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Unformatted text preview: TA KEV Name Student ID Number '- LAB Sec. # ; TA: ; Lab day/time: Toby Allen . Spring 2005 CHEMISTRY 2B FINAL EXAM Instructions: CLOSED BOOK EXAM! No books, notes, or additional scrap paper are permitted. All information required is contained on the exam. Place all work in the space provided. If you require additional space, use the extra sheet of the exam and refer to its page number in your response. A scientific calculator may be used (if it is a programmable calculator, its memory must be cleared before the exam). (circle answer HERE) an m >> 999w >I> ww©w €00 UUUU .90 > 000 Q, 3 5 35:53.3 e3 0009 UUUUU G > O ‘U mm D (1) Read each question carefiJlly. (2) For Parts I and 11, there is no partial credit given and only answers marked on this cover page will be graded. Answer all guestions. (3) For part III you should choose 2 problems out of 3 to be graded. You MUST indicate which 2 problems you would like graded. (4) The last 2 pages contain a periodic table and some usefiJl information. You may remove it for easy access. (5) If you finish early, RECHECK YOUR ANSWERS! @ \O @ >>>®> wwwmwwww "t'l'lt'fll'l'll'fll'fl U.C. Davis is an Honor Institution Possible Points m Part I (1-16) — 3 points each Part II (17—21)— 6points each — 3O /30 Part III (22) - 20 points ' .520: _ 00‘. Part 111 (23) — 20 points 7,0 ' ,20 T 120 Part III (24) - 20 points Re-grade bonus — 4 points Total Score - 122 points total ‘ L1 [122 O\ In ('D b: O 5" 5. > w :30 @ {Tl Part 11: int 38.3 63> waw 00 cu ml; 21. A B OD E Part III: 20 uoints each -m Name: Keg Final Exam (Page 2 of 12) Part 1: Multiple Choice, Concepts, (3 points each) Circle your answer on the cover sheet — No partial credit. ANSWER ALL QUESTIONS. 1. (3pts) A chemical equilibrium is A. the point at which both forward and reverse reactions stop. B. the point at which the forward reaction ceases. ©the point where concentrations of reactants and products become constant. D. the point at which all reactants have been converted to products. E. the point where as many moles of titrant as analyte have been added. 2. (3 pts) IfK> l andQ<Kthen A. the equilibrium favors reactants and the reaction will go towards reactants B. the equilibrium favors reactants and the reaction will go towards products C. the equilibrium favors products and the reaction will go towards reactants Q the equilibrium favors products and the reaction will go towards products E. the equilibrium favors products and the reaction is at equilibrium 3. (3 pts) Which of the following has the highest boiling point? A. pure water at sea level, B. pure water on top of Mount Everest, (9 sea water at sea level, D. sea water on top of Mount Everest, E. they’re all the same. 4. (3 pts) Which of these pairs is not in the correct order of acid strength? A. HI > HF ® H2803 > H2804 C. HClO4 > HC103 D. HCl > HCN E' :O: H l||.. || H—c—c—g—H > H—C —-c—-§—H H H H 5. (3 pts) The phase diagram for a pure substance is given below. The substance is stored in a container at 150 atm at 100°C. What happens if the container is cooled to room tem erature? 300 A. The liquid in the container freezes. 25° B. The liquid in the container vaporizes. 200 D. The solid in the container sublimes. Pressure, @ The vapor in the container liquefies. am 150 E. The solid in the container melts. 100 50 o 100 200 300 400 Temperature, K Name: “6% Final Exam (Page 3 of 12) 6. (3pts) An azeotropic mixture is a A. mixture of two or more substances where boiling point cannot be determined. B. solution of two or more substances present in the same amounts in the liquid phase. © liquid mixture of two or more substances in which the vapor has the same composition as the liquid. D. liquid mixture that can only boil in the tropical areas of Asia. E. solution of two or more substances that cannot be made to freeze. 7. (3pts) The following unit cell is body-centered cubic. The coordination number is A. 6 B. 3 C. 2 @ 8 E. depends on adjacent unit cells. 8. (3 pts) Why is the freezing of water at -10°C a spontaneous process? A. The process is exothermic. B. The entropy of the water decreases because of increased order. C. The internal energy of the system decreases. @ The surroundings absorbs heat enough to increase the total entropy. E. Because that’s the way it is. 9. (3 pts) Solid Mg(OH)2 can be precipitated from its solution by A. addition of strong acid B. addition of methanol CH3OH C. addition of a complexing agent (D addition of sodium hydroxide E. addition of weak acid 10. (3 pts) Which of the following processes will lead to a decrease in the entropy of the system? ® N2(g)+3H2 (g) :‘— 2NH3(g) B. H200) v——> H20(g) c. CaCO3(s) ..—.\ CaO(s)+C02(g) D. NH4NO3(s) +H20(l) :: NH.+ (aq)+N03' (aq) E. C02(S) x73- C02(g) 11. (3 pts) Why is ammonia very soluble in water? A. The molecules are of similar size. . B. The molecules experience van der Waals interactions. © Both molecules can hydrogen bond. D. Both molecules have hydrogen atoms. E. It has a dipole moment. Name: 146‘, Final Exam (Page 4 of 12) 12. (3 pts) Some devices that are sold as “weather forecast instruments” contain cobalt(II)chloride‘ and are based on the following equilibrium: CoC12(s) + 6 H20(g) ti CoC12-6H20(s) Purple P1nk What color will such an indicator show if rain is imminent? A. Pale blue B Purple D. Transparent 13. (3 pts) Which of the following pairs of compounds will form an ideal solution? A. water (H20) and hexane (C6H14) B. acetone (CH3COCH3) and CHCl3 C. water and 64 anol (CH3CH20H) D. hexane 33‘) and ethanol (CH3CH20H) ® hexane (C6 ' 14) and nonane (C9H20) 14. (3 pts) Consider a crystal with hexagonal closest packing. Where is the octahedral hole @B or C)? 15. (3 pts) Le Chatelier’s principle A. states that natural processes are spontaneous. B. states that a system completely recovers from any imposed change. @states that a system partially offsets change. D. states that a system remains at equilibrium during change. E. is hard to pronounce. 16. (3 pts) A reaction has AH<O and AS>0. A. It is spontaneous for high T. B. It is spontaneous for low T. C. It is non-spontaneous for all T. (D. It is spontaneous for all T. E. This is not possible. Name: léé'fl Final Exam (Page 5 of 12) Part 11: Multiple Choice, Short Calculations (6 points each) Circle your answer on the cover sheet— N 0 partial credit. ANSWER ALL QUESTIONS. All required data can be found on last 2 pages. 17. (6 pts) If the enthalpy of vaporization of water at its normal boiling point is 40.7 kJ/mol what is the entropy of vaporization? A’Smp 95513.:40'7 x [OBS/mo) 3‘73 . (S— K {£811 S/r'wl/IL A. 0.407 J/mol/K, B. 4.070 J/mol/K, C. 407.0 J/mol/K, D.O~lO?J/mol/K, @1091 J/mol/K 18. (6 pts) What is the molar solubility of silver sulfate? A3;SQ,,(3\ g3 gaff (Q2) + 30,3242) 2,5 5 AF: CZSQKS 3 4‘53 2 l‘S—xioqy— =5 5.: 3/ #5340 4‘ = [I’S‘Kto final/L .1.55x10'5mo1/L, B.1.22x10‘7mol/L, C.8.66x10‘8m01/L, D. 1.5x10'14mol/L. 19. (6 pts) Using the enthalpy of vaporization of water, find the vapor pressure of water at room em erature. t p 7T= mum, R: (er/m 1:298'151L f’ = 7 x _.___,, ’ at: = age—+— :2: ': “3567 5) P; _— (5.01571 A) latm 0.028 atm C) 0.996atm D) 1.1x10'69atm E) 0.10atm. Name: Kéfi Final Exam (Page 6 of 12) 20. (6 pts) Determine the AH" for the reaction: COC12(g) : CO(g) + C12(g) given that at 99.8°C, Kp = 6.7x10'9, and at 395 °c, Kp = 4.4x10'2. ’5: 371494 K.= Mme“l T7,: 668‘ISK K2: (P‘valO’Z 4th 19:: 1%? —# 4‘ K L t (5319’: __ ((AKZ/Kt f .. 83w: Slmol/IL AQ‘IHOAOI ’2 AH = -———-—-" ' “—‘eT—ff—T—‘f— ’i’: —’TL‘(— < 6635M. 372%1 : i~toxto537mol (l0 lcS/mol A. -252 kJ/mol @110 kJ/mol C. 126 kJ/mol D. +11.2 kJ/mol E. -32.2 kJ/mol 21. (6 pts) A system consists of IL of He gas at 2 atm pressure. The applied pressure is decreased suddenly and isothermally to 1 atm. The applied pressure is then increased suddenly and isothermally back to 2 atm to restore the original volume. How much work is done on the system? S'TAkéli PI , chlfl, VI= [1. Pp: loJ-m/ Vp= 2L. CPV=nQ75v vxl/PB‘ Wc‘PEleav c Pew}: ICJTVI :2) Vs/l=“l Liam ’ av: IL— A. none B. —1 L.atm ©+1 L.atm D. +2 L.atm E. -2 L.atm Name: lééfl Final Exam (Page 7 of 12) Part 111: Long Answer: CHOOSE 2 OF THESE 3 PROBLEMS! Show all work - in space grovided— Partial credit. Indicate on cover which questions to grade! Required data can be found on last 2 pages. (20 points each) 22. (20 pts) A. What volume of 1.0 mol/L NaOH must be added to 25.00 mL of 0.10 mol/L acetic acid (HC2H302) to reach the equivalence point? {laud r: 2§voOmL x 040”: Z-S‘O mmol 2*5—Ommol ’- A Vm‘ W “ 2‘50““ “P9 B. What is the concentration of acetate ions at the equivalence point? Viola! = 25700 + 2:5”0 == 2’7~3’“OmL _ 2 “O I l, . CCZ’HZOIJ = 5 mm/275’onL: O '09! M (#183 C. Write the hydrolysis reaction for acetate ions in water and compute the equilibrium constant. Cszof + H10 :5 HCLHJOI + OH‘ ("Z-MSW ——l ’+ I” ' . — -* (O 3 K‘K ( ~ 87Kl0‘75 “—— C N§3 D. Find the concentration of hydroxide ions at the equivalence point? wwmzwafl ow ‘ WT o ‘09 l O 0 CH6: --x + DC +>c éQ ox Oct ( '1 >L >C Kb : (”CZ 50H: :9— X'1 = $1596 KID" lo C‘l‘Pi3\ (cargo: coca-ac x c; ,/0 ‘0‘?! xs‘flxw’m 2) ’_ 6 :7 @H‘J =>c :- ‘7« H x10 M (3‘)}3\ B. What is the pH at the equivalence point? pea: 403 and = 5/37 Name: i<é>l Final Exam (Page 8 of 12) 23. (20 pts) A. Write the net reaction for the formation of complex ions from silver chloride and aqueous ammonia and compute the equilibrium constant for this reaction. AgCl(s) :2 A + aq) + C1(aq) l<sf:l 8H0 +aq)+2NH3(aq> : 1Ag<NH3)21+ 141= .. “(to 'cm + ZNH3(°2\ :2 CA 01,195+ +0 ( ’5') j ZF K: “‘10le = 2’88xlO—3 @433 B. What is the molar solubility of silver chloride in 0.10M ammonia solution? 53. (,iCs\ + ZNH3(%>:3 @3€~m\3&3*<—l?ca o o WT 0.10 rt CM: —25 +5 +5 5Q 0104/8 S 5 + -— z _ CM H33 (010 rzs\ (5p15\ S ,_ 3 , o 05“}? T 7 /- 288MB (7— 93\ “ML 7 0104: F 14340—25} -;§ 5 = S 39340 3 " O' [0.77" S i? 9 l‘l0775 _. 539%“? ,_ +g6x10'3mol/L (3,013) C. What is this solubility expressed as g/ 100mL solution? V_\ . NolaFWcSS(AjCl\ -; 107-9 41 1?.sz 2 1 #3 JSJ/mal ( pH => Soerr = <9 Sé >404 MOl/LK (wary/mo; O 069 j/tOOmL (ck/7'17”) Ciptfi Name: Kéfl .‘ Final Exam (Page 9 of 12) 24. (20 pts) Given the following reaction N2(g) + 3X2(g) ‘——‘ 2NX3(g) AHf = 0 0 -43 kJ/mol S° = 192 210 172 J/mol/K. A. What is AG° (in kJ/mol) at 700K? C W AH° z 2'6.qu 2’376 kSInoi if” A50: 20sz -— 10:2— zcuox = any I/mo] Cipfl (.39): A H° , “TaSO = -526 LIE/moi ~7001< (— negro“ tax/not} 27-:- ' ‘ m ‘ ' B. What is the equilibrium constant at 700K? ________+ 29'? 6 K3] 0’ CZPJQ (or: ’- T I A4 {L f < .. 2. +7HOS-S/mo I /Cg-3w&3/mov/I< x 700 A =7 l<= e:— ACf/ILT = e -— r C; , ~ 2 _ 4'2. _. ’ e 7 , 2 as IO {Laws} C. What is the equilibrium partial pressure of NX3(g) at 700K if N2 and X2 are mixed in equal proportions with partial pressures of 1 atm? NszD—k 3&(33 E": ZNX3(ii 1:417 I am l at!“ O CH’CT ~9Q _39( «F221 éq (I ~1qu qezgcan (pa can _ K: L 2 @701 a: (ac-2. = 281940" (w) K1,, 3% 0—3;} C\~Zx\3 ,5 9c: L63“ x(o’(° «J71 => [fa/3 32);: f~3l>rlodocdfl (2.6913 D. Umwrfl/fle stmt‘th‘ PWCFM_ If (02', OF N ((3 f5 WW‘ 1%,, 4 lain, Pan. 7: hi" 642g '2 O'cix S‘lelo’m: ‘P7X add—(OW P I? «asmo‘m ‘ ~17 (3 m 71>Q:__L3_3 : , 3_ 2 27,ng P . RAM, 81 (K I AC: AGO radeQ - ,q . «3 —— (4 ‘L'ZS’V‘O -— ‘ kg .315 to tcs/mol/K x700 (9% ’ 2%3 6 #10, A? 37 (F K End of Exam mm aS/mi — 141—738 «cs/moi ‘ 43 EM, (4300 afiPaNflNE'OUS, Name: Final Exam (Page 10 of 12) Extra page — if needed you must refer to this page number. Final Exam (Page 11 of 12) Name. Useful Eguations and Data Sheet Conversions and Constants. 1atm= 760 Torr= 760 mmHg— — 101325 Pa= 1013.25 kPa= 1. 01325 bar; 0°C= 273.15K= 32°F 1 cal = 4.184 J; R = 8.3145 J / mol / K = 0.08206 L atm Imol / K; NA = 6.022 x 1023 mol‘1 d (H20) = 1.00 g / mL cp (H20) = 4.184 J / g °C AH°f(H20(l)) = -285.8 kJ/mol AHovap(H20) = 44.0 kJ/mol KKHZO) = 1.86 K kg/mol 10,0120) = 0.512 K kg/mol K5p(AgI)= 8.5x10'17 K5p(AgSO4)= 1.5x10‘14 Ksp(AgCl)=1.8x10‘10 Kf([Ag(NH3)2]+) = 1.6x107 Equations: PV=nRT; q = n AHO csp=q/(AT x mass); E= q / (AT x n) Cszgas; PA: XAP°A ATf=-ifoxm; ATb=ibexm 1r=iMRT, (RT)A"gas pH = —log[H3O+] pOH = —log[OH’] pH + pOH = 14 pH =pKa + log([A-]/[HA]) pH = pKa+10gEL£Ij]j Kw= [H30+] [OH‘] =Ka x Kb pK =—log[K] ~14» : 1Q AU=q+w w=—PextAV AH=AU+PAV AH,“ :21: vAH AH° f(products) — Z v, AH 2 (reactants) AS,“ =2 v S 0 (products)— 2 v S ° (reactants) ASzr=AH" 1n§1=_ AH ———i AG°=AH°-TAS° AG=AG°+RT1nQ 7?, K1 R T2 T [2 ax2+bx+c=0 a x=:_l_)_i_3b__—_4_a_c_/o AG0=-RT1nK K: e—CKT/KT a Name: Final Exam (Page 12 of 12) Ionization Ionization Equilibrium Constant K Acid K“ a Iodic acid 111103 + Hzom—A 6-— 6130" +10; 1.6 x 10" Chkorous acid 16:102 + H20 4—— 113 0+ + 00; 1.1 x 1.0—2 Chioroacetic acid 111221120102 + H20 6—------3 1130+ + czuzclo2 1.4 x 10'3 Nitrous acid 111102 + H20 m H30 + NO; 7.2 x 10'“ Hydrofluoric acid HF 6» H20 #113041 + F’ 6.6 x 10—4 Fannie acid HCH02 + H20 =2 11:13”1 + cm; 1.8 x 10-4 Benzoic acid 110711502 + H20 .=——* 1130+ + 0,1150; 6.3 x 10-5 Hydrazoic acid HN3 4» H20 g 1130* + N3“ 1.9 X 10"5 Acetic acid 11(3211302 + H20 : 1130+ + C2H302" 1.8 x 10*5 Hypochiorous acid H001 + H20 # H30* + CC? 2.9 X 10“3 Hydrocyanic acid 11cm + H20 =2 H30: «1 CN" 6.2 x 10“10 Phenol HOCQHS + H20 .— 1131)" + C6H50" 1.0 x 10"“) Hydrogen peroxide 8202 + 1420““3 6—— [7130* + 1102 1.8 X 10”2 Base Kb =‘ Diethylamine (C2H5)2NH + H20 # (C2H5)2NH2+ + 0H" 6.9 x 10*4 Eiiiyiamine 911516112 + H20 =—-*-—— CZHSNH; + OH“ 4.3 x 10‘4 Ammonia NH3 + H20 Si NH4+ + OR“ 1.8 X 10"5 Hydroxylaminc HUM-12 + H20 # HONH3 + + 0H" 9.1 X 10‘9 Pyridine CSHSN + H20 6... CSHSNH" + OH” 1.5 X 10"9 Aniline (3611511112 + H20 6—— 661151116; «1 on" 7.4 x 10*“) 35.21 21 23 Se Ti 06 4466 4767 93,94 40 Zr .91 91 2? 92.91 107 166 M1 Hs 59) 12691 60 61 Nd Pm 13:9 14081 UPOS 144.2 1,145) 89 Ac Pu (223') 232 0 231a C' 238 D (93"?) (2441 194?) W748 /"""—VI“B‘—"‘—\ 2 16 Cu Zn Ga Ge 63.55 621.39 69.72 7254 47 Fe Co Ni 66.66 58.93 6669 54 Sn 6 X6 116.7 121.5 127.6 131.3 Tc Ru «'98) 101.1 102.9 106.4 1079 112.4 1148 56 71 6V4 Ba Re Ir Pb 1325.9 1373 17:0 17:17 18089 183.8 166.2 19052 192.2 1961 1970 2072 '38 L03 1223) 12826) 2L62’1 (1561) {20323 110 Uun (271) 67 Eu Gd Tb Ho Er Tm 152.0 151.3 1589 162?!) 164.9 161.3 1669 1730 Cm Cf 12471 1217) 1251) Uuu Uub 12721 1277) Periodic Table of the Elements 1 1 1.1 Uuq 12991 112 [M 101 Es Md (2‘12) 1257'“) (258) 16 VIA 0 16 .00 1s 6 s C Ar sop 97 32 07 3.5 96 34 65 :16 5681' Kr 74 .92 76.33 79. 90 Ram T _. 35 P0 At 1209) 1210 1 02 (2591 ...
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