KEY CHEMISTRY 2C. Section A FINAL EXAM

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1 thelifecurve.com KEY Last Name First Name Lab Sec. # ; TA: ; Lab day/time: Dr. Toupadakis Fall 00 Instructions: CHEMISTRY C Section A 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 9 pages total. () Read each question carefully. () For Part I, II and III there is no partial credit. There is partial credit for part 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 # 0 5 ( points each) / 50 # 69 (6 points each) / 4 # 30 (0 points total) / 0 # 3 (06 points total) / 06 # 3 (0 points total) / 0 # 33 (0 points total) / 0 # 34 (08 points total) / 08 # 35 (0 points total) / 0 # 36 (0 points total) / 0 # 37 ( points total) / # 38 (4 points total) / 4 Total Score (64) /64 or % Circle one 0. a b c d e 0. 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 0. a b c d e. a b c d e. a b c d e 3. a b c d e 4. a b c d e 5. a b c d e 6. a b c d e 7. a b c d e 8. a b c d e 9. a b c d e 0. a b c d e. a b c d e. a b c d e 3. a b c d e 4. a b c d e 5. a b c d e 6. a b c d e 7. a b c d e 8. a b c d e 9. a b c d e

2 Part I: Concepts ( points each) No partial credit is available. The reaction: Cl + CH =CH ClCH CH Cl is an example of: (a) a polymerization reaction (b) a substitution reaction (c) an elimination reaction (d) an isomerization reaction (e) an addition reaction. A nucleus is held together by (a) an electric force (b) a gravitational force (c) a nuclear force (d) a magnetic force (e) none of the above 3. P O 5 reacts with water. The product is: (a) P 4 (b) H 5 P 3 O 0 (c) H 3 PO (d) H 3 PO 3 (e) H 3 PO 4 4. If stands for the crystal field splitting and P for the electron pairing energy, how many unpaired electrons does a d 5 octahedral complex have if > P? (a) 5 (b) (c) (d) 3 (e) 4 5. The hybridization of the carbon atom marked with the asterisk (*) and the total # of and bonds for the structure below will be: (a) sp (b) sp + 3 (c) sp + 3 (d) sp + 3 (e) sp What is the coordination number and oxidation number of the metal ion in Pd(en)Cl? (a) CN = 3, ON = + (b) CN = 3, ON = +4 (c) CN = 3, ON = + (d) CN = 4, ON = 0 (e) CN = 4, ON = + O C *

3 3 7. What is the correct IUPAC name for [Co(NH 3 ) 5 Br]SO 4? (a) Bromopentaamminecobalt(III) sulfate (b) Pentaamminebromocobalt(II) sulfate (c) Pentaamminebromocobalt(III) sulfide (d) Pentaamminebromocobalt(III) sulfate (e) Pentaamminebromocobaltate(III) sulfate 8. As the atomic number of an element INCREASES, the ratio of neutrons to protons in stable nuclei (a) does not relate to stability. (b) increases. (c) stays the same. (d) decreases. (e) none of the above. 9. Choose the CORRECT statement: (a) Mass defect of a nucleus is the gain in mass that occurs when isolated protons and neutrons combine to form the nucleus. (b) Binding energy of a nucleus is the mass defect converted into energy that is gained during the nuclear reaction. (c) Binding energy of a nucleus is the mass of the nucleus converted to energy. (d) Mass defect of a nucleus is the loss in mass that occurs when isolated protons and neutrons combine to form the nucleus. (e) None of the above 0. Identify the species X in the following nuclear equation: (a) proton (b) neutron (c) particle (d) electron (e) none of the above 4 H X He. What is the IUPAC name for the compound? (a) (b) (c) (d) (e) 4,5dimethylhexane isopropylpentane methyl3propylbutane,3,4,5,6pentamethylheptane Octane

4 4. Choose the INCORRECT statement: (a) Both, lighter and heavier elements are less stable than midweight elements near iron56. (b) In nuclear fission heavy elements gain stability and release energy by fragmenting to yield midweight elements. (c) In nuclear fusion light elements gain stability and release energy by fusing together. (d) In a supercritical fission process, on average, less than one neutron causes another fission event. (e) None of the above 3. Hypophosphorous acid H 3 PO is a (a) monoprotic acid (b) diprotic acid (c) triprotic acid (d) strong acid (e) None of the above 4. Assume that the mechanism of a reaction has steps. The CORRECT statement is: (a) There are transition states. (b) There are intermediates. (c) There are 3 activation energies. (d) The overall order of the reaction is. (e) The reaction is bimolecular. 5. In the following electrochemical cell what is the oxidizing agent and how many electrons are transferred? Zn(s) Zn + (aq) Fe 3+ (aq), Fe + (aq) Pt(s) (a) Zn +, e (b) Zn, e (c) Fe 3+, e (d) Fe +, e (e) Pt, 0e 6. Consider the chemical equation: H O + H O H 3 O + + OH The number of moles of electrons transferred in the reaction represented by this chemical equation is:

5 5 (a) 0 (b) (c) (d) 3 (e) 4 7. Write the formula of tetraaquadichlorochromium(iii) chloride. (a) Cl[Cr(H O) 4 Cl ] (b) [Cr(H O) 4 Cl]Cl (c) [Cr(H O) 4 Cl]Cl (d) [Cr(H O) 4 Cl ]Cl (e) [(H O) 4 Cl Cr]Cl 8. Choose the CORRECT statement: (a) the rate constant increases as the temperature increases and the activation energy increases. (b) the rate constant increases as the temperature decreases and the activation energy decreases. (c) the rate constant increases as the temperature increases and the activation energy decreases. (d) the rate constant increases as the temperature decreases and the activation energy increases. (e) None of the above. 9. Determine the number of different isomers for the complex, M(en)F I. When trying to determine the number of isomers, keep the (en) group in the same position in all of your drawings. M stands for the metal ion and en for ethylenediamine. (a) 4 (b) 3 (c) (d) (e) 5 0. From the following elements, pick the transition metal. (a) Se (b) Re (c) Rn (d) Ra (e) Al. The coordination compounds [Co(NO )(NH 3 ) 5 ]SO 4 and [Co(ONO)(NH 3 ) 5 ]SO 4 are: (a) Ionization isomers (b) Linkage isomers (c) Geometric isomers (d) Coordination isomers (e) The same chemical compound. Which of the following is FALSE? (a) A fuel cell is a closed system. (b) A sacrificial anode can be used to prevent corrosion. (c) A primary cell battery cannot be recharged. (d) As a galvanic cell operates its potential for electrical work decreases. (e) None of the above.

6 6 3. Choose the CORRECT statement: (a) A zeroorder reaction is a reaction with a rate constant equal to zero. (b) The higher the value of an exponent in a rate law, the less sensitive is the reaction rate to a concentration change of that component. (c) The rate constant does not depend on the presence of a catalyst. (d) Intermediates sometimes can be isolated. (e) None of the above 4. 3 Bi decays with emission of an particle. The resulting nuclide is unstable and emits a particle. What is the final nuclide formed? (a) 09 Pb (b) 8 He (c) 3 Be (d) 4 Li (e) C 5. In the coordination compound [M(CN) (en) ]Br, the charge of the metal and its coordination number, respectively are: (a) +, 3 (b) +, 3 (c) +, 6 (d) +, 4 (e) +4, 6

7 7 Part II: Short Calculations (6 points each) No partial credit is available 6. A current of.00 A passing for 5.00 h through a molten metal chloride MCl x deposits. g of the metal. If the molar mass of the metal is 8.69 g/mol, then the value of x is: (a) (b) (c) 3 (d) 4 (e) 5 (5.00 h) 3600 h s.00 C s mol e C = mol e (. g M) mol M g M = 0.87 mol M mol e mol M = mol e mol M = e metal atom x = 7. The activation energy of a reaction is 53 kj/mol. Its rate doubles when the temperature increases by 0 o C? What was the temperature before it was increased? (a) (b) (c) (d) (e) 80 K 98 K 33 K 90 K 640 K k ln k E R T T ln J/mol J / mol K (x 0 ) K x K x = 98

8 8 8. Calculate the maximum kinetic energy of the beta particle emitted in the radioactive decay of 6 He. Assume that the beta particle has maximum energy when no other emission is involved. Use the conversion, amu = 93.5 MeV. Isotopic mass of 6 He = amu, Isotopic mass of 6 Li = 6.05 amu. (a).67 MeV (b) 5.85 MeV (c) 7.49 MeV (d) 3.5 MeV (e) 9.5 MeV Loss of mass = (mass of 6 He) (mass of 6 Li) = ( amu) (6.05 amu) = amu Energy equivalent = (93.5 MeV/amu)( amu) = 3.5 MeV 9. The general formula of the series of compounds whose molecule contains two double bonds, one ring and an ether group is: (a) (b) (c) (d) (e) C n H n O C n H n+ O C n H n+ O C n H n4 O C n H n6 O Saturated hydrocarbon, C n H n+ Two double bonds, C n H n+4 or C n H n One ring, C n H n or C n H n4 One ether group, C n H n4 O Therefore: C n H n4 O

9 9 Part III. Short Answer Fill in the blanks with the appropriate answers 30. [0 points possibleeach is worth pts] Identify the following molecules as chiral or achiral. Also, designate the molecule as R, S, fac, mer, cis, trans, or none of the above. Molecule Chiral (optically active) or Achiral (optically inactive)? Designation H C H C H H N H C N NH Co N Br H Cl CH Chiral Cis Achiral NONE B A A M A B B Achiral Fac OH H HOOC C CH 3 Chiral S CH 3 H 3 C H 3 C C CH 3 Achiral NONE

10 0 3. [6 points possibleeach is worth pts] Draw the following: trans,dibromocyclobutane Br H H Br CO mertribromotricarbonylferrate(ii) ion Br Br Fe CO CO Br 3chlorohexene Cl 3. [0 points possibleeach is worth pts] Identify the correct functional group with the molecule listed. Molecule Functional Group O Ether NH NH Amine CHO C H H Aldehyde CH 3 O Ketone NH Amide O

11 33. (0 ptseach is worth pt) Fill in the following table. Coordination number Metal d electrons Metal oxidation number Can it be chiral? Possible geometries [Cr(H O) 4 Cl ] + 6 d 3 +3 No octahedral [Cu(NH 3 ) 4 ] + 4 d 9 + no Sq. Planer or Tetrahedral 34. (8 pts totaleach is worth pts) Consider the following molecule: A B C HC C O a. ( Pts) What is the hybridization of atom A? sp b. ( Pts) What is the hybridization of atom B? sp 3 c. ( Pts) What is the hybridization of atom C? sp d. (4 Pts) What type of atomic orbitals overlap to form the C=O double bond? A. bond. C (? ) and O (? ) B. bond C (? ) and O (? ) A. bond. C ( sp ) and O ( p ) B. bond C ( p ) and O ( p )

12 35. [0 points each formula is worth pt] Fill in the blanks with the appropriate chemical formula(s). Sodium oxide + H O NaOH Sodium peroxide + H O NaOH + H O Potasium superoxide + H O KOH + H O + O P 4 O 0 + H O H 3 PO 4 Li + H O LiOH + H Na + Cl NaCl

13 3 Part IV. Long Answer Calculations Show your work for partial credit. Put answers in the spaces provided. 36. [0 Points] Consider a reaction that is represented by the following general chemical equation: A(g) + B(g) C(g) + 3 B(g) AT00F From the following data: 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 [A] = 3.0x0 M and [B] =.0x0 3 Μ. d) Could this reaction be a onestep reaction? Explain. Experiment Initial [A], M Initial [B], M Initial Rate (M/s) a) Overall reaction order is: Experiment and R [A] Experiment and 3 R [B] 0 R [A] R = k [A] Therefore: Overall reaction order = b) Rate constant k is: From experiment : R = k [A] = k(0.0) k = 0.50 M s c) Rate of reaction is: R = k [A] R = (0.50)( 3.0x0 ) Rate = 4.5x0 4 M/s d) No. If this reaction was a onestep reaction then its rate law expression would be: R = [A][B]. Because it is different from the experimentally determined rate law, the reaction is a multistep reaction.

14 4 37. ( pts) Consider the reaction described by the following chemical equation: AT00F NO + Br NOBr During the reaction an intermediate is detected, and the rate law determined by experiment is: Rate = k [NO] [Br ]. Could this reaction be a onestep reaction? Explain why.. For this reaction the following mechanism has been suggested: k Step : NO + Br NOBr Fast equilibrium k k Step : NOBr + NO NOBr Slow step a. Demonstrate by showing your work that this mechanism is acceptable because it satisfies the two requirements. b. Besides the fact that this mechanism satisfies the two requirements, what other additional support is there according to the data given in this problem? Explain briefly... Even though the rate law determined by experiment agrees with a onestep reaction mechanism, the detection of an intermediate drops this possibility because by definition a onestep reaction mechanism does not involve any intermediates. a. First requirement The sum of the steps must give the balanced equation of the reaction. Step : NO + Br NOBr k k Step : NOBr + NO NOBr NO + Br + NOBr + NO NOBr + NOBr Overall chemical equation: NO + Br NOBr

15 5 Second requirement The mechanism has to agree with the experimentally determined rate law. Overall rate = rate of slow step = rate of nd step = k [NO][NOBr ] First step is fast equilibrium Rate (forward) = Rate (reverse) k [NO][Br ] = k [NOBr ] [NOBr ] = (k /k )[NO][Br ] Overall rate = k [NO][NOBr ] = k [NO](k /k )[NO][Br ] = (k /k )k [NO] [Br ] = k[no] [Br ] Where k = (k /k )k b. The detection of an intermediate. 38. (4 points) Consider the reaction represented by the following chemical equation: Cr(s) + 3H O(l) + 3OCl (aq) Cr 3+ (aq) + 3Cl (aq) + 6OH (aq) 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.

16 6 A. B. (a) Anode half reaction Cr (s) Cr 3+ (aq) + 3 e E o = V (b) Cathode half reaction OCl (aq) + H O(l) + e Cl (aq) + OH (aq) E o = +0.94V (c) E o for the cell Cr (s) Cr 3+ (aq) + 6 e E o = V 3OCl (aq) + 3H O(l) + 6e 3Cl (aq) + 6OH (aq) E o = +0.94V Cr(s) + 3H O(l) + 3OCl (aq) Cr 3+ (aq) + 3Cl (aq) + 6OH (aq) E o cell = (+0.74 V) + (+0.94V) = +.68 V (d) Cell notation Cr(s) Cr 3+ (aq) OH (aq), OCl (aq), Cl (aq), Pt(s)

17 7 Potentially Useful Information: (You may remove this page for ease of access) Constants: R = J / mol K N A = 6.0 x 0 3 atm = 760 torr c = 3.00 x 0 8 m/s R = 0.08 L atm / mol K h = 6.66 x 0 34 J s nm = 0 9 m L = 000 cm 3 F = 96,485 C/ mol e amu =.6605x0 7 kg g = 6.0 x 0 3 amu MeV =.6 x 0 3 J Equations: Slope = Δy/Δx T K = To C To F 3 =.8 To C G H T S G G RT ln Q G RT ln K RT eq k Ae Ea PV rate k nrt ln k E R 0 k[ A] k [ A] t kt [ A] 0 T T h c ΔE = Δm c t [ 0 A] k rate k[ A] ln[ A] t kt ln[ A] 0 t k rate k[ A] [ A] t kt [ A] 0 t k[ A] 0 E cell = E ox + E red o o Ecell Ecell log Q and Ecell log Keq at 5 o C. n n C A ch arg e( C) current ( A) xtime( s) mol e 96, 485 C s o G nfe cell G nfe cell Rate of decay = N where N = # of remaining atoms ln N t = ln N o t 60 r v 430 o b 580 y g

18 8 Potentially Useful Information: (You may remove this page for ease of access) Electrode Half Reaction E /volts Cr O 7 (aq) + 4H + (aq) + 6e Cr 3+ (aq) + 7H O(l) +.33 O (g) + 4H + (aq) + 4e H O(l) +.3 Pt + (aq) + e Pt (s) +.0 Br (l) + e Br (aq) +.09 OCl (aq) + H O(l) + e Cl (aq) + OH (aq) Ag + (aq) + e Ag(s) I (s) + e I (aq) Cu + (aq) + e Cu(s) Fe 3+ (aq) + e Fe + (aq) Sn 4+ (aq) + e Sn + (aq) +0.5 H + (aq) + e H (g) 0.00 Sn + (aq) + e Sn (s) 0.4 V 3+ (aq) + e V + (aq) 0.6 Cr 3+ (aq) + e Cr + (aq) 0.44 Fe + (aq) + e Fe (s) 0.44 Cr 3+ (aq) + 3 e Cr (s) 0.74 Zn + (aq) + e Zn (s) 0.76 Cr + (aq) + e Cr (s).90 Mn + (aq) + e Mn(s).8 V + (aq) + e V (s).9 Al 3+ (aq) + 3e Al (s).66 Mg + (aq) + e Mg (s).36 Li + (aq) + e Li (s) Spectrochemical Series CN > CO > NO > en > NH 3 > SCN > H O > ONO > ox > OH > F > SCN > Cl > Br > I Visible Region of the Electromagnetic Spectrum Violet Blue Green Yellow Orange Red 400 nm nm

19 9 Potentially Useful Information: (You may remove this page for ease of access) Periodic Table Key H Atomic Number Symbol Atomic Mass Electronegativity He Li Be B C N O F Ne 0.8 Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc (98).9 44 Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po (09).0 85 At (0). 86 Rn () 87 Fr (3) Ra (6) Lr (60) 04 Unq 05 Unp 06 Unh 07 Uns 09 Une 57 La Ce Pr Nd Pm (45).3 6 Sm Eu Gd Tb Dy Ho Er Tm Yb Ac (7). 90 Th Pa (3).5 9 U Np (37) Pu (44).8 95 Am (43).3 96 Cm (47).3 97 Bk (47).3 98 Cf (5).3 99 Es (5).3 00 Fm (57).3 0 Md (58).3 0 No (59).3

KEY CHEMISTRY 2C. Section B FINAL EXAM

KEY CHEMISTRY 2C. Section B FINAL EXAM 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