ap05_frq_chemistry_b

ap05_frq_chemistry_b - AP® Chemistry 2005 Free-Response...

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Unformatted text preview: AP® Chemistry 2005 Free-Response Questions Form B The College Board: Connecting Students to College Success The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 4,700 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three and a half million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns. Copyright © 2005 by College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board. Admitted Class Evaluation Service, CollegeEd, Connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark of the College Entrance Examination Board and National Merit Scholarship Corporation. Other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at: http://www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program and Pre-AP: apcentral.collegeboard.com. 22 8 9 2 39 56 Ba 55 Cs *La 57 88.91 Ti Hf 72 91.22 Zr 40 47.90 V Ta 73 92.91 Nb 41 50.94 Cr W 74 95.94 Mo 42 52.00 Mn Re 75 (98) Tc 43 54.938 †Actinide Series *Lanthanide Series Db (262) Rf †Ac 105 Pr 58 Ce Nd 60 (263) Sg 106 Pa Th U 92 Np 93 (145) Pm 61 (262) Bh 107 232.04 231.04 238.03 237.05 91 90 140.12 140.91 144.24 59 (261) 104 89 226.02 227.03 Ra Fr (223) 88 87 132.91 137.33 138.91 178.49 180.95 183.85 186.21 Sr 87.62 Rb Y 38 85.47 37 44.96 (244) Pu 94 150.4 Sm 62 (265) Hs 108 190.2 Os 76 101.1 Ru 44 55.85 Fe Ni Pd 46 58.69 Cu Ag 47 63.55 Zn Cd 48 65.39 S 16 50 Sn 49 In Sb 51 53 I 52 Te Xe 54 Kr 83.80 Br 79.90 Se 78.96 As 74.92 Ge 36 72.59 Ga 69.72 Ar 18 35.453 39.948 Cl 17 35 34 33 30.974 32 28.09 26.98 31 32.06 P 15 Si 14 Al 13 19.00 16.00 Au 79 Hg 80 Tl 81 Gd 64 (269) § 110 Tb 65 (272) § 111 Dy 66 (277) § 112 207.2 Pb 82 208.98 Bi 83 Ho 67 Er 68 Tm 69 §Not yet named 195.08 196.97 200.59 204.38 Pt 78 Yb 70 (209) Po 84 Lu 71 (210) At 85 (243) Am 95 (247) Cm 96 (247) Bk 97 (251) Cf 98 (252) Es 99 (257) Fm 100 (258) Md 101 (259) No 102 (260) Lr 103 151.97 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97 Eu 63 (266) Mt 109 192.2 Ir 77 (222) Rn 86 102.91 106.42 107.87 112.41 114.82 118.71 121.75 127.60 126.91 131.29 Rh 45 58.93 Co Sc 30 Ca 29 40.08 28 K 27 39.10 26 20 19 25 24.30 22.99 24 Mg Na 23 12 Ne 20.179 F 11 10.811 12.011 14.007 O 9.012 6.941 N Be Li C 10 B 4.0026 6 3 5 1.0079 He 21 7 2 4 PERIODIC TABLE OF THE ELEMENTS H 1 DO NOT DETACH FROM BOOK. INFORMATION IN THE TABLE BELOW AND IN THE TABLES ON PAGES 3-5 MAY BE USEFUL IN ANSWERING THE QUESTIONS IN THIS SECTION OF THE EXAMINATION. STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25° C E °( V ) Half-reaction F2 ( g ) + 2 e - Æ Co + e Au3+ + 3 e Cl 2 (g ) + 2 e O 2 (g ) + 4 H + + 4 e Br2 (l ) + 2 e - Æ Co Æ A u(s ) Æ 2 Cl - Æ 2 H 2 O(l ) - 3+ 2 F2+ Æ 2 Br - 2 Hg 2+ + 2 e - Æ Hg 2+ + 2 e Ag + + e - Hg 2 2+ Æ H g (l ) Æ A g (s ) Æ 2 H g (l ) Æ Fe 2+ I 2 (s ) + 2 e Cu + + e - Æ 2 I- Æ C u(s ) - Hg 2 Fe + 2e 2+ 3+ +e - - Æ C u(s ) - Æ Cu + Sn 4+ + 2 e S(s ) + 2 H + + 2 e 2 H+ + 2 e- Æ Sn 2+ Æ H 2 S(g ) Cu Cu + 2e 2+ 2+ +e Æ H2 (g) + 2e - Æ P b(s ) + 2e - Æ S n(s ) + 2e - Æ N i(s ) + 2e - Æ C o(s ) + 2e - Æ C d(s ) Æ Cr 2+ Fe 2+ + 2 e - Æ F e(s ) - Æ C r(s ) - Æ Æ Z n(s ) H 2 ( g ) + 2 OH - Æ M n(s ) Æ A l(s ) Æ B e(s ) Æ M g(s ) Æ Na(s ) Æ Æ Ca(s ) S r(s ) Æ B a(s ) Æ R b(s ) Pb 2+ Sn 2+ Ni 2+ Co 2+ Cd 2+ Cr Cr 3+ 3+ +e - + 3e Zn + 2 e 2 H 2 O(l ) + 2 e 2+ Mn Al 2+ 3+ Be + 3e 2+ Na + e Ca Sr 2+ Ba 2+ - + 2e + 2+ - + 2e 2+ Mg + 2e + 2e - - - - + 2e Rb + e - - + 2e + - - Æ K (s ) + - Æ C s(s ) Li + e - Æ Li(s ) K +e + Cs + e + 3 2.87 1.82 1.50 1.36 1.23 1.07 0.92 0.85 0.80 0.79 0.77 0.53 0.52 0.34 0.15 0.15 0.14 0.00 – 0.13 – 0.14 – 0.25 – 0.28 – 0.40 – 0.41 – 0.44 – 0.74 – 0.76 – 0.83 – 1.18 – 1.66 – 1.70 – 2.37 – 2.71 – 2.87 – 2.89 – 2.90 – 2.92 – 2.92 – 2.92 – 3.05 ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS E v l p ATOMIC STRUCTURE E = hv c = lv h l= p = mu mu -2.178 ¥ 10 -18 En = joule n2 Planck’s constant, h = 6.63 ¥ 10 -34 J s Boltzmann’s constant, k = 1.38 ¥ 10 -23 J K -1 [OH − ] [ HB + ] [B] Avogadro’s number = 6.022 ¥ 10 23 mol -1 Electron charge, e = -1.602 ¥ 10 -19 coulomb K w = [OH − ] [H + ] = 1.0 × 10 −14 @ 25 C = Ka × K b 1 electron volt per atom = 96.5 kJ mol -1 pH = − log [ H + ], pOH = − log [OH − ] 14 = pH + pOH pH = pKa + log Equilibrium Constants Ka (weak acid) Kb (weak base) [A − ] [ HA ] [HB + ] [ B] pKa = − log Ka , pKb = − log Kb K w (water) K p (gas pressure) pOH = pKb + log K p = Kc ( RT ) Dn Kc (molar concentrations) , S = standard entropy where D n = moles product gas − moles reactant gas H = standard enthalpy G = standard free energy THERMOCHEMISTRY/KINETICS DS = DH = DG =  S products -  S reactants  DHf products -  DH f reactants  DGf products -  DGf reactants E T n m q c Cp DG = DH - TD S = - RT ln K = -2.303 RT log K = -n E t 1 A - ln A t = = = = = = = standard reduction potential temperature moles mass heat specific heat capacity molar heat capacity at constant pressure E a = activation energy k = rate constant A = frequency factor DG = DG + RT ln Q = DG + 2.303 RT log Q q = mcDT DH Cp = DT ln A u = velocity n = principal quantum number m = mass energy frequency wavelength momentum Speed of light, c = 3.0 ¥ 108 m s -1 EQUILIBRIUM [H + ] [A − ] Ka = [HA ] Kb = = = = = 0 Faraday's constant, = - kt = 96,500 coulombs per mole of electrons Gas constant, R = 8.31 J mol -1 K -1 1 = kt A0 = 0.0821 L atm mol -1 K -1 = 8.31 volt coulomb mol -1 K -1 - Ea 1 ln k = + ln A RT ej 4 GASES, LIQUIDS, AND SOLUTIONS P V T n D m u PA = Ptotal ¥ X A , where X A = moles A total moles Ptotal = PA + PB + PC + ... m n= M K = C + 273 PV1 P2V2 1 = T1 T2 m D= V 3kT 3 RT = urms = M m 1 KE per molecule = mu 2 2 3 KE per mole = RT 2 M2 r1 = M1 r2 molarity, M = moles solute per liter solution molality = moles solute per kilogram solvent DT f = iK f ¥ molality DTb = iK b ¥ molality p = iMRT A = abc pressure volume temperature number of moles density mass velocity = = = = = = = root-mean-square speed kinetic energy rate of effusion molar mass osmotic pressure van't Hoff factor molal freezing-point depression constant Kb A a b c Q I q t Ê n2 a ˆ Á P + 2 ˜ (V - nb ) = nRT Ë V¯ = = = = = = = urms KE r M p i Kf PV = nRT = = = = = = = = = molal boiling-point elevation constant absorbance molar absorptivity path length concentration reaction quotient current (amperes) charge (coulombs) time (seconds) E = standard reduction potential K = equilibrium constant Gas constant, R = 8.31 J mol -1 K -1 OXIDATION-REDUCTION; ELECTROCHEMISTRY = 0.0821 L atm mol -1 K -1 Q= [ C ] c [ D] d [ A ] a [ B] b = 8.31 volt coulomb mol -1 K -1 , where a A + b B Æ c C + d D Boltzmann' s constant, k = 1.38 ¥ 10 -23 J K -1 K f for H 2 O = 1.86 K kg mol -1 q I= t E cell = E cell log K = Kb for H 2 O = 0.512 K kg mol -1 0.0592 RT ln Q = E cell log Q @ 25 C n n 1 atm = 760 mm Hg = 760 torr STP = 0.000 C and 1.000 atm nE 0.0592 Faraday' s constant, 5 = 96,500 coulombs per mole of electrons 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) CHEMISTRY Section II (Total time—90 minutes) Part A Time—40 minutes YOU MAY USE YOUR CALCULATOR FOR PART A. CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. Attention should be paid to significant figures. Be sure to write all your answers to the questions on the lined pages following each question in this booklet. Do NOT write your answers on the lavender insert. Answer Question 1 below. The Section II score weighting for this question is 20 percent. Ka = [H3O+ ][OCl - ] = 3.2 ¥ 10 -8 [HOCl] 1. Hypochlorous acid, HOCl, is a weak acid in water. The Ka expression for HOCl is shown above. (a) Write a chemical equation showing how HOCl behaves as an acid in water. (b) Calculate the pH of a 0.175 M solution of HOCl. (c) Write the net ionic equation for the reaction between the weak acid HOCl(aq) and the strong base NaOH(aq). (d) In an experiment, 20.00 mL of 0.175 M HOCl(aq) is placed in a flask and titrated with 6.55 mL of 0.435 M NaOH(aq). (i) Calculate the number of moles of NaOH(aq) added. (ii) Calculate [H3O+] in the flask after the NaOH(aq) has been added. (iii) Calculate [OH−] in the flask after the NaOH(aq) has been added. Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE. 6 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) Answer EITHER Question 2 below OR Question 3 printed on pages 8 and 9. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 20 percent. 2. Water was electrolyzed, as shown in the diagram above, for 5.61 minutes using a constant current of 0.513 ampere. A small amount of nonreactive electrolyte was added to the container before the electrolysis began. The temperature was 298 K and the atmospheric pressure was 1.00 atm. (a) Write the balanced equation for the half reaction that took place at the anode. (b) Calculate the amount of electric charge, in coulombs, that passed through the solution. (c) Why is the volume of O2(g) collected different from the volume of H2(g) collected, as shown in the diagram? (d) Calculate the number of moles of H2(g) produced during the electrolysis. (e) Calculate the volume, in liters, at 298 K and 1.00 atm of dry H2(g) produced during the electrolysis. (f) After the hydrolysis reaction was over, the vertical position of the tube containing the collected H2(g) was adjusted until the water levels inside and outside the tube were the same, as shown in the diagram below. The volume of gas in the tube was measured under these conditions of 298 K and 1.00 atm, and its volume was greater than the volume calculated in part (e). Explain. Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE. 7 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) X → 2Y + Z 3. The decomposition of gas X to produce gases Y and Z is represented by the equation above. In a certain experiment, the reaction took place in a 5.00 L flask at 428 K. Data from this experiment were used to produce the information in the table below, which is plotted in the graphs that follow. Time (minutes) 0 10. 20. 30. 50. 70. 100. [X] (mol L− 1) 0.00633 0.00520 0.00427 0.00349 0.00236 0.00160 0.000900 ln [X] −5.062 −5.259 −5.456 −5.658 − 6.049 − 6.438 −7.013 [X] − 1 (L mol −1 ) 158 192 234 287 424 625 1,110 Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE. 8 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) (a) How many moles of X were initially in the flask? (b) How many molecules of Y were produced in the first 20. minutes of the reaction? (c) What is the order of this reaction with respect to X ? Justify your answer. (d) Write the rate law for this reaction. (e) Calculate the specific rate constant for this reaction. Specify units. (f) Calculate the concentration of X in the flask after a total of 150. minutes of reaction. STOP If you finish before time is called, you may check your work on this part only. Do not turn to the other part of the test until you are told to do so. Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 9 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) CHEMISTRY Part B Time—50 minutes NO CALCULATORS MAY BE USED FOR PART B. Answer Question 4 below. The Section II score weighting for this question is 15 percent. 4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described below. Answers to more than five choices will not be graded. In all cases, a reaction occurs. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You need not balance the equations. Example: A strip of magnesium is added to a solution of silver nitrate. (a) A solution of potassium carbonate is added to a solution of strontium chloride. (b) Propene is burned in air. (c) Excess ammonia is added to a solution of zinc nitrate. (d) Ethanoic acid (acetic acid) is added to a solution of barium hydroxide. (e) A small piece of potassium is added to water. (f) Powdered iron metal is strongly heated with powdered sulfur. (g) A solution of sodium fluoride is added to a solution of hydrochloric acid. (h) A strip of lead metal is added to a solution of silver nitrate. Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE. 10 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses. Answer BOTH Question 5 below AND Question 6 printed on page 12. Both of these questions will be graded. The Section II score weighting for these questions is 30 percent (15 percent each). 2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 22 H2O(l) → 2 KAl(SO4)2 12H2O(s) + 3 H2(g) 5. In an experiment, a student synthesizes alum, KAl(SO4)2 12H2O(s), by reacting aluminum metal with potassium hydroxide and sulfuric acid, as represented in the balanced equation above. (a) In order to synthesize alum, the student must prepare a 5.0 M solution of sulfuric acid. Describe the procedure for preparing 50.0 mL of 5.0 M H2SO4 using any of the chemicals and equipment listed below. Indicate specific amounts and equipment where appropriate. 10.0 M H2SO4 Distilled water 100 mL graduated cylinder 100 mL beaker 50.0 mL volumetric flask 50.0 mL buret 25.0 mL pipet 50 mL beaker (b) Calculate the minimum volume of 5.0 M H2SO4 that the student must use to react completely with 2.7 g of aluminum metal. (c) As the reaction solution cools, alum crystals precipitate. The student filters the mixture and dries the crystals, then measures their mass. (i) If the student weighs the crystals before they are completely dry, would the calculated percent yield be greater than, less than, or equal to the actual percent yield? Explain. (ii) Cooling the reaction solution in an ice bath improves the percent yield obtained. Explain. (d) The student heats crystals of pure alum, KAl(SO4)2 12H2O(s), in an open crucible to a constant mass. The mass of the sample after heating is less than the mass before heating. Explain. Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE. 11 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) 6. Consider two containers of volume 1.0 L at 298 K, as shown above. One container holds 0.10 mol N2(g) and the other holds 0.10 mol H2(g). The average kinetic energy of the N2(g) molecules is 6.2 × 10−21 J . Assume that the N2(g) and the H2(g) exhibit ideal behavior. (a) Is the pressure in the container holding the H2(g) less than, greater than, or equal to the pressure in the container holding the N2(g) ? Justify your answer. (b) What is the average kinetic energy of the H2(g) molecules? (c) The molecules of which gas, N2 or H2 , have the greater average speed? Justify your answer. (d) What change could be made that would decrease the average kinetic energy of the N2(g) molecules in the container? (e) If the volume of the container holding the H2(g) was decreased to 0.50 L at 298 K, what would be the change in each of the following variables? In each case, justify your answer. (i) The pressure within the container (ii) The average speed of the H2(g) molecules Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE. 12 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) Answer EITHER Question 7 below OR Question 8 printed on page 14. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 15 percent. 7. Answer the following questions about thermodynamics. Substance H2(g) C(s) Combustion Reaction H2(g) + Enthalpy of Combustion, ∆H comb , at 298 K (kJ mol−1) 1 O (g) → H2O(l) 22 −290 C(s) + O2(g) → CO2(g) −390 −730 CH3OH(l) (a) In the empty box in the table above, write a balanced chemical equation for the complete combustion of one mole of CH3OH(l). Assume products are in their standard states at 298 K. Coefficients do not need to be whole numbers. (b) On the basis of your answer to part (a) and the information in the table, determine the enthalpy change for the reaction C(s) + H2(g) + H2O(l) → CH3OH(l) . (c) Write the balanced chemical equation that shows the reaction that is used to determine the enthalpy of formation for one mole of CH3OH(l) . (d) Predict the sign of ∆S° for the combustion of H2(g). Explain your reasoning. (e) On the basis of bond energies, explain why the combustion of H2(g) is exothermic. Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). GO ON TO THE NEXT PAGE. 13 2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) 8. Use principles of atomic structure, bonding, and intermolecular forces to answer the following questions. Your responses must include specific information about all substances referred to in each part. (a) Draw a complete Lewis electron-dot structure for the CS2 molecule. Include all valence electrons in your structure. (b) The carbon-to-sulfur bond length in CS2 is 160 picometers. Is the carbon-to-selenium bond length in CSe2 expected to be greater than, less than, or equal to this value? Justify your answer. (c) The bond energy of the carbon-to-sulfur bond in CS2 is 577 kJ mol−1. Is the bond energy of the carbon-toselenium bond in CSe2 expected to be greater than, less than, or equal to this value? Justify your answer. (d) The complete structural formulas of propane, C3H8 , and methanoic acid, HCOOH, are shown above. In the table below, write the type(s) of intermolecular attractive force(s) that occur in each substance. Substance Boiling Point Propane 229 K Methanoic acid Intermolecular Attractive Force(s) 374 K (e) Use principles of intermolecular attractive forces to explain why methanoic acid has a higher boiling point than propane. END OF EXAM Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 14 ...
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This note was uploaded on 10/01/2009 for the course OC 9876 taught by Professor Dq during the Spring '09 term at UC Merced.

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