CHEM105b - Exam 3 (2008, Spring)a

CHEM105b - Exam 3 (2008, Spring)a - Chemistry 105b Name Dr....

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Unformatted text preview: Chemistry 105b Name Dr. Qin Last 3 digits of SID# : , Spring 2008 Lab T.A.'s Name Hour Exam 3 Lab Day & Time First Letter of Family Name uestion Score rader oints _— _-- _—_ _—— —_ _—_ —_ __ Instructions: 1. There are a total of 10 pages. Count the pages before you begin. 2. You must use black or blue ink. (No pencil, no whiteout, no erasable ink.) 3. If necessary, please continue your solutions on the back of the preceding page (facing you). A CADEMIC INTEGRITY (Please sign below): certify that I have observed all the rules of ACADEMIC INTEGRITY while taking this exam. Question 1: (30 points) Choose one correct answer for each question below. (10 items; 3 points each) (I) For a solution equimolar in HCN and NaCN, which statement is false? a) This is an example of the common ion effect. The [H+] is larger than it would be if only the HCN was in solution. c) The [H+] is equal to the K3. (1) Addition of more NaCN will shift the acid dissociation equilibrium of HCN to the left. e) Addition of NaOH will increase [CN‘] and decrease [HCN]. (2) Which of the following solutions will be the best buffer at a pH of 9.26? (Kgl for HC2H302 is 1.8 x 105, Kb for NH3 is 1.8 x 105). a) 0.10 M HC2H302 and 0.10 M Na C2H302 b) 5.0 M HC2H302 and 5.0 M Na C2H302 c) 0.10 M NH3 and 0.10 M NH4C1 5.0 M NH3 and 5.0 M NH4C1 e) 5.0 M HC2H302 and 5.0 M NH3 (3) Which of the following solid salts is more soluble in 1.0 M H+ than in pure water? a) NaCl CaCO3 \c) KCl d) AgCl e) KNO3 (4) The correct mathematical expression for finding the molar solubility (S) of Sn(OH)2 is: a) 252 = Ksp b) 253 = Ksp c) 10885 =K Sp 3 : 4s KSp 8 e) S3 = KSp (5) A weak acid HA has a pKa = 4.74. Consider the titration of 50.0 mL of 0.10 M HA with 0.10 M NaOH. When 25.0 mL of NaOH is added, the pH of the solution is a) 2.87 b) 3.50 @ 4.74 d) 8.72 e) 11.96 (6) Which of the following resu1t(s) in an increase in the entropy of the system? I. Br2(l) —> Br2(g) II. NaBr(s) —> Na+(aq) + Br—(aq) 111. 02(298 K) —> 02(373 K) IV. NH3(1 atm, 298 K) —> NH3(3 atm, 298 K) a) I b) 11, IV c) I, III, IV a I, II, III, IV , II, III (7) Which of the following statements is always true for a spontaneous process? a) ASS),S > 0 b) ASsurr > 0 @ ASuniv > 0 d) AGSyS > 0 e) AHSys > 0 (8) Which of the following statements is always true for a chemical reaction at equilibrium. a) AGO = 0 AG = 0 c) AGO = -RT 1n (Q) (1) AG = -RT In (K) e) none of the above (9) Which statement is true? a) For a given process, ASsurr and ASSys always have opposite signs. b) AH° is zero for a chemical reaction at constant temperature. @ Machines always waste some energy. (1) As long as the disorder of the system is increasing, a process will be spontaneous e) If Assn” < —AS sys, the process is at equilibrium. (10) Which of the following is true for exothermic processes? a) ASsurr < 0 b) ASsurr = ~AHsys/T c) ASsurr = 0 d) ASsurr > 0 two of these Question 2 (14 points) A 1.0—liter solution contains 0.25 M HF and 0.60 M NaF (Ka for HF is 7.2 x 10—4) (a) What is the pH of this solution? H? :33 Wit ’ I Cuflo : (um, 0:30“; Mariam-40M (b) If one adds 0.30 liter of 0.20 M KOH to the solution, how much does the pH change? k0” {\QW’LMQC‘LQ />( moles 0.? Ix: otgomx 030 L :. 0.0L, moles GQEM flflutm K? zwfim :Hfl'o :. L02KM><mL~ofiQ mafiédfl/lelb I ’ e 0lele Emlo : m + opt, mom/(M, Nam 1?]; (WM ' ’ - 0.509 , [01° " mil, ' PM“ * £03 OWL stion 3 (6 points) For the process CHCl3(s) —-> CHCl3(l), AH° = 9.20 kJ/mol and AS° = 43.9 K{ eriQ; LY‘: M, M Que J /mol/K. What is the melting point of chloroform at standard state? WC math?" m was 24 50 AQ :— O AHO’ “’50 3 ° ell ll: 43" Question 4 (12 points) Given the following KSp values PbCrO4 KSp = 2.0 310‘“ Pb(OH)2 KSp = 1.2 x 10—15 (a) Which substance has a higher solubility in pure water? Show detailed work. (Continue on the back of this page if needed) _._> 2+ 7“ a H, got (5) < PL («p + We“ (ts) . i D I f‘ I X? “A / (7 a 0‘ 9‘ 0 6M 56 x \ 1 (X j/t"l’:o‘l\ / a [04 K FM Haw 0W“ 7< 2% 4 f N 6W“ O< (MM Io’ ) fi 2 v? 2’ 1 './§ Ks? 2'- Cszl] COW] 1 9<~ (190m )4‘ x o -1,” A§§W_ 2% width go 17H (0 ...— 2x 1 w: 00 (IXt’OqYJ-Y’X'R“) 24%;: NW oc : m >< Io" moi/L. =9 mm Sow/:7 Wale: /o"*/zx:isw3:ois'/a at. _ 7Amfwe Malawi 1m his/W WW?me 6 (b) Without a calculation, predict which substance will decrease its solubility as the pH of the solution increases? Clearly state your rationale. (continue on the back of this page if needed) 95 Crows) :3 9191*(«p -t— aroma?) Q PMWL g...) r29” Mg) + .10 M' (a?) @ PM? CM]? ,@ will mfi % Mt rglol’lly, Solubili‘E/é Question 5 (12 points) For a protein, the unfolding process can be represted as folded protein (F) : unfolded protein (U) (a) The equilibrium constant K for the unfolding was measured as a function of temperature (in Kelvin). A graph of In K versus 1/ T for this reaction gives a straight line with a slope of —l .352 x 104 and an intercept of 14.51. Based on this data, calculate A6” for this unfolding at 50 °C. Assume AH0 and AS0 are temperature independEt. A H 0 6 Sq j; r ...——-——-— 4’ a a m K Q, 5520f: -; v ‘3’: I, . '3be W K 4' AW : L552. an wak 1'. HZWIo 3/09: '3 l/IEGYMVE :2 fig.” 3.4.5! 435°: (craft >< 32;“? '—'— IL°~l7 J/mofl/K p 0 a T o , T5362 337/3 K AQ :: 0H ’ OS ) Li. Acf t. l.lLWXl0$~3L3YlZ°~\71:HVZXW j/mofl (b) Now compare unfolding of protein A and protein B at standard |nK Figure 1 |nK Figure 2 state. At the folded state, both proteins have only 1 configuration (i.e., l microstate). At the unfolded state, protein A has 2100 configurations, while protein B has 250 configurations. Based on this information, which of the figure on the right can not be true? Clearly state your rationale. autism, g: : g: c o m 1 Wiaals’mie, g: > S8“ become A W W UK A 16 W'fibwrwnj. AgA2§A «g: > gg’ga 24$; C‘ng 43.5.“: 1;) :AWcWJC {VA 7 8 "76” K $15.2 WM, be {7146. protein A .5.“ w... “‘5...” a..." m. .,,_‘ protein B protein A N Question 6 (12 points) For the dissociation process CH3C02H(aq) : H+(aq) + CH3C02—(aq) at 25°C, Ka = 1.8 x 10—5 (a) What is AG° at 25°C? AGO: 'K'lflwk T We :LMWVBNC 2: WM (b) A solution has [CH3C02H] = 0.10 M, [H+] = 2.0 x 1(r8 M, and [CH3C02—] = 0.010 M. What is AG at 25°C for the dissociation process? Cwlco’] CM] 0.0! xwwo' L //———-“ : W :2 J ‘2 comm] 0 (9: e103(10'? Question 7 (14 points) Given rxn #1 H2(g) + 02(g) :2 H202(g) K; = 2.3 x 106 at 600 K rm #2 2H2(g) + 02(g) :2 2H20(g) K; = 1.8 x 1037 at 600 K Calculate AG° for H20(g) + (1/2)02(g) =———-‘— H202(g) at 600 K. (Xn m ML + 0,, g2 HLO,’ 435?. gm Ms? M My 22 how Wioz 12:02 «km/O :3. 041,01 0 Useful Physical Constants and Equations Tc = TK - 273.15 R = 8.314 J K'1 mol'1 Kw = 1.0x10'l4 at 25°C PH =PKa + 10g10( [A_] / [HAD Thermodynamics ASuniv = ASsys‘i'ssurr Assun’ = ‘ AHsys AG = AH - TAS AG°reaction = Z np AGf0 (products) - Z nrAGfo (reactants) AS°remion = 2 11p 8" (products) - 2 nr 8" (reactants) AG = AG°+ RT 1n (Q) At equilibrium, AG = 0 therefore, AG" = - RT In (K) and MK =~AH°/RT +AS°/R 11 ...
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CHEM105b - Exam 3 (2008, Spring)a - Chemistry 105b Name Dr....

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