CHM 131 - Exam 3 - 2005 Key

CHM 131 - Exam 3 - 2005 Key - UNIVERSITY 0 F ROCHESTER...

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Unformatted text preview: UNIVERSITY 0 F COLLEGEOFARTSANDSCIENCE ____________ ROCHESTER DEWNWW Chemistry 131 — Preliminary Exam III 22 November 2005 8:00 am. to 9:30 am. Name: Hfl L ID #: 20d Workshop Meeting Time (day /hour /room or bldg) (2 pts. Extra credit) This exam consists of six (6) questions, one extra credit problem, and 2 pages of potentially useful information. Please check BEFORE you begin to make sure that you have a complete exam. Please do all your work on the pages provided. You may use the backside of pages for additional work, but please tell us that you are doing so (i.e. an arrow with a note in clear English saying "more work on back"). SHOW ALL WORK. GIVE UNITS for all answers that require them. Partial credit can only be given for those answers for which work is provided. CIRCLE YOUR ANSWER. We ask this so that we don't have to interpret what you intended as your final answer. If you make any 5% approximations you will n_q_t receive credit if the approximation is not valid. Also if you need to use the quadratic formula, you need to SET UP THE CALCULATION and INSERT THE RELEVANT NUMBERS INTO THE FORMULA. Credit will L0} be given for answers that appear out of nowhere due to a fancy calculator. GOOD LUCK! Prob. W.s.' ;1(30) §2(20) §3(25) §4(25)' 5(25)'% 6(25) ?E.C.(5) Total E 1 ; I i 5 i Score a University of Rochester Rochester, New York 14627-0216 1) Consider the following 6 solutions: 0.1 MNaF wB 0.1MHC1 5A 0.1MKOH $8 0.1MCH3NH2W5 0.1MKC1 /\/ 0.1MNH4N02 WA a) (3 pts.) List the solutions or solutions that have a pH < 7.0. O~l/V\ HC\ )O‘lM Wi’lq/VOL b) (3 pts.) List the solutions or solutions that have a pH > 7.0. mm NQF, CHM/+47.) K c) (3 pts.) List the solution with the lowest pH. 04 /v\ Hm d) (3 pts.) List the solutions or solutions that have a pH = 7.0. o . \ /\/\ KQ e) (3 pts.) List the solution with the highest pOH o \ \ /V\ HQ Consider the following endothermic reaction at 373 K: heat + 2 NOBr(g) S 2 NO(g) + Br2(g) f) (3 pts) Circle the way the equilibrium position would shift if the temperature were lowered. @ RIGHT NO CHANGE NOT ENOUGH INFORMATION g) (3 pts.) Indicate how the value of the equilibrium constant K would change if the temperature were lowered. INCREASE NO CHANGE NOT ENOUGH INFORMATION h) (3 pts.) If the Kc for this reaction is 2 at 373 K, and the value of the reaction quotient Q = 0.2, circle the way the reaction would shift to reach equilibrium. LEFT NO CHANGE NOT ENOUGH INFORMATION i) (3 pts.) If you added more NO to the flask at constant temperature and volrmre, circle the answer that best describes the effect on the equilibrium amount of Br2? Br; increases Br; decreases Brz stays the same NOT ENOUGH INFORMATION j) (3 pts.) How would the equilibrium amount (say # of moi) of Br; change if the volume of the flask were increased at constant temperature? Brz decreases Brz stays the same NOT ENOUGH INFORMATION 2) a) (10 pts) A solution is made by dissolving 18.4 g of HCl (molar mass 36.46) in water and resulting in a total volume of 662 mL. Calculate the pH of this solution. ,v \ 22L : .505 mol Ha ammoh O! a 1% 13% X games 0 0.561L g Hzfgm]5f§d :7 C Hf} 3 0.7é3/V\ 1 PH: —703 CHTA :— § {0743 _ b) (5 pts) Calculate the pH of a . x a(OH)2 solution. 153w} ma Baolh K @0\ OH _—_ A 70” /V1 OH” . Mo\ BaO‘HL //K 1/ 2570745 91%.00 9) PH: [‘MO'FOH S 3) You make up a solution of nitrous acid (HNOz, molar mass 47.01) by adding 0.470 g to water with a total volume of 1.00 L. a) (5 pts) Write a balanced net ionic equation for the dissociation of nitrous acid 1n water. Hamlet )+ HLOCZB :fi H601; H‘ W0:@?) an i F M H/l/O’[email protected]?\ A [email protected]\ ‘lfl/VOL (a7\ b) (5 pts) Write an expression for the equilibrium constant for this reaction. KO: CW3 13/07:} firm/07:3 c) (15 pts) What 15 the pH of this solution at equilibrium? [HA/0%), :(0 WOS/Hm 3/“ l: 0 Ole m l L weigeC/‘i’ H79 au’i’OW/M Lia/'hOfl) CHM/01,1 *[H :33 4” E/WJX I at \ «o C a? >< >( E o.o\—>< >< >< Ki: 1L ,_. Lisa/5‘1 51ml; @T W/oléK L 0th ”L/ (L ’4 >< ,‘WD éwc/g/DK 0P ><+797pr790 :0 x:— (7557011 ~ Hem”) +C9X’m/a’é) )/3\ X: (7?76fli'7-’7//03)/z rt” K::—/— @674 373'/05/9~/ /7€05 4) At 1000 K, a sample of pure N02 gas decomposes according to: 2N02 (g) S 02 (g) + 2N0 (g) b) (5 pts) The value of the equilibrium constant Kp is 158 at 1000 K. What is the value of the equilibrium constant in terms of concentrations (Kc )? [Cult/V011. ‘- (WLH P01, 4:; i- :(IS‘X) ,1! C725 QM” (ea/M)?“ ’ Butt IZT accompany P/KT : 7 c) (10 pts) At eguilibrium the partial pressure of the 02 is 0.25 am. Calculate the pressure of the NO and the N02 in the mixture. HINT: No ICE table necessary A , we, Know 1 Mol 0L» :rflo\ /l/O Parm€&\ K a: Web”, (9 ca cd‘m Via—(‘3 W47 $17 d) (5 pts) If at the same temperature as parts (b) and (c) (1000 K) and at eguilibrium the partial pressure of the is 0.50 ‘ alue of KP? Hint: No calculations necessary. 5) Carbon dioxide (molar mass 44.00), when heated in the presence of solid atomic carbon (graphite) (molar mass 12.00), forms carbon monoxide (molar mass 28.00) until an equilibrium is reached. a) (5 pts) Write a balanced chemical equation for this reaction. Cow + €07,629: 160%} b) (10 pts) You put excess graphite into a closed container with 0.2 mol of carbon dioxide and notice that at equilibrium, the aver e molar mass of the gases in the container was 35.00 g/mol. (Note: The (weighted) average molar mass is the sum of the product of the mole fraction of each gas and its molar mass) Calculate the mole fractions of the carbon dioxide and the carbon momde 35: 9<¢0M¢o+ >6“,le —~ X6018 +WOQO 0\\$O Xw'l‘ Kay,” /l 0" fie :J XéoL :7 35:; ll~><cOL) aw L/LOQo :7 5522843797.”?sz— Zléxwt or OCng7//ézot7375’ 7660 1"7//é: cz//4 _, 0 512531540 c) (10 pts) What is the value of Kp if the total pressure in the container is 11 atm? vafi got/EOE: (P XCO)L :fi' 74; PT><CO 6) 0.100 mol of ammonium chloride (NH4C1) and 0.100 mol of sodium cyanide (NaCN) is placed into water with a total volume of 1.00 L. Using a special CHM 131 name-scope you look inside the solution and notice that there are 4 chemical reactions occurring in the flask. a) (5 its—ZWrite chemical equations for the 4 chemical reactions occurring in the flask. 6 ®1Haot4 Hsoléflt'ofilq l or HvOQZ) Q’HEYH‘OH—éfl /‘/ H%[email protected]‘13+Hv0d l :1 EWM/Hséflof ”Habit; Hgf)+~/I/#3éf) g 07¢ch) Jr Hflép :3 Wéz)l‘0rlé7\ @ NHatheflJr art/(42% : HWé‘fl ”VH3 (a?) b) (5 pts) Calculate the values of the equilibrium constants for all 4 equations from part (a). (Hint: K for a reaction which can be written as the sum of several other chemical equations is given by the product of the equilibrium constants for each reaction in the sum.) (9 KW: [Hijéji’i'3: [‘0 75/7 @ Z: RCCKJHOJN ll“ Rchl‘ibA 3 1L @ KQZ 3:»: %:5475b ' RCA/arse 5C Roam“ KM, /.o' ”IL , ,, 54%,: Ka KL/KW . / @ Kb: / —» .27; _ Z6 /0 A Q‘ér/ '§)(§.é'/pr’6) flow #1 .. W flwe d) (15 PtS) What is the pH ofthe solution? L1 ‘ Rims-f fZSTulfl\P/7:L\j gmi AOMROCi‘QS an) C’VHLle Z (CA/”30 : O‘l/Vl Kt? S MAS our. ML;- gen/K :f Hé/I/ +/z/#3 K a CHCA/I/g/HSB 1'/ I OI! O" “'0 \IO ’ \ c, ~>( xx +>( 1‘K <0" JZWqu) E OJ~K ott—x >< >< : Q70 :._>_<__._ O\i‘-X 0r X 20675“ or Xrot075~ofi$x orx WSK 2mm” or fl Wit/154M475]? flow/14 44/“36: C/l/Hlf}: a I Po. 0L1? : 0306714 PH Aol‘tr‘mmccx b7 QXA 3 [/ar <51" K) i A _ «— J m... was —,~ CGH 3am; : Com.“ CCfl/‘J COlOSll ‘ otter/l0 £0111: ‘73? 3 9%:— I‘7100 " POH: at; :gi—i Extra Credit Problem (5 pts) The sketch on the far left represents the [OI-1'] present in an ammonia solution of molarity M. If the concentration of ammonia in the solution is suddenly quadrupled, which sketch on the right best represents the resulting [OH'] in solution? Please back up your answer with a calculation (assume the 5% rule holds). Answers without calculations will receive very no credit. j: /V\ 1 (DO “/0 WW} Q ~>< X K : XL K K /l/l'—'>( E /V\"‘>< ’L 530»ch M77>< 50 Kb/K j: of XOZW 31% ma LM/k Slrmle 8+5” haUé,’ Xe: Kt’i/Vl : 2W: 2K0 X 3 3: OH’ {A box Kg : 1K0: 66H; (m EOK Potentially Useful Information Avogadro's Number, N = 6.02214 x 1023 mol" e/m = 4/7588 x 10“ C/kg 2X where X = element symbol, A = mass number, Z = charge number Average atomic mass = X aiMi ; where ai is the abundance of isotope i having mass Mi Metals tend to lose electrons to become cations, M+,M2+, etc. Halides tend to gain electrons to become anions, X" The mass of N atoms equals the atomic mass in grams M = molarity = moles of solute/liters of solution number of moles of atom X = [sample mass (g)/molecular mass (g/mol)][# mol of X/mol of molecules] Number of moles = sample mass/molecular mass The Gas constant R = 0.08206 L-atm/(mol-K) = 8.3145 J/(mol—K) latm = 760 mm Hg = 760 Torr = 101,325 pascal For IDEAL gases: PV=nRT , STP has P = 1 atm and T = 273 K = 0°C For 1 mol of gas at STP: V = nRT/P = 22.42 L Dalton's law of partial pressures states P = 2 P, —b -: Vbz — 4ac 2a For the equation ax2 + bx + c = 0, the solution is x = Law of mass action: jA + kB 2 KC + mD, K = [C]! [D]‘”/([A]j [B]k) THE FOLLOWING ARE VALID AT 25 °C Name Perchloric acid >>>1 Ba(OH)2 >>>1 H droxide Nitric Acid >>>1 CHsNHz H drochloric Acid Sulfuric acid >> >1 NH3 Hydrogen sulfate 1.2 x 10' Pyridine 1.7 x 10‘ 1011 HF H drofluoric acid 7.2 x 104- C6H5NH2 Aniline 3.8 x 10"”— HNOz Nitrous acid ‘ — E E HC2H302 Acetic acid HOCl H ochlorous acid The following equations work all of the time: [H+][0H1 = 1.0x10'” = w; pH = Jogthfi ; pH + pOH =14 [H41— Mfl These equations only work some of the time: 1/2 pH = pKa; pH = (ka[HA]o) ; pH = (PKal + PK32)/2 ; PH = PKa + 10g10([BaSC]/[Acid]) 21 22 23 So Ti V 44.96 47.88 50.94 39 40 41 Y Zr Nb M 88.91 91.22 92.91 95.94 57 72 73 74 La Hf Ta 1 38.9 1 78.5 1 80.9 89 Ac 58 59 60 Ce P Nd 140.1 140.9 144.2 92 24 Cr 42 o (227) l’ 90 91 Th Pa U 232.0 (231) 238 0 ...
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