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CHM 131 - Exam3 - 2004 Key

CHM 131 - Exam3 - 2004 Key - UNIVERSITY OF ROCHESTER ~ WWW...

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Unformatted text preview: UNIVERSITY OF COLLEGEOFARTSANDSCIENCE ROCHESTER ~ WWW... Chemistry 131 — Preliminary Exam III 18 November 2004 8:00 am. to 9:30 am. Name: Eagi’ This exam consists of six (6) questions, one extra credit problem, and 2pages'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_9_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 y_o_t be given for answers that appear out of nowhere due to a fancy calculator. GOOD LUCK! ' viziProbTE W.S. 1(30) 22(20) §3(25) §4(25)' 5(25)"; 6(25) {110(5) g Total , , ‘ ' 3 ' 1 ' 1 ' ‘: ' z ‘ 5‘ ‘ Score _ E : ‘ E g L ‘a x i i i 1 ‘, i x v University of Rochester Rochester, New York 14627-0216 , ~ 1) Solid sodium bicarbonate in a closed vesSel establishes the following equilibrium at 1000 K: Heat + >2NaHC03 (s) S Na2C03 (s) + H20 (3) + C02 (g) ' ‘ a) (3 pts.) Circle which way the equilibrium position would shift if CO2 were removed. LEFT 1 NO CHANGE NOT ENOUGH INFORMATION . b) (3 pts.) Circle which way the equilibrium position would shift if solid Na2C03 were removed. LEFT RIGHT NOT ENOUGH INFORMATION c) (3 pts.) Circle which way the equilibrium position would shifi if heat were added. LEFT NO CHANGE NOT ENOUGH INFORMATION d) (3 pts.) What would the value of the equilibrium constant K be if the amount of C02 in the vessel were doubled (at 1000 K)? K/2 @ NOT ENOUGH INFORMATION of mol) ‘of H20 change if the amount of 2K e) (3 pts.) How would the Equilibrium amount (say # CO2 in the vessel were doubled (at 1000 K)? H2O increases H2O stays the same NOT ENOUGH INFORMATION 1) (3 pts) Which solution has the highest % dissociation? (55 M — i.e. pure) H20 1.0 M HF 1.0 M HCN 1.0 M NH3 1.0 M NaO g) (3 pts) Which acid has the highest % dissociation? (55 M — i.e. pure) H20 1.0 M HCN 1.0 M NH3 1.0 M NaOH h) (3 pts) Circle the solution with the highest concentration of [IF] ions. 0.1 M NaNOz (55.5 M — i.e. pure) H2O 0.1 M NaNOs i) (3 pts) Circle the solution with the highest pH. 0.1 M NaNO2 (55.5 M — i.e. pure) H2O 0.1 M NaN03 0.1 M NH4N03 j) (3 pts.) Label the following solutions as SA (Strongly acidic), SB (Strongly Basic), WA (Weakly acidic), WB (Weakly Basic), or N (Neutral) 0.1 MNaN03 0.1 M K2804 0.1 M H2SO4 0.1 M CH3NH30C1 0.1 MNH3 /\/ W5 SA we W8 , N2, and H2 at equilibrium at 300 K. 2) A reaction vessel contains NH3 25 M, [N2] = 0.11 M, and [Hz]= 1.91M. The concentrations are [NI-13] = 0. given by: a) (5 pts) Calculate Kc for the synthesis of ammonia if the reaction is N2 (8) + 3H2 (8) 3 2NH3 (3) Kc: CM H3931 (AAA E HA3 value of Kp for the synthesis of ammonia for the above reaction: b) (10 pts) Calculate 1.5. @149 1%}: 724—315 ,0~!\>\ swag: :a: fix: 03W— ; aqsej . , (PMs/H31 ., Ms W 3 gen?- " KO“ W "<P~Ll(PHL3LfiT)1 P P’T le—T 3) You are fascinated by some of the ingredients in your toothpaste and so you decide to make up a solution sodium fluoride. ‘ ' a) (5 pts) Write a balanced net ionic equation for the hydrolysis (i.e. reaction with water) of sodiumfluoride. _ f, __, + HZOQO 2:3. Hrfiaz') + OH 82”) b) (5 pts) Write an expression for the equilibrium constant for this reaction. K _ [to VJIH 21,3?2/ KL: X /800-K: Tw><¥cr H F) " 7‘ D- 70” 7X1 ‘ -— i i 5/) [0155qu X44 %\0 $0 i\3C/’/& :5. $00 ,5 my/o V 4) For'the‘ reaction: C(s) + 2 H2 (3) ‘3 CH4 (g), Kp = 0.263 at 1000 °C. alue of the reaction quotient Q when 0.100 mol a) (5 pts) Write a numerical expression for the v action vessel. (You can evaluate the CH4 and 5.00 mol of C(s) are placed in a 4.16 L re expression if you like) ' . P 'L (PM ... ©\\\L0t0‘5’9~\00273 : 3‘5, <1th 10 QCHH 7’ "'7 ‘11 l L b) (20 pts) Calculate the total pressure when 0.100 mol CH4 and 5.00 mol of C(s) are brought into equilibrium at 1000 °C in a 4.16 L reaction vessel. Rom L0“ Wt Knew Paul: 16' Iod’m Kégalr ZHL :‘fCHLr >(2. ,. Pfesswe I O 16“] of; CH, 43% C 2x *X N“ E 9-K 251% X: /]W gingk X 1‘5 177 AQQCAGHEA7O 970.05%) ‘ \A/Q +3.},{Q flai' (00+ 5) A solution contains a mixture of 0.5 M HF, 0.2 M HOCl, and 0,1 M HCN. a) (10 pts) Calculate the [if] in this solution. ' Moo 3 60:65 are t . 0d f P (rim H’ F > P W“ W ”19% Hugging cat} HF :2 if +6): » eiudihm‘um + ’I: -5 O ' , C 3X +Y 'l’X / XL 5: f". [1170 "1 X: M¢\H Kg ’ O—EfX/X 0 \J)’ P“ L. y Otdler Z 0‘03? ' 6.5 was?) b) (5 pts) Calculate the pH of this solution. /~) 9H :. ”)03 CH-YB :3 c) (10 pts) Calculate e [CN'] in this solution. goo—5 is Ci‘EFmvwo’l L7: ,_ 'I\ —» cA/Cqflfi- HGT) \OI/L’V— é‘HTJ is AQ—WMCAQL £7 HF>AC$$OLra+rom —— WC CQA Seq ““5 wl We, ICE Ckar‘i" ,__ ' +‘ V kit/U f C/U 17H X; malw‘ 01% 3: o \ 0 col? {y , o ‘ PW L C, ->< +‘>( +7< E o.t»>< >( otOVHK //0 KC LKVQOHWA K4400]? at‘nce/ K4 : 43W) ‘3 50 0‘1 *‘X' Smct \\ r/O f_._,_,,,,.///:’5 W; (whim 5%» s hastens, Weiwmjeox A’s/him «'4 (’14sz H'Z/l/ o\\‘$sa¢\afi0fly 6) Note: You need to make sure that you keep in mind the water autoionization (if ' appropriate) for this problem. _ a) (5 pts) Calculate the pOH and the pH of a 2.0 x 10 ‘3 M solution of NaOH. ”3AA Mach: 10 ‘/0'3/V\ 0W (av deomlfl’rg bx? 3/633 == - .7 2POH‘) b) (20 pts) Calculate the pOH and the pH of a 2.0 x 10 ”7 M SOluti on OgNaOH, . . Am ‘ "‘ 70’ /l/\ LA/‘m (j A )0 /y\ fi—‘om (he, NaOH W 531 @Hhs /D“0 T ,g' 2W0 Neg) 5m QCCOUJY“ ‘QDF 344W [email protected]\1/OHCQ7\ 30 flats Problem ”(km/G733 . YObL can 5:4. +1.04%”: LOW) of LOH’B V H—> an} I @4372“ g‘0 (/57COH;> —‘ KW ,,_ O ’ UNSC Z ‘(xafmu q flow 3 9W a . . I [(7% 77—7 ,fiMc—fffoo’f smce W 3 : 9W £on )6 o I ,— HLO ,7 H‘FOH X; m; OH :1: 1.0957 3.0757 Consume} L c/ Kw — L016 H : (IN/51x ( 3 o 75 —><\ 1‘ 3070 M+>< 411070 X» X13 ’10 v5 >(+ ;.O/o "‘1 so , 2m; ‘gofmwlq ’>< - LAO/673: (170/07) -m\¢z.o,a«) R X: 9.0707: 239/07 Hm We, +314»: 44,35... 9\ F690? 5\'ncc X4 3.070 ”I ~7 COMB: 3075 «X; laévo /Vl 11L (”ask f3 fie Samfi a5 (1) ~\ /./——\ {30%; G 62 PM 7.3? 5‘ E Extra" Credit Problemi(5_ pts) Ammonium carbamate decomposes as follows: NH4C02NH2 (s) '3 2NH3. (g) + (mg) 9 ' in a flask it is found that at 40 °C the total pressure of all ' ' ' 0.363 aim. Calculate the equilibrium constant KP. From fie; €345 l 9V; GET 52:44—ij Vd‘T’A’Q, (LC Same 9w l/P‘; (PLOL\<P~H3\1 A152 PT:O“Eé$: @L’f’PA/Hl‘ :Llfiom (Bop-l :LI-CPCOLv f. (3&2ng afm P60} :— OL%3 /5 ; Oklaiagw‘ BOP/,5"— ‘9 _'- KP , L1 (om/133: ”ZOO/yo 7/ KP I Potentially Useful Information , e/m = -1/7588 x 1011 C/kg Avogadro's Number, N = 6.02214 x 10 mol' ’ 2X where X = element symbol, A = mass number, Z = charge number Average atomic mass = 2 ai i ; 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 om X = [sample mass (g)/molecu1ar mass (g/mol)] [# mol of X/mol of number of moles of at Number of moles = sample mass/molecular mass molecules] The Gas constant R = 0.08206 L—atm/(mol-K) = 8.3145 J/(mol-K) 760 Torr = 1 atm ’ STPhasP=1atmandT=273K=0°C nRT/P = 22.42 L Dalton's law of partial pressures states -—b :1: 'Jbz — 4ac For the equation ax2 + bx + c = O, the solution is x = 2 a For IDEAL gases: PV=nRT , .4 P =2 Pi For 1 mol of gas at STP: V = Law of mass action: jA + kB Z [C + mD, K = [C]’ [D]""/([A]j [B]k) EEK—“EE- HSO4' Hydrogen sulfate 1.2 x 10' 1.7 x 10' ion , E!- H drofluoric acid m- m- IE'IM_ 4-8 x10 _—_ m m EGE-lf—m [mm-m 10 10. B81 12. C01 14. N01 16.00 19.00 20.18 13 14 15 17 18 . AI Si CI Ar _ 26 98 28. 09 30. P97 35. 45 39. 95 19 20 - 26 27 28 29 30 31 32 33 K Ca Sc ' . ' Mn Fe Ni Cu Ga Ge As Se Kr 39.10 40. 08 54.94 55.85 58.93 58.69 63. 55 69. 72 72. 59 74. 92 78. 96 79. Br9O 83.80 33 41 42 45 47 49 50 Sr Nb Mo Ru Rh Pd Ag In Sn Xe 85R. 47 87. 62 88Y.91 92. 91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121. 8 127.6 126.9 131.3 55 57 72 73 74 75 76 77 78 79 81 82 83 85 Cs Ta Re Os Ir Au Tl Pb Bi Po A1 Rn 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) Ra Fr (223) 226 (227) 58 59 61 65 69 70 71 Ce Pr Nd Pm Gd Tb Ear Tm Yb Lu 140.1 140.9 1442 1(45) 157.3 158.9 162.5 164.9 1673 168.9 173.0 175.0 96 97 98 99 101 102 103 Cf E 90 9U2 9: 94 Th Pu Cm Bk Fm Md N0 Lr 232.0 (23381) 238.0 (2N3?!) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) ...
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