FINAL EXAM KEY. Professor Buhro. ID Number:

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1 FINAL EXAM Chemistry 465 KEY 10 May 011 Professor Buhro KEY Signature KEY 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. Potentially useful information, formulas, values, etc., are provided on the last pages of this exam, which you may detach for convenience. THIRD MIDTERM COMPREHENSIVE FINAL 1. (0 pts) 6. (15 pts). (10 pts) 7. (15 pts) 3. (15 pts) 8. (15 pts) 4. (0 pts) 9. (5 pts) 5. (15 pts) 10. (10 pts) Subtotal (80 pts) Subtotal (80 pts) Grand Total (160 pts) Final Semester Grade 1

2 1. 0 pts. Imagine that an iron (Fe) rod was polished to a high gloss using a 3-µm diamond powder (polishing grit). Please calculate the approximate tensile strength of the rod. Please assume that the stiffness of Fe is 165 GPa, the specific surface energy of Fe is 1.5 J/m, the work of fracture is J/m, and the Fe-Fe bond distance is.48 Å = m. Please also assume that the lengths of the surface cracks are comparable to the grit size of the diamond powder employed. Calculate the tensile load required for crack propagation from surface cracks having lengths of 3 µm: WE σ= g 1 / 5Nm 9 N m m 11 N = = = 10 GPa 6 π ( 3 10 m) m - pts. for wrong value 1 / (8 pts.) This value is much larger than (0.10)E, which is a crude approximation of the theoretical cleavage stress σ c, therefore the Fe rod should exhibit its theoretical strength (4 pts): γ σ c = surf E d 1 / - pts. for wrong value or incorrect substitution Nm 9 N m m 10 = = Pa = 8.8 GPa m - pts. for wrong value 1 / (8 pts.)

3 . 10 total pts. This question concerns experiments with mica conducted by E. Orowan, described in the Gordon text. Mica is a silicate (Si-O) mineral having a layered crystal structure (shown schematically below) somewhat analogous to the layered structure of MoS, and to a lesser extent graphite. The two-dimensional silicate (Si-O) layers in mica are constructed of covalent-bond networks, but adjacent silicate layers are not covalently bonded to one another. Instead, adjacent silicate layers are held together by very weak electrostatic interactions with intervening K + cations. Consequently, mica has physical properties similar to those of graphite and MoS the layers slide easily and separate easily. Mica can be obtained in macroscopically large sheets. edge view of the covalently bonded K + K + silicate-network layers, which have K + K + a low density of negative charges; here the layer planes are perpendicular to the page Orowan measured the tensile strengths of macroscopic mica sheets having different widths, as shown below. In the diagrams below the silicate layers in the mica crystal structure are parallel to the macroscopic sheets, and to the plane of the page. narrower mica sheet clamps wider mica sheet Mica sheets narrower than the width of the clamps used in the tensile test exhibited tensile strengths of 170 MN/m. However, mica sheets wider than the width of the clamps used in the tensile test exhibited tensile strengths of 3,100 MN/m. (a) 05 pts. Please describe the geometry of crack formation in the narrower mica sheets; that is, where did cracks initiate and in which direction did they propagate? You may wish to show a sketch in support of your answer. Cracks initiate at the left or right edges (3 pts.) and propagate inward toward the center ( pts.) 3

4 . (b) 05 pts. Please explain the apparently greater tensile strength of the wider mica sheets. Your answer should identify the effect or effects responsible for the enhancement in tensile strength, and should account for the tolerance of high stresses on the front and back surfaces of the sheets. The left and right edges in the wider sheets are unstressed by the grips; therefore surface cracks on these edges do not propagate ( pts.), as they do not exceed the critical crack length. Propagation of cracks initiated upon the front and back surfaces are stopped by the weak-interface mechanism (3 pts.) up to high imposed stresses (3100 MN/m ) total pts. (a) 07 pts. Please briefly describe experimental evidence demonstrating that elastic strain in metals is due to the stretching (straining) of chemical bonds. As a test piece is macroscopically strained, the microscopic strain (elongation) occurring in the chemical bonds may be monitored by XRD (4 pts.). The amount of strain measured macroscopically and microscopically agree closely (3 pts.), indicating that elastic strain is due to the stretching (or compression) of bonds. (b) 08 pts. Please briefly describe experimental evidence demonstrating that plastic strain in metals is not entirely due to the stretching (straining) of chemical bonds. Please identify the other phenomenon or phenomena that contribute to plastic strain. After plastic strain the material does not return to its original dimensions, which it would do if only the stretching of bonds was occurring (4 pts.). The permanent deformation requires the breaking and remaking of bonds (0 pts.). The process is known as slip or shearing (4 pts. for either name). 4

5 4. 0 total pts. Griffith s calculation of theoretical cleavage stress (σ c ) assumes that strain energy may be exchanged for the surface energies of the newly generated cleavage surfaces. In turn, surface energy is related to the bond-dissociation energies (BDEs) of the bonds cleaved at the cleavage interface. Let us make the crude assumption that each bond cleaved generates dangling (broken) bonds on the cleavage surfaces. Therefore, please assume that the sum of the energies of all the bonds cleaved is exactly equal to the total surface energy of the new surfaces. (a) 10 pts. Use this assumption to calculate the specific surface energy (γ surf ) of iron (Fe) in J/m. Fe has a bcc crystal structure with a =.87 Å = m. The Fe-Fe BDE = 156 kj/(mol of bonds). Please assume that cleavage occurs on a (001) plane in the crystal and that four bonds are cleaved per Fe atom on the cleavage interface. Hints: Determine the areal density of Fe atoms on a (001) plane per m. Note that Avogadro s number is N A = Remember that cleavage generates two cleavage surfaces each having an area equal to the area of cleavage interface. surf 156 kj 1000 J 1 mol bonds 4 bonds 1 Fe 1 interface area m γ (kj / mol) = 3 mol bonds kj bonds Fe. surface area m = 6. 3 J/m 10 ( m) (10 pts. for correctly setting up the expression above. Minus pts. for each missing or incorrect factor. No points given for the correct numerical answer, as that is given in part b below.) (b) 04 pts. The specific surface energy (γ surf ) of Fe is actually about 1 J/m, or about one-sixth (1/6) of the value you determined above in part (a). Please briefly explain why the actual specific surface energy is smaller than the value you calculated according to the assumptions outlined above. The surface reconstructs by pairing up some of the dangling bonds, thus remaking some bonds and regaining some of the lost bonding energy (0 pts.). Consequently, there are fewer remaining dangling bonds than the assumption above predicts (4 pts.), and consequently the specific surface energy is less than predicted. 5

6 4. (c) 06 pts. The theoretical cleavage stress (σ c ) generally overestimates the actual tensile strengths of materials by factors of around 10. What incorrect assumption is made by this theoretical model that produces the large discrepancy between the theoretical and actual strengths? Please explain your answer in no more than a few sentences. The incorrect assumption is that applied stress is uniformly distributed over all the bonds (6 pts.), or that every bond experiences the same average stress (6 pts.). Or, the model fails to account for stress concentration (6 pts.), or for the unequal distribution of stress over the chemical bonds (6 pts.) total pts. The following questions concern Griffith s interest in the mechanical properties of glass. (a) 05 pts. Griffith seemed to presume that glass was full of microscopic cracks, but we now know that the important weakening defects are on the surface. Please briefly describe three experimental observations confirming that the cracks responsible for weakening glass are primarily on the surface. Glass or glass fiber is weakened by surface abrasion or surface contact. Glass or glass fiber is strengthened by chemical etching, which removes the surface (and surface damage). The number of surface cracks anti-correlates with the measured strength of the glass or glass fiber. ( pts. each for the first two correct responses, and 1 pt. for the third.) (b) 05 pts. Please briefly describe a strategy for visualizing the surface cracks on glass that was used prior to the emergence of electron microscopes. Surface cracks may be decorated by depositing metallic (Na) crystals upon them from the vapor phase (5 pts.). They may be visualized in an optical microscope after decoration. (c) 05 pts. How is it possible to obtain glass specimens in which the tensile strengths approach the theoretical cleavage stress of glass? Glass specimens drawn as progressively thinner fibers exhibit near-theoretical strengths. 6

7 6. 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 χ(ce) = 1.1 and χ(o) = ln σ CeO 1.75 CeO 1.9 1/T (a) 05 pts. The ln σ vs. T 1 data for both compounds give linear plots. What conclusion may be drawn from the linearity in these plots? Please explain briefly. Both plots are single, straight lines, indicating that a single conductivity mechanism operates in each case. (b) 05 pts. Please identify the conductivity mechanism or mechanisms active in CeO -x and provide a brief justification for your answer. The compound is has high ionic character with cations and anions of nearly equal size (1 pt.), and easily becomes oxygen deficient. Thus, anion or oxideion ( pts.) vacancy conduction ( pts.) operates. (c) 05 pts. The plots above differ greatly (quantitatively) from one another. Please explain the physical or chemical origin of this difference. CeO 1.75 has a higher oxide-vacancy concentration, and consequently shows higher ionic conductivity than does CeO

8 7. 15 total pts. (a) 5 pts. The unit cell for the rock-salt (NaCl) structure is shown below. The smaller cations are black and the larger anions are gray. Note that an anion is positioned at the origin of the unit cell. Please resketch the cell after moving a cation to the origin, using the frame on the lower right. -1 pt. each missing ion, up to a max. of - (b) 5 pts. The unit cell for the zinc blende (ZnS) structure is shown below. The smaller cations are black and the larger anions are gray. Note that an anion is positioned at the origin of the unit cell. Please resketch the cell after moving a cation to the origin, using the frame on the lower right. -1 pt. each missing ion, up to a max. of - Anions may occupy either T + or T sites (alternating octants must be occupied). 8

9 7. (c) 5 pts. Although many crystal structures have anti relatives, such as fluorite and antifluorite, other crystal structures do not. Please explain why there are no anti-rock-salt, anti-zinc-blende, or anti-cscl structures. The anti structures in these cases do not exist because the structures generated by exchanging the cation and anion positions are identical to the original structures total pts. Five common crystal structures exhibited by simple ionic compounds are represented by the unit cells below. Please identify these structures by writing the common name or prototypical chemical formula under each. NiAs wurtzite CsCl (3 pts each) Rutile (TiO ) 9 Fluorite (CaF ) or antifluorite (Na O)

10 9. 5 total pts. (a) 16 pts. The XRD pattern below was obtained from a metal possessing an hcp crystal structure. The reflections contained within the region shown are the 10, 100, 101, and 00. The 00 reflection is indexed as shown. Please index the remaining reflections, and calculate the values of the lattice parameters a and c. Be sure to clearly show your quantitative work. (λ = 1.54 Å) 101 Hint: for hexagonal systems, ( pts) d hkl 4 = 3a h + k + hk ( ) l + c 1/ SOLN: λ 1.54 Å d 100 = = =.15 Å sinθ sin (1.0 ) (4 pts) 1/ Å = = a 3a a =.48 Å (3 pts) d 00 = 1.54 Å sin (17.9 ) =.51 Å (4 pts).51 Å = c c = c = 5.0 Å (3 pts) 10

11 9. (b) 09 pts. The XRD patterns for three metallic specimens are shown below. These specimens exhibit common metallic crystal structures. Please write the name or abbreviation for the structure exhibited by each in the spaces provided. fcc or ccp (3 pts) hcp (3 pts) bcc (3 pts) 11

12 total pts. (a) 05 pts. Several mechanisms exist that dramatically increase the work of fracture (W) in materials. Some of these are naturally occurring, and some are artificially designed by humans. Please list five (5) such mechanisms in the spaces below by writing an identifying name or phrase. Do not provide extensive definitions. i. Dislocation generation and/or motion ii. Weak-interface mechanism iii. Helix elongation (in wood) iv. Crazing v. Fiber pull-out Fiber debonding Transformation toughening (1 pt. each for any combination of these answers) (b) 05 pts. Structures are not as strong as the materials they are made from for the same reason that materials are not as strong as the chemical bonds they are composed of. What is this reason? Please give your answer in exactly two words, in the blanks below. Stress Concentration 1

FINAL EXAM Chemistry May 2011 Professor Buhro. ID Number:

FINAL EXAM Chemistry May 2011 Professor Buhro. ID Number: FINAL EXAM Chemistry 465 10 May 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