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_ap05_sg_chemistry_46574 - AP® Chemistry 2005 Scoring...

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Unformatted text preview: AP® Chemistry 2005 Scoring Guidelines 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. AP® CHEMISTRY 2005 SCORING GUIDELINES Question 1 HC3H5O2(aq) C3H5O2–(aq) + H+(aq) Ka = 1.34 × 10 – 5 Propanoic acid, HC3H5O2, ionizes in water according to the equation above. (a) Write the equilibrium-constant expression for the reaction. Ka = [H + ][C3 H5O 2 − ] [HC3 H5O 2 ] One point is earned for the correct equilibrium expression. Notes: Correct expression without Ka earns 1 point. Entering the value of Ka is acceptable. Charges must be correct to earn 1 point. (b) Calculate the pH of a 0.265 M solution of propanoic acid. HC3H5O2(aq) I C E 0.265 –x 0.265 – x C3H5O2–(aq) + H+(aq) 0 +x +x ~0 +x +x One point is earned for recognizing that [H+] and [C3H5O2− ] have the same value in the equilibrium expression. [H + ][C3 H5O 2 − ] ( x)( x) Ka = = [HC3H5O 2 ] (0.265 − x) Assume that 0.265 – x ≈ 0.265, x2 0.265 –5 (1.34 × 10 )(0.265) = x 2 One point is earned for calculating [H+]. then 1.34 × 10–5 = One point is earned for calculating the correct pH. 3.55 × 10– 6 = x 2 x = [H+] = 1.88 × 10−3 M pH = – log [H+] = – log (1.88 × 10−3) pH = 2.725 Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 2 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 1 (continued) (c) A 0.496 g sample of sodium propanoate, NaC3H5O2 , is added to a 50.0 mL sample of a 0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the following. (i) The concentration of the propanoate ion, C3H5O2−(aq) in the solution mol NaC3H5O2 = 0.496 g NaC3H5O2 × 1 mol NaC3 H5O 2 96.0 g NaC3H5O 2 One point is earned for calculating the number of moles of NaC3H5O2. mol NaC3H5O2 = 5.17 × 10−3 mol NaC3H5O2 = mol C3H5O2− [C3H5O2−] = mol C3H5O 2 − 5.17 × 10− 3 mol C3 H5O 2 − = = 0.103 M volume of solution 0.050 L One point is earned for the molarity of the solution. (ii) The concentration of the H+(aq) ion in the solution HC3H5O2(aq) C3H5O2–(aq) + H+(aq) I C E 0.103 +x 0.103 + x 0.265 –x 0.265 – x Ka = ~0 +x +x [H + ][C3 H5O 2 − ] ( x)(0.103 + x) = (0.265 − x) [HC3H5O 2 ] One point is earned for calculating the value of [H+]. Assume that 0.103 + x ≈ 0.103 and 0.265 − x ≈ 0.265 ( x)(0.103) 0.265 0.265 = 3.45 × 10−5 M x = [H+] = (1.34 × 10−5) × 0.103 Ka = 1.34 × 10−5 = The methanoate ion, HCO2−(aq) , reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation. HCO2H(aq) + OH–(aq) HCO2−(aq) + H2O(l) (d) Given that [OH−] is 4.18 × 10−6 M in a 0.309 M solution of sodium methanoate, calculate each of the following. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 3 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 1 (continued) (i) The value of Kb for the methanoate ion, HCO2−(aq) HCO2H + OH–(aq) HCO2−(aq) + H2O(l) I C E 0.309 –x 0.309 – x - 0 +x +x ~0 +x +x x = [OH–] = 4.18 × 10−6 M Kb = [OH − ][HCO 2 H] [HCO 2 − ] = (4.18 × 10− 6 )2 ( x)( x) = (0.309 − x) (0.309 − x) One point is earned for substituting 4.18 × 10−6 for both [OH−] and [HCO2H], and for calculating the value of Kb. x is very small (4.18 × 10−6 M), therefore 0.309 – x ≈ 0.309 Kb = (4.18 × 10− 6 )2 = 5.65 × 10−11 0.309 (ii) The value of Ka for methanoic acid, HCO2H Kw = Ka × Kb Ka = Kw 1.00 × 10−14 = Kb 5.65 × 10−11 One point is earned for calculating a value of Ka from the value of Kb determined in part (d)(i). Ka = 1.77 × 10−4 (e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer. Ka for propanoic acid is 1.34 × 10–5, and Ka for methanoic acid is 1.77 × 10– 4. For acids, the larger the value of Ka , the greater the strength; therefore methanoic acid is the stronger acid because 1.77 × 10– 4 > 1.34 × 10–5. One point is earned for the correct choice and explanation based on the Ka calculated for methanoic acid in part (d)(ii). Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 4 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 2 Answer the following questions about a pure compound that contains only carbon, hydrogen, and oxygen. (a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and 0.3370 g of H2O(g). (i) Calculate the individual masses of C, H, and O in the 0.7549 g sample. ⎛ 1 mol CO 2 ⎞ ⎛ 1 mol C ⎞ ⎛ 12.01 g C ⎞ massC = 1.9061 g CO2 × ⎜ ⎟×⎜ ⎟×⎜ ⎟ ⎝ 44.01 g CO 2 ⎠ ⎝ 1 mol CO 2 ⎠ ⎝ 1 mol C ⎠ = 0.5202 g C ⎛ 1 mol H 2 O ⎞ ⎛ 2 mol H ⎞ ⎛ 1.008 g H ⎞ massH = 0.3370 g H2O × ⎜ ⎟×⎜ ⎟×⎜ ⎟ ⎝ 18.016 g H 2 O ⎠ ⎝ 1 mol H 2 O ⎠ ⎝ 1 mol H ⎠ = 0.03771 g H Three points total are earned: One point each for the masses (or moles) of C, H, and O. massO = 0.7549 g – 0.5202 g – 0.03771 g = 0.1970 g O (ii) Determine the empirical formula for the compound. ⎛ 1 mol C ⎞ nC = 0.5202 g C × ⎜ ⎟ = 0.04331 mol C ⎝ 12.01 g C ⎠ ⎛ 1 mol H ⎞ nH = 0.03771 g H × ⎜ ⎟ = 0.03741 mol H ⎝ 1.008 g H ⎠ ⎛ 1 mol O ⎞ nO = 0.1970 g O × ⎜ ⎟ = 0.01231 mol O ⎝ 16.00 g O ⎠ ⎛ 0.04331 mol C ⎞ ⎛ 0.03741 mol H ⎞ ⎛ 0.01231 mol O ⎞ ⎜ ⎟:⎜ ⎟:⎜ ⎟ 0.01231 ⎠ ⎝ 0.01231 ⎠ ⎝ 0.01231 ⎠ ⎝ One point is earned for the number of moles of C, H, and O.* One point is earned for the empirical formula.* 3.518 mol C : 3.039 mol H : 1.000 mol O *based on the three masses as determined in part (a)(i) above The empirical formula is C7H6O2 . (b) A 0.5246 g sample of the compound was dissolved in 10.0012 g of lauric acid, and it was determined that the freezing point of the lauric acid was lowered by 1.68°C. The value of Kf of lauric acid is 3.90°C m–1. Assume that the compound does not dissociate in lauric acid. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 5 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 2 (continued) (i) Calculate the molality of the compound dissolved in the lauric acid. ∆Tf = i × Kf × molality ( i = 1 since compound does not dissociate) ∆T f 1.68°C molality = = = 0.431 molal Kf 3.90°C m −1 One point is earned for the correct molality. (ii) Calculate the molar mass of the compound from the information provided. 0.431 molal = 0.431 mol compound 1 kg lauric acid ⎛ 1 kg ⎞ ⎛ 0.431 mol compound ⎞ ncompound = 10.0012 g lauric acid × ⎜ ⎟×⎜ ⎟ ⎝ 1,000 g ⎠ ⎝ 1 kg lauric acid ⎠ ncompound = 0.00431 mol 0.5246 g compound molar mass of compound = = 122 g mol–1 0.00431 mol compound One point is earned for the molar mass of the compound. (c) Without doing any calculations, explain how to determine the molecular formula of the compound based on the answers to parts (a)(ii) and (b)(ii). The molar mass should be divided by (or compared to) the empirical mass to obtain a whole number. Each subscript in the empirical formula is multiplied by this whole number. One point is earned for the explanation. (d) Further tests indicate that a 0.10 M aqueous solution of the compound has a pH of 2.6. Identify the organic functional group that accounts for this pH. Since an aqueous solution of the compound is acidic, the compound must be an organic acid. The functional group in an organic acid is the carboxyl group – COOH. One point is earned for identifying the carboxyl group. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 6 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 3 Answer the following questions related to the kinetics of chemical reactions. − ⎯⎯ I−(aq) + ClO−(aq) ⎯OH → IO−(aq) + Cl−(aq) Iodide ion, I−, is oxidized to hypoiodite ion, IO−, by hypochlorite, ClO−, in basic solution according to the equation above. Three initial-rate experiments were conducted; the results are shown in the following table. Experiment [I−] (mol L−1) [ClO−] (mol L−1) 1 0.017 0.015 Initial Rate of Formation of IO− (mol L−1 s−1) 0.156 2 0.052 0.015 0.476 3 0.016 0.061 0.596 (a) Determine the order of the reaction with respect to each reactant listed below. Show your work. (i) I−(aq) From experiments 1 and 2: y x rate2 k[I− ]2 [ClO − ]2 = y x rate1 k[I− ]1 [ClO − ]1 One point is earned for the correct order of the reaction with respect to I−, with justification. 0.476 k (0.052) x (0.015) y = 0.156 k (0.017) x (0.015) y 3.05 = (0.052) x (0.017) x = 3.1x , therefore x = 1, The reaction is first order with respect to I−. (ii) ClO−(aq) From experiments 1 and 3: x y rate3 k[I − ]3 [ClO − ]3 = x rate1 k[I − ]1 [ClO − ]1y 0.596 k (0.016) x (0.061) y k (0.016)1 (0.061) y = = 0.156 k (0.017) x (0.015) y k (0.017)1 (0.015) y 3.82 = (0.94) One point is earned for the correct order of the reaction with respect to ClO−, with justification. (0.061) y (0.015) y 4.06 = 4.1 y, so y = 1, The reaction is first order with respect to ClO−. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 7 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 3 (continued) (b) For the reaction (i) write the rate law that is consistent with the calculations in part (a); One point is earned for the correct rate law based on exponents as determined in part (a). rate = k [I−]1 [ClO−]1 (ii) calculate the value of the specific rate constant, k, and specify units. k= k= rate −1 [I ] [ClO − ]1 −1 0.156 mol L s One point is earned for the value of k. −1 One point is earned for the correct units (consistent with orders found). −1 (0.017 mol L )(0.015 mol L−1 ) k = 610 L mol−1 s−1 (or 610 M −1 s−1) The catalyzed decomposition of hydrogen peroxide, H2O2(aq), is represented by the following equation. catalyst → 2 H2O2(aq) ⎯⎯⎯⎯ 2 H2O(l) + O2(g) The kinetics of the decomposition reaction were studied and the analysis of the results show that it is a firstorder reaction. Some of the experimental data are shown in the table below. [H2O2] (mol L−1) Time (minutes) 1.00 0.78 0.61 0.0 5.0 10.0 (c) During the analysis of the data, the graph below was produced. Time (minutes) Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 8 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 3 (continued) (i) Label the vertical axis of the graph. ln[H2O2] (or log[H2O2 ]) One point is earned for the y-axis label. (ii) What are the units of the rate constant, k, for the decomposition of H2O2(aq) ? minutes−1 (or sec−1) One point is earned for the correct units for k. (iii) On the graph, draw the line that represents the plot of the uncatalyzed first-order decomposition of 1.00 M H2O2(aq). Two points are earned for all three features (same origin, straight line, smaller negative slope), or one point for any two features. The line should have the same origin, be a straight line, and have a smaller negative slope. uncatalyzed ln[H2O2] Time (minutes) Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 9 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 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. General Scoring: Three points are earned for each reaction: 1 point for correct reactant(s) and 2 points for correct product(s). (a) A strip of zinc is placed in a solution of nickel(II) nitrate. Zn + Ni2+ → Zn2+ + Ni (b) Solid aluminum hydroxide is added to a concentrated solution of potassium hydroxide. Al(OH)3 + OH− → Al(OH)4− (c) Ethyne (acetylene) is burned in air. C2H2 + O2 → CO2 + H2O (d) Solid calcium carbonate is added to a solution of ethanoic (acetic) acid. CaCO3 + HC2H3O2 → Ca2+ + C2H3O2− + CO2 + H2O (e) Lithium metal is strongly heated in nitrogen gas. Li + N2 → Li3N (f) Boron trifluoride gas is added to ammonia gas. BF3 + NH3 → BF3NH3 Note: F3BNH3 also acceptable as a product (g) Sulfur trioxide gas is bubbled into a solution of sodium hydroxide. SO3 + OH− → SO42− + H2O Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 10 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 4 (continued) (h) Equal volumes of 0.1 M solutions of lead(II) nitrate and magnesium iodide are combined. Pb2+ + I− → PbI2 Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 11 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 5 Answer the following questions that relate to laboratory observations and procedures. (a) An unknown gas is one of three possible gases: nitrogen, hydrogen, or oxygen. For each of the three possibilities, describe the result expected when the gas is tested using a glowing splint (a wooden stick with one end that has been ignited and extinguished, but still contains hot, glowing, partially burned wood). Nitrogen: When the glowing splint is inserted into the gas sample, the glowing splint will be extinguished. Hydrogen: When the glowing splint is inserted into the gas sample, a popping sound (explosion) can be heard. One point is earned for each description. Oxygen: When the glowing splint is inserted into the gas sample, the splint will glow brighter or reignite. (b) The following three mixtures have been prepared: CaO plus water, SiO2 plus water, and CO2 plus water. For each mixture, predict whether the pH is less than 7, equal to 7, or greater than 7. Justify your answers. CaO plus water: The pH of the solution will be greater than 7. CaO in water forms the base Ca(OH)2 (or metal oxides are basic, or basic anhydrides). SiO2 plus water: The pH of the solution will be equal to 7. SiO2 is insoluble in water, so there would not be a change in the pH of the mixture. One point is earned for each description. CO2 plus water: The pH of the solution will be less than 7. CO2 in water forms the acid H2CO3 (or nonmetal oxides are acidic, or acidic anhydrides). (c) Each of three beakers contains a 0.1 M solution of one of the following solutes: potassium chloride, silver nitrate, or sodium sulfide. The three beakers are labeled randomly as solution 1, solution 2, and solution 3. Shown below is a partially completed table of observations made of the results of combining small amounts of different pairs of the solutions. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 12 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 5 (continued) Solution 1 Solution 2 Solution 1 Solution 3 black precipitate no reaction Solution 2 Solution 3 (i) Write the chemical formula of the black precipitate. The black precipitate is Ag2S . One point is earned for the correct formula. (ii) Describe the expected results of mixing solution 1 with solution 3. A precipitate will be produced when the two solutions are mixed. (iii) One point is earned for the correct observation. Identify each of the solutions 1, 2, and 3. Solution 1 is silver nitrate. Solution 2 is sodium sulfide. Solution 3 is potassium chloride. One point is earned for the correct identification of all three solutions. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 13 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 6 Answer the following questions that relate to chemical bonding. (a) In the boxes provided, draw the complete Lewis structure (electron-dot diagram) for each of the three molecules represented below. PF5 CF4 SF4 One point is earned for each correct complete Lewis structure. See diagrams above. One point is deducted when structures are correct but nonbonding electrons around F atoms are missing. (b) On the basis of the Lewis structures drawn above, answer the following questions about the particular molecule indicated. (i) What is the F–C–F bond angle in CF4 ? One point is earned for the correct bond angle. 109.5° (or within range 109°–110°) The bond angle given in this part must be consistent with the Lewis structure drawn in part (a). (ii) What is the hybridization of the valence orbitals of P in PF5 ? One point is earned for the correct hybridization. dsp3 The hybridization given in this part must be consistent with the Lewis structure drawn in part (a). (iii) What is the geometric shape formed by the atoms in SF4 ? One point is earned for the correct molecular geometry. Seesaw (or distorted tetrahedron or asymmetrical tetrahedron) The molecular geometry given in this part must be consistent with the Lewis structure drawn in part (a). Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 14 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 6 (continued) (c) Two Lewis structures can be drawn for the OPF3 molecule, as shown below. (i) How many sigma bonds and how many pi bonds are in structure 1 ? One point is earned for the correct number of sigma bonds. One point is earned for the correct number of pi bonds. 4 sigma bonds and 1 pi bond (ii) Which one of the two structures best represents a molecule of OPF3 ? Justify your answer in terms of formal charge. Structure 1 is the better structure because all of its atoms have a formal charge of zero. One point is earned for choosing the correct structure and either (1) indicating that the formal charge is zero on P or O , or (2) showing the calculation for formal charge. P: 5 – 5 – 0 = 0 F: 7 – 1 – 6 = 0 O: 6 – 2 – 4 = 0 Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 15 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 7 Use principles of atomic structure, bonding, and/or intermolecular forces to respond to each of the following. Your responses must include specific information about all substances referred to in each question. (a) At a pressure of 1 atm, the boiling point of NH3(l) is 240 K, whereas the boiling point of NF3(l) is 144 K. (i) Identify the intermolecular forces(s) in each substance. NH3 has dispersion forces and hydrogen-bonding forces. NF3 has dispersion forces and dipole-dipole forces. (Credit earned for hydrogen-bonding and dipole-dipole forces) One point is earned for the correct intermolecular attractive forces for both NH3 and NF3. (ii) Account for the difference in the boiling points of the substances. The higher boiling point for NH3 is due to the greater strength of the hydrogen-bonding intermolecular attractive forces among NH3 molecules compared to that of the dipole-dipole attractive forces among NF3 molecules. One point is earned for correctly identifying NH3 as having stronger intermolecular forces than NF3. (b) The melting point of KCl(s) is 776°C, whereas the melting point of NaCl(s) is 801°C. (i) Identify the type of bonding in each substance. Both KCl and NaCl have ionic bonds. One point is earned for naming ionic bonds as the bonds in KCl and NaCl. (ii) Account for the difference in the melting points of the substances. The difference in the melting points is due to the different strengths of ionic bonding in the substances. The charges on the cations and anions are the same in both compounds, therefore the relative size of the ions is the determining factor. Since Na+ has a smaller ionic radius than K+, the lattice energy of NaCl is higher than that of KCl. Thus more energy is required to overcome the ionic forces in solid NaCl than in solid KCl, and NaCl has the higher melting point. One point is earned for a correct explanation of the cause of the difference in melting points of KCl and NaCl. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 16 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 7 (continued) (c) As shown in the table below, the first ionization energies of Si , P, and Cl show a trend. Element First Ionization Energy (kJ mol−1) Si 786 P 1,012 Cl 1,251 (i) For each of the three elements, identify the quantum level (e.g., n = 1, n = 2, etc.) of the valence electrons in the atom. One point is earned for the principal quantum level for all three elements. The valence electron is located in the n = 3 level for all three atoms. (ii) Explain the reasons for the trend in first ionization energies. Because the valence electrons in all three elements are shielded by the same number of inner core electrons and the nuclear charge increases going from Si to P to Cl, the valence electrons feel an increasing attraction to the nucleus going from Si to P to Cl. Valence electrons having a greater attraction to the nucleus, as in Cl , will be more difficult to remove, so Cl has the highest ionization energy. P has the second highest ionization energy, and Si has the lowest ionization energy. One point is earned for explaining that greater ionization energy is due to increased nuclear charge. Note: Explanations of the trend on the basis of effective nuclear charge are acceptable. (d) A certain element has two stable isotopes. The mass of one of the isotopes is 62.93 amu and the mass of the other isotope is 64.93 amu. (i) Identify the element. Justify your answer. Copper. The relative average atomic mass is between the two isotopic masses given. One point is earned for the element and the explanation. (ii) Which isotope is more abundant? Justify your answer. The isotope with mass 62.93 amu must be more abundant because its mass is closer to 63.55 amu (the relative weighted average atomic mass for copper) than is the mass of the other isotope. One point is earned for the correct choice and explanation. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 17 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 8 AgNO3(s) → Ag+(aq) + NO3−(aq) The dissolving of AgNO3(s) in pure water is represented by the equation above. (a) Is ∆G for the dissolving of AgNO3(s) positive, negative, or zero? Justify your answer. ∆G for the dissolving of AgNO3(s) is negative. Because AgNO3(s) is known to be soluble in water, the solution process must be spontaneous, therefore ∆G is negative. One point is earned for the correct sign of ∆G and a correct explanation. (b) Is ∆S for the dissolving of AgNO3(s) positive, negative, or zero? Justify your answer. ∆S is positive because the solid reactant AgNO3(s) is more ordered than the aqueous ion products, Ag+(aq) and NO3− (aq). One point is earned for the correct sign of ∆S and a correct explanation. (c) The solubility of AgNO3(s) increases with increasing temperature. (i) What is the sign of ∆H for the dissolving process? Justify your answer. The sign of ∆H must be positive for the solubility of AgNO3 to increase with increasing temperature. Solubility is an equilibrium process, and since increasing temperature (accomplished by adding heat) shifts the equilibrium towards the products side in the chemical equation, heat must be absorbed during the solution process. Therefore, the solution process is endothermic, and ∆H > 0 . One point is earned for the correct sign of ∆H and a correct explanation. (ii) Is the answer you gave in part (a) consistent with your answers to parts (b) and (c)(i) ? Explain. Yes. Although ∆H is positive, ∆S is also positive; thus ∆G can be negative because the value of the T∆S term in the equation ∆G = ∆H − T∆S is positive and can be greater than the value of the ∆H term. A positive number minus a greater positive number yields a negative number for the value of ∆G . One point is earned for the correct sign and a correct explanation. The compound NaI dissolves in pure water according to the equation NaI(s) → Na+(aq) + I−(aq). Some of the information in the table of standard reduction potentials given below may be useful in answering the questions that follow. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 18 AP® CHEMISTRY 2005 SCORING GUIDELINES Question 8 (continued) Half-reaction E˚ (V) O2(g) + 4 H+ + 4 e− → 2 H2O(l) 1.23 I2(s) + 2 e− → 2 I− 0.53 2 H2O(l) + 2 e− → H2(g) + 2 OH− –0.83 Na+ + e−→ Na(s) –2.71 (d) An electric current is applied to a 1.0 M NaI solution. (i) Write the balanced oxidation half-reaction for the reaction that takes place. 2 I− → I2(s) + 2 e− One point is earned for the correct half-reaction. (ii) Write the balanced reduction half-reaction for the reaction that takes place. One point is earned for the correct half-reaction. 2 H2O(l) + 2 e− → H2(g) + OH− (iii) Which reaction takes place at the anode, the oxidation reaction or the reduction reaction? The oxidation half-reaction occurs at the anode. One point is earned for the correct choice. (iv) All electrolysis reactions have the same sign for ∆G°. Is the sign positive or negative? Justify your answer. The sign of ∆G for all electrolysis reactions is positive. Because electrolysis reactions are non-spontaneous, energy in the form of applied electrical current (electrical work) must be applied to make the reaction occur. One point is earned for the correct sign of ∆G and a correct explanation. Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 19 ...
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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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