Chapter 4 What is a binary phase diagram? Phase diagrams
Chapter 4 What is a binary phase diagram? Phase diagrams are diagrammatic representations of the phases present in a system under specified equilibrium conditions, most often composition, temperature and pressure. Binary systems contain two components, for example, Fe + C, NaNbO3 + LiNbO3, Pb + Sn and binary phase diagrams display the equilibrium phase relations for such systems. in the diagram. A single phase will be represented by a volume Phase boundaries form two-dimensional surfaces in the representation, and three phases will co-exist along a line. As most experiments are carried out at atmospheric pressure, a planar diagram, using temperature and composition as variables is usually sufficient. phase diagrams. These sections are called isobaric A point in such a binary phase diagram defines the temperature and composition of the system. A single phase will be represented by a planar area in the diagram, phase boundaries are represented by lines and three phases will coexist at a point. What is a eutectic point? A eutectic point in any system is characterised by the co-existence of three phases, one liquid and two solids that can only be in equilibrium at one temperature and composition, at a fixed pressure. point. The eutectic point is therefore an invariant The reaction that occurs on cooling through a eutectic point, the eutectic transformation: L solid + solid is an invariant reaction. The reaction is reversible, and on heating: solid + solid L The liquidus in the neighbourhood of a eutectic has a characteristic shape, meeting the solidus at the eutectic point. The eutectic composition, which is the composition at which the eutectic point is found, solidifies at the lowest temperature in the system, the eutectic temperature. What is the difference between carbon steel and cast iron? Plain carbon steels are alloys of carbon and iron only and contain less than about 2 wt % carbon although commercial steels rarely contain much more than 1.4 wt % C. They can be heated to give a homogeneous austenite solid solution. In this condition, they can readily be worked or formed. When the carbon content is greater than about 2 wt % and less than about 5 wt % the material cannot be heated to give a homogeneous solid solution and is called cast iron. At all temperatures below the eutectic temperature of 1148C the solid is a mixture of austenite and cementite or ferrite and cementite. The effect of this is that the materials are hard, brittle, and resist deformation. The material can be cast into the desired shape but not worked in the same way as steels. Commercial cast irons rarely contain much more than about 4.5 wt % C. Quick quiz 1. c; 2. b; 3. c; 4. b; 5. a; 6. a; 7. b; 8. c; 9. b; 10. c; 11. a; 12. b; 13. a; 14. b; 15. c; 16. c; 17. a; 18. b; 19. b; 20. c; 21. c; 22. c; 23. a; 24. b; 25. a; 26. a; 27. a; 28. b. Calculations and questions 1. 73.6 at %. 2. 73.5 g. 3. 64.3 at %. 4. 63.6 at % Sn; 36.4 at % Pb. 5. Ti2Al. 6. a. 33.2 at %; b. ~1322C; c. ~53 wt % Ni, ~55 at % Ni; d. ~1360C; e. ~74 wt % Ni, ~76 at % Ni. 7. a. solid + liquid; b. ~64 wt %; c. ~36 wt %. 8. a. 71 wt % liquid, 29 wt % solid; b. 73 vol % liquid, 27 vol % solid. 9. a. 40.1 wt %; b. ~74 mol % Cr2O3; c. ~18 % liquid; d. ~2215C; e. ~80 mol % Cr2O3. 10. a. ~2150C; b. ~59 mol % Cr2O3; c. ~2090C; D. ~12.5 mol % Cr2O3. 11. a. 30 mol % Cr2O3; b. no liquid; c. 100 % solid. 12. a. ~36 mol % Cr2O3; b. ~17 mol % Cr2O3; c. 30 % liquid, 70 % solid. 13. a. 3186C, Re, 2334C, Ru; b. 55.9 g Re, 44.1 g Ru; c. solid , Ru70Re30; d. liquid, Ru70Re30. 14. a. ~42 at % Ru;b. ~63 at % Ru; ~86 % liquid, ~14 % solid. 16. a 47.0 g BeO, 53.0 g Y2O3; b. ~46 %; c. BeO; d. ~54 %; e. ~37.5 mol % BeO. 17. a. BeO + Y2O3; b. BeO + Y2O3; c. 20 mol % ½ (Y2O3) : 80 mol % BeO. 18. a. 27.6 g Sn + 72.4 g Pb; b. liquid; c. 40 at % Sn : 60 at % Pb; d. 100 % liquid. 19. a. + liquid; b. ~ 22 at % Sn, liquid ~ 46 at % Sn; c. ~25 %, liquid ~ 75 %. 20. a. + (solid); b. , 9 at % Sn, ~99 at % Sn; c. ~65.5 %, ~ 34.5 %. 21. a. (austenite); 100% ; 1.5 wt % C. 22. a. (austenite) + cementite; b. ~91.2, cementite ~8.8 %; c. ~1 wt % C, cementite, 6.70 wt % C. 23. a. ~2.3 wt % C; b. ~1 % liquid; c. ~0.9 at % C; d. ~86 %. 24. a. WO2 + ZrO2 + W18O49; b. ~33.3 % ZrO2, ~20.5 % WO2, ~46.3 % W 18O49. 25. a. W 18O49 + ZrW2O8 + W 20O58; b. W 18O49 ~19 %, ZrW 2O8, ~30 %, W 20O58 ~51 %. Solutions 1 A copper – nickel alloy is made up with 28 grams of copper in 100 g alloy. What is the atom % Ni in the alloy? 28 g of copper contains (28 / molar mass) x NA atoms = (28 / 63.546) x NA atoms 72 g of nickel contains (72 / molar mass) x NA atoms = (72 / 58.693) x NA atoms atom % Ni = [(72 / 58.693) x NA ] / [(28 / 63.546) + (72 / 58.693)] x NA = (72 / 58.693) / [(28 / 63.546) + (72 / 58.693)] = 73.6 at % 2 A solid solution of aluminium oxide, (Al2O3), and chromium oxide, (Cr2O3), has a composition Al0.70Cr1.30O3. What mass of Cr2O3 needs to be weighed out to prepare 100 grams of sample? Al0.70Cr1.30O3 = 0.35 Al2O3 + 0.65 Cr2O3 We need 0.35 moles Al2O3 = 0.35 x molar mass Al2O3 g = 0.35 x 101.964 g = 35.69 g and 0.65 moles Cr2O3 = 0.35 x molar mass Al2O3 g = 0.65 x 151.992 g = 98.79 g The total mass = 134.45 g To prepare 100 g of mixture we need (35.69 / 134.45) x 100 = 26.54 g Al2O3 and (98.79 / 134.45) x 100 = 73.45 g Cr2O3 3 A copper – zinc alloy is made up with 35 g Cu in 100 g alloy (35 wt %). What is the atom % Zn in the alloy? 35 g of copper contains (35 / molar mass) x NA atoms = (28 / 63.546) x NA atoms 65 g of zinc contains (65 / molar mass) x NA atoms = (65 / 65.39) x NA atoms atom % Zn = (65 / 65.39) x NA ] / [(28 / 63.546) + ((65 / 65.39))] x NA = (72 / 58.693) / [(28 / 63.546) + (72 / 58.693)] = 64.3 at % 4 A solder contains 50 wt % Sn and 50 wt % Pb. What are the atom % of Sn and Pb in the solder? The solder contains 50 g tin and 50 g lead. 50 g of tin contains (50 / molar mass) x NA atoms = (50 / 118.71) x NA atoms 50 g of lead contains (50 / molar mass) x NA atoms = (50 / 207.2) x NA atoms The atom % tin = [(50 / 118.71) x 100] / [(50 / 118.71) + (50 / 207.2)] = 63.6 at % The atom % lead = [[(50 / 118.71) x 100] / [(50 / 118.71) + (50 / 207.2)] = 36.4 at % 5 An intermetallic compound in the Ti – Al system is found at 78 wt % Ti and 22 wt % Al. What is the approximate formula of the compound? The intermetallic compound contains 78 g Ti and 22 g Al. The number of moles of Ti = 78 / molar mass = 78 / 47.88 = 1.629 moles The number of moles of Al = 22 / molar mass = 22 / 26.98 = 0.815 moles The composition = Ti1.629Al0.815 = Ti2Al 6 An equilibrium sample of a copper – nickel alloy with a composition of 65 wt % Ni is prepared. With reference to the Cu – Ni phase diagram (Figures 4.6 and 4.8): a. What is the atom % of copper present in the alloy? b. On heating the alloy from room temperature, at what temperature does liquid first appear? c. What is the composition of the liquid? d. At what temperature does the solid disappear? e. What is the composition of the final solid? Note: use a large photocopy of the figure; results will be approximate. (a) 65 wt % Ni = 65 g Ni + 35 g Cu 65 g of Ni contains (65 / molar mass) x NA atoms = (65 / 58.693) x NA atoms 35 g of Cu contains (35 / molar mass) x NA atoms = (35 / 63.546) x NA atoms atom % Cu = [(35 / 63.546) x NA ] / (35 / 63.546) + (65 / 58.693)] x NA = [(35 / 63.546) x 100 ] / [(35 / 63.546) + (65 / 58.693)] = 33.2 at % (b) 1322C (c) 53 wt % Ni; 55 at % Ni (d) 1360C (e) 74 wt % Ni; 76 at % Ni 7 For the sample in the previous question, the alloy is held at a temperature of 1340C. a. What phase(s) are present? b. How much solid is present, if any? c. How much liquid is present, if any? (a) Solid + liquid (b) Using the lever rule and measuring distances instead of compositions on an enlarged photocopy, as the scale is linear: % solid 9/14 x 100 % 64% (c) Using the lever rule and measuring distances on an enlarged photocopy instead of compositions, as the scale is linear: % solid 5/14 x 100 % 36% 8 With reference to the Cu – Ni phase diagram (Figures 4.6 and 4.8), an equilibrium sample of a copper – nickel alloy with a composition of 37 wt % Ni is held at a temperature of 1250 C. a. What are the amounts of solid and liquid present? b. If the density of the solid is 8.96 x 10 3 kg m-3, and the density of the liquid is 90% of that of the solid, calculate the volume % of the solid and liquid present. (a) Using the enlarged figure and the lever rule: % liquid [(47 – 37) / 47 – 33)] x 100 71% liquid % solid [(37 – 33) / 47 – 33)] x 100 29% liquid (b) volume = weight / density volume of liquid = [71 / (0.9 x 8.96 x 103)] = 0.0088 m3 volume of solid = [29 / (8.96 x 103)] = 0.00324 m3 volume % liquid = 0.0088 x 100 / (0.0088 + 0.00324) = 73 % volume % solid = 0.00324 x 100 / (0.0088 + 0.00324) = 27 % 9 An equilibrium sample of composition 50 mole % Cr2O3 is prepared using the phase diagram of the Al2O3 - Cr2O3, system (Figure 4.21). a. What is the weight % of Al2O3 present? b. The sample is held at 2200 C. What is the composition of the solid phase present? c. How much liquid phase is present? d. At what temperature will the last of the solid disappear? e. What will the composition of this solid be? Note: use a large photocopy of the figure; results will be approximate. (a) 50 mol % Cr2O3 contains 1 mole of Al2O3 and 1 mole of Cr2O3. The weight % = molar mass Al2O3 x 100 / molar mass Al2O3 + molar mass Cr2O3 = (101.964 x 100) / (101.964 + 151.992) = 40.1 wt % (b) 74 mol % Cr2O3 (c) amount of liquid (lever rule) 5.4 / 30 = 18 % (d) the last solid disappears at 2215C (e) the last solid has a composition 80 mol % Cr2O3 10 With reference to (Figure 4.21), a sample of composition 30 mole % Cr2O3 is held at 2300C and then slowly cooled. a. At what temperature does solid first appear? b. What is the composition of the solid. c. At what temperature does the liquid finally disappear? d. What is the composition of the last drop of liquid? Note: use a large photocopy of the figure; results will be approximate. (a) 2150C (b) 59 mol % Cr2O3 (c) 2090C (d) 12.5 mol % Cr2O3 11 With reference to (Figure 4.21), a sample of composition 30 mole % Cr2O3 is held at 2080C. a. What is the composition of any solid phase present? b. What is the composition of any liquid phase present? c. How much of each phase is present? (a) The composition of the solid is 30 mol % Cr2O3 (b) No liquid present (c) 100% solid is present 12 With reference to (Figure 4.21), a sample of composition 30 mole % Cr2O3 is held at 2100C. a. What is the composition of any solid phase present? b. What is the composition of any liquid phase present? c. How much of each phase is present? Note: use a large photocopy of the figure; results will be approximate. (a) The composition of the solid is 36 mol % Cr2O3 (b) The composition of the liquid is 17 mol % Cr2O3 (c) The amount of liquid (lever rule) is 6 / 20 = 30 % The amount of solid (lever rule) is 14 / 20 = 70 % 13 The phase diagram of the rhuthenium, (Ru), rhenium (Re), system is given in Figure 4.22. a. What are the melting points of pure Re and pure Ru? b. A sample of composition 70 at % Ru is made up. What weights have to be added to prepare 100 grams of sample? c. This alloy is held at a temperature of 2200C. What phases are present and what are their compositions? d. The alloy is held at 3000C. What phases are present and what are their compositions? Note: use a large photocopy of the figure; results will be approximate. (a) Re: 3186C; Ru: 2334C (b) The composition of the alloy is Ru0.7Re0.3. Following the method in Q2: mass of Ru = 0.7 x molar mass Ru = 0.7 x 101.07 = 70.749 g mass of Re = 0.7 x molar mass Re = 0.3 x 186.21 = 55.863 g For 100 g of alloy we need 70.749 x 100 / [70.479 + 55.863] = 55.9 g (c) At 2200C there is only solid , composition Ru0.7Re0.3. (d) At 3000C there is only liquid, composition Ru0.7Re0.3. 14 With reference to (Figure 4.22), a sample of composition 60 at % Ru is made up and held at 2700C. a. What is the composition of any solid phase present? b. What is the composition of any liquid phase present? c. How much of each phase is present? Note: use a large photocopy of the figure; results will be approximate. (a) The composition of the solid is 42 at % Ru (b) The composition of the liquid is 63 at % Ru (c) The amount of liquid (lever rule) is 18 / 21 = 86 % The amount of solid (lever rule) is 3 / 21 = 14 % 15 The phase diagram of the BeO – Y2O3 system is given in Figure 4.23. Explain why this figure differs from that of the lead – tin system, Figure 4.9. Why does the composition axis us ½ [Y2O3] instead of Y2O3? BeO and Y2O3 are oxides with totally different chemistry. The atomic sizes of the two metal atoms are also quite different. This means that: No solid solutions are expected The vast difference in the chemical properties results in a deep eutectic, as a liquid containing the two species will find it more difficult to crystallize easily into two phases at the same time. The composition axis is chosen to keep he number of metal atoms constant at 1 across the composition range. 16 With respect to Figure 4.23, a composition is made up with 80 mol % BeO and held at 2000C. a. What weights of the components are needed to make 100 g of sample? b. How much solid is present? c. What is the composition of the solid? d. How much liquid is present? e. What is the composition of the liquid? Note: use a large photocopy of the figure; results will be approximate. (a) Following the method in Q2: mass of BeO = 0.8 x molar mass BeO = 0.8 x 25.01 = 20.008 g mass of Y2O3 = 0.2 x ½ molar mass Y2O3 = 0.1 x 225.812 = 22.581 g For 100 g of alloy we need 20.008 x 100 / [20.008 + 22.581] = 47 g BeO and 22.581 x 100 / [20.008 + 22.581] = 53 g Y2O3 (b) The amount of solid (lever rule) is 17.5 / 38 = 46% (c) BeO (d) The amount of liquid (lever rule) is 20.5 / 38 = 54% (e) 37.5 mol % BeO 17 With respect to the sample in the previous question, the material is cooled to 1400C. a. What phases are present? b. What are the compositions of the phases? c. What are the proportions of each phase present? Note: use a large photocopy of the figure; results will be approximate. (a) BeO + Y2O3 (b) BeO and Y2O3 (c) 80 mol % BeO; 20 mol % ½ Y2O3 i.e 80 mol % BeO; 10 mol % Y2O3 18 With respect to the Pb – Sn phase diagram (Figures 4.9 – 4.12), a sample is made up with 40 at % Sn. a. What weights of lead and tin are needed to make 100 g solid? b. What phase(s) are present when the sample is held at 300C? c. What are the compositions of the phase(s)? d. How much of each phase is present at 300C? Note: use a large photocopy of the figure; results will be approximate. (a) Following the method in Q2: mass of Sn = 0.4 x molar mass Sn = 0.4 x 118.71 = 47.484 g mass of Pb = 0.6 x molar mass Pb = 0.6 x 207.2 = 124.32 g For 100 g of alloy we need 47.484 x 100 / [47.484 + 124.32] = 27.6 g Sn and 124.32 x 100 / [47.484 + 124.32] = 72.4 g Pb (b) Only liquid (c) Sn4Pb6 (d) 100% liquid 19 With respect to the Pb – Sn phase diagram (Figures 4.9 – 4.12), a sample made up with 40 at % Sn is cooled slowly to 250C. a. What phases are present at 250C b. What is the composition of each phase? c. How much of each phase is present? Note: use a large photocopy of the figure; results will be approximate. (a) solid + liquid (L) (b) 22at% Sn; L 46 at % Sn (tie-lines) (c) The amount of (lever rule) is 46 - 40 / 46 -22 = 25 % The amount of liquid (lever rule) is 40 - 22 / 46 – 22 = 75 % Note: at a glance these figures suggest that the total number of atoms of Sn is 22/100 + 46/100 = 68/200 = 34 at %. this is clearly wrong as the alloy is made up to be 40 at %. the correct figure is given by: 25% of the sample has a composition of 22 at % Sn 75 % of the sample has a composition of 46 at % Sn The total content is 0.22 x 0.25 + 0.75 x 46 = 40 at % Sn 20 With respect to the Pb – Sn phase diagram (Figures 4.9 – 4.12), the sample made up with 40 at % Sn is cooled further to 100C. a. What phases are present at 100C? b. What is the composition of each phase? c. How much of each phase is present? Note: use a large photocopy of the figure; results will be approximate. (a) solid + solid (b) 9 at % Sn; 99 at % Sn (tie-line) (c) The amount of (lever rule) is 99 – 40 / 99 – 9 = 65.5 % The amount of (lever rule) is 40 - 9 / 99 – 9 = 34.5 % 21 With respect to the Fe – C phase diagram (Figure 4.15), an alloy with a composition of 1.5 wt % C is homogenised by heating for a long period at 1000C. a. What phase(s) are present? b. How much of each phase is present? c. What are the compositions of the phase(s)? Note: use a large photocopy of the figure; results will be approximate. (a) Austenite () (b) 100 % (c) 1.5 wt % C 22 With respect to the Fe – C phase diagram (Figures 4.15), an alloy with a composition of 1.5 wt % C is homogenised by heating for a long period at 1000C, is subsequently cooled slowly to 800C. a. What phase(s) are present? b. How much of each phase is present c. What are the compositions of the phase(s)? Note: use a large photocopy of the figure; results will be approximate. (a) Austenite () and cementite (Fe3C). (b) The amount of (lever rule) is 6.7 – 1.5 / 6.7 – 1 = 91.2 % The amount of Fe3C (lever rule) is 1.5 - 1 / 6.7 – 1 = 8.8 % (c) 1 wt % C (tie-line); Fe3C = 6.70 (constant) 23 With respect to the Fe – C phase diagram (Figures 4.15), an alloy with a composition of 5 at % C is homogenised by heating for a long period at 1350C. a. What is the composition of the liquid phase present? b. How much of the liquid phase is present? c. What is the composition of the solid phase present? d. How much solid phase is present? Note: use a large photocopy of the figure; results will be approximate. To use the diagram convert 5 at % C to wt %. The overall composition is given by Fe95C5. To obtain the weights follow the method in Q2: mass of C = 5 x molar mass C = 5 x 12.01 = 60.05 g mass of Fe = 95 x molar mass Fe = 95 x 55.847 = 5305.465 g wt % C = 60.05 x 100 / (60.05 + 5305.465) = 1.1 wt % (a) 2.3 wt % (b) The amount of liquid phase (lever rule) is 1.1 – 0.9 / 2.3 – 0.9 = 14 % (c) solid 0.9 wt % C (tie-line) (d) The amount of solid phase (lever rule) is 2.3 – 1.1 / 2.3 – 0.9 = 86 % 24 With respect to the phase diagram of the WO3 – WO2 – ZrO2 system (Figure 4.17), a sample is made up of an equimolar mixture of WO3, WO2 and ZrO2, (1:1:1) and heated at 1100C to equilibrium. a. What phases are present? b. How much of each phase is present? Note: use a large photocopy of the figure; results will be approximate. (a) The 1:1:1 composition is at the centre of the triangle, marked P. The phases are WO2, ZrO2, W 18O49. (b) amount of ZrO2 = aP / aA 29 / 87 = 33.3 % amount of WO2 = bP / bB 15 / 73 = 20.5 % amount of W 18O49 = cP / cB 31.5 / 68 = 46.3 % Note: these figures do not total 100 % but are within the accuracy of the measurements on the figure. 25 With respect to the phase diagram of the WO3 – WO2 – ZrO2 system (Figure 4.17), a sample is made up of a mixture of 80 mol % WO3, 10 mol % WO2 and 10 mol % ZrO2, (8:1:1) and heated at 1100C to equilibrium. a. What phases are present? b. How much of each phase is present? Note: use a large photocopy of the figure; results will be approximate. (a) From the figure, the 80 mol % WO3, 10 mol % WO2 and 10 mol % ZrO2, (8:1:1) composition is located at P. The phases present are W 18O49, ZrW 2O8, W 20O58. (b) amount of ZrW 2O8= aP / aA 8.5 / 28.5 = 30 % amount of W 20O58 = bP / bB 10 / 20 = 50 % amount of W 18O49 = cP / cB 3.5 / 18.5 = 19 % Note: these figures do not total 100 % but are within the accuracy of the measurements on the figure. 26. According to the SiO2 – MgO – Al2O3 phase diagram (Figure 4.20), what phases may be present in steatite ceramics and cordierite ceramics? The composition triangles on the phase diagram show that steatite lies almost completely in the MgSiO3 – SiO2 – Mg2Al4Si5O18 triangle, so steatites will consist mainly of enstatite, silica and cordierite. The cordierite phase falls into three different triangles and cordierites will contain mixtures of phases (i) silica, mullite and cordierite, (ii) silica, enstatire and cordierite or (iii) enstatite, mullite and cordierite.
© Copyright 2024