ap03_sg_chemistry_b_26429

ap03_sg_chemistry_b_26429 - AP® Chemistry 2003 Scoring...

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Unformatted text preview: AP® Chemistry 2003 Scoring Guidelines Form B The materials included in these files are intended for use by AP teachers for course and exam preparation; permission for any other use must be sought from the Advanced Placement Program®. Teachers may reproduce them, in whole or in part, in limited quantities for noncommercial, face-to-face teaching purposes. This permission does not apply to any third-party copyrights contained herein. This material may not be mass distributed, electronically or otherwise. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here. These materials were produced by Educational Testing Service® (ETS®), which develops and administers the examinations of the Advanced Placement Program for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, and their programs, services, and employment policies are guided by that principle. The College Board is a national nonprofit membership association whose mission is to prepare, inspire, and connect students to college and opportunity. Founded in 1900, the association is composed of more than 4,300 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 22,000 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 equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. For further information, visit www.collegeboard.com Copyright © 2003 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. AP Central is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark jointly owned by the College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service. Other products and services may be trademarks of their respective owners. For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 1 Total Score 10 points 2 HI(g) ↔ H2(g) + I2(g) 1. After a 1.0 mole sample of HI(g) is placed into an evacuated 1.0 L container at 700. K, the reaction represented above occurs. The concentration of HI(g) as a function of time is shown below. (a) Write the expression for the equilibrium constant, Kc , for the reaction. Kc = [H2][I2] [HI]2 1 point for correct expression (b) What is [HI] at equilibrium? From the graph, [HI]eq is 0.80 M 1 point for equilibrium [HI] Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 2 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 1 (cont’d.) (c) Determine the equilibrium concentrations of H2(g) and I2(g). 2 HI(g) → H2(g) + I2(g) I C E 1.0 M −0.20 M 0.80 M 0 +0.10 M 0.10 M 1 point for stoichiometric relationship between HI reacting and H2(g) or I2(g) forming 0 +0.10 M 0.10 M 1 point for [H2]eq and [I2]eq [I2] = [H2] = 0.10 M (d) On the graph above, make a sketch that shows how the concentration of H2(g) changes as a function of time. From the graph, [H2]eq is 0.10 M The curve should have the following characteristics: - start at 0 M; - increase to 0.1 M; - reach equilibrium at the same time [HI] reaches equilibrium 1 point for any two characteristics 2 points for all three characteristics (e) Calculate the value of the following equilibrium constants for the reaction at 700. K. (i) Kc Kc = [H2][I2] [0.10][0.10] = = 0.016 [0.80]2 [HI]2 1 point for correct substitution (must agree with parts (b) and (c)) (ii) Kp Kp = Kc = 0.016 The number of moles on the product side is equal to the number of moles on the reactant side Kp = Kc(RT)∆n ∆n = 2 – 2 = 0 Kp = Kc(RT)0 Kp = Kc 1 point for Kp = Kc (with verification) Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 3 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 1 (cont’d.) (f) At 1,000 K, the value of Kc for the reaction is 2.6 × 10 −2. In an experiment, 0.75 mole of HI(g), 0.10 mole of H2(g) , and 0.50 mol of I2(g) are placed in a 1.0 L container and allowed to reach equilibrium at 1,000 K. Determine whether the equilibrium concentration of HI(g) will be greater than, equal to, or less than the initial concentration of HI(g) . Justify your answer. [H2][I2] [0.10][0.50] = 8.9 × 10−2 2= [HI] [0.75]2 Kc = 2.6 × 10−2 1 point for calculating Q and comparing to Kc Q > Kc 1 point for predicting correct change in [HI] Q= To establish equilibrium, the numerator must decrease and the denominator must increase. Therefore, [HI] will increase. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 4 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 2 Total Score 10 points 2. Answer the following questions that relate to chemical reactions. (a) Iron(III) oxide can be reduced with carbon monoxide according to the following equation. Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) A 16.2 L sample of CO(g) at 1.50 atm and 200.°C is combined with 15.39 g of Fe2O3(s). (i) How many moles of CO(g) are available for the reaction? PV = nRT PV nCO = RT = (1.50 atm) (16.2 L) = 0.626 mol CO L·atm 0.0821 mol·K (473 K) 1 point for correct substitution 1 point for answer (ii) What is the limiting reactant for the reaction? Justify your answer with calculations. nFe 2O3 æ 1 mol Fe2O3 ö = 15.39 g Fe2O3 ç159.7 g Fe O ÷ = 0.0964 mol Fe2O3 nCO required è 2 3ø 3 mol CO ö = 0.0964 mol Fe2O3 æ ç1 mol Fe O ÷ = 0.289 mol 2 3ø è CO required to completely react with 0.0964 mol Fe2O3 0.626 mol CO are available, so CO is in excess and Fe2O3 is limiting. OR nFe 2O3 required æ1 mol Fe2O3 ö = 0.626 mol CO ç 3 mol CO ÷ = 0.209 mol è ø 0.209 mol Fe2O3 corresponds to 33.4 g Fe2O3 (the amount of Fe2O3 required to completely react with 0.626 mol CO) 1 point for moles of CO or Fe2O3 required 1 point for correct conclusion NOTE: Answer must be consistent with moles of CO calculated in part (a) 0.0964 mol of Fe2O3 is available, so there is not enough Fe2O3 to completely react with 0.626 mol CO. Therefore, Fe2O3 is the limiting reactant. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 5 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 2 (cont’d.) (iii) How many moles of Fe(s) are formed in the reaction? 2 mol Fe ö nFe = 0.0964 mol Fe2O3 æ ç1 mol Fe O ÷ = 0.193 mol Fe produced 2 3ø è If 0.626 mol CO is used as the limiting reactant, then the number of moles of Fe formed is: 2 mol Fe ö nFe = 0.626 mol CO æ = 0.417 mol Fe è3 mol COø 1 point for correct number of moles of Fe formed Note: Values must be consistent with answer in part (a) (ii). (b) In a reaction vessel, 0.600 mol of Ba(NO3)2(s) and 0.300 mol of H3PO4(aq) are combined with deionized water to a final volume of 2.00 L. The reaction represented below occurs. 3 Ba(NO3)2(aq) + 2 H3PO4(aq) → Ba3(PO4)2(s) + 6 HNO3(aq) (i) Calculate the mass of Ba3(PO4)2(s) formed. æ 2 mol H3PO4 ö nH3PO4 = 0.600 mol Ba(NO3)2ç 3 mol Ba(NO ) ÷ = 0.400mol H3PO4 è 3 2ø required to completely react with 0.600 mol Ba(NO3)2 . There is 0.300 mol H3PO4 available. Therefore, H3PO4 is the limiting reactant. 1 point for determining the limiting reactant æ1 mol Ba3(PO4)2ö ÷× è 2 mol H3PO4 ø massBa3(PO4)2 = 0.300 mol H3PO4ç æ602 g Ba3(PO4)2 ö ç1 mol Ba (PO ) ÷ = 90.3 g Ba3(PO4)2 3 4 2ø è If Ba(NO3)2 is used as the limiting reactant: æ1 mol Ba3(PO4)2ö massBa3(PO4)2 = 0.600 mol Ba(NO3)2 ç ÷ è 3 mol Ba(NO3)2 ø æ602 g Ba3(PO4)2 ö ç1 mol Ba (PO ) ÷ = 120. g Ba3(PO4)2 3 4 2ø è 1 point for determining the correct mass of Ba3(PO4)2 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 6 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 2 (cont’d.) (ii) Calculate the pH of the resulting solution. æ 6 mol HNO3 ö nHNO = 0.300 mol H3PO4 ç ÷ = 0.900 mol HNO3 3 è2 mol H3PO4ø 0.900 mol HNO3 = 0.45 M [HNO3] = 2.0 L Since all the H3PO4 has reacted, the only acid in the solution is HNO3. Since HNO3 is a strong acid it completely dissociates. pH = −log[H+] = −log(0.45) = 0.35 If Ba(NO3)2 is used as the limiting reactant æ 6 mol HNO3 ö nHNO = 0.600 mol Ba(NO3)2 ç ÷ = 1.2 mol HNO3 3 è3 mol Ba(NO3)2ø 1.2 mol HNO3 = 0.60 M [HNO3] = 2.0 L pH = −log[H+] = −log(0.60) = 0.22 1 point for number of moles of H+ 1 point for calculation of [H+] and pH Must be consistent with part (b) (i) (iii) What is the concentration, in mol L−1, of the nitrate ion, NO3−(aq), after the reaction reaches completion? The final concentration of NO3− must be the same as the initial concentration. æ 2 mol NO3− ö ÷ = 1.2 mol NO3− nNO − = 0.600 mol Ba(NO3)2 ç 3 è1 mol Ba(NO3)2ø [NO3−] = − 1.2 mol NO3 = 0.60 M NO3− 2.0 L 1 point for concentration of NO3− Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 7 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 3 Total Score 10 points 3. In an experiment, a sample of an unknown, pure gaseous hydrocarbon was analyzed. Results showed that the sample contained 6.000 g of carbon and 1.344 g of hydrogen. (a) Determine the empirical formula of the hydrocarbon. 1 mol C ö = 0.5000 mol C è12.00 g Cø 1 mol H ö nH = 1.344 g H æ = 1.333 mol H è1.008 g Hø 1 point for number of moles of carbon and number of moles of hydrogen 1 mol C : 2.667 mol H 1 point for ratio of moles of carbon to moles of hydrogen nC = 6.000 g C æ 0.5000 mol C 1.333 mol H : 0.5000 0.5000 3 (1 mol C : 2.667 mol H) = 3 mol C : 8.000 mol H 1 point for correct formula The empirical formula is C3H8 (b) The density of the hydrocarbon at 25°C and 1.09 atm is 1.96 g L−1. (i) Calculate the molar mass of the hydrocarbon. PV = nRT grams PV = molar mass RT grams RT molar mass = VP molar mass = 1.96 g L−1 1 point for correct substitution and 1 point for answer OR L·atm 0.0821 mol·K 298 K 1.09 atm 1 point for calculation and 1 point for units molar mass = 44.0 g mol−1 1.96 g L−1 × 22.4 L mol −1 = 43.9 g mol −1 (1 point maximum) (ii) Determine the molecular formula of the hydrocarbon. Empirical mass × n = molar mass Empirical mass for C3H8 is 44 g mol−1 44 g mol−1 × n = 44 g mol−1 Þ n = 1, so the molecular formula is the same as the empirical formula, C3H8 1 point for reporting correct formula with verification Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 8 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 3 (cont’d.) In another experiment, liquid heptane, C7H16(l), is completely combusted to produce CO2(g) and H2O(l), as represented by the following equation. C7H16(l) + 11 O2(g) → 7 CO2(g) + 8 H2O(l) o The heat of combustion, ∆ H comb , for one mole of C7H16(l) is −4.85 × 103 kJ. (c) Using the information in the table below, calculate the value of ∆ H fo for C7H16(l) in kJ mol−1. Compound ∆ H fo (kJ mol−1) CO2(g) H2O(l) −393.5 −285.8 ∆ H fo = ∑ ∆ H fo (products) − ∑ ∆ H fo (reactants) = 7 ∆ H fo (CO2) + 8 ∆ H fo (H2O) − [ ∆ H fo (C7H16) + 11 ∆ H fo (O2)] 1 point for correct coefficients kJ kJ kJ kJ −4,850 mol = 7(−393.5 )+ 8(−285.8 mol ) − [ ∆ H fo (C7H16) + 11 (0 mol) ] 1 point for the mol correct substitution kJ kJ kJ −4,850 mol = −2,754 mol −2,286mol − ∆ H fo (C7H16) into ∆ H fo equation kJ ∆ H fo (C7H16) = −191 mol (d) A 0.0108 mol sample of C7H16(l) is combusted in a bomb calorimeter. (i) Calculate the amount of heat released to the calorimeter. æ −4850 kJ ö qreleased = 0.0108 mol C7H16 ç1 mol C H ÷ è = 52.4 kJ of heat released 1 point for the amount of heat released 7 16ø (ii) Given that the total heat capacity of the calorimeter is 9.273 kJ °C−1, calculate the temperature change of the calorimeter. Q = Cp ∆T 52.4 kJ = 9.273 kJ °C−1 × ∆T 52.4 kJ ö ∆T = æ ç9.273 kJ °C−1÷ = 5.65°C è ø ∆T = −(−52.4 kJ)/(9.273 kJ )°C−1 = +5.65°C 1 point for the correct change in temperature (Must be consistent with answer in part (d)(i)) Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 9 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 4 Total Score 15 points 4. (a) Hot hydrogen gas is passed over heated copper(II) oxide solid. H2 + CuO → Cu + H2O Note: Cu2O is an acceptable product (b) Solid sodium hydride is added to water. NaH + H2O → Na+ + OH− + H2 (c) Propanone is burned in air. C3H6O + O2 → CO2 + H2O Note: CO is an acceptable product (d) A solution of lead(II) nitrate is added to a solution of potassium sulfate. Pb2+ + SO42− → PbSO4 (e) Ammonia gas is mixed with hydrogen chloride gas. HCl + NH3 → NH4Cl Note: 1 product point for NH4+ + Cl− ; 2 points (total) for NH3 + H+ → NH4+ (f) Sulfur trioxide gas is bubbled into water. SO3 + H2O → H+ + HSO4− (g) Excess concentrated potassium hydroxide solution is added to a solution of nickel(II) chloride. OH− + Ni2+ → Ni(OH)42− Note: Ni(OH)2 , Ni(OH)3−, Ni(OH)53−, and Ni(OH)64− are acceptable products (h) Solid sodium acetate is added to 1.0 M hydrobromic acid. NaC2H3O2 + H+ → HC2H3O2 + Na+ Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 10 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 5 Total Score 10 points 5. Oxalic acid, H2C2O4 , is a primary standard used to determine the concentration of potassium permanganate, KMnO4 , in solution. The equation for the reaction is as follows. 2 KMnO4(aq) + 5 H2C2O4(aq) + 3 H2SO4 → 2 MnSO4(aq) + 10 CO2(g) + 8 H2O(l) + K2SO4(aq) A student dissolves a sample of oxalic acid in a flask with 30 mL of water and 2.00 mL of 3.00 M H2SO4. The KMnO4 solution of unknown concentration is in a 25.0 mL buret. In the titration, the KMnO4 solution is added to the solution containing the oxalic acid. (a) What chemical species is being oxidized in the reaction? H2C2O4 is the substance being oxidized. The half-reaction is: H2C2O4(aq) → 2 CO2(g) + 2 H+(aq) + 2 e− 1 point for identifying H2C2O4 or C2O42− as species oxidized OR The oxidation state of carbon changes from +3 to +4 in CO2. (b) What substance indicates the observable endpoint of the titration? Describe the observation that shows the endpoint has been reached. In the reaction, the purple KMnO4 solution in the buret is added to the colorless solution in the flask. KMnO4 reacts with H2C2O4 upon addition, so the purple KMnO4 color disappears as it is added to the solution in the flask that contains unreacted H2C2O4 . 1 point for identifying KMnO4 as reacting species that indicates the endpoint As soon as all the H2C2O4 has reacted (endpoint), the KMnO4 is in excess and the solution in the flask will turn pink (pink is the color produced when the more concentrated purple KMnO4 solution in the buret is diluted in the solution in the flask). 1 point for indicating color change is from colorless to pink at the endpoint (c) What data must be collected in the titration in order to determine the molar concentration of the unknown KMnO4 solution? The mass of oxalic acid, the initial volume of the KMnO4 solution in the buret, and the final volume of the KMnO4 solution in the buret 1 point for the mass of oxalic acid 1 point for the initial and final volume or for saying the change in volume of KMnO4 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 11 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 5 (cont’d.) (d) Without doing any calculations, explain how to determine the molarity of the unknown KMnO4 solution. Determine the moles of oxalic acid (by dividing the mass of oxalic acid measured/weighed, by the molar mass of oxalic acid). 1 point for determining moles of H2C2O4 Use the stoichiometric ratio of the amount (2 mol) of KMnO4 to amount (5 mol) of H2C2O4 from the balanced chemical equation to convert from amount (in moles) of H2C2O4 to amount (in moles) of KMnO4. 1 point for using correct stoichiometric factor Divide the amount (in moles) of KMnO4 by the volume, expressed in liters, of KMnO4 needed to reach the endpoint. 1 point for dividing moles of KMnO4 by liters of KMnO4 solution (e) How would the calculated concentration of the KMnO4 solution be affected if 40 mL of water was added to the oxalic acid initially instead of 30 mL? Explain your reasoning. There would be no effect on the concentration of the KMnO4 solution. We are only interested in the moles of oxalic acid. Since it is a solid, the moles of oxalic acid are calculated from the mass of oxalic acid. The volume of water used to dissolve the oxalic acid is independent of the moles of oxalic acid. 1 point for effect 1 point for explanation Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 12 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 6 Total Score 9 points 6. Answer the following questions about electrochemistry. (a) Several different electrochemical cells can be constructed using the materials shown below. Write the balanced net-ionic equation for the reaction that occurs in the cell that would have the greatest o positive value of Ecell . 1 point for selection of correct two redox couples Al(s) → Al3+(aq) + 3 e− Cu2+(aq) + 2 e − → Cu(s) 1 point for correctly balanced net ionic equation 2 Al(s) + 3 Cu2+(aq) → 2 Al3+(aq) + 3 Cu(s) (b) Calculate the standard cell potential, E°cell , for the reaction written in part (a). Al3+(aq) + 3 e − → Al(s) Cu2+(aq) + 2 e − → Cu(s) E° = −1.66 V E° = +0.34 V o o o Ecell = Ecathode − Eanode = +0.34 V −(−1.66 V) = +2.00 V o 1 point for correct Ecell (Must be consistent with part (a)) Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 13 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 6 (cont’d.) (c) A cell is constructed based on the reaction in part (a) above. Label the metal used for the anode on the cell shown in the figure below. The metal is aluminum solid. 1 point for correct metal (Must be consistent with part (a)) (d) Of the compounds NaOH , CuS , and NaNO3 , which one is appropriate to use in a salt bridge? Briefly explain your answer, and for each of the other compounds, include a reason why it is not appropriate. NaOH is not appropriate. The anion, OH−, would migrate towards the anode. The OH− would react with the Al3+ ion in solution. CuS is not appropriate. It is insoluble in water, so no ions would be available to migrate to the anode and cathode compartment to balance the charge. 1 point for correctly indicating whether each compound is appropriate, along with an explanation (3 points total) NaNO3 is appropriate. It is soluble in water, and neither the cation nor the anion will react with the ions in the anode or cathode compartment. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 14 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 6 (cont’d.) (e) Another standard cell is based on the following reaction. Zn + Pb2+ → Zn2+ + Pb If the concentration of Zn2+ is decreased from 1.0 M to 0.25 M, what effect does this have on the cell potential? Justify your answer. [Zn2+]ö è[Pb2+]ø o Ecell = Ecell – 0.059 ln æ [Zn2+]ö < 1, therefore è[Pb2+]ø If [Zn2+] is reduced, then the ratio æ [Zn2+] lnæ 2+ ö < 0. Thus Ecell increases (becomes more positive). è[Pb ]ø 1 point for correctly indicating how Ecell is affected 1 point for explanation in terms of Nernst equation and Q Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 15 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 7 Total Score 8 points 7. Account for the following observations using principles of atomic structure and/or chemical bonding. In each part, your answer must include specific information about both substances. (a) The Ca2+ and Cl− ions are isoelectronic, but their radii are not the same. Which ion has the larger radius? Explain. Both Ca2+ and Cl− ions have 18 electrons. Their electron configuration is 1s22s22p63s23p6. However, they differ by the number of protons in the nucleus. Calcium has 20 protons and chlorine has 17 protons. 1 point for indicating that chloride ion has the larger ionic radius The valence electrons are shielded by the same number of electrons in each ion (10), so the effective nuclear charge (ENC) experienced by the valence electrons in Ca2+ is +10 and for Cl− the ENC is +7. The valence electrons in Cl− experience a smaller attraction to the nucleus due to the smaller nuclear charge, so Cl− has the larger ionic radius. (The same argument is acceptable when comparing the total number of protons versus total number of electrons for each ion.) 1 point for correct explanation (b) Carbon and lead are in the same group of elements, but carbon is classified as a nonmetal and lead is classified as a metal. Binary compounds of carbon exhibit covalent character (property of a nonmetallic element), whereas binary compounds of lead exhibit ionic character (property of a metallic element). OR Oxides of carbon, when dissolved in water, are acidic (property of a nonmetallic element), whereas oxides of lead, when added to water, are basic (property of a metallic element). OR Carbon is a poor thermal conductor (property of a nonmetallic element), whereas lead is a very good thermal conductor (property of a metallic element). 1 point each for indicating the characteristic of each element and the difference in behavior exhibited by the element, and then relating the behavior to a metal or nonmetal Note: Students may use other examples where the chemical or physical properties of carbon and lead differ to distinguish between the two elements.) Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 16 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 7 (cont’d.) (c) Compounds containing Kr have been synthesized, but there are no known compounds that contain He. Helium has a filled shell (the first shell), so does not tend to lose or gain electrons. Therefore, helium does not react. 1 point for filled shell for He Krypton, while having filled 4s and 4p sublevels, has empty 4d and 4f sublevels. These empty orbitals affect the reactivity of Kr. 1 point for indicating presence of empty d orbitals in Kr Note: Also acceptable is a comparison of the ionization energies of helium, and krypton and then the justification for krypton being more reactive. (d) The first ionization energy of Be is 900 kJ mol−1, but the first ionization energy of B is 800 kJ mol−1. The electron configuration for Be is 1s22s2, whereas the electron configuration for B is 1s22s22p1. The first electron removed in boron is in a 2p subshell, which is higher in energy than the 2s subshell, from which the first electron is removed in beryllium. The higher in energy the subshell containing the electron to be removed (ionized), the lower the ionization energy. 1 point for indicating the difference in the subshell where the first electron is removed for each element 1 point for associating higher energy sublevel with lower ionization energy Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 17 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 8 Total Score 8 points 8. The decay of the radioisotope I-131 was studied in a laboratory. I-131 is known to decay by beta ( −0 e ) emission. 1 (a) Write a balanced nuclear equation for the decay of I-131. 131 53 I Note: “β” for 0 −1 e → 131 54 Xe + 0 −1 e 1 point for correct equation is acceptable (b) What is the source of the beta particle emitted from the nucleus? A neutron spontaneously decays to an electron and a proton. 1 point for identifying a neutron as the source of the beta emission The radioactivity of a sample of I-131 was measured. The data collected are plotted on the graph below. (c) Determine the half-life, t1/2 , of I-131 using the graph above. The half-life is 8 days. That is the time required for the disintegration rate to fall from 16,000 to one-half its initial value, 8,000. 1 point for half-life Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 18 AP® CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 8 (cont’d.) (d) The data can be used to show that the decay of I-131 is a first-order reaction, as indicated on the graph below. (i) Label the vertical axis of the graph above. The label on the y-axis should be ln or log one of the following: disintegrations or moles or atoms or [I-131] or disintegration rate. 1 point for correct label on y-axis (ii) What are the units of the rate constant, k , for the decay reaction? From the graph, the units on the rate constant are days−1 (Units of time−1 is acceptable) 1 point for correct units (iii) Explain how the half-life of I-131 can be calculated using the slope of the line plotted on the graph. The slope of the line is −k. The slope is negative, so k is a positive number. The half-life can then be calculated using the 0.693 relationship t1/2 = k. 1 point for indicating slope is k 1 point for half-life equation (d) Compare the value of the half-life of I-131 at 25°C to its value at 50°C. The half-life will be the same at the different temperatures. The half-life of a nuclear decay process is independent of temperature. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 19 1 point ...
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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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