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samplefinal_bradforth04_key - Chem 105b E e Pom" Final...

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Unformatted text preview: Chem 105b E e Pom"? Final Exam Print Name Last, First Wednesday, May 5, 2004 Last 4 digits Soc Sec. # Professor Stephen Bradforth TA‘s Name Lab Day & time: _____—__— Questions Points 1-3 15 _— 4 _— 5 5 ' 6 2. Mn 7 25 _ 8 25 9 20 10 25 l 1 30 12 Total 200 _— Please Sign Below: I certify that I have observed all the rules of Academic Integrity while taking this Signature: INSTRUCTIONS 1. You must show your work to receive credit. For the long answer questions (problems 7-12) you must show all working for full credit. State and justify any approximations you make. 2. If necessary, please continue your solutions on the back of the preceding page (facing you). 3. YOU MUST Use black ink. 4. There are 12 problems on 15 numbered pages (plus a piece of graph paper. Please count them before you begin. For the long answer questions (problems 7 —12) you must show all working for full credit. State and justify any approximations you make. The following data may be of use: TC = TK - 273.115 1 R = 8.314 J K_ mol- = 0.08206 L atm moi" NA = 6.02 x 1023 Integrated Rate Laws: First Order: [A] = [A]0 exp (- kt) Second Order: 1/ [A] = 1/ [A]0 + kt The Arrhenius Equation k = A exp(—EA/RT) lnk=lnA— (EA/RT) Acid — Base Equ ilibria: 14 1.0 x 10' @25°C + = - 10g10[H30 ] - 10g10[0H-] - 1<ng0 Ka The Henderson-Hasselbalch Equation: pH = pKa + logto([A']/[HA] ) Thermodynamics: asum-v: ASSys + A3511” ASS“1T — - AH/T AGofeaCtion = 2 np AGfo (products) - 2 n1. AGfO (reactants) AG = AH-TAS AGO = -RTan Electrochemistry: AG" = - n F E0 Faraday Constant (F) = 96,485 C / mole of electrons E = E0 — (0.0592 In) log10(Q) Transition metals: Spectrochemical Series CN' > N02' > en >NH3 > H20 > OH' > F' > C1' > Br' >1" The solution of the quadratic equation a x2 + b x + c = 0 is given by: -b‘i V(b2 - 4ac) a 1| 2a You will require the data on following pages: $.25“: mutt—«5.39 NEON an— E E m on 6.“... ow mm 8.3 _m w. mPZmEmAm mm? m0 m—Amfih UHQOHMm—m TABLE OF K“ VALUES Name Chlorous acid Hydrofluoric acid Nitrous acid Formic acid Lactic acid Benzoic acid Acetic acid Hypochlorous acid Hypobromous acid Hydrocyanic acid Boric acid Ammonium ion Phenol Hypoiodous acid Carbonic acid Sulfuric acid Sulfurous acid Oxalic acid Ascorbic acid TABLE OF K1, VALUES Name Ammonia Methylamine Ethylamine Pyridine Aniline Formula Km. HCIOZ 12 x 10—2 HP 1.2 x 1041 HNO: 4.0 x 10-4 HC02H 1.8 x 10-4 HCaHsOa 1.38 x 10-4 Hal-1502 6.4 x 10-5 HC2H302 1.8 x 10-5 HClO 3.5 x 10—8 HBTO 2 x 10-9 HCN 6.2 x 10-10 H303 5.8x 10-10 NHJ 5.6 x 10-10 l-IOCeHs 1.6 x 10—10 H10 2 x 10-1] Hzcog 4.3 x 10-7 H2304 Large H2503 1.5 x 10—2 H2C204 6.5 x 10-2 HzcsHGOa 7.9 x 10-5 Formula K1, NH: 1.8 x 10—5 CH3NH2 4.38 x 10—4 €szan 5.6 x 10-4 CsHsN 5.6 x 10-9 CsHsNH2 3.8 X 10—]0 5.6x10‘“ 1.2x10—2 1.0><10—7 6.1 ><10—5 1.6x10‘12 Standard reduction Potentials as 25°C (298 K) for Many Common Half-Reactions Half-Reaction F; + 2.2- a 2F' Ag2+ + e‘ -> Ag+ Co3++ e' —-’ Co2+ H202 + 2H+ + 29,- -+ 21120 Ce4++ e' —’ 033+ P1302 + 4H+ + 5042- + Ze' —> PbSO4 + 21420 23' + 211+ + 104' —’ 103‘ + H20 Au3+ + 36' —> Au Pboz + 4H+ + 26 —> sz+ + 21120 C12 + 26' —> 2C1' Crzoqz- + mm + 6c' —> 2Cr3+ + 71120 02 + 4H+ 46‘ -+ 21120 103- + 6H+ + 52' —> 1€12 + 31120 Brz + 20' —> ZBr' v02+ + 2H+ + e' —> v02+ + H20 AuC14' + 36' -’ Au 4- 4C1" NO3' + 4H+ + 3c" —> N0 + 21120 C102 + c' —»c102' 2Hg2+ + 26' -> Hg22+ Ag” + c' —> Ag Hggz+ + 2c" -9 2Hg F63+ + e" —> F€2+ 02 + 2H+ + 2c' —> H202 Mn04' + e' -> Mn042' 12 + 26' -» 21' Cu+ + e' -+ C0 8 °(V) 2.87 1.99 1.82 1.78 1.70 1.69 1.60 1.50 1.46 1.36 1.33 1.23 1.20 1.09 1.00 0.99 0.96 0.954 0.91 0.80 0.80 0.77 0.68 0.56 0.54 0.52 Half-Reaction 02 + 21120 + 40’ —> 4011‘ C021” + 20' -> Cu HggClg + 26‘ —> 2Hg + 2C1' AgCl + e'l—r Ag + C1' 5042- + 411+ + 2e' —-> H2503 + H20 Cu2+ + e‘—> Cu+ 2H+ + 23' —> H; Fe3+ + 36' —* Fe Pb2+ + 29,- —> Pb Sn2+ + 20' —> Sn Ni2+ + 2e' —> Ni PbSO4 + 2e- —+ Pb + 5042- Cd2+ + 2e' —+ Cd Fe2+ + 2e" —3- Fe C13+ + e' —r Cr2+ Cr3+ + 3e' —-> Cr 2112+ + 26' -> Zn 21120 + 29: —> H2 + ZOH' Mn2+ + 26' —> Mn A13+ + 3:5 —> A] H; + 2e' —) 2H' Mg2+ + 26' -> Mg La3+ + 3e' —> La Na+ + e' -> Na Ca2+ + 2e‘ -+ Ca Ba2+ + 26' -9 Ba K+ + e‘ —> K 1.1” + e" —> Li 8 °(V) 0.40 0.34 0.34 0.22 0.20 0.16 0.00 0.03 -0.136 0.14 023 -0.35 -0.40 -0.44 -0.50 -0.73 -0.76 -0.83 . —1.18 -1.66 -2.23 _2.37 —2.37 —2.71 -2.76 -2.90 -2.92 -305 Question 1) (5 points) At 100°C the equilibrium constant for the reaction COC12(g) = C0(g) + C12(g) has the value 0ch = 2.19 x 10“". Are the following mixtures (1) and (ii) of coca, co, and c1; at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (0 [c0012] = 5.00 x 10'2 M; [c0] = 3.31 x 10‘6 M; [(312] = 3.31 x 10'6 M (ii) [0001;] = 3.50 x 10'3 M; [c0] = 1.11 x 10'5 M; [012] = 3.25 x 10'6 M (a) (i) is at equilibrium and (ii) is at equilibrium (b) (i) is at equilibrium and (ii) is not at equilibrium, left to right @ (i) is at equilibrium and (ii) is not at equilibrium, right to left (d) (1') is not at equilibrium, right to left, and (ii) is at equilibrium (e) (i) is not at equilibrium, left to right, and (ii) is at equilibrium Question 2) (5 points) Arrange the following 0.10 M solutions in order of most acidic to most basic. (a) HCl, KNOz, KC], NH4C1, KOH (b) NH4C1, HCl, KCl, KOH, KNO; (0) HCI, KCl, KN02,N}I4C1, KOH .HCl, NH4C1, KCl, KNOZ, KOH (e) KNOz, HCl, KCl, KOH, NH4C1 (f) None of the above I Question 3) (5 points) Which of the following will lower the activation energy for a reaction? (3) increasing the concentrations of reactants (b) raising the temperature of the reaction @ adding a suitable catalyst (d) all of these (e) there is no way to lower the activation energy of a reaction. ye & WWULE 53919291 R'MMF . as 5 meat \(‘r Short answer questions -9Q Pariwel Wt may ha ‘5 Obv’tm what W mama-L . Question 4) (10 points) For the reaction 2802(g)+02(g)=2803(g) AH°=- 198kJ the equilibrium constant K = 2.5 x 10‘0 at 500K. (3) What is Kp for this reaction? Ideal???) F5037“ 95°: (2“): , Ch} ‘ (3:0 K? r r50: ‘90; “' Ch: (cu-)1— CUJRT) Ch: coL RT Zwa‘ : K/ RT (—éosznouu (b) An equilibrium mixture of SO; (g), 02 (g) and 803 (g) is formed in a reaction A vessel. What will happen in each of the following situations: . . . . . - 6 l X \08 (1) The pressure on the reaction vessel 15 increased (by compressmg the vessEl}. _ ' 7:5- Efi’mlthnm 5W3 k’ Lea—L}- (ii) The vessel is allowed to cool to 298 K. Eiwltgnwm Slack-g R‘CTHT Question 5) (5 points) Draw all geometrical isomers for the following compound: CH3CH=CHCH2CH=C(CH3)2 ' H CH3, \ / Q -—~ c. ‘\ H / CH1 m CH3 \ . / C, :2 C. / \ Q43 H H ..._ 0‘3 / C 2:: Ci CH CH 045 / I 3 -: C. C \ / H H Question 6) (20 points) Carefiilly read each of the following statements. Mark each one as true or false, and for each statement you mark as false, amend the sentence to provide the most informative, chemically correct, statement. (Example: If the statement read, "USC is a university located in San Diego", the best revision would be "USC is a university located in Los Angeles", and not "USC is not located in San Diego") amt me... (a) A peptide is formed by a condensation of the W group and carboxylic acid functional groups present in amino acids. .— erLSEi (b) At the boiling point of a liquid, M F e I Al" :TAS .n 9f Asian-“(z 0 35' AG =0 P‘L-SE Air «one v 4’ a l W? i j (c) In the galvanic cell Zn(s) | Zn2+(aq) || Pb2+(aq) |Pb(s), the lead electrode gets heavier as the cell runs down. E (d) If NaClO4 is dissolved in water, the water is substantially cooled. As this process is spontaneous, AHsystem for this process must be negative, as ASHM is positive. a Qm‘Q—r AHSKAJ‘W J 63 Wd T5353;b“> WEI (e) For a diprotic acid K,. is always guaflé’r than K32. iAHi err Kal< Kn, ‘Ww A [El (f) Elemental copper has an electron configuration 452 3d9. 44‘ 3d ‘° Fame (g) Concentrated HCl acts to solubilize AgCl precipitate by forming a complex anion. (h) By speeding up the forward reaction rate constant, a catalyst changes the equilibrium amount of product that can be made by the reaction it catalyzes. ” E’Hflr —— alto PM! up baskwoggk ”02%! PALM? "' SO Eiwltbnm M ACDNGAQLOL OF (1%)? kcw W (i) The ionization potential generally i s down Group II in the periodic 3&5- . because the positive charge on the n Moreasing. {OD/mi Ann m ‘ . . _ , I! dams. «Jaws wow— We 1% PW _ _ =95 LAW 3.ng W (k) In the titration of a weak base w1th a strong ac1d th d pomt 15 pH 7. low l’km Of SM bow. wad-c #7er Own} F19 Q9- ____. Dr“ 0mg] M Mai“: 51W SEA—(meat- ___...—- Long Questions — show all your working for full credit. Question 7) (25 points) The following (unbalanced) redox reaction is used in acidic solution in the Breathalyzer test to determine the level of alcohol in the blood: 0 n+(aq) + Cr2072' (aq) + cnacnzon (aq) —> Cr3+ (aq) + CHg-Cf/ (aq) + H20 (I) H The organic reduction half reaction is not in standard tables. It can be written as: CHrq + 2H+ + 2 e' 9 CH3CH20H E“ = + 0.22 V (a) In this reaction ethanol is oxidized to another organic molecule - what type of fimctional group does it contain? was) Awe/w of (b) Identify the elements undergoing changes in oxidation state and indicate the initial and final oxidation numbers for these elements. - CPLOZI'LP Cr 3" )Pts) Cr +Q, .-—--v + 3 o ~cn1pu — Cl—H Car-50m O 1 +1 (c) Write down the complete balanced reaction. o / g[ matron > chl _________________._.._...—_._.___.—.________________ 1+ //0 304 3011034 + Cry}? 814* 7 20 + 3 013 c _ a pm) i pts) Question 7 continued) ((1) What is the electrochemical cell voltage for the reaction at stande conditions? CH3CH19H > CH3 (0H + 2H*+ 1e“ mu’laré 95+ GL0; 1' ————-> Cr3+ *— 0.22. (e) If the blood alcohol level of a legally intoxicated driver is sufficient to produce an ethanol concentration of 1 X 10' M in the breathalyzer, what would be the measured cell voltage if all other species are present at standard state concentrations? (')(3 E: E0 _ OOSSL “TALE? (03mm) )3 {Ar—£2 Ln ——. amt —. 0% \/ I l Question 8) (25 points) In most areas of the United States fluoride at about 1 ppm (ppm = part per million) is added to drinking water. This practice has long been controversial but is endorsed by the American Dental Association. Fluoride ions in drinking water convert the hydroxyapatite, Ca5(PO4)3OH of teeth into fluoroapatite, Ca5(PO4)3F. The K51, of the two compounds are 1.0 x 10'36 and 1.0 x 10‘60 respectively. The solubility equilibria to consider are Ca5(PO4)3OH(s) == 5 Ca1+ (aq) + 3 P043' (aq) + OH' (aq) C35(P04)3F(S) = s Ca2+ (aq) + 3 P03 (aq) + F (aq) (a) Write down the equilibrium constant expressions for each reaction '1’“) (n K: ECQUJSEPQEJEEOWJ Oil K: [Cw/“ii? [POWB'TSCFH] (b) What are the molar solubilities of each substance in water? i pts) ‘B{ Ca 50300 3 OH mp : Ix was __ merges-33 [ou']:(§x)s(31)3x ‘-' 3%315 JCS Cog £00030“ :3 Scout + aflofi'a— cu * - E2 MALE. '51 31 DC..— 9 fl -: 2 ' 8 X '0". S M :27 SoWiUb “1 7C “2 8%3’13 Fix Co: [90.01 F ‘5 4.0 .. at, = '— PS l K“... git-55 gtO D 9”].Oxlo'c‘0_ GlXib—xN @ SOL-UL {K}: 1: 844,375 "" ' Question 8 continued) (0) Speculate as to why fluoride is added to water. 139 Mag jlkwag H Wei—tr [we Sw'Ecug \Weg PM r; W) _ , _ , Wad b3 WW5) £MO~M¢ZI WM ‘5 LE 85W g Wejflzr (53 3 Wit-(rs 05 Mag/115K993 » .r (d) From the two K5], values, and using AG? (OH' (aq)) = -157 kJ/mol and ANS META AGfO (F‘ (aq)) = -279 kJ/mol, compute the difl’erence in AGfo for solid l 7 hydroxyapatite and fluoroapatite. S— !“ ’V’L lpts) AGO: * RTLn Ks? @233,“ AG0 RY Rack“; «3ka dLSSo\\/v8 LY +logkj/mut. AG“: —-' (83m) (198) L“(l.0xio*1‘) @ AG°( - _ Ca (900 F 5:” Sea” APO? + F I '1' '5 F) : SAG“) (can) + zbgf (Pof’)—+ AGOCF’Z ‘— AGOdLSSQUnkE/i (4‘) AG (924900300 = 53G? (039+ ant—f (F0q9§+ Imam") SW 0 8 o . ,° G 0(F: - F — ’6 a 2-..- G dt (I - Acting—I) 4. A I 2 AC; (€11,000; ) AC): CC $0033 0“) é magma) = 3&2-107- 279"” ‘5 AGO} C 9130900: Question 9) (20 points) One common type of reaction is between alcohols and carboxylic acids. Let us consider the reaction between methanol, CH30H, and acetic acid, CH3COOH. (a) What type of alcohol is CH30H'? pts) 'pRUWR-K‘f C \o) (b) Write down the complete chemical reaction between these two reactants, showing the relevant reactive fimctionalities in full. What class of reaction is this? 3 pts) 0 O H \t CllgtiH Ho 3% c— 013 -————a 043 O we— CH3 (c) A certain polyester has the repeating unit ‘{—|C|—.—|CI—OWCH2‘.—CH2—O%; Identify the monomers used to make the polyester. [0 pts) 0 t O Question 10) (25 points) Neutral hydroxyl radical is a highly reactive species (a free radical) formed from ozone in the lower atmosphere over Los Angeles and by radiation induced damage in cells. In the following elementary reaction OH reacts with chloride ions: OH(aq)+ Cl'(aq) '9 CIOH' (aq) This reaction can occur at the Southern California coast with sea spray from the Pacific Ocean or biochemically in the cell where the concentration of chloride ions is also high. (a) Write down the rate law for the reaction. , ptS) (b) To evaluate the rate constant a set of experiments is run with the C1“ concentration in large excess. Under these conditions, write an algebraic expression for the variation of [OH] with time. i pts} A \m 'E' ii!!! ___n_mmm.w_m. . ..,_ -, _ “NEW... ___m__.=__ 1 “m .9: m ‘iIIH-III-I- “¥I M — L “ll 5 fl - I-LVI .._._.....==E- =:=:=EEEEE L ‘ V V _Illu 'E: _I I—III- — IIIII .— -— 1-I' IIIII E5 Do not detach thi‘sipiece of graph paper. m. m _. W mm— H. ...... . u... .. . . 2.. _. _. __ _m . “ __.____ _. m .i . H \ . n .I ! m ._ m n _ ._ an _. __m n _“ 1:” .. ._ , _-.. -_ .u __"-.;...:__._ ...._m.._. ”f -- - _ .. _ _ _ .u m an a: H . .. g .. r. .m . .. nu. u -- I E “ I I l- n I _ =. u .m . . Em "— .= __ n .. -.nn_ H _n. n .— __. Kim“: .H m ._ "E 1. n —m- m =- _. n .— an: n n. n .n n . _. n n .n _. nun—n... _.: n .1 u .u n.“ "n.__m_____. _m _m. .. “ _n___m mu_ __. _ _u. ”mu_.__.u __M__ a, .\ W a W A ~ Question 10 continued) The following data was collected for [OH] as a function of time with [C1] = 0.5 M. t (seconds) [OH] lmol L" (A to H 3 0 1.00x104 ‘9 2‘0 1.0x10‘9 0.74><10‘4 _g.gn 0.40x10“ - 10, I21} -— I’L . 2 09 (c) Find the value for the rate constant, k, you included in your rate law for part (a). Make sure to heck for and include units I n . .0 I , . -. in (SW- ‘Q : oSXm‘Br‘ _ W I ‘3 —| 8 -\ -1 lawn, a: 0'3“? S = (moms Coda 0.8M ‘l Ml“: Mi Wamplx/ M (M W loop 03351 Vow/y \Za hwjgcm. Question 11) (30 points) The following scenario actually came up in an experiment in my laser laboratory this semester. We wanted to do some laser spectroscopy on the CN(aq) anion, but because it acts as a weak base in water (Kb= 1.6 X 105), a solution cannot be made that does not contain significant [OH']. (3) Calculate the {OH‘] and pOH of a 1.00 M solution of NaCN in water. WM loam CMGJW ptS) K . it a) QLAQ i 2 m 991-; l,€><lo"'§ by lmxfl j 1:: LL ”JD—S: 0.00% M=EDI .3 . "I i i M AVE LPDKIO :Qifiolal/ PD”: ~io<jm (00$!) -; ll HE‘ 1 Unfortunately, 0H'(aq) also absorbs the laser light! The concentration you calculated in (a) makes the experiment impossible to conduct. A suggestion was made to titrate the solution against a strong acid to reduce the [OH‘] while leaving a large enough [CDT] to do the laser experiment. .._-u—n_ Consider the following situation: 0 (b) 100.0 mi. of a 1.00 t 5 solution of NaCN, a weak base, is titrated against 1.00 M HCl. Write the reac o ; u' -_ on, and calculate the concentrations of all the anions in solution, Cl', CN and l - a (jar OH' present after 25.0 mL of HCl has been added. W] . - CB ptsfit‘i—‘E’h‘ (N _ —1- H u ———-4> HW + Q\ Ifiifl OJ mol ZSMLxloom ..._ __ 2 0-013 as. 0 Dana o. 0.01? wt 0.015% NM £9. Posh ._ 0.0;»: "h, EIMEV— I %‘%§- —‘ 047.5 (*6_”03 _,_«---::.:—‘ Q .8; 3; ac. pts) 5 pts) Question 11 continued) (c) The probability of absorption of light (the absorbance) by a particular species is given by a simple relationship known as Beer’s Law (you have encountered this in the CHEM 105B laboratory) ‘ Absorbance = a I c From this relationship we can see if the titration was successful and now allows us to perform a study on CN. The e for each anion at the laser wavelength are: 2(01-1') = 1000 M“ cm'], s(cr) = 16 M“ cm“1 and 3(CN’) = 0.3 M" cm". If the solutions are tested in a cell with l = 1 cm, calculate which anionic species absorbs the most light before the titration (with concentrations in part (a)) w which absorbs the most light after the titration (with concentrations you computed in part (b)). Show all your working. LEM” L a C A“,- 203(08} (Lcfl .; C). 8 Act/r :L-"O)(\°°°)(O‘OOKQ; 4 d———— Oughbm AFTeYL Pier =( ‘-°)(0-X\Ko.g); 0&8 onr — (LOX Iomiltw‘xw“): 0mg ’- Aq" 2 0-0) (Ab) (0.2,): 3-7— C (d) Was the titration successful in making CN the main absorber? Explain your answer. What safety considerations might you alert the student to before starting this experiment. ND —— 9325 CV absw/brrvvm‘rg ’04, las’kriigbl’. (SW-“MM V1 Mex/8 Obie/lam L3 OH”) 13 Question 12) (25 points) A compound (X) contains one mole of platinum, one mole of thiocyanate (SCN'), one mole of chloride, and three moles of ammonia. The compound was subjected to the following tests: (X) + H+ we no reaction (X) + Ag+ ——-> 1 mole of AgCl (X) + Fe+3 ”6 no reaction or color change (Hint: Fe+3 + SCN' would give a red solution) (a) Write the molecular formula for the compound. [Pt C “WA 3 (3%)] CL W) (b) How many d electrons does the platinum metal have? Ptzj- _ 0‘3 % pts} 14 ptS) : pts) Question 12 continued) (0) Further experiments show that the compound is diamagnetic. Draw the correct d orbital splitting diagram for the compound with the correct number of electrons in each orbital. What is the shape of the compound? t5 m3 anew/W when“? O‘TWN? _ ' 5:0 WW . 4t 4% (d) Draw all possible isomers of this compound. WRouG! _". NH»; l H3 N _ Pk _______ g C M 20 Week \ 1- TM NH!) fit—O toad" ;‘ ‘ NHg _\- l s M I H314 _. m/ __ we: SON” WW l l J' ““213 NW3 ...
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