MATSCI 204 Thermodynamics and Phase Equilibria Winter Chapter #5 Practice problems

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MATSCI 204 Thermodynamics and Phase Equilibria Winter 2013 Chapter #5 Practice problems Problem 4 a-assuming that you are cooling an equimolar liquid Au-Bi solution reversibly from 1200 C, describe

1 MATSCI 204 Thermodynamics and Phase Equilibria Winter 2013 Chapter #5 Practice problems Problem 4 a-assuming that you are cooling an equimolar liquid Au-Bi solution reversibly from 1200 C, describe the sequence of events (i.e. phases that form and disappear at what temperatures and all phase present as you cool) until you reach 0 C. b-sketch and label Gibbs free energy vs. composition diagrams for the Au-Bi system at 371 C, 247 C and 241 C. 1

2 Solution a- At 1200 C, a 50% at Au-Bi solution is homogeneous. As we cool, the first phase transformation occurs at 530 C, where pure Au is in equilibrium with the liquid. The two-phase equilibrium persists until 371 C, where Au disappears and the liquid is in equilibrium with a Au 3 Bi intermetallic compound. At 241 C, the system incurs in a triple point, where Au 3 Bi, liquid Au-Bi solution and Bi are in equilibrium. Upon further cooling until 116 C, the system is in a two-phase Au 3 Bi/Bi field. At T<116 C, the intermetallic disappears and the system is a mixture of Au and Bi. b- At 371 C, three phases are in equilibrium: Au, Au 3 Bi and the liquid: Notice that all solids are line compounds At 247 C, the phases in equilibrium are Au, Au 3 Bi, the liquid and Bi: 2

3 At 241 C, there is three-phase coexistence at x Bi >0.33 3

4 Problem: PtSi is an ohmic contact used in Si microelectronic devices. It is one of several silicide compounds that can form in the Pt-Si binary system (see attached). Note: the dashed line near XSi = 0 is the estimated solubility of Si in the Pt-Si fcc alloy. a) Sketch the activity of Pt as a function of composition at 1000 C. Label your sketch fully and be as quantitative as possible. b) Assume that the liquid behaves as a regular solution with an enthalpy of mixing of 1000J/mol at XSi = 0.5. The heat of fusion of Si is 50.2 kj/mol and that of Pt is 19.7 kj/mol(tm,pt = 1768 C). Determine the Gibbs free energy change at 1210 C for the reaction: Pt(l) + Si(l) = PtSi(s). 4

5 Solution Plot of activity of Pt as a function of 1000 o C. atomic % Si L+ Pt 2 Si L+ Si fcc Pt(Si) L fcc +L L L a Pt L+ PtSi Ideal solution N.B. Activity of Pt drops abruptly to xsi 0. b) We are asked to calculate G 0S PtSi " G 0L 0L Pt " G Si From the phase T = 1210 o C, PtSi(s) undergoes congruent melting. 5

6 This is equivalent to 1 2 "G PtSi L mix = "G mix at x=0.5 with the consistent choice of reference L states (as discussed in class), where "G mix is the free-energy of mixing of the liquid Pt-Si solution. The factor or ½ comes from the fact that at x=0.5, we have 0.5 moles of Pt and 0.5 moles of Si, which form 0.5 moles of PtSi. We choose pure solid Pt and Si as reference states: "G PtSi mix = G 0S PtSi # G 0S Pt # G 0S Si is the free-energy of formation of PtSi from pure solid components. L "G mix = x Si G L 0S ( Si # G Si ) + x Pt G L 0S ( Pt # G Pt ) At the congruent melting point: 1 2 G 0S PtSi " G 0S " G 0S ( Pt ) = x Si Si G L 0S ( Si " G Si ) + x Pt G L 0S ( Pt " G Pt ) (1) The liquid solution is a regular solution with a given "H mix, therefore: x Si G L 0L Si " G L 0L Si " G Pt ( ) ( ) + x Pt ( G Pt ) = #H mix + RT x Si ln x Si + x Pt ln x Pt We now introduce G 0S Pt and G 0S Si : x Si G L Si " G 0S + #G 0L $S ( Si ) + x Si Pt G L Pt " G 0S + #G 0L $S Pt Si Plugging this result in Eq. 1: 1 2 G 0S PtSi " G 0S " G 0S 0L ( Pt ) = #H Si mix + RT( x Si ln x Si + x Pt ln x Pt ) " x Si #G $S Si " x Pt #G Pt Moreover, 1 2 G 0S PtSi " G 0S " G 0S ( Pt ) = 1 Si 2 G 0S PtSi " G 0L " G 0L ( Pt ) " #G 0L $S Pt " #G 0L $S Si Si 2 2 Since at the congruent melting point, x Si =x Pt =0.5, We have: 1 2 G 0S PtSi " G 0L " G 0L ( Pt ) " #G 0L $S Pt " #G 0L $S Si = Si 2 2 0L = #H mix + RT( 0.5ln ln0.5) " 0.5#G $S 0L $S Si " 0.5#G Pt As a result, the free-energy of Pt L +Si L =PtSi S is: G 0S PtSi " G 0L Pt " G 0L Si = 2 [#H mix + RT( 0.5ln ln0.5) ]= kj/mol ( ) = #H mix + RT x Si ln x Si + x Pt ln x Pt ( ) 0L $S (2) 6

7 Problem 4 a) A homogeneous liquid of composition x SiO2 =0.6 is cooled slowly from 2500 C to rom temperature such that at any temperature during cooling the sample may be assumed to be at equilibrium. List the phases present during each stage of the cooling process and the temperature ranges over which they are observed. Be sure to indicate the temperatures at which phase boundaries are crossed during cooling of this sample and describe which new phases are fomed (and which old phases may disappear) at each of these temperatures. b) The kinetics of formation of Zircon, ZrSiO 2, are notoriously slow. Repeat question b, assuming that the cooling process is performed more quicky, such that zircon does not form. c) Sketch the variation of the activity of SiO 2 across the composition range from pure ZrO 2 to pure SiO 2 at 2300 C. Indicate regions of to phase and single phase stability and be quantitative in locating phase boundaries. 7

8 Solution Part a) Equilibrium cooling: ZrO 2 /SiO 2 system; X SiO2 = 0.6, T init = C. i) 2500 o C > T > 2336 o C Liquid(L) T = 2336 o C L L1 + L2 (phase separation) ii) 2336 o C > T > 2250 o C L1 + L2 T = 2250 o C L1 ZrO2 (tet) + L2 (monotectic rxn) iii) 2250 o C> T > 1687 o C ZrO2(tet) + L2 T = 1687 o C L2 ZrO2(tet)+SiO2(crist) (eutectic) iv) 1687 o C > T > 1676 o C ZrO2(tet) + SiO2 (crist) T = 1676 o C ZrO2(tet) + SiO2(crist) ZrSiO4(peritectoid) v) 1676 o C > T > 1470 o C ZrSiO4 + SiO2 (crist) T = 1470 o C SiO2(crist) SiO2(trid) (pseudomorphic trans.) vi) 1470 o C > T > 867 o C ZrSiO4 + SiO2 (trid) T = 867 o C SiO2(trid) SiO2(H-qtz) (pseudomorphic trans.) vii) 867 o C > T > 573 o C ZrSiO4 + SiO2 (H-qtz) T = 573 o C SiO2(H-qtz) SiO2(L-qtz) (pseudomorphic trans.) viii) 573 o C > T > 25 o C ZrSiO4 + SiO2 (L-qtz) Part b) Till T = 1676 oc, the phases present & the phase transformations are same as that of Part (a). Part c) 1676 o C > T > 1470 o C ZrO2 (tat) + SiO2 (crest) T = 1470 o C SiO2 (crist) SiO2 (trid) 1470 o C > T > 1170 o C ZrO2 (tet) + SiO2 (trid) T = 1170 o C ZrO2 (tet) ZrO2 (mon) 1170 o C > T > 867 o C ZrO2 (mon) + SiO2 (trid) T = 867 o C SiO2 (trid) SiO2 (H-quartz) 867 o C > T > 573 o C ZrO2 (mon) + SiO2 (H-quartz) T = 573 o C SiO2 (H-quartz) SiO2 (L-quartz) 573 o C > T > 25 o C ZrO2(mon) + SiO2 (L-quartz) At T = 2300 oc, the following phase fields exist. ZrO2 (cub) + L1 L1 0<x< <x<

9 L1 + L2 L <x<0.61 x>0.61 9

10 Problem 2: This problem will require the use of Mathematica or a similar type of software to solve a non-algebraic equation. There is an analytical solution, which Saahil posted on C2G. FeO and MnO form ideal liquid and solid solutions. Calculate the temperature at which equilibrium melting begins when the equimolar solid solution is heated. Calculate the composition of the first-formed liquid and the temperature at which equilibrium melting is complete. The heat of fusion of FeO is kj/mol at the melting point (1643 K) and that of MnO is 54.4 kj/mol at its melting point (2148 K). Throughout the problem, assume that the Turnbull approximation is valid. 10

11 Answer According to the problem, the phase diagram of the FeO-MnO system looks like this: We need to calculate x L, T 1 and T 2. At T 1, the solid solution having x S =0.5 is in equilibrium with the liquid solution having composition x L. Therefore, at T 1, the g vs. x curves of the liquid and solid solution have a common tangent at x S =0.5 and x L. Mathematically, this is equivalent to: "g S = "g L "x xs = 0.5 "x xl g L x L = g S x S = "g S "x xs = 0.5 ( ) 0.5 # x L where g S and g L are the g vs. x curves of the solid solution and the liquid solution. These two curves can be calculated but we must take care to use the same reference state for both solutions. At T 1, we choose pure liquid FeO and pure solid MnO as reference states. We obtain: 0L g L = xµ MnO + (1" x)µ 0L FeO + RT[ x ln(x) + (1" x)ln(1" x) ] and 0S = xµ MnO 0S g S = xµ MnO 0S = xµ MnO S $L [ ] + x#g MnO + (1" x)µ 0L FeO + RT x ln(x) + (1" x)ln(1" x) 0S + (1" x)µ FeO 0S + (1" x)µ FeO [ ] [ ] + (1" x)#g FeO + RT x ln(x) + (1" x)ln(1" x) + RT x ln(x) + (1" x)ln(1" x) L $S 11

12 S at T=T 1 we can use the Turnbull approximation to calculate "g #L MnO and "g FeO S #L S "g #L MnO = "h MnO T m,mno L "g #S FeO = "h S #L FeO T m,feo ( T m,mno $ T) ( T $ T m,mno ) S (you can verify that "g #L L MnO > 0 and "g #S FeO > 0 at T=T 1 ) S #L : Using the first condition (derivatives must be equal at x S =0.5 and x L ) we obtain: " #h S $L FeO T m,feo + % x ( T 1 " T m,feo ) = RT ln L (. - ' * 0 + #h S $L MnO T m,mno " T 1, & 1" x L )/ T m,mno ( ) because ln # 0.5 & % ( ) 0 $ 1" 0.5' we extract: T 1 = S #L "h FeO T m,feo S "h #L S #L FeO $ "h MnO $ "h S #L + % MnO + RT- ln' T m,mno, & x L (. * 0 1$ x L )/ The second condition gives: [ ] + x L #h MnO RT 1 x L ln(x L ) + (1" x L )ln(1" x L ) = RT ln(0.5) #h S $L FeO T m,feo S $L ( T 1 " T m,feo ) " #h FeO ( ) = T m,mno T m,mno " T 1 S $L T m,feo ( T 1 " T m,feo ) % ( 0.5 " x L ) These two equations can be put through Mathematica to find x L and T 1. I calculated T 1 as a function of x L with the first equation and then found the root of the second by plotting it. With this approximate method, I obtain 0.34<x L <0.35 and 1953K<T 1 <2015K. Numerical solution using Mathematica gives x=0.33 and T 1 =1888K (pretty close). Using the same procedure at T 2, we need to simultaneously solve: "g S = "g L "x xl = 0.5 "x xs g S x S = g L x L = "g L "x xl = 0.5 ( x S # 0.5) The procedure is identical and we obtain x S =0.66 and T 2 =1971K 12

13 Problem BaO is reacted with pure CO 2 to form BaCO 3. What is the maximum number of phases coexisting at equilibrium at an arbitrary temperature and pressure? 13

14 Answer The system has 2 independent components, at fixed P and T, the number of degrees of freedom is 2-P. Therefore, the maximum number of phases at fixed (and arbitrary) P and T is 2. Another way of understanding this is to realize that at an arbitrary T, the equilibrium dissociation pressure of BaCO 3 is fixed. The only way to have 3 phases coexisting is to have the pressure equal the dissociation pressure, which is not arbitrary. 14

15 Problem The Al-Zn fcc solid solution contains a miscibility gap below a critical temperature T C. The Gibbs free-energy of the solution is represented by the equation: G = x Al G 0 Al + x Zn G 0 Zn + RT x Al ln x Al (J/mol) Calculate T C and the associated composition x Zn 0. & ( 4000' [ ( ) + x Zn ln( x Zn )] + x Al x Zn ( 13180x Al x Zn )% 1" T # $ 15

16 Answer The critical temperature is defined by " 2 #g mix "x Zn 2 = 0 and " 3 #g mix "x Zn 3 =0. ( ) = RT[ x Al ln( x Al ) + x Zn ln( x Zn )] + x Al x Zn 13180x Al x Zn "g mix = G # x Al G 0 0 Al + x Zn G Zn ' ) 4000( ( )& 1# T The first derivative is: "#g + mix % x = RT ln Zn (. % - ' * 0 + 1$ T ( 2 ' *( $ 33474x "x Zn, & 1$ x Zn )/ & 4000) Zn x Zn ) The second derivative is: " 2 #g mix % 2 = RT 1 1 ( + ' + * + 1$ T. 2-0 ( $33474x "x Zn & x Zn 1$ x Zn ), 4000/ Zn x Zn ) = 0 which can be rearranged in: # 1 & ) RT% ( " T, 2 +.("33474x $ x Zn ( 1" x Zn )' * Zn x Zn ) = 0 or equivalently: 4000RT 4000 " T = ( " x Zn)x Zn ( 1" x Zn ) (1) The third derivative is: " 3 #g % mix 3 = RT $ 1 2 "x Zn x + 1 ( ' * &' Zn ( 1$ x Zn ) 2 )* + + 1$ T. - 0 ( 21342) = 0, 4000/ which can be rearranged as: 4000RT 4000 " T = 21342x 2 Zn( 1" x Zn ) 2 (2) 1" 2x Zn By equating (1) and (2), we finally obtain: 64026x 2 Zn "109672x Zn = 0 whose two roots are x Zn =1.315 and x Zn =0.398, which is obviously the physical one. Putting this value back into (1), we obtain T C =610.1K $ % 16

17 Problem 3 Note the attached binary phase diagram for the system Ga-As. 1- For a composition x As =0.8 that is heated to 1600K, list the phases present during each stage of slow cooling to room temperature, including the temperatures associated with any phase changes and the As compositions of phases present at such temperatures. 2- Sketch the variations of the activity of Ga as a function of composition at 1200K. Label your sketch fully and be quantitative when possible. 3- A stoichiometric GaAs single crystal is heated at 1200K in a chamber that contains P Ga =1.7x10-8 atm. Use the following information (equilibrium boiling point of Ga=2477K; vaporization enthalpy of Ga: 256kJ/mol) to determine whether Ga will be deposited on the wafer surface under these conditions. (Hint: compare P Ga in the chamber with the equilibrium pressure of Ga over GaAs. Recall that the equilibrium pressure of a component in an alloy is related to its activity) 17

18 Answer K>T>1335K homogeneous liquid phase present (x As =0.8) T=1335K: solidification of GaAs L->L +GaAs where L is richer in As than GaAs 1335K>T>1083K: As concentration in L increases up to x As =0.98. GaAs has constant composition T=1083K: eutectic solidification of the remaining L liquid: L ->GaAs+As 1083K>T>300K: GaAs and pure As are present in 2-phase equilibrium Ga will be deposited on the wafer if the P Ga in the chamber exceeds the equilibrium pressure of Ga over GaAs. The activity of Ga in GaAs is: GaAs a Ga = P Ga where P 0 0 Ga is the pressure of Ga over pure liquid Ga at 1200K. P Ga Using the Clausius Clapeyron equation (or equivalently the Turnbull approximation), we have 0 ln( P Ga ) = " #H $ vap,ga 1" T ' RT & % T ) hence P 0 Ga =1.8x10-6 atm. vap ( According to the plot obtained in question 2, the activity of Ga on the G-rich side of GaAs is ~ Therefore, P GaAs Ga ~1.66x10-6 atm. The equilibrium pressure of Ga over GaAs is orders of magnitude higher than the pressure of Ga in the chamber: Ga deposition will not occur. 18

Problem 3 Consider the attached Au-Sn phase diagram: What is the only compound that melts congruently? Find: 4 peritectics, 2 eutectics, 1 peritectoid and 1 eutectoid.

19 Problem 3 Consider the attached Au-Sn phase diagram: What is the only compound that melts congruently? Find: 4 peritectics, 2 eutectics, 1 peritectoid and 1 eutectoid. For each one write down the reaction that occurs when the triple point is reached from higher temperatures as well as the reaction temperature and composition of the phases involved. Which Au-Sn compound is the only one that can be cast directly from the melt? 19

20 Answer AuSn melts congruently. Only compositions of liquids and solid solutions will be stated. Peritectics (there are 4 C Au+L -> Au 3.6 Sn 0.4 compositions: Au is 5% Sn in 95% Au; L is 79% Au and 21% C Au 3.6 Sn 0.4 +L -> Au 0.85 Sn 0.15 compositions: L is 78% Au and 22% C AuSn+L ->AuSn 2 compositions: L is 29% Au and 71% C AuSn 2 +L -> AuSn 4 compositions: L is 12% Au and 88% Sn C L -> Au 0.85 Sn AuSn compositions: L is 71% Au and 29% Sn; Au 0.85 Sn 0.15 is 82% Au and 18% Sn (i.e. it s a solid solution of Au 0.85 Sn 0.15 and excess C L -> AuSn 4 + Sn compositions: L is 6% Au and 94% Sn C Au 0.85 Sn AuSn -> Au 5 Sn compositions: Au 0.85 Sn 0.15 is 86% Au and 14% Sn (it s nonstoichiometric with an excess of Au) C AuSn 4 -> Sn + AuSn 2 AuSn has a congruent melting point and therefore can be cast directly from the melt. 20

21 Problem Consider the attached phase diagram of the binary system Ni/Al. List all the eutectics, peritectics, eutectoids, peritectoids and congruent melting points. In your list, write the reaction that occurs upon cooling and the reaction temperature. 21

22 Answer Congruent: 1638 C: L->NiAl Eutectics: 640 C: L->Al+NiAl 3 ; 1385 C: L->Ni 3 Al+Ni Peritectics: 1133 C: L+NiAl->Ni 2 Al 3 ; 1395 C: L+NiAl->Ni 3 Al; 854 C: L+Ni 2 Al 3 ->NiAl 3 Peritectoid: 700 C: NiAl+Ni 3 Al->Ni 5 Al 3 22

23 Problem 4 Consider the following phase diagram of a binary system A/B. You can consider AB as a line compound. We will work at the T=1300K isotherm 1- First assume that all solutions (solid and liquid) are ideal. - plot the activity of A as a function of x B. Be quantitative wherever possible. (pick the equilibrium state of A at T=1300K as the reference state) - calculate the activity of A in AB when AB is in equilibrium with the A-rich liquid solution and with the B-rich liquid solution as well. - estimate the molar melting enthalpy of pure A Δh m at its melting point. Explain the approximations you make. 2- The terminal B-rich solution now obeys the dilute solution model for solvent and solute throughout the whole composition range at 1300K. Henry s coefficient for A is Furthermore the liquid solution in equilibrium with the terminal B-rich solid is modeled as a regular solution with a molar enthalpy of mixing of the form "h mix = #x A x B. - calculate α. - what s the activity of A in AB when AB is in equilibrium with the B-rich liquid solution (i.e. right side of the AB line compound)? 3- The liquid in equilibrium with the terminal A-rich solid is not an ideal solution either. Assume it s a regular solution and the Δg mix at x B =0.15 is -7 kj/mol. 23

24 Furthermore, the activity coefficient of B at x B =0.15 is equal to 0.4. What is the activity of A in AB when it s in equilibrium with the A-rich liquid solution (i.e. left side of the AB line compound)? 24

25 Answer 1 With the stated reference state and using all the information (i.e. that all solutions are ideal), we conclude that the a vs. x lines for the two liquid solutions are both on the diagonal, which fixes all points: As a result, the activity of A in AB on the A-rich side is 0.85 and that of A on the B-rich side is The Δh m of A is obtained from the intercept of the activity of A in the B-rich terminal solid solution with the y-axis at x B =0 (or x A =1). Due to the choice of reference states, the $ intercept is exp " #g m ' A & ). Since the solution is ideal, we conclude that: % RT ( % a A = x A " exp # $g m ( A ' *. And since a A =0.3 at x A =0.1, we get & RT ) $ "g m A = #RT ln a ' A & ) = #RT ln( 3) % x A ( Further, using the Turnbull approximation, since we are at T<T m of A, we can write: $ #RTT m ln a ' A & ) x "h m % A = A (. (T # T m ) Plugging in the numbers, you get Δh m ~143 kj/mol 25

26 2- The solution on the B-rich side is not ideal anymore but regular with Δh mix =αx A x B. Henry s coefficient gives us the activity of A at x A =0.1, which is equal to that of A at x A =0.3 (2-phase equilibrium): a A =K.a A (ideal)=0.25*0.3= At x A =0.3, A in the liquid phase follows the regular solution model with Δh mix =αx A x B, we have (see notes): $ a A = "x A = exp #x 2 ' B & ) x A therefore " = RT % RT ( x ln # a & A 2 % ( using a A =0.075 and x A =0.3, we obtain B $ x A ' α~-30.5kj/mol. $ At the other end of the B-rich solution composition, a A = "x A = exp #x 2 ' B & ) x A with % RT ( x A =0.45 and x B =0.55. Plugging in the numbers, we obtain a A ~0.2 (γ~0.43). 3- Now the A-rich solution is a regular solution but there is no indication that Δh mix is proportional to x A x B! On the other hand, "g XS mix = "g mix # RT( x A ln x A + x B ln x B ) = RT( x A ln$ A + x B ln$ B ) You have all the data to obtain γ A =0.9, from which you get a A =0.85*γ A =

MATSCI 204 Thermodynamics and Phase Equilibria Winter Chapter #4 Practice problems

MATSCI 204 Thermodynamics and Phase Equilibria Winter 2013 Chapter #4 Practice problems Problem: 1-Show that for any extensive property Ω of a binary system A-B: d ( "# ) "# B, = "# + 1$ x B 2- If "# has