Chem271_S10_x2_key

Chem271_S10_x2_key - Chemistry 271, Section 22xx YOur Name:...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemistry 271, Section 22xx YOur Name: if; g, Prof. Jason Kahn " University of Maryland, College Park Your SID #: General Chemistry and Energetics Your Section #: Exam II (100 points total) April 7, 2010 You have 52 minutes for this exam. Exams written in pencil or erasable ink will not be reugraded under any circumstances. Explanations should be concise and clear. I have given you more space than you should need. There is extra space on the last page if you need it. You will need a calculator for this exam. No other study aids or materials are permitted. Generous partial credit will be given, i.e., if you don’t know, guess. Useful Equations: K. = [Ham/[HA] pH = —10g<[Hn> Kb = [HA][HO”]/[A‘] Kw = [n+1 [Ho-1 pH = 13K. + log [A‘]/[HA] AG = —n3‘AE R = 0.08206 L-atm/mole K 0 0C = 273.15 K aneq = —AH°/(RT) + ASO/R AS—q/TZO R: 8.314 J/mole K: 1.987 cal/moleK 521can AG: AH—TAS 1: = L(] —v2/c2)’é fr" : 96500 C(oulomb)/mole 1 V : 1 J/C AG" 2 — RflnKeq W = NI/(H nil) I’ll/I10 = exp[—(Ei—E0)/k1] TM 2 AH°/[AS°+ R1n(CT/4)] 2.303RT/5" = 0.0592 Volts at 25 °C E = E" — 2.303(RT/n5‘7flong Chemical standard state: 1 M solutes, pure liquids, 1 atm gases Biochemical standard state: pH 7, all species in the ionic form found at pH 7 Honor Pledoe: At the end of the examination time. please write out the followincr sentence and sign it, or talk to me about it: “I pledge on my honor that l have not given or received any unauthorized assistance on this examination." Chemistr 271. section 22xx Exam ll 4/7/2010 21'? 1. 112 pts) Multiple choice: Circle the single best answer for each Question (A; 4 pts) The “predominant configuration” is a. The microstate that has the most uniform distribution of energy. b. The microstate that corresponds to a Boltzmann distribution of particles. 4}! ((9 A set of mierostates that is more like! y to be observed than any other possible configuration, at equilibrium. (1. Only observed when systems are small enough so molecules can be enumerated. 6. None of the above. (B; 4 pts) The Joule experiment on the expansion of an ideal gas into a vacuum showed that a. The entropy change for the isothermal expansion of an ideal gas is negative. *‘1 The enthalpy change for the isothermal expansion of an ideal gas is zero. c. The free energy change for any expansion of an ideal gas is negative. d. The heat transfer q for expansion of an ideal gas into vacuum is positive. e. None of the above. (C; 4 pts) State functions are useful in thermodynamics because a. Crashing them gets us into the news. + L( The value of a state function for a system is independent of the path taken to arrive there. 0. The use of state functions allows us to avoid doing any work. d. They are independent of temperature. 6. They are independent of concentration. 2. 28 ts Short-answer nestions (a; 6 pts) Briefly describe what a DNA microarray is and how it is used. — A M452“ Qua/7 if a Elf/l1 M SufFMJL ma. micale olifl‘ww'f D‘V‘i @ MAW Lou/t FYI/EFL WM dv" May-(’M/ fMOA_ z W M iW {Mi/ti-MA UCLLAKNJ‘ (affix/7‘ S’flgf“ v’ "_‘ "i W ’pi‘ziu’c W l4 Sc:wa para. 6..» {Ma .3 WW is it... M? M ’il/4 {KW Lyéflr‘i’wvfflh {73 “VA S/c‘l‘ i} 1445131an- I . U” M uJ flu, Cm/[tsiphk if; L SW Mi'xbrgj."(. Vii-43') a". I“. or. renal. W w v? I, Scoreforthe page Chemistr 271 section 22xx Exam ll 4/7/2010 3/7 (b; 5 pts) What is the difference between a voltaic cell and an electrolytic cell? What metal is obtained from its oxide using an industrially important electrolytic cell? —~ A W: U0 H61 a“ Mtg; q SWnfifl/LJ €(eaflndLamLo-A roubfn.‘ ‘lu [iambic uyéul Lauri/m W-Curm—f, ham; 4-...“ @ Jig/192(th ceU W M “CA/M Veltln?‘ Faun/(A941 off-Vt « nan-9pr Wat/151% @-— Aiummum (Hail’HcmmH') Mr amidst/a, mycwwoflmgfla (c; 8 pts) Give an example of an exothermic ordering reaction. Briefly explain why exothermicity per 36 makes a reaction thermodynamically favorable, but on the other hand explain why at high enough temperature an exothermic ordering reaction will become nonspontaneous. @" pr»ch «gill? * {6M W— Wait” #6624? ' ’— gafik/Q/Mtl'CIr? mu», fiA My! “JAR/I M +4 {it 6’9 “wwwb-QL’” M Mchch L- @ .._ {mm-76‘ Tl A61: +15” (fawn, 55‘ ;; nféfhvfi W i} W {L4 (MM; Mama: Ik L‘— __~_ Ski/M?” KW “5 TT/Eoiwrr‘} WED (d;4pts)W a 18 required to escri earnicr e f’asWfil’efiffiEfiE"? @9’ [ ~ TL: Mm, Vela-.3, “raw, use/kw, M Lama rm 34 m’” @b’“% WW MAL-:1“ L: swififl. (17,1 E W a ocfijo/"EJ 5a «,7 414/22. HIV, (e; 5 pts) Define a “high energy bond” or “high energy molecule” and give an example. A Lrgbr LAM“)? LW-fil M“ W. 1% @ 01h lie/YUM :i'v flint W “10% 5h,“ Nay-n»! my); +7, bunk} butt 3‘ CMwa l‘D Willa Mn HEM. @ PVT?) yi/VTSFIIVthAyd/r‘w, {LVMtutc‘g/Jrl J, Fez/087A)?“ Score for the page—gig 1/._ a? W‘" Chemistry 271, section 22xx Exam “4/7/2010 4/7 3. (22 pts} Applied thermodynamics DNA hybridization is described by the equilibrium W + C :2 W'C, with equilibrium constant Khyb. We assume that AH" = 40000 cal/mole and AS“ = 7185 calimole K. Recall that the Tm for DNA melting is defined as the temperature at which 0. = ‘/2, i.e. half of the total concentration CT of strands are in double—stranded form. Thus [WC] 2 (CTX l/é)/2 2 CH4 (we divide by 2 because 32 Molar W-C has a strand concentration of 2y M), [WI : (CTX 1/2)]2 : CT/4, and [CI : C714. Therefore, the equilibrium constant Km 2 [W-CJ/[W][CJ = 4/CT at T = Tm. For C,: 2.5 uM, the _ v - ; o 1:45]: C or o-osCT thelmodynamic values above give Tm — 55 C. [W 4’ . We want to measure the temperature T.,_05 at which 0!. = 0.05. / common ___- (a; 4 pts) What is Km in terms of Crwhen 0L 2 0.05, so IWCI : CT>< 0. 5’2“? |( a fLUCJ CT 10.0g/L- (Ll 0'1” LYL a- __.4~"'—'*-m_., vs _ ..____________' = ’H [Wj [C] FCCTX 035b,) L I“ CT . 0.953 CT @ e: (b; '12 pts) The van’t Hoff equation we used in class is ln{Km) = (-AH°/R)(UT) + (AS‘U’R). First, plug in the expression for Khyb at Tm for Km and “Tm” for T. Second, write a similar equation by plugging in the expression for KW, at Tom for Km] and “T0705” for T. Subtract one equation from the other and solve to obtain an expression for T035 in terms of Tm and AH". Give a numerical .wer for TOM. l (t :1 ~41? 1 +930 ._/ [m‘q‘y'il 35_[\ + in [L‘l’ @(l) {L " ' *(ig’rfigj‘kflflddo h ) TMMK Ll 2";th Lye—550 - l J! @ih < /CT’B Try- K * 328K liolgvtd .lll ,u WL of F _ " on <0 ai— ow, +2 - Ll CT“ _. Aldo .L __ J... -———~ __ - .——----~ -— _‘ _._ FL @ in C1— OJH3 (L TM Tans + 9/ , [C MC. K. AW“ 04” [M Too! [Q Chemist ' 27] section 22m Exam 11.4/7/2010 5/7 (c; 6 pts) As you know, oligonucleotide Tm tends to increase with length. In detecting SNPs (singlek fl} nucleotide polymorphisms or singleibase pair differences between people’s genome sequences), we use ...—-— h bridization to oil onucleotides that are as short as ossible but no shorter B ed on o t' izi . . Y g p : ~ - GEM 'W ' (pg bridization signaland specificity,explgj temfft’.p SWdC’JE-féf‘fll ‘ ZA I) ._~ Tin. sLurl'er lb: it?” I W , (Alfie/M bah/Van Carr/M GA’J that.»er L’f.y."Mk lie all?) 17 ‘l‘W [Lu-c will 44¢ W “A ghlfl‘r 'l‘. [guil— 5 flat (lth C? {be ski/l" 3'44 “MA cf. Ry h g, 7—- "I. 13L jig/Am 10/479 “mt” +31 $ij 3 H. weal‘JEMV Tfi’e} 7, m r —' ’ 7m with 7 7‘1 £7“an ' 4. 22 ts Thermod nami tamer» few I, . _" :3; l . _ Witt Mam ’\ A; The Second Law of Thermodynamics states that the entropy of the universe indrease/ for any spontaneous Cf E? 34:. ‘3? 1M. Liv/L441 process: ASSySm + assumundmgs a 0. We derived the “master relation” ASSN,“ — qsystem/T 2 0 to convert themj 2nd Law to a form that refers only to the system. Emma ~ » > (a; 6 pts) What restriction does the master relation require for the nature of the interaction between the . _ 1'» ‘ ‘system and the surroundings? Briefly and qualitatively ex 1 in the orio'n f the sec nd term’s r . __ farm M 4 MUM . a «r « - Er ~ farts son, m: @725 WI! ‘ dependence on “UT”. 1) 5M pmwm [We h f \ , ' L" 175 Cat/1% 055W}: an; ' l (a enmfcdtrjyka ,- dz " TLK 5ygl‘5’ln Camran- tKCLga? [Mnflx 5" r ‘ / was Wale/fiat M “Q Wt - MA “MW rtfil@fis T T” l" “W "" We ‘3’ a aw m. if g M Maw. 72% WW" 6/525 0(t7mCV‘Z};L€/_Zgl tar... al doom/aw! syskm MM aw ct r \ I H W V‘s-1'2: t ( * #35ij m6: ‘ - 'U’ltzfi (b; 3 pts) For a reversible process carri d out at constant T and? we er} a e to replace the master relation with the equation AS — AH/T 2 which motivated us to define the Gibbs free energy G = H — TS How . . . -- . - 9 Via .rjw‘iitlmfls ex JMM " does this give us a sump e ondition for spontaneity. } Kg . 5 J» A 53 ' M H” > #153435" 2:: team ,. M- at: Chemistr 271 section 22m Exam ll 4/7/2010 6/7 We showed that GA 2 G"A + RTlnPA describes the molar free energy of substance A. (c; 9 pts) For the reaction A <> 2 B, use this equation to calculate the free energy change AG for converting a mole of A to 2 moles of B. At equilibrium we know that AG : 0 for any infinitesimal change. Use this to demonstrate that a ratio of partial pressures must be a constant, KP. M“: 4561 (Us) ~ 6(a) P @ 4361 —= Magrarinra)— (a; HQTM a?) lgag—G’: +ZaTiMP5 “[ZTIM&4PA W _ Ag,“ + (21121.“ Pg—lnpfi) A a at @156» 2 as +tt‘ntr‘il= AGOt'ZTW 2. O: AGO+¢TM(%> p6?- (30L was mmuri‘ld '- .- T ln<"“ OHM ‘2 50 c a f”) f .. Pa r WW m» heater; ((1; 4 pts) We see that even if the standard state molar free energy of substance B is more positive than that of substance A, there will still be some B present at equilibrium. Give a qualitative argument based on N QMWQ, a la? («Gee/{7% gamma/i4. T4 We cmm/FWJNfifml-DB) ’lL-e/LMQEW'F I i Jwtlm microstates that rationalizes this conclusion. Score for the page 5 Chemistry 271. section 22):): Exam [1. 4/7/2010 7/7 5. (16 pts) Electrochemistrg Consider the reduction of the iodate anion 103‘ to iodide, 1'. (a; 8 pts) Balance the half-reaction in acidic solution. Its Standard reduction potential E"red = 1.085 ‘V. Why . . . . ... {t 7“ f is the 1odate anlon a relatively good oxrdrzrng agent? if ' ~ / “it u» to: A; 1-.2 r“ .. .— :5-7 rg-‘rgi/mug/wni -,/ Mfffixfiwii': :5 J "—9 I (12'. I ;'-'."’r,j‘:._; GcM’E/l IZJM 0: $03’A'I 4-3/9” % :11? __ -H ‘4 oil/M H: 2130;4— é HJF—e 2r +3HLO “mm a new ow: rem ta .9 a) W? Th “we. in: i an Icy“ Cc +35“ - Ft ‘3 ’ ' 4/J u ri.‘\ /"1;.x ..‘. 1' @ gfccf/‘m:fim ?gfie(40W—:\:— CUE-“M e"'V> fled (-94—- n- Wm at S‘iVW‘a, Wm? [vial W—I-OKQ/kg—c w ¢{S(, The reduction of ferric iron to ferrous iron 18 described by Fe+3 + e‘ —> Fe”, E°red = 0.771 V. (b; 8 pts) What is the net reaction if 10371— are in one half cell and Pea/Fe+2 in the other, all at standard conditions? What is Bow”? /r Io; tug/rt +ée"—9 I'lk BHLo 5:“ :+(.o:5\/ he” re“ —a Fe +7“ 130,”: : +O‘H’IV @pfid‘e/K ltd. Rf} WWI-x pg *1...) (—1” re” EQMGX : ~6.'r’5/l\/ Sail:ng deem 6 P{*7—_;épe+3+6€ Few:..- jam, orientate »i m ‘~v r. J _ f‘ b P .— = osV~a’+?I\/~031L{V + ECG“ 8 "‘\e--'I“"'_" . Z Scorefor the page—LE ‘3 ii-"H'i’j/r- ‘ 5‘: t7? (,1 x’ i“ 7 I?/"»"' i ...
View Full Document

Page1 / 7

Chem271_S10_x2_key - Chemistry 271, Section 22xx YOur Name:...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online