CHEM 08 (Spring-008) Exam. (05 pts) Name: --------------------------------------------------------------------------, CLID # -------------------------------- LAST NAME, First (Circle the alphabet segment of your LAST NAME): A-E F-L N-P Q-Z Please answer the following questions: Part I: Multiple Choices (8 pts: @ pts each). Circle the ONE best answer:. Which of the following statements is FALSE: a) Reactions with a negative H rxn and a positive S rxn are product-favored at all temperatures. b) The entropy of a pure, perfect crystal is zero at 0 K. c) A sample of pure I vapor has higher entropy than pure solid I (both at room temperature). d) Reactions with positive H rxn and a positive S rxn can never be product-favored.. Which of the following reactions is most likely to have the most negative change in entropy? a) CaCO (s) CaO (s) + CO (g) b) NH (g) N (g) + H (g) c) O (g) + N (g) NO (g) d) NH (g) + HCl (g) NH Cl (s). Calculate the ph of a solution that is 0.5 M HC H O (acetic acid) and 0.5 M NaC H O (sodium acetate), K a (HC H O ) =.8 x 0-5 a).89 b) 9. c).75 d) 8.8. The solubility of CaSO at 5 C is listed as 00. mg/00 ml (CaSO = 6 g/mol). What is the K sp of CaSO? a).0 x 0 - b). x 0-6 c).0 d). x 0-5. Pb(OH) is a relatively insoluble base. A saturated solution has a ph of 9.0. Calculate the Ksp for Pb(OH). What is the predicted value of K sp for Pb(OH)? a).9 x 0-6 b) 5.8 x 0-6 c).5 x 0-6 d) 5.0 x 0-6 6. Which of the following sulfides has the least molar solubility? CdS (K sp = 8 x 0-8 ), CoS (K sp = 8 x 0 - ), CuS (K sp = 6 x 0-7 ) and ZnS (K sp = x 0-5 ): a) CuS b) ZnS c) CoS d) CdS 7. For the process: O (g) O (g), H = +98 kj. What would be predicted for the sign of S rxn and the conditions under which this reaction would be spontaneous? S rxn Spontaneous ------------------------------------------------------------------------------------------------------------------- a) positive at low temperature only b) positive at high temperature only c) negative at high temperature only d) negative at low temperature only
8. Given the following standard molar entropies, calculate S for the reaction: CH (g) + O (g) CO (g) + H O (g) S (J/mol.K) 86. 05.0.6 88.7 a) 5. b). c) -. d) -9. 9. A buffer may be prepared by mixing a weak acid with a roughly equivalent amount of strong base. Which of the acids shown below would be the best for the preparation of a buffer with a ph of 9.00? a) chlorous acid, HClO; K a =.8 x 0-8 b) formic acid, HCO H; K a =.9 x 0 - c) dihydrogen phosphate ion, H PO -, K a = 6. x 0-8 d) ammonium ion, NH + ; K a = 5.6 x 0-0 0. Examine the titration curve shown. Which of the following titrations most likely could it represent? a) HCl by NaOH b) HCl by aqueous NH c) NaOH by HCl d) Aqueous NH by HCl. Use the graph shown below to answer the following questions: In the titration of 5 ml of 0. M HC H O (acetic acid) by 0. M KOH. Which of the following indicators that can be used to detect the equivalence point for this titration (circle it)? Indicator color change ph range Thymol blue red yellow..8 Bromophenol blue yellow blue.0.6 Methyl red red yellow. 6. Bromothymol blue yellow blue 6.0 7.6 Cresol blue yellow red 7. 8.8 Phenopththalein colorless pink 8. 0.0. Calculate the ph of a solution that is formed by mixing 5.0 ml of 0.0 M HCl with 0.0 ml of 0.5 M NaOH? a).0 b) 7.00 c).0 d).08 Part II. ( pts): Fill in the blanks: a) Write the solubility product expression for Ag CrO : K sp = --------------------------------------- b) Standard molar entropies of pure substances are always ---------------------- at 0 K. c) The conditions that would apply to a reaction that is spontaneous at room temperature: Gº = ------------------------ and K ---------------
Part III (5 pts) Calculations: Show all work for full credit. Please express all answers with proper units and correct number of significant figures.. (8 pts) For the following reaction at 98 K: Al O (s) + H (g) Al (s) + H O (g) Al O (s) H (g) Al (s) H O (g) H f (kj/mol) -669.8 0 0 -.8 S (J/K.mol) 5.0 0.6 8. 88.8 a) Calculate G in kilojoules G rxn (kj) b) Is the calculated entropy change consistent with what you expected? Why? Yes because: No because: c) Calculate the equilibrium constant for the reaction at 98 K. K =. (9 pts) Will a precipitate form when 00. ml of 8.0 x 0 - M Pb(NO ) is added to 00. ml of 5.0 x 0 - M Na SO? (K sp for PbSO = 6. x 0-7 ). Explain your answer. No, Q (9 x 0-8 ) < Ksp
. (8 pts) In the titration of 5.0 ml of 0.0 M formic acid, HCO H (Ka =.8 x 0 - ) with 0.00 M NaOH solution: a) What is the ph of HCO H solution before the titration begins? b) What is the ph at the mid point of the titration? c) What is the ph at the equivalence point?
You will have 60 minutes to complete this exam. The exam has pages plus a reference page. When you are told to do so, tear off the Periodic cover sheet and use as required during exam. Constants & Useful Equation : 5 6 7 H.0 Li 6.9 Na.99 9 K 9. 7 Rb 85.7 55 Cs.9 87 Fr () Be 9.0 Mg.0 0 Ca 0.08 8 Sr 87.6 56 Ba 7. 88 Ra 6.0 PREFERENCE SHEET FOR CHEM 08 Exam Spring 008 Gas Constant: R = 8. J/K.mol = 8.5 x 0 - kj/k.mol = 0.08 L atm/mol.k K = 7 + C Periodic Table of the Elements 5 6 7 8 9 0 Sc Ti V Cr Mn Fe Co Ni Cu Zn.96 7.88 50.9 5.00 5.9 55.85 58.9 58.69 6.55 65.8 9 Y 88.9 57 La 8.9 89 Ac 7.0 atm = 760 torr = 760 mmhg d (density) = m)/v ln (P /P ) = ( H vap /R){(T -T )/T T } C g = k.p g n i = P i /P t PV = nrt π = MRT T = i. Km. P = χ.p o 0 Zr 9. 7 Hf 78.5 0 Rf (6) Nb 9.9 7 Ta 8.0 05 Db (6) Mo 95.9 7 W 8.8 06 Sg (6) Tc (98) 75 Re 86. 07 Bh (6) Ru 0. 76 Os 90. 08 Hs (65) 5 Rh 0.9 77 Ir 9. 09 Mt (66) 6 Pd 06. 78 Pt 95. 0 Uun (69) 7 Ag 07.9 79 Au 97.0 Uuu (7) 8 Cd. 80 Hg 00.6 Uub (77) 5 6 7 8 9 B C N O F 0.8.0.0 6.00 9.00 Al 6.98 Ga 69.7 9 In.8 8 Tl 0. Si 8.08 Ge 7.59 50 Sn 8.7 8 Pb 07. 5 P 0.97 As 7.9 5 Sb.8 8 Bi 09.0 6 S.06 Se 78.96 5 Te 7.6 8 Po (09) 7 Cl 5.5 5 Br 79.90 5 I 6.9 85 At (0) 58 59 60 6 6 6 6 65 66 67 68 69 70 7 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 0. 0.9. (5) 50. 5.0 57. 58.9 6.5 6.9 67. 68.9 7.0 75.0 90 Th.0 9 Pa.0 9 U 8.0 9 Np 7.0 9 Pu () 95 Am () 96 Cm (7) P = χ.p o Rate = k[a] ln [A] o = kt ln[a] = - kt + ln[a] o t / = ln /k = 0.69/k [A] Rate = k[a] /[A] - /[A] o = kt t / = /k[a] o Rate = k [A] = - kt + [A] o t / = [A] o /k k = Ae -Ea/RT ln k = (-E a /R)(/T) + ln A ln (k /k ) = (E a /R){(T -T )/T T } K p = K c (0.08 T) n G rxn = G rxn + RTlnK 97 Bk (7) 98 Cf (5) 99 Es (5) 00 Fm (57) 0 Md (58) 0 No (59) 0 Lr (60) He.00 0 Ne 0.8 8 Ar 9.95 6 Kr 8.80 5 Xe. 86 Rn () 5