MATSCI 204 THERMODYNAMICS AND PHASE EQUILIBRIA Winter 2016 Final Exam You get no credit for answers that are not clearly justified Exam strategy: the points given for every problem are related to how important the topic is not necessarily to how difficult the problem is so look at all the questions first to optimize your exam time! Please write neatly and clearly
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Problem 1 (20 points) 2 points each T/F answer. If F, 1 point for the F, 1 points for the correct explanation)
Answer the following questions with True or False and write a 1-line explanation of why you think the statement is false (you don’t need to write anything if the statement is true ). Write the explanation on this sheet and don’t forget to turn it in!
T/F
The energy of the Universe can never change
T T/F
The entropy of the Universe can only increase for any spontaneous process
T
T/F A bigger thermodynamic driving force for a transformation always translates into a faster process
F: a faster process relates to kinetics, which are not necessarily linked to thermodynamic driving forces
T/F
In a binary binary system system at an arbitrary arbitrary P, there there can never be more than 3 phases in equilibrium
T (NB: arbitrary P means that P is fixed at an arbitrary value so that’s a lost degree de gree of freedom)
T/F
Diffusion always occurs down a thermodynamic activity gradient
T ––––––––––––––––––––––––––––––––––––––––––––––––––––– –––––––––––––––––––––––––– –––––––––––––––––––––––––––––––––––––––––––––– –––––––––––––––––––
T/F
The decomposition pressure of an oxide MO depends only on the strength of the M-O bonds, i.e. the oxidation enthalpy
F: it also depends on the entropy of the reaction
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T/F
The chemical potential of a species depends on the reference state chosen
F: the chemical potential is a physical quantity that’s independent of the reference state, it’s measure requires a reference state
T/F
The activity of a species depends on the reference state chosen
T
T/F
Miscible species at constant (P,T) always display a negative enthalpy of mixing
F: they can have a positive enthalpy of mixing because the entropy of mixing is always positive
T/F
Miscible species at constant (P,T) always display a negative Gibbs free-energy of mixing
T ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
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Problem 2 (40 points-10 points each)
The Salleo solution is a special type of non-ideal binary solution. It is regular and exhibits a mixing enthalpy of the form: !hmix ! X 2 (1 " X ) . So that we all use the same symbols, we will =
assume that the elements forming the solution are A and B and the mole fraction in the equation above is that of B. 1. What is the entropy of mixing of the Salleo solution? 2. Find an equation relating the activity coefficients of the two components to composition. 3. We measure Henry’s coefficient of A at room temperature and it is equal to 1.8. What is the heat of mixing at X=0.5? 4. What is Henry’s coefficient at 500°C (assuming that the mixture still obeys the Salleo solution model)?
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Answer: 1. Regular solution=mixing enthalpy is ideal: !S mix = " R[ X ln( X ) + (1 " X )ln(1 " X )]
2. If the solution is regular, the only contribution to non-ideality is the mixing enthalpy. We have: RT ln(! A )
=
! H A, mix
.
The partial molar enthalpy of mixing is obtained from the !hmix using the tangent construction found in the notes. With our naming convention: ! H B, mix ( X ) = !hmix ( X ) + ! H A, mix ( X )
=
!hmix ( X ) "
d !hmix dX d !hmix
(1 " X ) X
dX
I obtain:
( ) RT ln (! ) RT ln ! B
A
2! X 1 ! X
=
! X
2
2
( ) ( 2 X ! 1)
=
3. Henry’s coefficient K for A is the limit of "A when X tends to 1. Hence:
K A
=
! ! $ & " RT %
exp #
and as a result we extract #=1.47 kJ/mol. At X=0.5, the heat of mixing is 183 J/mol. 4. In a regular solution, the heat of mixing cannot depend on T (p. 168 of the notes), therefore # is independent of T:
K A
=
! ! $ & with T=500°C, I get K A=1.26. " RT %
exp #
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Problem 3 (60 points)
Calvin, an eager PhD student in Stanford is itching to create a new startup company, and having just finished taking Thermodynamics, is now well versed in reading phase diagrams. When examining the binary phase diagram of Gallium and Arsenic, he notices that GaAs is the only compound that forms between the two elements, and has a very narrow region of stability. Calvin realizes that he may have a new way to grow GaAs crystals quickly and at low cost, the key to commercial success. Calvin realizes that the synthesis’ kinetics will be much more favorable if at least one of the precursors is in the liquid phase . Since Ga has a low melting point, Calvin decides to synthesize a large GaAs boule by taking a large block of arsenic, immersing it in an excess of molten gallium (to ensure that the entire block of arsenic is reacted), and waiting long enough for the system to reach equilibrium. a) What is the temperature range in which Calvin could have made his GaAs boules? (5 points)
Leaning on his newly made boule of GaAs, Calvin proudly tells his colleagues about his new method of making GaAs wafers when Hobbes walks in sipping from a can of soda, and asks Calvin what kind of electrical contacts he plans on using. Calvin proudly responds that he will use Al, since it is earth abundant, and relatively cheap to come by. Hobbes chuckles, putting his can down on the boule, and says, “I thought you just finished studying thermodynamics. Didn’t you learn anything about activity?” b) Using the Ga-As phase diagram below, plot the activity of gallium at 300K as a function of XGa. Make sure to state explicitly your reference state, as well as the value of activity and composition at every point where the activity changes. (10 points) c) What is activity of Ga in Calvin’s boule? Justify your answer to get credit.(5 points)
Hobbes asks Calvin to look-up the phase diagram of Ga-Al to try and determine what would happen if Al contacts were deposited on the boule. d) In the Ga-Al binary system, plot the activity of gallium at 300K as a function of X Ga. As before, make sure to explicitly state your reference state, as well as the value of activity and composition at every point where the activity changes. Justify all the assumptions you make. (10 points)
e) What is the activity of Ga in the aluminum contacts immediately upon deposition (i.e. before any reaction or equilibration)? What do you conclude should happen over time? (10 points)
“So, what’s the problem?” asks Calvin. Smiling, Hobbes replies, “I’ll show you why it matters.” Hobbes reaches out to pick up his soda that had been sitting on the boule, and as his fingers close
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around the can, the aluminum crumbles into pieces. “You see,” Hobbes explains, “Ga can attack Al and break apart its grain boundaries, totally destroying the contacts that you want to put down on your new solar cells.” Hobbes walks away smugly, while Calvin distraughtly cries out, “Hey, by leaching out Ga you decomposed part of my GaAs boule and now I will have toxic As in my lab!” f) Is Calvin right, did the boule decompose and he will be poisoned? (10 points) After Hobbes walks away, Calvin thinks for a while longer, and realizes that Hobbes’s argument might not be relevant for him. After all, he was planning on using his high quality wafers in a solar concentrator, where the temperatures are around 800K. Frantically, he looks around for a pad and paper to redo his calculations at a higher temperature. g) Recalculate the activity plots for the higher temperature case. Remember to include your chosen reference state, as well as the activity and composition values at the points where the activity changes. Does Calvin get a different answer at high temperature? (5 points) h) How could Calvin have changed the synthesis process to avoid having such reactive Ga in his GaAs? (5 points) Data: Assume all solutions are ideal Enthalpy of fusion of gallium is 5.6 kJ/mol Phase diagrams on the next page
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Answer: a) The synthesis Calvin envisions can be done anywhere from 303-1083 K b) This should be a step function: from x=0 to .5, it’s in equilibrium with solid arsenic, so the activity is 0, and from .5 to 1, it’s in equilibrium with solid gallium so the activity is 1.
c) The circled point is the activity of Gallium in the boule. Because the boule was formed from an excess of Gallium, you expect the boule to be slightly gallium richer, thus choosing the upper bound of the steep drop at the line compound. d) The Ga in Al line is a little more complicated because there is a Eutectic at 299.3 K, and the melting point of Ga is 303 K. I choose solid Ga as the reference because it makes my life easier. Thus, you expect a flat line until the transition to liquid at ~.98, then a drop along the liquid Ga line, (at a slope of exp(5600(3/300)/2494.2) until you hit the transition to two phases (~.97), where you get a flat line until ~.08. Then, the activity should drop to zero along a straight line.
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Zooming in near the XGa=1, you are crossing a pure Ga and a narrow Ga-L 2-phase coexistence because 300K is between the eutectic temperature and the melting point of Ga:
e) The activity of Ga in Al solid is zero, and is thus quite a bit lower than it is in the boule. The gallium should diffuse from the wafer into the aluminum until the activity of Ga in both the GaAs boule and the Al1-xGax phase is the same. f) No, Calvin is wrong. In order to get enough gallium to leave the wafer to begin forming pure As solid (which is what Calvin is presumably concerned about), the activity of gallium in GaAs would have to drop to zero, which would be thermodynamically impossible as the activity of the gallium in Al will be greater than zero for all compositions. g) Higher temperature will not affect the result very much. The reference chosen should be liquid Ga. There should be a barely perceptible drop on the GaAs activity just at x=1 because the liquid gallium phase is stable there for more than just a line compound. The GaAl diagram should drop linearly for the first 70%, then horizontally until around 6%. The slope of the activity in the Al1-xGax phase should be the same as it was in part e. This is checkable by comparing the activity at the point at which the sloped region begins.
h) Three options that I can think of: 1. Run the synthesis at a higher temperature, where you are able to have two liquids mixed with a seed crystal to ensure that you can carefully control the stoichiometry. In that case, you can just put more arsenic than gallium in the
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melt. It’s true that the arsenic’s activity will be higher, then, but then you won’t have to worry about the aluminum problem, at least. 2. Keep adding solid blocks of arsenic until the entire pool of gallium is used up. The last one that you add will not be fully converted to GaAs, and will likely be junk, but the others will not have an excess of gallium. 3. Start at the exact right composition (GaAs) in the liquid and freeze the congruent solid.
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"#$%&'( )
(60 points)
Consider the Pt-Si phase diagram below.
1. In your blue book, going from left to right, to the best of your knowledge list all eutectics, peritectics, eutectoids, peritectoids and congruent melting points. For each reaction write the phases involved and write the reaction that occurs as you cool the system. If according to this phase diagram, there are reaction that violate the phase rule, indicate them as well. (10 points) 2. Sketch the activity of Si as a function of Si content at T=1019°C and T=974°C (for simplicity assume that the activity is always proportional to XSi). (20 points) 3. Draw a sketch of the g vs. XSi curves at T=974°C across the whole compositional range. (10 points) 4. Sketch a heating curve (i.e. T vs. time when a constant heat flux is applied) at XSi=10 at. %. For simplicity, use only straight lines. Be as quantitative as possible about relative slopes and the temperatures where the slopes change. (20 points)
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Answer 1. I count 11 horizontal lines, which correspond to 3-phase equilibria or violations of the phase rule. This is a good way to ensure we don’t miss anything. I have:
Congruent melting points: 1769°C: Pt(l)->Pt(s) 1105°C: Pt2Si(l)-> Pt2Si(s)-HT 1232°C: PtSi(l)->PtSi(s) 1414°C: Si(l)->Si(s) Eutectics: 851°C: L->Pt3Si-HT+Pt 1008°C: L->Pt6Si5+Pt2Si 974°C: L->PtSi+Si Peritectics: 874°C: L+Pt12Si5->Pt3Si 1004°C: L+Pt2Si-HT->Pt12Si5 1019°C: L+PtSi->Pt6Si5 Eutectoids: none that can be distinguished from phase rule violations Peritectoids: 842°C: Pt+Pt3Si-HT->Pt25Si7-RT 870°C: Pt3Si-HT+Pt12Si5-HT->Pt5Si2 Violations (4 phases intersecting a line): 440°C, 770°C, 745°C Including the 4-phase intersecting points, I have 11 special temperatures. *+
At 1019°C we have:
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At 974°C, we have:
3. Here is a plausible set of g vs. X curves. Important aspects are the common tangents and ensure that the phase combinations shown in the phase diagram always correspond to the lower Gibbs free energy section in the g vs. X plot.
4. To obtain the heating curve, we follow the phase behavior along a vertical line at X Si=0.1
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• • •
•
•
•
From RT to 842°C, the slope is given by the inverse of the specific heat of the solids. To cross the peritectoid temperature, all Pt 25Si7 must react, this leads to a plateau. From 842°C to 851°C, Pt and Pt 3Si coexist, the slope is given by the inverse specific heat of the two solids (no reaction as they are both line compounds). At 851°C, there is a eutectic reaction: all Pt 3Si must disappear, which leads to another plateau. From 851°C and 1640°C, Pt and L coexist. The slope is shallower than the previous segments because to increase the temperature Pt and L must be heated and some Pt must melt too. The last solid is found at 1640°C, beyond which there is only liquid and the slope is steeper than the previous segment because no latent heat of melting is involved.
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Problem 5 (50 points)
ZrO2 is a very stable refractory oxide, which means that it has a very negative heat of formation (i.e. strong Zr-O bonds) and a high melting point. There are no other oxides in the Zr/ZrO 2 phase diagram. A slab of ZrO 2 is put in contact with a slab of Zr. The two materials are then sealed in an ampoule and heated until equilibrium is reached. At 1800K, the measured solubility of O 2 in the solid Zr equiliubrated with ZrO 2 is 15 parts-permillion. You can certainly consider it a dilute solution of O 2 in Zr. We are initially interested in the properties of this solid solution. 1. Using your intuition and the information above: do you expect Henry’s coefficient for O2 in Zr to be large or small? (10 points) 2. Calculate Henry’s coefficient at 1800K. (10 points) 3. The measurement is repeated at 1850K and it is found that Henry’s coefficient doubled. Calculate the molar heat of mixing of O 2 in Zr (i.e. the enthalpy of mixing 1 mole of O 2 in Zr, not that of making 1 mole of Zr/O2 solid solution). (10 points) 4. The slab of ZrO2 is replaced by a film epitaxially grown on an inert substrate that puts ZrO2 in a state of compression. For simplicity, the mechanical state is considered equivalent to a hydrostatic pressure of 200 MPa: a. Using your intuition, do you predict that the maximum solubility of O2 in Zr will increase or decrease and why? (5 points) b. Calculate the equlibrium solubility of O 2 in Zr under these new conditions (T=1800K). (10 points) 5. Explain in a short paragraph why chips of equilibrated Zr/ZrO2 are excellent oxygen getters when dropped in molten metal. (5 points) Data: At 298K: !h°F (ZrO2)=-1080 kJ/mol S°(ZrO2)=50.3 J/K.mol S°(Zr)=39.2 J/K.mol S°(O2)=205 J/K.mol Assume that !Cp for the formation reaction of ZrO 2 is ~0 VZrO2=2.16x10 -5 m3/mol (assume the compressibility is negligible)
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Answer 1. Considering that ZrO 2 is a very stable oxide, this means that Zr has a very strong affinity for O. As a result, I would expect that Henry’s coefficient would be very small.
2. To calculate Henry’s coefficient of O 2 in Zr, we need to calculate its activity and compare it to the expected activity for an ideal solution. Because the solid solution of Zr and O 2 is equilibrated with ZrO 2, the oxygen pressure over the solid solution must be the same as the decomposition pressure of ZrO 2. With the data given, we can calculate !h0298 and !S0298. I get !h0298=-1080 kJ/mol and !S0298=-194 J/K.mol. We need to calculate !g01800 in order to obtain the equilibrium oxygen pressure. If the !Cp of the formation reaction is ~0, we can assume that !h0T and !S0T are independent of temperature. As a result, !g0T= !h0298-T !S0298. For T=1800K, I get !g0T=-731 kJ/mol.
!P $ The decomposition pressure is given by: # O02 & " P %eq
=
! 'gT 0 $ -22 exp # & " which gives PO2~6.2x10 " RT %
atm. (I told you it was a very stable oxide!). Using pure oxygen at 1 atm as the reference -22 state, the activity of O2 in the oxide and in the metal is 6.2x10 . -6 If the solution of O2 in Zr was ideal, its activity would be equal to 15x10 using the same reference state of pure O2 at 1 atm (the pressure of O 2 at equilibrium would be equal to the mole fraction times the pressure of pure oxygen in its chosen reference state). As a -22 -6 -17 result, Henry’s coefficient K is 6.2x10 /15x10 =4x10 , a very small number indeed! 3. In the dilute solution limit, Henry’s coefficient is the activity coefficient. The dependence of the activity coefficient on temperature is given by:
( ) ! (1 / T ) ! ln !
" H mix =
R
# In the dilute
solution limit, the total molar enthalpy of mixing is only equal to the partial molar enthalpy of mixing of the solute because the solvent is ideal. Hence we obtain
( ) , which is the molar heat of mixing we are looking for. Using the data " (1 / T ) ! ln (! ) provided (" is Henry’s coefficient), we have =-384 kJ/mol. $ -46164 and ! H ! (1 / T ) RT ln ( K ) where K is Henry’s coefficient since a You can also use the equation ! H , ! H mix
=
" ln !
R
O2 , mix
O2 mix
=
dilute solution is also regular. You get a different result (my fault, I did not run this check when I set-up the problem) and you get full credit. 4. a- Applying pressure increases the chemical potential of ZrO2 solid and as a result I expect the partial pressure of oxygen (and therefore the equilibrium solubility of O2 in Zr) to increase. b- Hydrostatic pressure changes the activity of ZrO2 according to the following equation: 200 MPa
(
RT ln a ZrO
2
)
=
!
v ZrO dP ~ v ZrO 2
2
("P ) # Plugging in the numbers, I get aZrO2~1.33.
0.1 MPa
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" PO % $ ' #P & 2
0
Since !gT 0 RT ln =
a ZrO
, we get as a new equilibrium pressure
2
!P $ # & "P % O2
0
200 MPa
=
a ZrO
2
!P $ # & "P % O2
0
-22
~8.2x10 atm. Henry’s coefficient in Zr does not depend
0.1 MPa
on the pressure condition in ZrO2, therefore the same multiplication factor applies to the equilibrium solubility of O2 in Zr: XO2~20 ppm. 5. Because the activity of O2 is so low in these chips, every oxygen molecule that encounters them will be captures and used to transform Zr into ZrO2 and maintain the low oxygen activity mandated by the 2-phase equilibrium. They cease to work as getters when all the Zr is consumed and only ZrO2 is left because in this case there is no 2-phase equilibrium anymore.
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Problem 6 (50 points)
I’d like to protect Fe from oxidation at 1000°C with a sacrificial amount of Cr dissolved into it, so that Cr oxidizes first leaving Fe intact. Henry’s coefficient of Cr in Fe is 0.1. 1. What is the minimum mole fraction of Cr I need to dissolve in Fe? (10 points)
Use the Ellingham diagram to answer the following questions (briefly justify each one to get full credit): 2. 1 mole of Cr2O3 and 1 mole of Al2O3 are sealed in a closed, evacuated container at 1200°C? a. What is the equilibrium state of the system (name phases present and oxygen pressure)? (10 points) b. Explain why it is unphysical to not observe any Cr2O3 at equilibrium. (5 points) 3. What CO/CO2 ratio oxidizes Ni at 800°C? (10 points) 4. Will Ti reduce water to H2 at 500°C? Calculate the ratio of H2 to H2O pressures at equilibrium. (10 points) 5. What is the equilibrium pressure of oxygen at the melting point of Mn? (5 points)
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Answers %# Pure Cr is below all the Fe lines, which means it oxidizes more easily. Diluting it in Cr
reduces its activity and makes it less reactive. The minimum amount of Cr is the mole fraction that provides an activity such that the equilibrium pressure of Cr is equal to that of Fe. Any more Cr will oxidize preferentially compared to Fe. 0 For the Cr oxidation at 1000°C, !g =~-530 kJ/mol. 2/3
So we have for the Cr oxidation equilibrium:
K Cr
aCr O =
2
3
! P $ 4/3 # 0 &a "P % O2
=
! 530,000 $ & " R ( 1273 %
exp # '
Cr
We need to plug in the lowest O2 pressure that would oxidize Fe in order to find the Cr activity that would make Cr oxidize at that pressure rather than Fe. From the Ellingham -15 diagram, that’s oxidation to FeO, which occurs at PO2~10 atm. Assuming that the activity -6 of Cr 2O3 is 1 (pure oxide), I get a Cr =8.7x10 . Since aCr =K.XCr , We get XCr =87 ppm. &# The oxides will start decomposing in the vacuum. The decomposition pressure of Cr 2O3 is
higher than Al2O3 therefore the final state will be Al 2O3, Cr 2O3 in equilibrium with Cr at the -18 Cr 2O3 decomposition pressure (~10 atm.). The reason why the pressure is not fixed below the Cr 2O3 decomposition pressure is that for this happen, a ll the Cr 2O3 would have to decompose entirely: you can calculate the volume necessary to have 1.5 moles of O2 at -18 pressures below 10 atm… '# Reading from the scale in the Ellingham diagram, I have a CO/CO2 ratio of ~0.02#
4. If we look at the H2 to H2O ratio in equilibrium with Ti at 500°C, we get something between 12 13 10 and 10 , so yes, Ti certainly reduces H2O! 0 The ratio of pressures is obtained by calculating !g from the Ellingham diagram since the scale does not allow to read it exactly. 0 The reaction Ti+2H2O->TiO2+2H2 can be split as Ti oxidation ( !g 500=~-775 kJ/mol) and water splitting ((!g0500=+400 kJ/mol), so for the reaction above we have (!g0TOT=-375
! P H 2 $2 # 0& "P % kJ/mol). Using the equation 2 ! P H 2O $ # 0 & " P %
=
0 ! (gTOT $ 12 exp # ' & , we obtain PH2/PH2O~4.7x10 . " RT %
5 Reading from the diagram I get ~10-19 atm.
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