Chem 1312H Final Exam Key

Chem 1312H Final Exam Key - K2 CHEM 1312H/1412 FINAL...

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Unformatted text preview: K2): CHEM 1312H/1412 FINAL EXAM (PART 2) Spring Semester, 2008 Print Name Signature Instructions 1. Palt 2 contains 7 written-response questions. Answer these questions in the space provided below each question. The value of each question is indicated in the exam booklet. . There is a one hour and 45 minute (105 minute) time limit on Part 2. All exam papers must be handed in promptly when time expires. 3. This exam is subject to the UGA Honor Code. You may not receive assistance from any source during the exam. Also, you may not discuss the contents of the exam with other students in the course who have not yet taken the exam. 4. If you have a question about the exam, raise your hand and Dr. Kutal will come to you. l\.) A Periodic Table and other potentially useful information are provided in a supplementary handout. 1. Below are the structures of four complex ions (ox is the bidentate oxalate anion). 00 II II 0X = C—C | I ‘O O" (i) Identify which, if any, of these complex ions are isomers (geometric or optical), which, if any, are identical (have identical structures), and which, if any, are distinctly different. (25 pts) OH2 0H2 7 2 W W “4 W HBN _/ / NH3 (alor (lo) 3.5 a jeoneifx'cal OH2 0H2 isomer (MC _ (a) (b) (C) is disc-Fermi 4/m Hie NH3 NH3 [005 CH? H70 W” 7 ‘ 2 Z C0"r 0x OH2 H20 _/ 0H2 NH3 (c) (d) (ii) What is the overall charge on complex (a)? How many 3d electrons does the metal in complex (a) possess? If complex (a) is low spin, how many unpaired electrons does it possess? (15 pts) 0 CO3+ has 0 (lb 6 [90+f‘an [Omit] ura‘l‘i‘an) so all elec+rcms pour M low-er enegy t1 Ofloflml’c :1 no uA‘oakeJ Q lé‘cl‘rm} :63; t2. 2. Enkephalins are pentapeptides, short biopolymers that are synthesized by humans to control pain. Enkephalins bind to certain receptors in brain cells, which are also known to bind morphine and heroin. One enkephalin results from the condensation of amino acids in the sequence tyrosine-glycine-glycine-phenylalanine-methionine. (a) Draw the structure of this pentapeptide. (25 pts) H i‘ i i‘ HZN—(IJ—COZH HzN—(ID—COZH HgN—CIZ—COQH HzN—(lJ—COZH CH2 H CH2 cHz Glycine CI;H2 i Phenylalanine CH3 OH ' Methionine Tyrosine I l cit). H H CH»- 9“ CH2. l :9 Ll 0 CH3 (b) Predict the sign of the entropy change, AS, for the reaction that produces the enkephalin from the five amino acid molecules. Explain your prediction. (10 pts) ine reaction invdlv 5 lirxlcm iv e‘i’l‘er We amino aCs'ci 6 J 5 via (or/“firm of Wade 10ml; (wings) Ma evast/‘on 0% a Water molQleC- H 0 This cocdn/xxah’m (‘Qac‘i‘t'on doef noi' charge the if!" AME—U 0i; for‘i‘fcle} ‘ 5 a,{\“z\o 0“le q i + q [ta/0 Niewles - 50 A; 5km {A be close w 12W- 14 omit/Ni), 05m; 5+ he 4: if Ml ne sci-We *3 “1an He decrease In +(‘msla‘b‘mal YMMV” 0‘4 H9 [09 Chain proquxceci- 3 I D) In hexane solution the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and (CH3)AuPH3 according to the following mechanism. in Step 1 (CH3)3AuPH3 :2 (CH3)3Au + PH3 (fast) 42-: Step2 (CH3)3Au i19- C2H6 + (CH3)Au (slow) 123 Step 3 (CH3)Au + PH3 ——> (CH3)AuPH3 (fast) (a) What is the balanced equation for the overall reaction? (10 pts) LCH3)_3Au PH} ‘P (CH3)QM PH} 1- C.sz (b) Identify all intermediates in the reaction. (10 pts) (CH3); 9%, LCH3)'AL4\ 0A3 PH} are intermediate; (c) What is the molecularity of each elementary step in the mechanism? (10 pts) S+ef> \ l (unimolecularl we? 2 5+2? 3 l " 2 L tzmolewlor] (d) What is the predicted rate law for this mechanism? (15 pts) Cal-e= Iii-1 I: (CH3)3ALA] Ham Sl—efl L. I: (CH3); Au PHJJ : g“! l: (CH3); AVJLPHJJ ‘ Au 50 L (“bk Ad : (*1!) [PH 3 3 in“; rm : film/E) [(Cll5)3 Au 9H,] ,5; CC cm); Audi‘s} C 9 H3 3 C 1° H33 (e) How would adding PH3 to the solution affect the reaction rate? EXplain. (5 pts) The (sale {5 {Aversely Profaft'mnal 1‘0 CPPBJ/ 50 5&de 9H3 to 1‘4“? $0l‘Al‘i‘m Would deCl‘eA’UQ were, FeaC't‘z‘o/l Pod-e. 4. Consider the following equilibrium. 2 N0(g) + 02(g) :2 2 N02(g) A reaction mixture at 175 K initially contains 522 torr of NO and 421 torr of 02. At equilibrium the total pressure of the reaction mixture is 748 torr. Calculate the equilibrium constant, Kp, at this temperature. Note: 760 torr = 1 atm. (50 pts) 1 Nolb) l‘ Oztj) : 1 1 51z+ocr L121+arr O C ~1)( ~74 llx E (52202.20 Hll‘fi 1x +011"! frefwre L521_2x) + + 5 'l' (Mm 0d" 81‘,“ I X: lag farr 50 air ejuilarium PM“ 2[l0\$)= 390 +drr: 0.513a+m 7" 2. K91 PNOZ t (0-fl3) : PM: 2‘ P09, Lo.|WL{)1(0-2°1'1) ' / Metallic elements are essential components of many important enzymes operating within the human body. Carbonic anhydrase, which contains Zn2+, is responsible for rapidly interconveiting dissolved C02 and HCO3'. The zinc in carbonic anhydrase is coordinated to three nitrogen—containing groups and a water molecule. The action of the enzyme can be attributed to the greater acidity of the coordinated water molecule compared to bulk (uncoordinated) ’water molecules present in solution. Explain this greater acidity in terms of the factors that influence acid strength. (35 pts) In a complex ion, the transition metal is an electron pair acceptor. a Lewis acid: the ligand is an electron pair donor, a Lewis base. In carbonic anhydrase, the Zn2+ ion withdraws electron density from the O atom of water. The electronegative oxygen atom compensates by withdrawing electron-density from the 0—H bond. The 0-H bond is polarized and H becomes more ionizable, more acidic than in the bulk solvent. This is similar to the effect of an electronegative central atom in an oxyacid such as H2804. 6. (a) Which aqueous solution from the right hand column exhibits the property in the left hand column? (20 pts) _————_-———— Property Solution 1. lowest electrical (a) 0.10 m KCl(aq) conductivity 2. lowest boiling (b) 0.15 m C12H22011(aq) point 3. highest vapor pressure (C) 0.10 m CH 3COOl-l(aq) of water at 25 °C 4. lowest freezing point (d) 0.05 111 NaCl lOWGS‘f elQCWCCal COAduci'fi/C‘l‘y b lowesl- boi lt‘nj loaf/ML cl hijlxefir vafw Pressure d [owefi freezmj Poi/\‘l‘ a, (b) Calculate the freezing point of solution ((1). The freezing point depression constant for water is 1.86 °C/m. (20 pts) 0.05 m NoCl solwflon COA'l‘alAS 0.0: m NOT ml 0'03 m Cl") £0 +49 @4€PC+Il\ie (Rajah-l7 1:3 o.lUM K; ' m : L8e "C/m X 010% - 0.1% 00 Freebg IDUer : —- o [0) 0C 7. Consider the equilibria for dissolving the two salts, NaCl and AgCl, in water. NaCl(5) :2 Na+(aq) + Cl_(aq) AgC1(s) ‘2 Ag+(aq) + C1-<aq) (a) Calculate AG° at 298 K for each process. (15 pts) AG:(,m(NaCl) = (—26'1.9 kJ/mol) + (—13l.2 kJ/mol) — {—3810 kI/mol) A91 kJ/mol (+7711 kJ/mol) + (—l3l.2 kJ/mol) - (—109.70 kj,"mol) = +536 kl’mol AGsaoln(AgCI) (b) The two values should be very different. Does this result arise primarily from a difference in the enthalpy of solution or the entropy of solution? (15 pts) salt AIiscoln ASEOIn '— TAsgoln NaC] +3.6 kJ/mol +432 J/mol-K i129 kJ/mol AgC] +657 k],/mo] +343 J/n‘tol-K —lO.2 kJ/mol The entropy terms for the solution of the two salts are very similar. That seems sensi- ble because each solution process should lead to a similar increase in randomness as the salt dissolves, forming hydrated ions. In contrast, we see a very large difference in the enthalpy term for the solution of the two salts. The differ— ence in the values of A6301“ is dominated by the difference in the values of AHEUIH. (0) Calculate the solubility product constant, Ksp, for each salt at 298 K. (15 pts) (c) The solubility product, K5,, is the equilibrium constant for the solution process. As such, we can relate K51D directly to AGSOID by using Equation 1923: K5,, = ewacgoln/RT We can calculate the K5], values in the same way we applied Equation 19.23 in Sample Exercise 19.13. We use the AGEND values we obtained in part (a), remembering to con— vert them from to J/mol: NaCl: K5}, = [Na+(nq)][cr(nq)1 = e(“glnm'lts-“WZQBH = (#37 = 40 KS.“ 2 = (7‘ #155,600)?»(8.314)(298)] : (7- 22,4 : X 10"“) (d) How will the solubility of each salt change with an increase in temperature. Explain your reasoning. (5 pts) Semmse. A520“. is fofttive) Solmlfillty I‘M—WM“ wft’l’t focPeMmj #ngrmm owt‘y T0 the I‘_r\cc€0~3)’:) “drama-W) 05 {4Q cragorerm +0 00:0,”, 8 ...
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Chem 1312H Final Exam Key - K2 CHEM 1312H/1412 FINAL...

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