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_ap06_chemb_sg - AP® Chemistry 2006 Scoring Guidelines...

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Unformatted text preview: AP® Chemistry 2006 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. © 2006 The 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 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 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. AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 1 → C6H5COOH(s) ← C6H5COO– (aq) + H+(aq) Ka = 6.46 × 10 – 5 1. Benzoic acid, C6H5COOH, dissociates in water as shown in the equation above. A 25.0 mL sample of an aqueous solution of pure benzoic acid is titrated using standardized 0.150 M NaOH. (a) After addition of 15.0 mL of the 0.150 M NaOH, the pH of the resulting solution is 4.37. Calculate each of the following. (i) [H+] in the solution [H+] = 10− 4.37 M = 4.3 × 10− 5 M One point is earned for the correct answer. (ii) [OH −] in the solution [OH − ] = Kw + [H ] = 1.0 × 10−14 M 2 4.3 × 10 −5 M = 2.3 × 10−10 M One point is earned for the correct answer. (iii) The number of moles of NaOH added mol OH − = 0.0150 L × 0.150 mol L−1 = 2.25 × 10− 3 mol One point is earned for the correct answer. (iv) The number of moles of C6H5COO– (aq) in the solution mol OH − added = mol C6H5COO−(aq) generated, thus mol C6H5COO−(aq) in solution = 2.25 × 10− 3 mol One point is earned for the correct answer. (v) The number of moles of C6H5COOH in the solution Ka = [H + ][C6H 5COO - ] [C6 H5COOH] (4.3 [C6H5COOH] = ⇒ [C6H5COOH] = [H + ][C6H 5COO - ] Ka 2.25 10 -3 mol 10 -5 M ) 0.040 L -5 6.46 10 = 3.7 × 10−2 M One point is earned for the correct molarity. One point is earned for the correct answer. thus, mol C6H5COOH = (0.040 L)( 3.7 × 10−2 M) = 1.5 × 10−3 mol © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 2 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 1 (continued) Alternative solution for part (a)(v): pH = pKa + log [C6 H5 COO − ] [C6 H5 COOH] ⇒ pH − pKa = log [C6H5COO−] − log [C6H5COOH] ⇒ log [C6H5COOH] = log [C6H5COO−] − ( pH − pKa ) 2.25 × 10−3 mol ) − (4.37 − 4.190) 0.040 L = log ( = −1.25 − 0.18 = −1.43 ⇒ = 10−1.43 = 3.7 × 10−2 M [C6H5COOH] thus, mol C6H5COOH = ( 0.040 L )( 3.7 × 10−2 M ) = 1.5 × 10−3 mol (b) State whether the solution at the equivalence point of the titration is acidic, basic, or neutral. Explain your reasoning. At the equivalence point the solution is basic due to the presence of C6H5COO− (the conjugate base of the weak acid) that hydrolyzes to produce a basic solution as represented below. → C H COO− + H O ← C H COOH + OH− 65 2 One point is earned for the prediction and the explanation. 65 In a different titration, a 0.7529 g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in water and titrated with 0.150 M NaOH. The equivalence point is reached when 24.78 mL of the base solution is added. (c) Calculate each of the following. (i) The mass, in grams, of benzoic acid in the solid sample mol C6H5COOH = (0.02478 L) × (0.150 mol OH− L−1) × 1 mol C6 H 5COOH 1 mol OH − = 3.72 × 10− 3 mol C6H5COOH mass C6H5COOH = 3.72 × 10− 3 mol C6H5COOH × 122 g C6 H 5COOH 1 mol C6 H 5COOH One point is earned for the correct answer. = 0.453 g C6H5COOH © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 3 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 1 (continued) (ii) The mass percentage of benzoic acid in the solid sample mass % C6H5COOH = 0.453 g C6 H 5COOH × 100 0.7529 g One point is earned for the correct answer. = 60.2% © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 4 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 2 2. Answer the following questions about voltaic cells. (a) A voltaic cell is set up using Al /Al 3+ as one half-cell and Sn /Sn2+ as the other half-cell. The half-cells contain equal volumes of solutions and are at standard conditions. (i) Write the balanced net-ionic equation for the spontaneous cell reaction. 3 Sn2+ + 2 Al → 3 Sn + 2 Al 3+ One point is earned for the correct direction. One point is earned for the balanced net-ionic equation. (ii) Determine the value, in volts, of the standard potential, E°, for the spontaneous cell reaction. E° = − 0.14 V − (−1.66 V) = 1.52 V (or, 1.52 J C−1) One point is earned for the correct answer. (Potential must be positive.) (iii) Calculate the value of the standard free-energy change, ∆G°, for the spontaneous cell reaction. Include units with your answer. 96,500 C 6 mol e− × × (1.52 J C−1) ∆G° = − n F E° = − 1 mol 1 mol e − One point is earned for indicating the correct mol e− to mol reaction ratio. = − 8.80 × 105 J mol−1 (or – 880 kJ mol−1) One point is earned for the correct answer with correct units. (iv) If the cell operates until [Al 3+] is 1.08 M in the Al /Al 3+ half-cell, what is [Sn2+] in the Sn /Sn2+ half-cell? change in [Sn2+] = 3 mol Sn 2+ 0.08 mol Al3+ 0.12 mol Sn 2+ × = 1L 1L 2 mol Al3+ [Sn2+] = 1.00 mol L−1 – 0.12 mol L−1 = 0.88 mol L−1 One point is earned for the correct answer. (b) In another voltaic cell with Al /Al 3+ and Sn /Sn2+ half-cells, [Sn2+] is 0.010 M and [Al 3+] is 1.00 M. Calculate the value, in volts, of the cell potential, Ecell , at 25°C. Ecell Answers must be consistent with part (a)(i). One point is earned for the proper exponents. One point is earned for the correct substitution of concentrations. 0.0592 (1.00)2 = 1.52 V − log 6 (0.010)3 = 1.52 V − 0.0592 V = 1.46 V One point is earned for the correct answer. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 5 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 3 3. Answer the following questions about the thermodynamics of the reactions represented below. Reaction X: 1 1 → I ( s) + Cl (g) ← ICl(g) 22 22 ∆H f = 18 kJ mol –1, ∆S 298 = 78 J K–1 mol –1 Reaction Y: 1 1 → I 2 ( s) + Br (l) ← IBr(g) 2 22 ∆H f = 41 kJ mol –1, ∆S 298 = 124 J K–1 mol –1 (a) Is reaction X , represented above, spontaneous under standard conditions? Justify your answer with a calculation. ∆G° = ∆H° − T∆S° = (18 kJ mol−1) − ( 298 K)(0.078 kJ mol−1 K−1) = −5 kJ mol−1 Reaction is spontaneous because ∆G° < 0 . One point is earned for the correct value of ∆G°. One point is earned for a correct justification of spontaneity. (b) Calculate the value of the equilibrium constant, Keq , for reaction X at 25°C. ∆G° = −R T ln Keq ⇒ ln Keq = − ln Keq = − DG RT ( -5 kJ mol -1 )(103 J kJ -1 ) = 2.019 (8.31 J mol -1 K -1 )(298 K) One point is earned for the correct answer. Keq = e 2.019 = ( 7.5314) = 8 (c) What effect will an increase in temperature have on the equilibrium constant for reaction X ? Explain your answer. DH DS + RT R Since ∆H ° is positive, an increase in T will cause − ∆H °/RT to become a smaller negative number, therefore Keq will increase. OR ∆G° = −R T ln Keq = ∆H ° − T∆S° ⇒ ln Keq = - The reaction is endothermic (∆H = +18 kJ mol −1); an increase in temperature shifts the reaction to favor more products relative to the reactants, resulting in an increase in the value of Keq . One point is earned for the correct choice with a correct explanation. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 6 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 3 (continued) (d) Explain why the standard entropy change is greater for reaction Y than for reaction X . Both reaction X and reaction Y have solid iodine as a reactant, but the second reactant in reaction X is chlorine gas whereas the second reactant in reaction Y is liquid bromine. Liquids have lower entropies than gases, thus in reaction Y the reactants are more ordered (and have lower entropies) than in reaction X . The products of both reaction X and reaction Y have about the same disorder, so the change in entropy from reactants to products is greater in reaction Y than in reaction X. One point is earned for a correct explanation. (e) Above what temperature will the value of the equilibrium constant for reaction Y be greater than 1.0 ? Justify your answer with calculations. ∆G° = ∆H° − T∆S° Keq = 1 when ∆G° = 0 T= ∆H = ∆S ⇒ ⇒ T∆S° = ∆H° 41 kJ mol−1 0.124 kJ mol−1K −1 = 330 K One point is earned for ∆G° = 0. One point is earned for the correct temperature. So when T > 330 K, ∆G° < 0 kJ mol−1 ⇒ Keq > 1.0 (f) For the vaporization of solid iodine, I2(s) → I2(g), the value of DH 298 is 62 kJ mol –1. Using this information, calculate the value of DH 298 for the reaction represented below. → I2(g) + Cl2(g) ← 2 ICl(g) → I2(s) + Cl2(g) ← 2 ICl(g) → I ( g) ← I ( s ) DH 298 = 2 × 18 kJ mol−1 → I2(g) + Cl2(g) ← 2 ICl(g) DH 298 = − 26 kJ mol−1 DH 298 = − 62 kJ mol−1 ________________________________________________ 2 2 One point is earned for DH 298 of either the first or second equation. One point is earned for the correct sum of the DH 298 values. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 7 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 4 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 can be earned for each reaction: one point for the correct reactant(s) and two points for the correct product(s). (a) Solid calcium carbonate is strongly heated. CaCO3 → CaO + CO2 (b) A strip of magnesium metal is placed in a solution of iron(II) chloride. Mg + Fe2+ → Mg2+ + Fe (c) Boron trifluoride gas is mixed with ammonia gas. BF3 + NH3 → F3 BNH 3 (d) Excess concentrated hydrochloric acid is added to a solution of nickel(II) nitrate. Ni2+ + Cl− → [Ni Cl 4]2− (e) Solid ammonium chloride is added to a solution of potassium hydroxide. NH4Cl + OH− → NH3 + H2O + Cl− (f) Propanal is burned in air. CH3CH2CHO (or C3H6O) + O2 → CO2 + H2O (g) A strip of aluminum foil is placed in liquid bromine. Al + Br2 → AlBr3 (or Al2Br6) (h) Solid copper(II) sulfide is strongly heated in air. CuS + O2 → CuO + SO2 ( Cu and SO3 are also acceptable as products. ) © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 8 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 5 5. A student carries out an experiment to determine the equilibrium constant for a reaction by colorimetric (spectrophotometric) analysis. The production of the red-colored species FeSCN2+(aq) is monitored. (a) The optimum wavelength for the measurement of [FeSCN2+] must first be determined. The plot of absorbance, A , versus wavelength, λ , for FeSCN2+(aq) is given below. What is the optimum wavelength for this experiment? Justify your answer. The optimum wavelength is 450 nm because that is the wavelength of maximum absorbance by FeSCN2+(aq) . One point is earned for the correct answer with justification. (b) A calibration plot for the concentration of FeSCN2+(aq) is prepared at the optimum wavelength. The data below give the absorbances measured for a set of solutions of known concentration of FeSCN2+(aq). Concentration ( mol L− 1 ) 1.1 × 10 – 4 3.0 × 10 – 4 8.0 × 10 – 4 12 × 10 – 4 18 × 10 – 4 Absorbance 0.030 0.065 0.160 0.239 0.340 © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 9 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 5 (continued) (i) Draw a Beer’s law calibration plot of all the data on the grid below. Indicate the scale on the horizontal axis by labeling it with appropriate values. One point is earned for a straight-line plot. One point is earned for a correctly scaled horizontal axis. (ii) An FeSCN2+(aq) solution of unknown concentration has an absorbance of 0.300. Use the plot you drew in part (i) to determine the concentration, in moles per liter, of this solution. See plot in part (i). At A = 0.300 , [FeSCN2+] is approximately 16 × 10− 4 mol L−1. One point is earned for the correct answer. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 10 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 5 (continued) (c) The purpose of the experiment is to determine the equilibrium constant for the reaction represented below. → Fe3+(aq) + SCN–(aq) ← FeSCN2+(aq) (i) Write the equilibrium-constant expression for Kc . Kc = [FeSCN2 + ] [Fe3 + ][SCN - ] One point is earned for the correct expression. (ii) The student combines solutions of Fe(NO3)3 and KSCN to produce a solution in which the initial concentrations of Fe3+(aq) and SCN–(aq) are both 6.0 × 10 –3 M . The absorbance of this solution is measured, and the equilibrium FeSCN2+(aq) concentration is found to be 1.0 × 10 –3 M. Determine the value of Kc . Fe3+(aq) I C E 6.0 × 10−3 M −1.0 × 10−3 M 5.0 × 10−3 M + SCN−(aq) 6.0 × 10−3 M −1.0 × 10−3 M 5.0 × 10−3 M → ← FeSCN2+(aq) 0 +1.0 × 10−3 M +1.0 × 10−3 M One point is earned for the correct substitutions and the calculated value. -3 Kc = (5.0 1.0 10 10 -3 )(5.0 10 -3 ) One point is earned for the correct equilibrium concentration. = 40. (d) If the student’s equilibrium FeSCN2+(aq) solution of unknown concentration fades to a lighter color before the student measures its absorbance, will the calculated value of Kc be too high, too low, or unaffected? Justify your answer. The value of Kc will be too low; the lower absorbance reading indicates a lower [FeSCN2+] than actually existed before the fading occurred, so substitution of a lower [FeSCN2+] into the equilibrium expression will result in a lower value of Kc. One point is earned for the correct prediction. One point is earned for the correct justification. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 11 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 6 GeCl4 ICl4– SeCl4 ICl4+ 6. The species represented above all have the same number of chlorine atoms attached to the central atom. (a) Draw the Lewis structure (electron-dot diagram) of each of the four species. Show all valence electrons in your structures. One point is earned for each correct structure. (b) On the basis of the Lewis structures drawn in part (a), answer the following questions about the particular species indicated. (i) What is the Cl – Ge – Cl bond angle in GeCl4 ? 109.5° One point is earned for the correct angle. (ii) Is SeCl4 polar? Explain. Yes. The SeCl4 molecule is polar because the lone pair of nonbonding electrons in the valence shell of the selenium atom interacts with the bonding pairs of electrons, causing a spatial asymmetry of the dipole moments of the polar Se-Cl bonds. The result is a SeCl4 molecule with a net dipole moment. One point is earned for the correct answer. (iii) What is the hybridization of the I atom in ICl4– ? d2sp3 or sp3d2 One point is earned for the correct hybridization. (iv) What is the geometric shape formed by the atoms in ICl4+ ? See-saw (or distorted tetrahedral or disphenoidal) One point is earned for the correct shape. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 12 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 7 7. Account for each of the following observations in terms of atomic theory and/or quantum theory. (a) Atomic size decreases from Na to Cl in the periodic table. Across the periodic table from Na to Cl, the number of electrons in the s- and p- orbitals of the valence shell increases, as does the number of protons in the nucleus. The added electrons only partially shield the added protons, resulting in an increased effective nuclear charge. This results in a greater attraction for the electrons, drawing them closer to the nucleus, making the atom smaller. One point is earned for indicating the increase in nuclear charge. One point is earned for attributing the size decrease to the greater attraction of the nucleus for the electrons caused by the increase in nuclear charge. (b) Boron commonly forms molecules of the type BX3 . These molecules have a trigonal planar structure. Boron has three valence electrons, each of which can form a single covalent bond with X. The three single covalent bonds of the boron atom orient to minimize electron-pair interaction, resulting in bond angles of 120° and a trigonal planar structure. One point is earned for describing the valence electrons and the bonds. One point is earned for a correct VSEPR argument. (c) The first ionization energy of K is less than that of Na. 1 Both Na and K have an s valence-shell electron 1 ; K: [Ar] 4s1). The K atom configuration (Na: [Ne] 3s valence electron has a higher n quantum number, placing it farther from the nucleus than the Na atom valence electron. The greater distance results in less attraction to the nucleus. Because its valence electron is less attracted to its nucleus, the K atom has the lower ionization energy. One point is earned for the size explanation. One point is earned for describing the attraction to the nucleus. (d) Each element displays a unique gas-phase emission spectrum. Each element has a unique set of quantized energy states for its electrons (because of its unique nuclear charge and unique electron configuration). As the electrons of an element absorb quanta of energy, they change to higher energy states (are excited) – during de-excitation, energy is released as EM radiation as the electrons cascade to lower energy states. Each photon of the EM radiation is associated with a specific wavelength ( λ = hc/E ), a flux of which produces the lines of the emission spectrum. One point is earned for describing the quantized energy states and emission phenomenon. One point is earned for describing the effect of the uniqueness of the nucleus and/or electron configuration. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 13 AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 8 8. Use chemical and physical principles to account for each of the following. (a) An aluminum container filled with an aqueous solution of CuSO4 eventually developed a leak. Include a chemical equation with your answer. Al(s) + Cu2+(aq) → Al3+(aq) + Cu(s) Cu2+ has a higher reduction potential than does Al3+, which results in the oxidation and eventual disappearance of the Al metal (depending upon the amount of Cu2+ ). One point is earned for the correct equation (phases not required). One point is earned for the explanation of relative reactivity. (b) The inside of a metal container was cleaned with steam and immediately sealed. Later, the container imploded. The high temperature of the steam causes the air/water mixture in the container to be at an elevated temperature. When the container is sealed and the temperature decreases, the pressure of the residual gases decreases below the external pressure, causing the implosion. The decrease in pressure occurs because pressure is proportional to temperature and/or vapor pressure of water decrease with temperature, which means that condensation occurs upon cooling with a resultant pressure drop. One point is earned for explaining the implosion in terms of internal pressure decrease. One point is earned for the explanation of the change of pressure (either cause is accepted). (c) Skin feels cooler after rubbing alcohol has been applied to it. Rubbing alcohol evaporates rapidly. Evaporation is endothermic so heat energy is absorbed from the skin in the process, which causes the cooling sensation. One point is earned for reference to the volatility of the alcohol. One point is earned for discussing the endothermic nature of the process. (d) The redness and itching of the skin caused by ant bites (injections of methanoic acid, HCO2H) can be relieved by applying a paste made from water and baking soda (solid sodium hydrogen carbonate). Include a chemical equation with your answer. HCO2H + NaHCO3 → NaHCO2 + H2O + CO2 One point is earned for the equation. − Methanoic acid is neutralized by the HCO3 ion; with the neutralization of the acid; the redness and itching of the ant bites subside. One point is earned for the explanation. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for 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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