SECOND MIDTERM EXAM Chemistry April 2011 Professor Buhro

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1 SECOND MIDTERM EXAM Chemistry April 011 Professor Buhro Signature Print Name Clearly ID Number: Information. This is a closed-book exam; no books, notes, other students, other student exams, or any other resource materials may be consulted or examined during the exam period. Calculators are permitted. Partial credit will be given for partially correct reasoning in support of incorrect or correct final answers. Additional space for answers is provided at the end of this exam; please clearly label any answers you place there. Please find Potentially Useful Information attached as the last pages of this exam. 1. (15 pts). (0 pts) 3. (5 pts) 4. (0 pts) 5. (10 pts) 6. (10 pts) Total (100 pts) 1

2 1. 15 total pts. The compound having the nominal composition CeO has the fluorite crystal structure and exhibits good oxide-ion (O ) conductivity. This compound easily becomes oxygen deficient, and its formula is best described as CeO -x where x may have the values 0 < x < 0.5. Ionic conductivity data for CeO -x specimens having different compositions are plotted below. Please answer the questions below. For your reference the ionic radii are r(ce 3+ ) = 1.8 Å, r(ce 4+ ) = 1.11 Å, and r(o ) = 1.4 Å. The Pauling electronegativities are (Ce) = 1.1 and (O) = ln CeO 1.75 CeO 1.9 1/T (a) 05 pts. The ln vs. T 1 data for both compounds give linear plots. What conclusion may be drawn from the linearity in these plots? Please explain briefly. (b) 05 pts. Please identify the conductivity mechanism or mechanisms active in CeO -x and provide a brief justification for your answer. (c) 05 pts. The plots above differ greatly (quantitatively) from one another. Please explain the physical or chemical origin of this difference.

3 . 0 total pts. The following description of the alumina-silica (Al O 3 -SiO ) phase diagram dates back to the 1960s (it has since been revised). Al O 3 and SiO melt at 060 and 170 C, respectively. Only one compound forms from alumina and silica, having the composition 3Al O 3 SiO or Al 6 Si O 13. This compound is known as mullite and it melts congruently at 1850 C. Eutectic points occur at 5 mol % Al O 3 and 1595 C, and at 67 mol % Al O 3 and 1840 C. Please sketch this 1960s-era phase diagram in the space below, and label all regions. (The corrected alumina-silica phase diagram is found in F. J. Klug et al. J. Am. Ceram. Soc. 1987, 70, ) T (C) Al O 3 mol % SiO SiO 3

4 3. 5 total pts. Please consider the Ti Ni phase diagram shown below and provide the requested information: (a) 01 pt. Please identify each line compound on the diagram by writing its formula below. (b) 04 pts. Please identify each congruent melting point on the diagram by writing its temperature and the corresponding component composition below. (c) 05 pts. Please identify each phase that forms solid solutions by writing its (nominal) formula below. 4

5 (d) 03 pts. Please identify each eutectic point on the diagram by writing its temperature and the corresponding component composition below. (e) 03 pts. Please identify the peritectic point on the diagram by writing its temperature and the corresponding component composition below. Please indicate if this peritectic point is also an incongruent melting point. (f) 04 pts. Assume that a liquid having a component composition of Ti:Ni = 30:70 is cooled infinitely slowly from 1600 o C to 1000 o C. Please give the component composition, the phase composition, and the composition(s) of the phase(s) present in the specimen at 1000 o C. (g) 05 pts. Now assume that the same liquid (as in part f) having a component composition of Ti:Ni = 30:70 is cooled at a normal cooling rate from 1600 o C to 1000 o C. Please sketch the microstructure of the specimen at 1000 o C. Please be sure to clearly label all phases and microconstituents that are present. 5

6 4. 0 total pts. The Madelung constant A for the NaCl structure is Here you will calculate the Madelung constants for three small fragments of (clusters from) the NaCl structure in (a)- (c) below. Assume that the nearest cation-anion separations in these clusters are the same as in a NaCl crystal. Please show your work, and then answer the following questions. (a) 05 pts. Determine A for a cluster consisting of one anion and one cation: + A = (b) 05 pts. Determine A for a cluster consisting of two anions and two cations positioned at the vertices of a square: + + A = (c) 05 pts. Determine A for a cluster consisting of four anions and four cations positioned at the vertices of a cube: A = (d) 05 pts. Why is NaCl normally found as an extended, nonmolecular crystal lattice rather than as small molecular clusters like those shown above? Please explain briefly. 6

7 5. 10 total pts. The compound formed when graphite is fully intercalated with elemental K has the formula KC 8. The XRD pattern in the region of the 004 reflection ( = 16.56º) of KC 8 is shown below. KC 8 can be de-intercalated by heating, which expels and vaporizes the K. The XRD pattern undergoes changes as this process occurs, as described below. The experiment considered here was discontinued before all of the K was removed. Intensity (a.u.) Degrees -Theta (a) 03 pts. As the de-intercalation proceeded, the XRD peak at 16.46º gradually disappeared as another peak grew in to a maximum intensity at one of the positions indicated by the vertical lines. This second peak then gradually disappeared as another peak grew in to a maximum intensity at one of the other positions indicated. Eventually peaks had appeared at all the positions marked by the verticals. What process was responsible for this behavior? Please give your answer in a single sentence. (b) 03 pts. Which new peak was the first to appear and which the last to appear as deintercalation proceeded? Please identify these peaks by their values. (c) 04 pts. Please provide an appropriate stage label for each vertical in the space above it. 7

8 6. 10 total pts. A and B are metallic elements having fcc crystal structures and the lattice parameters a = 3.5 and 4.95 Å, respectively. A partial A-B phase diagram, for the temperature range below the solidus, is given below. Please determine the lattice parameter(s) of the phase or phases present at state point p. Be sure to show your work. T (C) p A mol % B B 8

9 Extra Work Space Please clearly label any work placed here for grading. 9

10 Potentially Useful Information Engel-Brewer rules: s, p electrons structure bcc hcp fcc 4 diamond (cubic) 1 Å = 10 8 cm = m Cu K radiation ( = 1.54 Å) n = dsin d hkl = a(h + k + l ) -1/ for cubic systems (a = b = c, = = = 90 o ) 1/ 4 l d hkl ( ) h k hk for hexagonal systems (a = b c, = = 90 o, = 10 o ) 3a c d hkl = (h /a + k /b + l /c ) -1/ for orthonormal (including tetragonal and orthorhombic) crystal systems ( = = = 90 o ); for tetragonal systems, a = b c for orthorhombic systems, a b c sin ( 1 )/sin ( ) = m 1 /m, where m = (h + k + l ) In an ideal hcp metal, in which all 1 nearest neighbors are equidistant, c/a = The area of a parallelogram is: b = base h = height area = b h h parallelepiped b The volume of a parallelepiped is: h base area = parallelogram area h = height parallelepiped volume = (base area) h 10

11 The volume of a sphere is: V sphere = r 3 Definitions of sine and cosine functions: sin = opposite side/hypotenuse (of a right triangle) cos = adjacent side/hypotenuse (of a right triangle) Lowest-angle reflections in XRD powder patterns: Primitive-cubic lattice 100 Primitive but non-cubic lattice 100, or 010, or 001 bcc lattice 110 fcc lattice 111 simple hcp-based structure 100 or 00 Reflections present (allowed by symmetry): primitive (P) all hkl may be present* body-centered (I) h + k + l = n (even)* face-centered (F) hkl are all odd or all even* A-centered (A) k + l = n* B-centered (B) h + l = n* C-centered (C) h + k = n* Systematic absences (extinctions): primitive (P) no absences required by lattice type* body-centered (I) h + k + l = odd* face-centered (F) 100, 110, 10, 11, (300, 1), 310, 30, 31, etc.* diamond (F) 100, 110, 00, 10, 11, (300, 1), 310,, 30, 31, etc. *Note well: additional absences may be present, depending on the space-group symmetry (rather than lattice symmetry) of specific, individual cases. radius ratio rules: r M /r X = CN of M = CN of M = CN of M = 8 Z 1Ze VCoul 4 0d U ( d) U 0 ANZ1Ze 4 d 0 ANZ 1Ze d NB n d 1 1 n Structure Madelung constant A NaCl 1.75 CsCl 1.76 Zinc blende Wurtzite CaF.5 Rutile (TiO ).41 CdI.36 Ion configuration Born exponent, n He 5 Ne 7 Ar, Cu + 9 Kr, Ag + 10 Xe, Au

12 e = C 4 0 = = C J -1 m -1 1 Å = 10 8 cm = m n e i ii 1 i E Aexp a RT Coordination-number ratio: For a compound M n X m, CN M /CN X = m/n d-spacings in intercalated crystals: d 00l = d 1 + (n 1)d 0 where: d 00l is the characteristic spacing, the translational period length, in the z direction (such as d 001 ) d 1 is the interlayer spacing in the stage-1 compound d 0 is the interlayer spacing in the empty (non-intercalated) host crystal n is the stage number 1

FIRST MIDTERM EXAM Chemistry March 2011 Professor Buhro

FIRST MIDTERM EXAM Chemistry March 2011 Professor Buhro FIRST MIDTERM EXAM Chemistry 465 1 March 2011 Professor Buhro Signature Print Name Clearly ID Number: Information. This is a closed-book exam; no books, notes, other students, other student exams, or any