ap07_chemistry_form_b_sgs_final_complete - AP® Chemistry...

Unformatted text preview: AP® Chemistry 2007 Scoring Guidelines 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 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,500 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. © 2007 The College Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, SAT, and the acorn logo are registered trademarks of the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. 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. AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 1 A sample of solid U3O8 is placed in a rigid 1.500 L flask. Chlorine gas, Cl2(g), is added, and the flask is heated to 862°C. The equation for the reaction that takes place and the equilibrium-constant expression for the reaction are given below. ( pUO2 Cl2 )3 ( pO2 ) 3 UO2Cl2(g) + O2(g) U3O8(s) + 3 Cl2(g) Kp = ( pCl2 )3 When the system is at equilibrium, the partial pressure of Cl2(g) is 1.007 atm and the partial pressure of UO2Cl2(g) is 9.734 × 10− 4 atm. (a) Calculate the partial pressure of O2(g) at equilibrium at 862°C. U3O8(s) + 3 Cl2(g) I C E --- 3 UO2Cl2(g) + O2(g) ? 0 1.007 atm 9.734 × 10− 4 atm UO2Cl2(g) × 0 9.734 × 10− 4 atm ? One point is earned for the correct answer. (1 mol O 2 ) = 3.245 × 10− 4 atm O2(g) (3 mol UO 2 Cl2 ) (b) Calculate the value of the equilibrium constant, Kp , for the system at 862°C. Kp = (pUO 2Cl2 )3 ( pO 2 ) ( pCl2 )3 = (9.734 × 10−4 )3 (3.245 × 10−4 ) (1.007)3 = 2.931 × 10−13 One point is earned for the correct substitution. One point is earned for the correct answer. (c) Calculate the Gibbs free-energy change, ΔG °, for the reaction at 862°C. ΔG ° = − RT ln Kp = (−8.31 J mol−1 K−1)( (862+273) K)(ln (2.931 × 10−13)) = 272,000 J mol−1 = 272 kJ mol−1 One point is earned for the correct setup. One point is earned for the correct answer with units. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 1 (continued) (d) State whether the entropy change, ΔS°, for the reaction at 862°C is positive, negative, or zero. Justify your answer. ΔS° is positive because four moles of gaseous products are produced from three moles of gaseous reactants. One point is earned for the correct explanation. (e) State whether the enthalpy change, ΔH°, for the reaction at 862°C is positive, negative, or zero. Justify your answer. Both ΔG° and ΔS° are positive, as determined in parts (c) and (d). Thus, ΔH° must be positive because ΔH° is the sum of two positive terms in the equation ΔH° = ΔG° + TΔS°. One point is earned for the correct sign. One point is earned for a correct explanation. (f) After a certain period of time, 1.000 mol of O2(g) is added to the mixture in the flask. Does the mass of U3O8(s) in the flask increase, decrease, or remain the same? Justify your answer. The mass of U3O8(s) will increase because the reaction is at equilibrium, and the addition of a product creates a “stress” on the product (right) side of the reaction. The reaction will then proceed from right to left to reestablish equilibrium so that some O2(g) is consumed (tending to relieve the stress) as more U3O8(s) is produced. One point is earned for a correct explanation. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 2 Answer the following problems about gases. (a) The average atomic mass of naturally occurring neon is 20.18 amu. There are two common isotopes of naturally occurring neon as indicated in the table below. Isotope Ne-20 Ne-22 Mass (amu) 19.99 21.99 (i) Using the information above, calculate the percent abundance of each isotope. Let x represent the natural abundance of Ne-20. 19.99 x + 21.99(1− x) 19.99 x + 21.99 − 21.99 x 19.99 x − 21.99 x −2 x x ⇒ percent abundances are: = = = = = 20.18 20.18 20.18 − 21.99 −1.81 0.905 One point is earned for the correct answer. Ne-20 = 90.5% Ne-22 = 9.5% (ii) Calculate the number of Ne-22 atoms in a 12.55 g sample of naturally occurring neon. 23 12.55 g Ne × 6.022 ×10 Ne-22 atoms 1 mol Ne 0.095 mol Ne-22 × × 1 mol Ne-22 20.18 g Ne 1 mol Ne = 3.6 × 1022 Ne-22 atoms One point is earned for the correct molar mass. One point is earned for the correct fraction of Ne-22 in Ne. One point is earned for the number of atoms. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 2 (continued) (b) A major line in the emission spectrum of neon corresponds to a frequency of 4.34 × 10−14 s−1. Calculate the wavelength, in nanometers, of light that corresponds to this line. c = λν λ= ⇒ λ= 8 3.0 × 10 m s −1 14 −1 4.34 × 10 s c ν × One point is earned for the correct setup. 1 nm 10− 9 m = 690 nm One point is earned for the answer. (c) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet (UV) radiation, as shown by the equation below. Ozone serves to block harmful ultraviolet radiation that comes from the Sun. ææ O3(g) æUV Æ O2(g) + O(g) A molecule of O3(g) absorbs a photon with a frequency of 1.00 × 1015 s−1. (i) How much energy, in joules, does the O3(g) molecule absorb per photon? E = hν = 6.63 × 10−34 J s × 1.00 × 1015 s−1 = 6.63 × 10 −19 J per photon One point is earned for the correct answer. (ii) The minimum energy needed to break an oxygen-oxygen bond in ozone is 387 kJ mol−1. Does a photon with a frequency of 1.00 × 1015 s−1 have enough energy to break this bond? Support your answer with a calculation. 6.63 × 10 −19 J 6.022 × 1023 photons 1 kJ × × 3 = 399 kJ mol −1 1 photon 1 mol 10 J 399 kJ mol−1 > 387 kJ mol−1, therefore the bond can be broken. One point is earned for calculating the energy. One point is earned for the comparison of bond energies. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 3 2 H2(g) + O2(g) → 2 H2O(l) In a hydrogen-oxygen fuel cell, energy is produced by the overall reaction represented above. (a) When the fuel cell operates at 25°C and 1.00 atm for 78.0 minutes, 0.0746 mol of O2(g) is consumed. Calculate the volume of H2(g) consumed during the same time period. Express your answer in liters measured at 25°C and 1.00 atm. (0.0746 mol O2) × V= n H 2 RT P = 2 mol H 2 = 0.149 mol H2 1 mol O2 One point is earned for calculation of moles of H 2 . (0.149 mol H2 )(0.0821 L atm mol −1 K −1 )(298 K) 1.00 atm One point is earned for substitution into PV = nRT . One point is earned for the answer. = 3.65 L H2 (b) Given that the fuel cell reaction takes place in an acidic medium, (i) write the two half reactions that occur as the cell operates, O2 + 4 H+ + 4 e− → 2 H2O One point is earned for each of the two half reactions. H2 → 2 H+ + 2 e− (ii) identify the half reaction that takes place at the cathode, and O2 + 4 H+ + 4 e− → 2 H2O One point is earned for either the equation of the correct half reaction, or for indicating “the reduction half reaction” if the correct equation is given in (b)(i) . (iii) determine the value of the standard potential, E°, of the cell. E° = 1.23V + 0.00 V = 1.23 V One point is earned for the standard potential. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 3 (continued) (c) Calculate the charge, in coulombs, that passes through the cell during the 78.0 minutes of operation as described in part (a). − (0.0746 mol O2 ) × 4 mol e 96, 500 C × = 2.88 × 104 C 1 mol O2 1 mol e− One point is earned for the stoichiometry. One point is earned for the answer. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 4 For each of the following three reactions, in part (i) write a balanced equation for the reaction 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 graded. (a) Solid ammonium carbonate decomposes as it is heated. (i) Balanced equation: (NH4)2CO3 → 2 NH3 + CO2 + H2O One point is earned for the correct reactant. Two points are earned for correct products. One point is earned for balancing mass and charge. (ii) Predict the algebraic sign of ΔS ° for the reaction. Explain your reasoning. The algebraic sign of ΔS ° for the reaction will be positive because one mole of solid (with relatively low entropy) is converted into four moles of gas (with much greater entropy). One point is earned for the correct answer. (b) Chlorine gas, an oxidizing agent, is bubbled into a solution of potassium bromide. (i) Balanced equation: Cl2 + 2 Br − → 2 Cl − + Br2 One point is earned for correct reactants. Two points are earned for correct products. One point is earned for balancing mass and charge. (ii) What is the oxidation number of chlorine before the reaction occurs? What is the oxidation number of chlorine after the reaction occurs? The oxidation number of chlorine is 0 before the reaction and −1 after the reaction. One point is earned for the correct answer. (c) A small piece of sodium is placed in a beaker of distilled water. (i) Balanced equation: 2 Na + 2 H2O → H2 + 2 Na+ + 2 OH− One point is earned for correct reactants. Two points are earned for correct products. One point is earned for balancing mass and charge. (ii) The reaction is exothermic, and sometimes small flames are observed as the sodium reacts with the water. Identify the product of the reaction that burns to produce the flames. It is the H2 gas that burns. One point is earned for the correct answer. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 5 Answer the following questions about laboratory solutions involving acids, bases, and buffer solutions. (a) Lactic acid, HC3H5O3, reacts with water to produce an acidic solution. Shown below are the complete Lewis structures of the reactants. In the space provided above, complete the equation by drawing the complete Lewis structures of the reaction products. One point is earned for each correct structure. (b) Choosing from the chemicals and equipment listed below, describe how to prepare 100.00 mL of a 1.00 M aqueous solution of NH4Cl (molar mass 53.5 g mol−1). Include specific amounts and equipment where appropriate. NH4Cl(s) Distilled water 50 mL buret 100 mL beaker 100 mL graduated cylinder 100 mL volumetric flask 100 mL pipet Balance mass of NH4Cl = (0.100 L)(1.00 mol L−1)(53.5 g mol−1) = 5.35 g NH4Cl 1. Measure out 5.35 g NH4Cl using the balance. One point is earned for the mass. 2. Use the 100 mL graduated cylinder to transfer approximately 25 mL of distilled water to the 100 mL volumetric flask. 3. Transfer the 5.35 g NH4Cl to the 100 mL volumetric flask. One point is earned for using a volumetric flask. 4. Continue to add distilled water to the volumetric flask while swirling the flask to dissolve the NH4Cl and remove all NH4Cl particles adhered to the walls. One point is earned for diluting to the mark. 5. Carefully add distilled water to the 100 mL volumetric flask until the bottom of the meniscus of the solution reaches the etched mark on the flask. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 5 (continued) (c) Two buffer solutions, each containing acetic acid and sodium acetate, are prepared. A student adds 0.10 mol of HCl to 1.0 L of each of these buffer solutions and to 1.0 L of distilled water. The table below shows the pH measurements made before and after the 0.10 mol of HCl is added. pH Before HCl Added 7.0 4.7 4.7 Distilled Water Buffer 1 Buffer 2 pH After HCl Added 1.0 2.7 4.3 (i) Write the balanced net-ionic equation for the reaction that takes place when the HCl is added to buffer 1 or buffer 2. C2H3O2− + H3O+ → HC2H3O2 + H2O One point is earned for the equation. (ii) Explain why the pH of buffer 1 is different from the pH of buffer 2 after 0.10 mol of HCl is added. Before the HCl was added, each buffer had the same pH and thus had the same [H+]. Because Ka for acetic acid is a constant, the ratio of [H+] to Ka must also be constant; this means that the ratio of [HC2H3O2 ] to [C2H3O2− ] is the same for both buffers, as shown by the following equation, derived from the equilibrium-constant expression for the dissociation of acetic acid. [HC2 H3O2 ] − [C2 H 3O 2 ] = [H + ] Ka After the addition of the H+, the ratio in buffer 1 must have been greater than the corresponding ratio in buffer 2, as evidenced by their respective pH values. Thus a greater proportion of the C2H3O2− in buffer 1 must have reacted with the added H+ compared to the proportion that reacted in buffer 2. The difference between these proportions means that the original concentrations of HC2H3O2 and C2H3O2− had to be smaller in buffer 1 than in buffer 2. One point is earned for a correct answer involving better buffering capacity or relative amount of base (acetate ion). © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 5 (continued) (iii) Explain why the pH of buffer 1 is the same as the pH of buffer 2 before 0.10 mol of HCl is added. Both buffer solutions have the same acid to conjugate-base mole ratio in the formula below. [H+] = Ka [HC2H3O2 ] ⎡ C2 H 3 O 2 − ⎤ ⎣ ⎦ One point is earned for the correct answer involving ratio of acid to base in the buffer. Therefore, the buffers have the same [H+] and pH. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 6 First Ionization Energy (kJ mol−1) 1,251 496 738 1,000 Element 1 Element 2 Element 3 Element 4 Second Ionization Energy (kJ mol−1) 2,300 4,560 1,450 2,250 Third Ionization Energy (kJ mol−1) 3,820 6,910 7,730 3,360 The table above shows the first three ionization energies for atoms of four elements from the third period of the periodic table. The elements are numbered randomly. Use the information in the table to answer the following questions. (a) Which element is most metallic in character? Explain your reasoning. Element 2. It has the lowest first-ionization energy. Metallic elements lose electron(s) when they become ions, and element 2 requires the least amount of energy to remove an electron. One point is earned for the identification. One point is earned for the justification. (b) Identify element 3. Explain your reasoning. Magnesium. Element 3 has low first and second ionization energies relative to the third ionization energy, indicating that the element has two valence electrons, which is true for magnesium. (The third ionization of element 3 is dramatically higher, indicating the removal of an electron from a noble gas core.) One point is earned for the identification. One point is earned for the justification. (c) Write the complete electron configuration for an atom of element 3. 1s2 2s2 2p6 3s2 One point is earned for the correct electron configuration. (d) What is the expected oxidation state for the most common ion of element 2? 1+ One point is earned for the correct oxidation state. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 6 (continued) (e) What is the chemical symbol for element 2? Na One point is earned for the correct symbol. (f) A neutral atom of which of the four elements has the smallest radius? Element 1 One point is earned for the correct identification of the element. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). ...
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