University f Waterl DEPARTMENT OF CHEMISTRY CHEM 13 Test # Wednesday, March 11, 009 This is test versin 001. Fill in vals 001 fr the Card Number (r Test Master) n yur cmputer answer card. Name (Print in ink!): First name Surname Student Number: Signature: Aids Allwed: Time Allwed: Electrnic Calculatr 50 minutes Instructins: 1. The instructrs names and sectin numbers are given belw. Circle yur instructr s name and sectin number nw. T. Leung Sectin 001 9:30 am R. Le Ry Sectin 00 10:30 am D. Gilbert Sectin 003 11:30 am G. Dmitrienk Sectin 00 1:30 pm. Print yur name and ther infrmatin in the tp right crner n the frnt f the cmputer card. 3. Fill in yur ID number, Sectin Number and yur test versin in the designated fields.. Use a pencil t fill in the vals n the cmputer card. Make sure any erasures are cmplete and clean. If yu need a pencil, raise yur hand and ask fr ne. 5. This test bklet has 1 questins and 7 pages (including this Cver Page and the Data Sheet). Check nw t ensure that n pages are missing. If pages are missing, raise yur hand and ask fr a new test bklet. 6. There is nly ne crrect answer fr each questin. Circle A, B, C, D r E n this bklet. When yu have finished, transfer yur answers t the cmputer card by filling in the apprpriate val (A,B,C,D r E) in the clumn fr that questin n the card. 7. Attempt all questins. N marks are deducted fr wrng answers. D all rugh wrk n this test bklet. Scrap paper is nt permitted. 8. At the end f the test, turn in this test bklet and yur cmputer card, and leave immediately. Nte: The last page is a DATA SHEET. Yu may tear it ff.
[1] In an equilibrium reactin, the fllwing cnditins culd exist: A) I and II I. Tw ppsing reactins ccur at equal rates. II. The cncentratins f prducts and f reactants are equal. III. The value f equilibrium cnstant depends n the initial cncentratins f reactants. IV. Temperature changes cause the equilibrium cnstant t change. Which f the abve statements are true? B) I and III C) II and III D) II and IV E) I and IV [] If K C is the equilibrium cnstant fr the fllwing reactin: A) 1 K C B) K C C) K C D) 1 KC E) K C AB(g) A (g) + B (g) then the equilibrium cnstant fr the reactin: A (g) + B (g) AB(g) is: [3] Cnsider the reactin: 3 Fe(s) + H O(g) Fe 3 O (s) + H (g) K C can be written in terms f K P as: A) K C = K P B) K C = K P (RT) 1 C) K C = K P RT D) K C = K P E) K C = K P
[] Fr the reactin: Cl (g) + H O(g) HCl(g) + O (g), K C = 1.5 10 9 at 500 K. If the initial cncentratins at 500 K are: [Cl ] = 0.50 ml L 1, [H O] = 0.0 ml L 1, [HCl] = 0.50 ml L 1, and [O ] = 0.015 ml L 1, then the fllwing will ccur. A) The reactin prceeds t the right. B) The reactin prceeds t the left. C) The reactin is already at equilibrium. D) The reactin vlume must be specified in rder t predict which directin the reactin will prceed. E) The value f K P at anther temperature (different frm 500 K) must be specified in rder t predict which directin the reactin will prceed at 500 K. [5] Cnsider the reactin: N O(g) + N H (g) 3 N (g) + H O(g). If 0.100 mle f N O and 0.50 mle f N H are placed in a 1.000 L cntainer at high temperature, and at equilibrium 0.060 mle f N O is present, what is K C fr this reactin? A). 10 B) 1.7 10 C) 0.0 D) 5.7 10 1 E). 10 3
[6] If the system: Cl (g) + H O(g) HCl(g) + O (g) [ΔH = 11. kj ml 1 ] was in equilibrium, which f the fllwing wuld increase the amunt f HCl? I. decreasing the vlume II. increasing the vlume III. adding a catalyst t speed up the frward reactin IV. increasing the temperature V. decreasing the temperature A) I and IV B) II and V C) II and IV D) I and III E) I and V [7] Cnsider the reactin: H (g) + I (g) HI(g) at a temperature fr which K C = 100. If the initial cncentratins are: [H ] = 0.0 M, [I ] = 0.0 M, and [HI] = 0.0 M, what is the equilibrium cncentratin f H? A) 0.067 M B) 0.1 M C) 0.07 M D) 0.0 M E) 3.06 M [8] Which f the fllwing represents a cnjugate acid-base pair? A) H PO 3 and PO B) HSO and SO 3 C) HNO 3 and NO 3 D) HCl and NaOH E) Nne f the abve
[9] What is the ph f a 0.0 ml L 1 slutin f ethylamine, C H 5 NH [K b =.3 10 ]? A) 11.97 B).03 C).37 D) 11.63 E) 7.00 [10] What is the ph f a.00 10 7 M slutin f HCl at 5 ºC? A) 7.30 B) 6.70 C) 6.6 D) 7.38 E) 7.00 [11] A 0.1 M slutin f which f the fllwing acids will have the highest ph? A) Acetic acid, HCOOH [K a = 1.8 10 5 ] B) Hypbrmus acid, HOBr [K a =.06 10 9 ] C) Hypchlrus acid, HOCl [K a = 3.5 10 8 ] D) Hydrfluric acid, HF [K a = 6.6 10 ] E) Hydridus acid, HOI [K a =.3 10 11 ] 5
[1] Cnsider fur separate slutins btained by disslving 0.10 mle f each f the fllwing salts in 1.0 L f pure water: 1. NaF. Na O 3. H N + Cl. NaCl The rder f decreasing ph (highest t lwest) fr these aqueus slutins is: A) > 3 > > 1 B) > > 1 > 3 C) > 1 > > 3 D) 1 > > > 3 E) > 1 > 3 > [13] The rder f decreasing acidity (highest t lwest) fr the fllwing cmpunds is: A) CHF COOH > CHF CH OH > CH BrCH OH > HBr B) CHF COOH > CH BrCH OH > HBr > CHF CH OH C) CHF CH OH > CHF COOH > CH BrCH OH > HBr D) HBr > CHF COOH > CHF CH OH > CH BrCH OH E) CHF COOH > HBr > CH BrCH OH > CHF CH OH [1] A diprtic acid H A has the fllwing acidity cnstants: K a1 = 1.0 10 8 and K a =.00 10 3 When this acid is disslved in pure water, which f the fllwing relatinships is true at equilibrium? A) [H 3 O + ] [H A] B) [H A] > [HA ] > [A ] C) [H 3 O + ] [A ] D) [HA ] [A ] E) [H 3 O + ] [HA ] 6
CHEM 13 (Winter 009) Test # 1 1A H 1.008 Li 6.91 Na.99 K 39.10 Rb 85.7 Cs 13.9 Fr (3) A Be 9.01 Mg.31 Ca 0.08 Sr 87.6 Ba 137.3 Ra 6 3 3B Sc.96 Y 88.91 (57-71) La-Lu (89-103) Ac-Lr B Ti 7.88 Zr 91. Hf 178.5 10 Rf 5 5B V 50.9 Nb 9.91 Ta 180.9 105 Db 6 6B Cr 5.00 M 95.9 W 183.9 106 Sg CHEM 13 DATA SHEET 7 7B Mn 5.9 Tc (98) Re 186. 107 Bh 8 Fe 55.85 Ru 101.1 Os 190. 108 Hs 9 8B C 58.93 Rh 10.9 Ir 19. 109 Mt 10 Ni 58.69 Pd 106. Pt 195.1 110 Uun 11 1B Cu 63.55 Ag 107.9 Au 197.0 111 Uuu 1 B Zn 65.38 Cd 11. Hg 00.6 11 Uub 13 3A B 10.81 Al 6.98 Ga 69.7 In 11.8 Tl 0. 113 Uut 1 A C 1.01 Si 8.09 Ge 7.59 Sn 118.7 Pb 07. 15 5A N 1.01 P 30.97 As 7.9 Sb 11.8 Bi 09.0 16 6A O 16.00 S 3.07 Se 78.96 Te 17.6 P (09) 17 7A F 19.00 Cl 35.5 Br 79.90 I 16.9 At (10) 18 8A He.003 Ne 0.18 Ar 39.95 Kr 83.80 Xe 131.3 Rn () K a and K b fr sme Acids and Bases Acid K a Acetic, CH 3 COOH 1.7 x 10-5 Chlracetic, CH ClCOOH 1.1 x 10-3 Ascrbic, H C 6 H 6 O 6 7.9 x 10-5 (K a1 ) - HC 6 H 6 O 6 1.6 x 10-1 (K a ) Benzic, HC 7 H 5 O 6.6 x 10-5 Frmic, HCOOH 1.8 x 10 - Hypbrmus, HOBr.06 x 10-9 Hypchlrus, HOCl 3.5 x 10-8 Hydrfluric, HF 6.6 x 10 - Hydrcyanic,HCN.93 x 10-10 Hydrgen sulfide, H S 1.0 x 10-7 (K a1 ) HS - 1. x 10-13 (K a ) Nitrus, HNO.5 x 10 - Phsphric, H 3 PO 7.1 x 10-3 (K a1 ) H PO - 6.3 x 10-8 (K a ) HPO -. x 10-13 (K a3 ) Sulfuric, H SO strng - HSO 1.0 x 10 - (K a ) Base K b Ammnia, NH 3 1.8 x 10-5 Aniline, C 6 H 5 NH.3 x 10-10 Trimethylamine N(CH 3 ) 3 6.3 x 10-5 Physical cnstants Cnversin factrs Avgadr cnstant, N A = 6.01 10 3 0 C 73.15 K Electrn charge, e = 1.60 10-19 C 1 atm = 101.35 kpa Faraday cnstant, F = 9685 C ml -1 = 760 trr Gas cnstant, R = 0.08058 atm L K -1 ml -1 1 pm = 1 10 1 m = 8.315 J K -1 ml -1 1 A = 1 ampere = 1 C s -1 = 8.315 kpa L K -1 ml -1 In prduct fr water, K w = 1.0 10-1 K sp fr Sme Salts AgCl 1.8x10-10 Ag 3 (PO ) 1.8x10-18 MnS 3.0x10-13 AgBr 5.0x10-13 Al(PO ) 9.8x10-1 ZnS.5x10-1 AgI 8.3x10-17 Ca 3 (PO ) 1.3x10-3 CdS 1.0x10-7 PbCl 1.7x10-5 Ni 3 (PO ).7x10-3 Ag S 1.6x10-9 AgOH 1.5x10-5 Ag (CrO ) 1.x10-1 BaCO 3.6x10-9 Sr(OH) 3.x10 - Sr(CrO ) 3.6x10-5 Li CO 3 1.7x10-3 Ca(OH) 6.5x10-6 Ba(CrO ) 1.x10-10 MnCO 3.x10-11 Mg(OH) 7.1x10-1 Mn(OH) 6.0x10-1 Al(OH) 3 1.3x10-33 K f fr Sme Cmplexes Fe(OH) 7.9x10-16 Cr(OH) 3 6.3x10-31 Ag(NH 3 ) + 1.6x10 7 Zn(OH).5x10-17 Fe(OH) 3 1.6x10-39 Ag(CN) 5.6x10 18 Zn(NH 3 ) +.1x10 8 Sme Standard Reductin Ptentials F (g) + e F (aq).889 V Fe 3+ (aq) + 3e Fe(s) 0.0 Au + (aq) + e Au(s) 1.691 Pb + (aq) + e Pb(s) 0.17 O (g) + H + (aq) + e H O(l) 1.9 Ni + (aq) + e Ni(s) 0.36 Br (l) + e Br (aq) 1.078 C + (aq) + e C(s) 0.8 Ag + (aq) + e Ag(s) 0.799 Tl + (aq) + e Tl(s) 0.336 Fe 3+ (aq) + e Fe + (aq) 0.77 Cd + (aq) + e Cd(s) 0.0 O (g) + H O(l) + e OH (aq) 0.01 Fe + (aq) + e Fe(s) 0.1 Cu + (aq) + e Cu(s) 0.339 Cr 3+ (aq) + 3e Cr(s) 0.7 AgCl(s) + e Ag(s) + Cl (aq) 0. Zn + (aq) + e Zn(s) 0.76 AgBr(s) + e Ag(s) + Br (aq) 0.073 Li + (aq) + e Li(s) 3.00 H + (aq) + e H (g) 0.0 Sme frmulae = A [ A] = [ A] e kt [ A] [ ] kt [ A] = [ ] + kt 1 1 A k = Ae E 1 1 ( k k 1) = ( T T1 ) a ln / R K = K ( RT ) Δn p c x = Ea RT - ± - a b b ac ΔH 1 1 ( K K1) = R ( T T1 ) = ( ) ln / E E 0.059 n lg Q (at 98 K)