thelifecurve.com KEY Last Name First Name Lab Sec. # ; TA: ; Lab day/time: Dr. Toupadakis Spring 2013 CHEMISTRY 2C Section B Instructions: FINAL EXAM CLOSED BOOK EXAM! No books, notes, or additional scrap paper are permitted. All information required is contained on the exam. Place all work in the space provided. If you require additional space, use the back of the exam. A scientific calculator may be used (if it is a programmable calculator, its memory must be cleared before the exam). This exam has 15 pages total and 28 problems. (1) Read each question carefully. (2) For Part I, and III there is no partial credit. There is partial credit for part II and IV. (3) The last three pages contain a periodic table and some useful information. You may remove them for easy access. (4) If you finish early, RECHECK YOUR ANSWERS! U.C. Davis is an Honor Institution Possible Points Earned Points # 01 16 (03 points each) / 48 # 17 (10 points) / 10 # 1819 (06 points each) / 12 # 20 (08 points total) / 08 # 21 (06 points total) / 06 # 22 (04 points total) / 04 # 23 (12 points total) / 12 # 24 (08 points total) / 08 # 25 (10 points total) / 10 # 26 (10 points total) / 10 # 27 (10 points total) / 10 # 28 (14 points total) / 14 Total Score /152 or % Circle one 01.a b c d e 02.a b c d e 03. a b c d e 04.a b c d e 05.a b c d e 06. a b c d e 07. a b c d e 08. a b c d e 09. a b c d e 10. a b c d e 11. a b c d e 12. a b c d e 13.a b c d e 14. a b c d e 15. a b c d e 16. a b c d e
TOUPADAKIS FINAL Spring 2013 CHEM 2C 2 of 15 Part I: Concepts (3 points each) No partial credit is available 1. The following reaction: is an example of: (a) An addition/hydrohalogenation reaction. (b) A substitution. (c) An isomerization reaction. (d) A condensation reaction. (e) An addition/hydrogenation reaction. 2. Choose the incorrect statement. Radioactive decay is: a) a firstorder process. b) a process whose rate depends on the chemical environment of the decaying nucleus. c) a process whose rate is pressure and temperature independent. d) a process with a low activation energy. e) a process whose rate is not affected by a change in temperature. 3. The hybridization of the terminal carbon atoms and the total number of σ bonds, π bonds and nonbonding pairs of electrons for the line structure below will be: O O (a) sp 3 6 σ 1π 3 nonbonding electron pairs (b) sp 2 11 σ 2π 5 nonbonding electron pairs (c) sp 24 σ 3π 5 nonbonding electron pairs (d) sp 3 16 σ 1π 4 nonbonding electron pairs (e) sp 13 σ 2π 4 nonbonding electron pairs 4. When gold metal, Au is oxidized with a 3:1 V/V mixture of concentrated hydrochloric acid and concentrated nitric acid (a) the nitrate ions act as the oxidizing agent and the chloride ions provide the driving force for the completion of the reaction. (b) hydronium ions act as the oxidizing agent (c) hydronium ions act as the reducing agent (d) the chloride ions act as the oxidizing agent and the nitrate ions provide the driving force for the completion of the reaction. (e) none of the above
TOUPADAKIS FINAL Spring 2013 CHEM 2C 3 of 15 5. How many unpaired electrons are there in [MnCl 6 ] 4? Note that Cl is a weak field ligand. (a) 0 (b) 1 (c) 2 (d) 4 (e) 5 6. The cation, Zn 2+ in water is colorless because: (a) Zn is a transition metal (b) Zn is a main group metal (c) Zn 2+ has a d 10 electron configuration (d) Zn 2+ has no d electrons (e) None of the above 7. The correct IUPAC name for Na 3 [Fe(CN) 5 CO] is: a) sodium pentacyanocarbonyliron(iii) b) pentacyanocarbonylferrate sodium (III) c) sodium carbonylpentacyanoferrate(ii) d) sodium pentacyanocarbonylferrate(ii) e) sodium carbonylpentacyanoferrate(iii) 8. What is the coordination number of Co in tris(ethylenediamine)cobalt(iii) ion? (a) 1 (b) 2 (c) 3 (d) 6 (e) 9 9. When a nucleus of an atom undergoes electron capture: a) the atomic number decreases by 1. b) the number of neutrons decreases by 1. c) the number of protons increases by 1. d) the atomic weight increases by 1. e) None of the above. 10. Choose which properties of the following molecule are correct: (a) It has an R configuration (b) It has an S configuration CH 2 CH 3 (c) Is not optically active (d) Has a plane of symmetry C CH 3 (e) None of the above H CH(CH 3 ) 2
TOUPADAKIS FINAL Spring 2013 CHEM 2C 4 of 15 11. Determine the number of chiral centers in the molecule below. (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 12. What type of emission is observed for the decay of C14 to N14? (a) Alpha (b) Beta (c) Gamma (d) Positron (e) Neutron 13. You are told that the potential energy diagram for a reaction mechanism has three peaks. Which of the following CANNOT be determined from the information above? (a) The number of elementary reactions involved in the mechanism. (b) The number of intermediates. (c) The number of transition states. (d) The ratedetermining step of the mechanism. (e) All of the above must be true. 14. How many moles of electrons are transferred per mole of oxidizing agent in the electrochemical cell below? (a) 5 (b) 4 (c) 3 (d) 2 (e) 1 Ag(s) Ag + (aq) MnO 4 (aq), Mn 2+ (aq) C(s) 15. Which of the following oxides is the MOST basic? (a) Al 2 O 3 (b) P 4 O 10 (c) CaO (d) CO 2 (e) BeO 16. Choose the INCORRECT statement: (a) Activated complexes cannot be isolated. (b) Intermediates are sometimes seen in the overall chemical equation. (c) The ratedetermining step always has the largest activation energy. (d) Most reactions occur in a multistep mechanism. (e) Catalysts speed up a reaction by decreasing the activation energy for a reaction.
TOUPADAKIS FINAL Spring 2013 CHEM 2C 5 of 15 Part II: Short Calculations 17. (10 points) Explain, (show with numbers) that pure iron solid does not have the tendency to be oxidized to Fe 2+ (aq) by pure water. Why then do we need to paint iron structures? Explain by considering oxygen as the oxidizing agent. In deed, iron solid does not have the tendency to be oxidized by water as the E o cell value shows. Fe(s) > Fe 2+ (aq) + 2e + 0.44 V 2H 2 O(l) + 2e > H 2 (g) + 2OH (aq) 0.83 V E o cell = 0.39 V. Thus the oxidation of iron is not thermodynamically favored. We need to paint iron structures because in natural water there is oxygen dissolved. Iron is oxidized by oxygen in the presence of water. Without water, pure oxygen cannot oxidize iron. Fe(s) > Fe 2+ (aq) + 2e + 0.44 V O 2 (g) + 4H + (aq) + 4e > 2 H 2 O(l) +1.23 V E o cell = + 1.67 V the oxidation of iron is thermodynamically favored. 18. (6 points) What is the change in mass (grams) when 2 moles of hydrogen atoms combine to form 1 mole of hydrogen molecules? 2H > H 2 ΔE = 436 kj Mass defect = Δm = ΔE/c 2 = [( 436 kj)(10 3 J/kJ)(1 kg m 2 s 2 /1 J)] /(3.00 x 10 8 m/s) 2 = 4.84 x 10 12 kg = 4.84x10 9 g.
TOUPADAKIS FINAL Spring 2013 CHEM 2C 6 of 15 19. (6 points) How much time, in minutes, is required to electroplate 4.4 mg of zinc from a zinc nitrate solution using a current of 0.50 A? ch arge( C) = current ( A) xtime( s) t = (0.0044 g Zn)(1 mol Zn/65.38 g Zn)(2 moles of e /1 mol Zn) *(96485 C/mol e )/0.50 A t = 25.97 s (1 min/60 s) = 0.43 minutes
TOUPADAKIS FINAL Spring 2013 CHEM 2C 7 of 15 Part III. Short Answer Fill in the blanks with the appropriate answers 20. [8 points possibleeach is worth 2 pts] Identify the following species as chiral or achiral. Also, designate them as, fac, mer, cis, or trans. Species Chiral (optically active) or Achiral (optically inactive)? Designation Achiral Cis Achiral trans Achiral fac Achiral trans 21. [6 points possibleeach is worth 3 pts] Draw the following: trans1,3dimethylcyclopentane cistetraaquadichlorochromium(iii) ion OH 2 Cl OH 2 Cr Cl OH 2 OH 2 +
TOUPADAKIS FINAL Spring 2013 CHEM 2C 8 of 15 22. [04 points possibleeach is worth 1 pt] Identify the correct functional group with the molecule listed. Molecule Functional Group Alcohol Aldehyde Carboxylic Acid Amide 23. (12 ptseach is worth 2pts) Consider the following molecule: a) How many chiral centers does it have? 2 b) How many different types of functional groups? Do not count the groups CH, CC, and the ring as functional groups. 3 c) How many functional groups total? Do not count the groups CH, CC, and the ring as functional groups. 4 d) How many sp 2 carbons? 2 e) How many secondary alcohol groups? 1 f) What is the formal charge of the carbon atom that the arrow is pointing to? 0
TOUPADAKIS FINAL Spring 2013 CHEM 2C 9 of 15 24. [8 points] Fill in the missing species below: 40 19 K + β 40 18 Ar 14 7 N + 1 0n 1 1H + 14 6C Part IV. Long Answer Calculations Show your work for partial credit. Put answers in the spaces provided. 25. (10 points) Calculate the fraction of the remaining nuclei after 100 halflife times. (Show your work) The fraction (f n ) of the remaining nuclei after n halflife times is (0.5) n Therefore: f 100 = 0.5 100 = 8x10 31
TOUPADAKIS FINAL Spring 2013 CHEM 2C 10 of 15 26. [10 Points] Consider the reaction that is represented by the following chemical equation: From the following data: 2NO(g) + Cl 2 (g) 2NOCl(g) Experiment Initial [NO], M Initial [Cl 2 ], M Initial Rate (M/min) 1 0.10 0.10 0.18 2 0.10 0.20 0.35 3 0.20 0.20 1.45 a) Determine the overall reaction order (explain or show your work). b) Determine the value and units of the rate constant. c) Determine the rate of the reaction for [NO] = 1.0 M and [Cl 2 ] = 2.0 Μ. d) Could this reaction be a onestep reaction? Explain. a) and b) Assume that the rate law is: R = k[no] x [Cl 2 ] y Experiment 1 0.18 = k[0.10] x [0.10] y Experiment 2 0.35 = k[0.10] x [0.20] y Experiment 3 x = 2 y = 1 k = 180 M 2 min 1 1.45 = k[0.20] x [0.20] y Overall reaction order is: x + y = 1 + 2 = 3 Rate constant k = 180 M 2 min 1 c) Rate of reaction is: R = k[no] 2 [Cl 2 ] R= (180 M 2 min 1 )(1.0) 2 (2.0) Rate = 360 M/min d) Could this reaction be a onestep reaction? Explain Yes. If this reaction was a onestep reaction then its rate law expression would be: R = k[no] 2 [Cl 2 ]. Because it would be the same as the experimentally determined rate law, the reaction could be a onestep reaction even though most unlikely as having molecularity equal to three.
TOUPADAKIS FINAL Spring 2013 CHEM 2C 11 of 15 27. (10 pts) A possible mechanism for the reaction: is the following: Br 2 + 2NO 2NOBr k 1 Br 2 + NO NOBr 2 k 1 NOBr 2 + NO k 2 2NOBr If the second step is rate determining, what is the rate law that corresponds to the above mechanism? Assume that the equilibrium state in step 1 is reached very fast. Rate rxn = R 2 = k 2 [NOBr 2 ][NO] Rate 1for = k 1 [Br 2 ][NO] and Rate 1rev = k 1 [NOBr 2 ] Equilibrium state: Rate 1for = Rate 1rev k 1 [Br 2 ][NO] = k 1 [NOBr 2 ] [NOBr 2 ] = (k 1 /k 1 )[Br 2 ][NO] Therefore: Rate rxn = k 2 [NOBr 2 ][NO] = k 2 (k 1 /k 1 )[Br 2 ][NO][NO] = (k 1 k 2 /k 1 )[Br 2 ] [NO] 2 = k[br 2 ] [NO] 2 Thus the rate law is: Rate rxn = k[br 2 ] [NO] 2 28. (14 points) Consider the reaction represented by the following chemical equation: 5Sn(s) + 16H + (aq) + 2MnO 4 (aq) > 5Sn 2+ (aq) + 2Mn 2+ (aq) + 8H 2 O(l) A. Make a drawing of a cell that uses this reaction and indicate on the diagram the following: (a) The anode and the cathode. (b) The direction in which the anions and cations migrate through the salt bridge. (c) The direction in which the electrons migrate through the external circuit. B. (a) Write the anode half reaction. (b) Write the cathode half reaction. (c) Calculate E o for the cell. (d) Write the cell notation.
TOUPADAKIS FINAL Spring 2013 CHEM 2C 12 of 15 A. B. (a) Anode half reaction Sn (s) > Sn 2+ (aq) + 2e E o = + 0.14 V (b) Cathode half reaction MnO 4 (aq) + 8H + (aq) + 5e > Mn 2+ (aq) + 4H 2 O(l) E o = + 1.51V (c) E o for the cell E o cell = (+ 0.14 V) + (+ 1.51 V) = +1.65 V (d) Cell notation Sn(s) Sn 2+ (aq) H + (aq), MnO 4 (aq), Mn 2+ (aq) Pt(s)
TOUPADAKIS FINAL Spring 2013 CHEM 2C 13 of 15 Potentially Useful Information: (You may remove this page for ease of access) Constants: R = 8.3145 J / mol K N A = 6.022 x 10 23 1 atm = 760 torr c = 3.00 x 10 8 m/s R = 0.0821 L atm / mol K h = 6.626 x 10 34 J s 1 nm = 10 9 m 1 L = 1000 cm 3 F = 96,485 C/ mol e 1 amu = 1.6605x10 27 kg 1 g = 6.02 x 10 23 amu 1 MeV = 1.6 x 10 13 J 1 J = 1 kg m 2 /s 2 Equations: Slope = Δy/Δx T K = To C + 273.15 G = H T S G = G + RT ln Q G = RT ln K eq k = Ae Ea RT PV = nrt k ln k 2 Eα 1 = R T2 T1 1 1 = hν c =ν λ ΔE = Δm c 2 [ [ A] t = kt + [ A] 0 0 rate = k A] = k t 1 = 2 [ 0 A] 2k rate = k[ A] ln[ A] t = kt + ln[ A] 0 t 1 = 2 0.693 k rate = k[ A] 2 1 [ A] t 1 = kt + [ A] 0 t 1 2 = 1 k[ A] 0 E cell = E ox + E red E cell o 0.0592 o 0.0592 = Ecell logq Ecell = log Keq n n at 25 o C o G = nfe cell G = nfe cell 1 C 1 A = ch arge( C) = current ( A) xtime( s) 1 mol e = 96, 485 C 1 s Rate of decay = λ N where N = # of remaining atoms ln N t = ln N o λt
TOUPADAKIS FINAL Spring 2013 CHEM 2C 14 of 15 Potentially Useful Information: (You may remove this page for ease of access) Electrode Half Reaction E /volts Co 3+ (aq) + e Co 2+ (aq) +1.82 MnO 4 (aq) + 8H + (aq) + 5e Mn 2+ (aq) + 4H 2 O(l) +1.51 Cr 2 O 2 7 (aq) + 14H + (aq) + 6e 2Cr 3+ (aq) + 7H 2 O(l) +3 O 2 (g) + 4H + (aq) + 4e 2 H 2 O(l) +1.23 Pt 2+ (aq) + 2 e Pt (s) +1.20 Br 2 (l) + 2e 2Br (aq) +1.09 OCl (aq) + H 2 O(l) + 2e Cl (aq) + 2OH (aq) +0.94 Ag + (aq) + e Ag(s) +0.80 I 2 (s) + 2e 2 I (aq) +0.54 Cu 2+ (aq) + 2e Cu(s) +0.34 Fe 3+ (aq) + 1 e Fe 2+ (aq) +0.77 Sn 4+ (aq) + 2e Sn 2+ (aq) +0.15 2H + (aq) + 2 e H2 (g) 0.00 Sn 2+ (aq) + 2 e Sn (s) 0.14 V 3+ (aq) + e V 2+ (aq) 0.26 Cr 3+ (aq) + e Cr 2+ (aq) 0.424 Fe 2+ (aq) + 2 e Fe (s) 0.44 Cr 3+ (aq) + 3 e Cr (s) 0.74 Zn 2+ (aq) + 2 e Zn (s) 0.76 2H 2 O(l) + 2e H 2 (g) + 2OH (aq) 0.83 Cr 2+ (aq) + 2 e Cr (s) 0.90 Mn 2+ (aq) + 2e Mn(s) 1.18 V 2+ (aq) + 2e V (s) 1.19 Al 3+ (aq) + 3e Al (s) 1.66 Mg 2+ (aq) + 2e Mg (s) 2.36 Li + (aq) + 1e Li (s) 3.040 Spectrochemical Series CN > CO > NO 2 > en > NH 3 > SCN > H 2 O > ONO > ox 2 > OH > F > SCN > Cl > Br > I Visible Region of the Electromagnetic Spectrum Violet Blue Green Yellow Orange Red 400 nm 475 510 570 590 650 nm
TOUPADAKIS FINAL Spring 2013 CHEM 2C 15 of 15 Potentially Useful Information: (You may remove this page for ease of access) Periodic Table Key 1 H 1.008 2.20 Atomic Number Symbol Atomic Mass Electronegativity 2 He 4.003 3 Li 6.941 0.98 4 Be 9.012 1.57 5 B 10.81 2.04 6 C 12.01 2.55 7 N 14.01 3.04 8 O 16.00 3.44 9 F 19.00 3.98 10 Ne 20.18 11 Na 22.99 0.93 12 Mg 24.31 1 13 Al 26.98 1.61 14 Si 28.09 1.90 15 P 30.97 2.19 16 S 32.06 2.58 17 Cl 35.45 3.16 18 Ar 39.95 19 K 39.10 0.82 20 Ca 40.08 1.00 21 Sc 44.96 6 22 Ti 47.90 1.54 23 V 50.94 1.63 24 Cr 52.00 1.66 25 Mn 54.94 1.55 26 Fe 55.85 1.83 27 Co 58.93 1.88 28 Ni 58.70 1.91 29 Cu 63.55 1.90 30 Zn 65.38 1.65 31 Ga 69.72 1.81 32 Ge 72.59 2.01 33 As 74.92 2.18 34 Se 78.96 2.55 35 Br 79.90 2.96 36 Kr 83.80 37 Rb 85.47 0.82 38 Sr 87.62 0.95 39 Y 88.91 1.22 40 Zr 91.22 3 41 Nb 92.91 1.6 42 Mo 95.94 2.16 43 Tc (98) 1.9 44 Ru 101.1 2.2 45 Rh 102.9 2.28 46 Pd 106.4 2.20 47 Ag 107.9 1.93 48 Cd 112.4 1.69 49 In 114.8 1.78 50 Sn 118.7 1.96 51 Sb 121.8 2.05 52 Te 127.6 2.1 53 I 126.9 2.66 54 Xe 13 55 Cs 132.9 0.79 56 Ba 137.3 0.89 71 Lu 175.0 1.27 72 Hf 178.5 73 Ta 180.9 1.5 74 W 183.9 2.36 75 Re 186.2 1.9 76 Os 190.2 2.2 77 Ir 192.2 2.20 78 Pt 195.1 2.28 79 Au 197.0 2.54 80 Hg 200.6 2.00 81 Tl 204.4 2.04 82 Pb 207.2 2.33 83 Bi 209.0 2.02 84 Po (209) 2.0 85 At (210) 2.2 86 Rn (222) 87 Fr (223) 0.7 88 Ra (226) 0.9 103 Lr (260) 104 Unq 105 Unp 106 Unh 107 Uns 109 Une 57 La 138.9 1.10 58 Ce 140.1 1.12 59 Pr 140.9 1.13 60 Nd 144.2 1.14 61 Pm (145) 1.13 62 Sm 150.4 1.17 63 Eu 152.0 1.2 64 Gd 157.3 1.20 65 Tb 158.9 1.2 66 Dy 162.5 1.22 67 Ho 164.9 1.23 68 Er 167.3 1.24 69 Tm 168.9 1.25 70 Yb 173.0 1.1 89 Ac (227) 1.1 90 Th 232.0 91 Pa (231) 1.5 92 U 238.0 8 93 Np (237) 6 94 Pu (244) 1.28 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259)