ap10_frq_chemistry_formb - AP® Chemistry 2010...

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Unformatted text preview: AP® Chemistry 2010 Free-Response Questions Form B The College Board 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 College Board is composed of more than 5,700 schools, colleges, universities and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,800 colleges through major programs and services in college readiness, college admission, guidance, assessment, financial aid and enrollment. Among its widely recognized programs are the SAT®, the PSAT/NMSQT®, the Advanced Placement Program® (AP®), SpringBoard® and ACCUPLACER®. 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. © 2010 The College Board. College Board, ACCUPLACER, Advanced Placement Program, AP, AP Central, SAT, SpringBoard and the acorn logo are registered trademarks of the College Board. Admitted Class Evaluation Service is a trademark owned by the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at: 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: apcentral.collegeboard.com. 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. -2- GO ON TO THE NEXT PAGE. STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25∞C Half-reaction E ∞(V) 2F 2+ F2 ( g ) + 2 e 3+ - - Æ Æ Æ Æ - 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 Co + e Au3+ + 3 e Cl2 (g ) + 2 e O2 (g ) + 4 H + 4 e + Co Au(s) 2 Cl 2 H 2 O(l ) 2 Br Hg2 2+ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Br2 (l ) + 2 e 2 Hg 2+ 2+ - + 2e - Hg + 2 e Ag+ + e Hg2 2+ + 2 e Fe + e I 2 (s) + 2 e 3+ - Hg(l ) Ag(s) 2 Hg(l ) Fe 2 I2+ Cu + e Cu2+ + 2 e Cu2+ + e Sn 4+ + 2 e S(s) + 2 H + + 2 e + - Cu(s) Cu(s) Cu+ Sn 2+ H 2 S(g ) H2 ( g) Pb(s) Sn(s) Ni(s) Co(s) Cd(s) Cr 2+ Fe(s) Cr(s) Zn(s) H 2 ( g) + 2 OH Mn(s) Al(s) Be(s) Mg(s) Na(s) Ca(s) Sr(s) Ba( s) Rb(s) K ( s) Cs(s) Li(s) -3- 2H + 2e + 2+ - Pb + 2 e Sn 2+ + 2 e Ni2+ + 2 e Co2+ + 2 e Cd 2+ + 2 e Cr 3+ + e Fe 2+ + 2 e Cr 3+ + 3 e Zn 2+ + 2 e 2 H 2 O(l ) + 2 e Mn + 2 e Al3+ + 3 e Be 2+ + 2 e Mg 2 + + 2 e Na + + e Ca 2+ + 2 e Sr 2+ + 2 e Ba 2+ + 2 e Rb+ + e K + + eCs+ + e Li+ + e 2+ - GO ON TO THE NEXT PAGE. ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS ATOMIC STRUCTURE E = hv c = lv h l= p = mu mu -2.178 ¥ 10 -18 En = joule n2 E v l p = = = = energy frequency wavelength momentum u = velocity n = principal quantum number m = mass Speed of light, c = 3.0 ¥ 108 m s -1 Planck’s constant, h = 6.63 ¥ 10 -34 J s Boltzmann’s constant, k = 1.38 ¥ 10 -23 J K -1 Avogadro’s number = 6.022 ¥ 1023 mol -1 -14 EQUILIBRIUM [H + ] [A - ] Ka = [HA] Kb = [OH - ][HB+ ] [B] + K w = [OH ][H ] = 1.0 ¥ 10 = K a ¥ Kb Electron charge, e = -1.602 ¥ 10 -19 coulomb @ 25 C 1 electron volt per atom = 96.5 kJ mol -1 Equilibrium Constants K a (weak acid) K b (weak base) K w (water) K p (gas pressure) K c (molar concentrations) S = standard entropy H = standard enthalpy G = standard free energy E T n m q c Cp pH = - log [H + ], pOH = - log[OH - ] 14 = pH + pOH pH = pK a + log pOH = pK b + log [A - ] [HA] [HB+ ] [B] pK a = - log K a , pK b = - log K b K p = K c ( RT ) Dn , where D n = moles product gas - moles reactant gas THERMOCHEMISTRY/KINETICS DS = DH DG Â S products -Â S reactants = Â DHf products -Â DH f reactants = Â DGf products -Â DGf reactants = - RT ln K = -2.303 RT log K = -n E DG = DH - T D S = = = = = = = standard reduction potential temperature moles mass heat specific heat capacity molar heat capacity at constant pressure Ea = activation energy k = rate constant A = frequency factor Faraday's constant, DG = DG + RT ln Q = DG + 2.303 RT log Q q = mc DT DH Cp = DT ln [ A ] t - ln [ A ]0 = - kt 1 1 = kt [A ]t [A ]0 = 96,500 coulombs per mole of electrons = 0.0821 L atm mol -1 K -1 = 8.31 volt coulomb mol -1 K -1 Gas constant, R = 8.31 J mol -1 K -1 ln k = - Ea 1 + ln A RT () -4- GO ON TO THE NEXT PAGE. GASES, LIQUIDS, AND SOLUTIONS PV = nRT Ê n2 a ˆ Á P + 2 ˜ (V - nb) = nRT Ë V¯ PA = Ptotal ¥ X A , where X A = Ptotal = PA + PB + PC + ... n= m M moles A total moles P V T n D m u = = = = = = = pressure volume temperature number of moles density mass velocity K = C + 273 PV1 PV 1 = 22 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 urms = root-mean-square speed KE r M p i Kf A a b c Q I q t = = = = = = = = = = = = = = kinetic energy rate of effusion molar mass osmotic pressure van't Hoff factor molal freezing -point depression constant absorbance molar absorptivity path length concentration reaction quotient current (amperes) charge (coulombs) time (seconds) K b = molal boiling -point elevation constant E = standard reduction potential K = equilibrium constant OXIDATION-REDUCTION; ELECTROCHEMISTRY Gas constant, R = 8.31 J mol -1 K -1 = 0.0821 L atm mol -1 K -1 = 8.31 volt coulomb mol -1 K -1 Boltzmann's constant, k = 1.38 ¥ 10 -23 J K -1 K f for H 2 O = 1.86 K kg mol -1 K b for H 2 O = 0.512 K kg mol -1 Q= I= [C] c [D] d [A] a [B] q t , where a A + b B Æ c C + d D b Ecell = Ecell log K = nE 0.0592 0.0592 RT ln Q = Ecell log Q @ 25 C n n 1 atm = 760 mm Hg = 760 torr STP = 0.00 C and 1.0 atm Faraday's constant, = 96,500 coulombs per mole of electrons -5- GO ON TO THE NEXT PAGE. 2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) CHEMISTRY Section II (Total time—95 minutes) Part A Time— 55 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 the goldenrod booklet. Do NOT write your answers on the lavender insert. Answer Questions 1, 2, and 3. The Section II score weighting for each question is 20 percent. 1. The compound butane, C4H10 , occurs in two isomeric forms, n-butane and isobutane (2-methyl propane). Both compounds exist as gases at 25°C and 1.0 atm. (a) Draw the structural formula of each of the isomers (include all atoms). Clearly label each structure. (b) On the basis of molecular structure, identify the isomer that has the higher boiling point. Justify your answer. The two isomers exist in equilibrium as represented by the equation below. Æ n-butane(g) ¨ isobutane(g) Kc = 2.5 at 25°C Suppose that a 0.010 mol sample of pure n-butane is placed in an evacuated 1.0 L rigid container at 25°C. (c) Write the expression for the equilibrium constant, Kc , for the reaction. (d) Calculate the initial pressure in the container when the n-butane is first introduced (before the reaction starts). (e) The n-butane reacts until equilibrium has been established at 25°C. (i) Calculate the total pressure in the container at equilibrium. Justify your answer. (ii) Calculate the molar concentration of each species at equilibrium. (iii) If the volume of the system is reduced to half of its original volume, what will be the new concentration of n-butane after equilibrium has been reestablished at 25°C ? Justify your answer. Suppose that in another experiment a 0.010 mol sample of pure isobutane is placed in an evacuated 1.0 L rigid container and allowed to come to equilibrium at 25°C. (f) Calculate the molar concentration of each species after equilibrium has been established. © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -6- 2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) 5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) 2. A galvanic cell and the balanced equation for the spontaneous cell reaction are shown above. The two reduction half-reactions for the overall reaction that occurs in the cell are shown in the table below. Half-Reaction Fe3+(aq) + e- Æ Fe2+(aq) MnO4-(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(l) E° (V) at 298 K + 0.77 +1.49 (a) On the diagram, clearly label the cathode. (b) Calculate the value of the standard potential, E°, for the spontaneous cell reaction. (c) How many moles of electrons are transferred when 1.0 mol of MnO4-(aq) is consumed in the overall cell reaction? (d) Calculate the value of the equilibrium constant, Keq , for the cell reaction at 25°C. Explain what the magnitude of Keq tells you about the extent of the reaction. Three solutions, one containing Fe2+(aq), one containing MnO4-(aq), and one containing H+(aq), are mixed in a beaker and allowed to react. The initial concentrations of the species in the mixture are 0.60 M Fe2+(aq) , 0.10 M MnO4-(aq) , and 1.0 M H+(aq) . (e) When the reaction mixture has come to equilibrium, which species has the higher concentration, Mn2+(aq) or MnO4-(aq)? Explain. (f) When the reaction mixture has come to equilibrium, what are the molar concentrations of Fe2+(aq) and Fe3+(aq)? © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -7- 2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) 3. A sample of ore containing the mineral tellurite, TeO2 , was dissolved in acid. The resulting solution was then reacted with a solution of K2Cr2O7 to form telluric acid, H2TeO4 . The unbalanced chemical equation for the reaction is given below. . . . TeO2(s) + . . . Cr2O72-(aq) + . . . H+(aq) Æ . . . H2TeO4(aq) + . . . Cr3+(aq) + . . . H2O(l) (a) Identify the molecule or ion that is being oxidized in the reaction. (b) Give the oxidation number of Cr in the Cr2O72-(aq) ion. (c) Balance the chemical equation given above by writing the correct lowest whole-number coefficients on the dotted lines. In the procedure described above, 46.00 mL of 0.03109 M K2Cr2O7 was added to the ore sample after it was dissolved in acid. When the chemical reaction had progressed as completely as possible, the amount of unreacted (excess) Cr2O72-(aq) was determined by titrating the solution with 0.110 M Fe(NO3)2 . The reaction that occurred during the titration is represented by the following balanced equation. 6 Fe2+(aq) + Cr2O72-(aq) + 14 H+(aq) Æ 2 Cr3+(aq) + 6 Fe3+(aq) + 7 H2O(l) A volume of 9.85 mL of 0.110 M Fe(NO3)2 was required to reach the equivalence point. (d) Calculate the number of moles of excess Cr2O72-(aq) that was titrated. (e) Calculate the number of moles of Cr2O72-(aq) that reacted with the tellurite. (f) Calculate the mass, in grams, of tellurite that was in the ore sample. 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. © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. -8- 2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) CHEMISTRY Part B Time— 40 minutes NO CALCULATORS MAY BE USED FOR PART B. Answer Question 4 below. The Section II score weighting for this question is 10 percent. 4. For each of the following three reactions, in part (i) write a balanced equation and in part (ii) answer the question about the reaction. In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may use the empty space at the bottom of the next page for scratch work, but only equations that are written in the answer boxes provided will be scored. (a) Solid copper(II) sulfate pentahydrate is gently heated. (i) Balanced equation: (ii) How many grams of water are present in 1.00 mol of copper(II) sulfate pentahydrate? ____________________________________________________________________________________ ____________________________________________________________________________________ © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -9- 2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) (b) Excess concentrated aqueous ammonia is added to a solution of nickel(II) nitrate, leading to the formation of a complex ion. (i) Balanced equation: (ii) Which of the reactants acts as a Lewis acid? _____________________________________________________________________________________ _____________________________________________________________________________________ (c) Methylamine (CH3NH2 ) is added to a solution of hydrochloric acid. (i) Balanced equation: (ii) Methylamine dissolves in water to form a solution. Indicate whether this solution is acidic, basic, or neutral. _____________________________________________________________________________________ _____________________________________________________________________________________ © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -10- 2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) Answer Question 5 and Question 6. The Section II score weighting for these questions is 15 percent each. Your responses to these questions will be scored 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. 5. A solution of 0.100 M HCl and a solution of 0.100 M NaOH are prepared. A 40.0 mL sample of one of the solutions is added to a beaker and then titrated with the other solution. A pH electrode is used to obtain the data that are plotted in the titration curve shown above. (a) Identify the solution that was initially added to the beaker. Explain your reasoning. (b) On the titration curve above, circle the point that corresponds to the equivalence point. (c) At the equivalence point, how many moles of titrant have been added? (d) The same titration is to be performed again, this time using an indicator. Use the information in the table below to select the best indicator for the titration. Explain your choice. Indicator Methyl violet Methyl red Alizarin yellow pH Range of Color Change 0 – 1.6 4–6 10 – 12 (e) What is the difference between the equivalence point of a titration and the end point of a titration? (f) On the grid provided on the next page, sketch the titration curve that would result if the solutions in the beaker and buret were reversed (i.e., if 40.0 mL of the solution used in the buret in the previous titration were titrated with the solution that was in the beaker). © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -11- 2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. GO ON TO THE NEXT PAGE. -12- 2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) H2(g) + Cl2(g) → 2 HCl(g) 6. The table below gives data for a reaction rate study of the reaction represented above. Experiment 1 2 3 Initial [H2 ] (mol L ) 0.00100 0.00200 0.00200 −1 Initial [Cl2] (mol L ) 0.000500 0.000500 0.000250 −1 Initial Rate of Formation of HCl (mol L−1 s−1) 1.82 × 10−12 3.64 × 10−12 1.82 × 10−12 (a) Determine the order of the reaction with respect to H2 and justify your answer. (b) Determine the order of the reaction with respect to Cl 2 and justify your answer. (c) Write the overall rate law for the reaction. (d) Write the units of the rate constant. (e) Predict the initial rate of the reaction if the initial concentration of H2 is 0.00300 mol L−1 and the initial concentration of Cl2 is 0.000500 mol L−1. The gas-phase decomposition of nitrous oxide has the following two-step mechanism. Step 1: N2O → N2 + O Step 2: O + N2O → N2 + O2 (f) Write the balanced equation for the overall reaction. (g) Is the oxygen atom, O, a catalyst for the reaction or is it an intermediate? Explain. (h) Identify the slower step in the mechanism if the rate law for the reaction was determined to be rate = k [N2O]. Justify your answer. STOP END OF EXAM © 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com. -13- ...
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This note was uploaded on 05/05/2011 for the course CHEM 504 taught by Professor John during the Fall '11 term at American College of Computer & Information Sciences.

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