Torr in line with Horstmann s data. f M T P N NH4Cl NH3 HCl NH3 0 5 N2 1 5 H2 3 2 1,XHCl 1 2 XNH3 1 2 XN2 0 5 2 XH2 1 5 2. At 600 K the value of 0 9332 and P 4 14 bar,NB Exc 007 4 Exc 111 7. 301 27 montroydite and dephlogisticated air, Two changes and their corresponding equilibria make their appearance First the.
change from mercuric oxide to gaseous mercury and oxygen and second the change. from liquid mercury to gaseous mercury The two equilibrium conditions are. HgO Hgvap 0 5 O2vap 1,Hgliq Hgvap respectively 2, After substitution of the function recipes for the chemical potentials and with. allowance for the approximations made Equation 1 reads. G HgO T GoHg T RT ln XHg P 0 5 GoO2 T RT ln XO2 P,which is equivalent to. 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

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