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Studen t P erform an ce Q & A : 2008 A P® C hem istry Free-R esponse Q uestions The following comments on the 2008 free-response questions for AP ® Chemistry were compiled by the Chief Reader, Eleanor Siebert of Mount St. Ma ry’s College in Los Angeles, California. They give an overview of each free-response question and of how students performed on the question, including typical student errors. General comment s regarding the skills and content that students frequently have the most problems with are incl uded. Some suggestions for improving student performance in these areas are also provided. Teach ers are encouraged to attend a College Board workshop to learn strategies for improvi ng student performance in specific areas. Question 1 W hat was the intent of this question? This question was designed to probe student unders tanding of gases and gaseous equilibria. Part (a) required students to write the expression for Kp. In part (b) students were asked to determine the number of moles of CO 2(g) given the volume, pressure, and temperature. This determination required the use of the ideal gas law. In part (c)(i) students had to select the correct data from the table and use Dalton’s law of partial pressures to determine the pressure of CO( g) at equilibrium. In part (c)(ii) students were asked to determine the value of Kp using equilibrium pressures. In part (d) they had to explain the effect of a catalyst on the total pressure of gases at equilibrium. In part (e) students were given a new set of initial conditions and asked to determine the direction th e reaction would proceed to reach equilibrium. How well did students perform on this question? Generally, students performed in the middle range on this question. The mean score was 4.17 out of 9 possible points. Students often earned points in parts (b ), (c)(ii), and (d). The most frequently earned point was in part (d), while the most frequently missed point s were for part (e). Students often exhibited a better conceptual knowledge of gases than ma thematical skills when using data. W hat were com m on student errors or om issions? Common student errors in part (a) included failing to express Kp in terms of pressure (typically substituting concentration) or copying the expression Kp = Kc(RT )!n from the “Advanced Placement Chemistry Equations and Constants” sheet in the exam booklet and attempting to use it. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.
In parts (b) and (c) students often failed to recognize the significance of initial versus equilibrium pressures for their calculations. Also, students often ma de algebraic errors. In part (c)(i) students were frequently confused about when to use the stoichio metric relationship between the gases given in the reaction. In part (e) students often did not use a calculation in the justification even though they were asked to do so; they also frequently used stoichiometric r easoning and/or Le Chatelier’s principle instead of comparing the reaction quotient, Q, to Kp . Based on your experience of student responses at the AP Reading, what m essage would you like to send to teachers that m ight help them to im prove the perform ance of their students on the exam ? " Teachers should ask students to write equilibri um expressions in their classroom assessments. They should also ask students to distinguish between a Kc expression and a Kp expression. The use of simulated lab data as a follow-up to labs may help students better distinguish between initial values of measurements and equilibrium values. " Teachers should work with students to help them develop a qualitative and quantitative ability to justify predictions. " Finally, teachers should stress to their students that they should read all of the parts of a question thoroughly and answer the question as it was written. Question 2 W hat was the intent of this question? This question assessed student knowledge and skills relati ng to gravimetric analysis, which is included in several of the laboratory experiments recommended in the AP Chemistry Course Description . In parts (a) through (c) students were asked to analyze and interp ret a data table. They had to explain how they correctly determined that all the water of hydratio n had been driven off from a sample of a hydrate; calculate an appropriate formula for the hydrate; a nd determine the effect of an error in laboratory procedure on the calculation of the mass of water released upon heating. Parts (d) and (e) required students to describe a quantitative laboratory procedur e to determine the mass of a precipitate from a mixture and then calculate the number of moles and percent by mass of a component of the mixture. How well did students perform on this question? Overall, students did reasonably well when answ ering this question. The mean score was 4.15 out of 10 possible points, and the distribution of scores was relatively even. Students were generally successful on parts (a) and (b)(i ), and responses that earned 1 to 2 points usually garnered them here. Part (b)(ii) proved a bit more challenging, as students frequently used an incorrect mass to calculate the number of moles of wa ter of hydration and determine the formula. In part (c) students usually recognized that the la boratory procedure error described would result in a calculation of too large a mass of water; however, many had difficulty pr oviding an appropriate justification and consequently failed to earn the point. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 2
In part (d) most students described a quantitative laboratory procedure for the isolation of a precipitate, but many did not elaborate sufficiently to earn full credit. While most students recognized filtration as the method of choice, they often did not include the necessary step of drying the precipitate prior to weighing, nor did they explain the need to determine the m ass of the precipitate by difference. A significant proportion of students misinterpreted the intent of th e question; these students attempted a mathematical explanation of the steps necessary to calculate the mass of AgCl from the data. In part (e)(i) students were generally successful with the calculation of the moles of MgCl 2; however, many used an incorrect value for the total mass of the sample in part (e)(ii) and so did not earn the final point. W hat were com m on student errors or om issions? (a) Students most frequently answered this part correctly. Common errors were: " A description or restatement of the data provided, without explanation " An unclear explanation of the term “constant mass” " A description of the data as having high accuracy and/or precision (b)(i) Students frequently answered this pa rt correctly as well. Common errors were: " Calculating an incorrect mass of water from th e data (students often reversed the water and MgCl 2) " Subtraction errors " Failing to round the final answer appropriate ly and reporting an answer with the wrong number of significant figures (b)(ii) Two points were available for this question, and students often earned one point for applying a mole ratio to end up with a hydrated formula. Common errors were: " Using an incorrect mass to determine the moles of MgCl 2 " Failing to recognize that a mole ratio was re quired and applying the number of moles of water to the formula " Misunderstanding the meaning of the term “f ormula” (some students tried to provide a balanced equation for a reaction between H2O and MgCl 2 instead) (c) Many students were able to correctly identify the effect of the error but unable to supply an appropriate justificati on. Common errors were: " Confusion between the terms “hydrate” and “water” " Not understanding that the hydrate was a solid and that water would not also “splash out” (d) This was the most difficult part of th e question for students. Common errors were: " Omitting at least one of the required three steps (filtering, drying, or we ighing the precipitate by difference) " Stating only the need to weigh or mass the precipitate " Not describing the steps in sufficient detail " Referencing the hydrate procedure from the first part of the question © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 3
(e)(i) Many students earned at least one poi nt in this part. Common errors were: " Miscalculation of the molar mass of AgCl " Failure to recognize the 1: 2 stoichiometric ratio between MgCl 2 and AgCl " Not answering the question and calculating the mass of MgCl 2 " Using an incorrect mass of the sample from another part of the question " Failing to round the final answer appropriately and reporting an answer with an incorrect number of significant figures (e)(ii) Many students answered this question correctly. Common errors were: " Using an incorrect value for the total mass of the sample " Trying to calculate a percent error rather than the percent by mass " Calculating the mass of MgCl 2 but not using it to calculate the percent by mass " Failing to round the final answer appropriately and reporting an answer with an incorrect number of significant figures Based on your experience of student responses at the AP Reading, what m essage would you like to send to teachers that m ight help them to im prove the perform ance of their students on the exam ? " Have students perform the recommended laboratory experiments. " Provide opportunities for students to engage in guided-inquiry experiments. Students need to practice the analysis of data collected in tabular form and also be able to recognize when data are valid. " Give students laboratory experiences in which techni que is important, as was the case in part (d). Students rarely mentioned that the precipitate s hould be washed and rinsed to remove the other soluble salts. " Remind students that significant figures are important in calculations, and review the rules for addition and subtraction. " Deemphasize the use of algorithms for numerical ca lculations. Students are often able to solve gravimetric stoichiometry problems but are unawa re of the intermediate values calculated. Question 3 W hat was the intent of this question? This question tested a diverse set of student skills. Parts (a) and (b) were intended to assess the ability of students to understand the relationship among standard reduction potentials of ha lf-reactions and the cell potential, and the relationship between the cell potential and the change in Gibbs free energy of the reaction. Part (c) assessed students’ ability to relate the change in entropy of the reaction to the phases of reactants and products given in the balanced equation. Parts (d), (e), and (f) required students to answer questions related to the kinetics of a different reactio n; calculate reaction orders from experimental data; write a rate law that was consistent with the orders ; and determine a rate constant. Those parts of the question were intended to assess the students’ unders tanding of kinetics and the meaning of reaction orders, and their ability to write and interpret a rate law. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 4
How well did students perform on this question? Because this question assessed a broad range of skills, the range of student performance was also quite broad. The mean score was 3.65 out of 9 possible poi nts. Surprisingly, the modal score was 0, with 18 percent of the responses failing to earn any points and another 6 percent of papers left completely blank. Those students who were able to earn points on the qu estion performed fairly well, and the distribution of scores from 1 through 9 was relatively even. Many stude nts earned all of the points available on parts (a) through (c) but earned no points on parts (d) through (f), or vice versa; this suggests that one or more of the topics had likely been omitted or covered superficially in the students’ courses. W hat were com m on student errors or om issions? In part (a) the most common answer was obtained by simply subtracting the values of E# given, obtaining +0.28 V rather than the correct +0.96 V. Because th e unit (V) was provided in the values, responses that omitted the unit still earned the point. In part (b) common errors included: " Identifying an incorrect number of electrons tr ansferred in the reaction, with common errors of n = 5 (sum of electrons in each half-reaction) and n = 22 (sum of coefficients in balanced equation) " Using E# for one of the half-reactions (answer carried down from part [a] rather than the cell potential given) " Giving incorrect or inconsistent units or omitting units altogether " Omitting the algebraic sign of !G# In part (c) many students attempted to determine the +/ $ sign of !S# from the relationship !G# = !H# $ T!S# and then use the value of !G# obtained in part (b). A small fraction of responses determined the incorrect sign of !S# by considering the total number of moles of products versus reactants and failing to look carefully at the phases. In part (d) many students performed well, but common errors included: " Incorrect algebra (e.g., 3n = 9 % n = 3 ) " Failure to provide justification In part (e) common errors included: " Omission of rate constant " Exponents that were inconsistent with values obtained in part (d) " Writing an equilibrium expression rather than a rate law In part (f) those students who wrote a well-formed rate law in part (e) also did well here. The most common errors included: " Failing to include units or including incorrect units (e.g., M s$1 ) " Substituting inconsistent value from the experimental data " Poor algebra and an inability to use scientific notation " Confusion between the rate of reaction and the rate constant, k © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 5
Based on your experience of student responses at the AP Reading, what m essage would you like to send to teachers that m ight help them to im prove the perform ance of their students on the exam ? " Student performance on tasks like part (a) might be improved by encouraging students to write an equation that relates the cell potential to those of the half-reactions. " Emphasize “ballpark” values and the physical mean ing of numerical quantities (e.g., students should recognize that $360,000 kJ mol $1 is an absurdly large value for the !G# of a chemical reaction). " Give students practice writing and interpreting rate laws, with particular emphasis on units of reaction rates, rate constants, rat es of formation, and concentrations. " Emphasize the precise use of symbols and notation (e.g., mol versus m versus M , k versus K, parentheses versus brackets, and the appropriate use of superscripts and subscripts). Question 4 W hat was the intent of this question? This question was intended to assess students’ ability to write both molecular and net-ionic equations and to recognize when each is appropriate. Various aspects of the question were intended to reinforce knowledge gleaned from the classroom and from experience in the laboratory. How well did students perform on this question? Students displayed a wide range of knowledge and sk ills in their responses to this question. The mean score was 6.81 out of 15 possible points. Scores covered the range from 0 to 15, with close to a perfect bell-shaped distribution curve. The most common scores were 4, 5, and 6, and there were relatively few blank papers. W hat were com m on student errors or om issions? Common student errors included: " Showing insoluble substances, or substances stat ed to be solid or gaseous, in ionized form " Showing gaseous forms as if in aqueous solution " Including spectator ions in the same form on both sides of the equation " Not balancing equations so that coefficien ts are in terms of lowest whole numbers " Not canceling reagents that appear on both sides of the equation " Confusing the terms “colorless” and “clear” (CuSO 4 , for example, forms a clear but colored solution) " Mistaking common formulas (e.g., writing “HCl 2” rather than “HCl,” or “2Cl” rather than “Cl 2”) " Using inexact language (e.g., “ phenolphthalein will show its basic color” rather than “phenolphthalein turns pink in basic solution”) " Reading the prompt inexactly (e.g., not writing the formula of a complex ion when directed to do so) " Balancing by stoichiometry of atoms but not by charge " Omitting an explanation or justification following an assertion in parts (a)(ii), (b)(ii), and (c)(ii) " Writing more than one answer in the provided answer box © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 6
" Placing charges within formulas (e.g., “H 1+Cl 1$ ” rather than “HCl”) " Adding inappropriate products just to “balance” the equation " Adding and omitting randomly the charges on species in the written reaction Based on your experience of student responses at the AP Reading, what m essage would you like to send to teachers that m ight help them to im prove the perform ance of their students on the exam ? " Give students practice with writing net-ionic equations. Emphasize fundamental knowledge and skills regarding equation writing, including elemental forms and common ions, especially polyatomic anions. Student responses often showed errors in the systematic presentation of chemical species, including soluble versus insol uble salts, strong versus weak acids or bases, and gaseous versus aqueous states. Some students al so had difficulty with the art of balancing equations, neglecting to conser ve either mass or charge. " Encourage students to write balanced net-ionic equa tions to describe their work in the laboratory. A student who has performed an acid-base titrati on with phenolphthalein as an indicator is unlikely to forget the characteristic pink color of the indicator as the soluti on becomes basic at the titration’s endpoint. Similarly, a student who has created a precipitate of Al(OH) 3 by adding strong base drop-wise to a solution containing aluminum(III) cation is apt to remember the excitement of redissolving that precipitate by c ontinuing to add drops of strong base to form a complex ion. A student who has oxidized HCl( g) with O2(g) in the lab is not following safe lab practices, but that student should have been taught that HCl is not ionized in the gaseous state. " Remind students to refer to the resources availa ble to them during the exam, specifically the “Periodic Table of the Elements” and “Standard Reduction Potentials in Aqueous Solution at 25°C” pages in the exam booklet, for memory prompts about the behavior of elements and common ions. " Provide students with opportunities to practice writing about chemistry. Precise language is important. For example, many responses for part (a )(ii) were unclear as to whether the added acid reacted with the complex or the hydroxide ions from the ionized NaOH . Many descriptions of LeChâtelier shifts or limiting reactants were vague. A good observation in the lab is not “it changed color” but specifically what the color chan ge was. It is not sufficient to say “the acid reacted with the ions.” What ions? What reacti on occurred? Instead, specify why the added acid affected the concentration of the complex ion. Pronouns should have unambiguous antecedents; sentences that might have made sense if written as “the acid reacts with the base and tends to neutralize the solution” were too often seen as a statement such as “it reacts with it and tends to neutralize it.” Question 5 W hat was the intent of this question? This question was designed to assess student understandi ng of the structure and properties of atoms and molecules. In parts (a) through (c) students had to demonstrate their understanding of ionization energy and provide explanations for its va riance among different atoms. In parts (d) through (f) students were © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 7
required to sketch Lewis electron-dot diagrams, iden tify molecular shape and hybridization, and predict molecular polarity. How well did students perform on this question? The mean score was 3.92 out of 9 possible points, with scores of 4 and 5 both being modal. This question had a wide bell-shaped distribution of scores. W hat were com m on student errors or om issions? In part (a) the correct chemical equation was seldom obtained. Some responses attempted to develop a mathematical equation involving the given value of the first ionization energy. In parts (b) and (c) the explanations that were to be made on the basis of nuclear charge and atomic size instead often involved discussions of periodic tre nds, electron configuration, or electronegativity. In part (d) most students drew correct Lewis elect ron-dot diagrams. Performance varied when students attempted to use these diagrams to reason out the sh ape, the central atom hybridization, and then molecular polarity. Based on your experience of student responses at the AP Reading, what m essage would you like to send to teachers that m ight help them to im prove the perform ance of their students on the exam ? " Students need to be able to discuss the factors de termining the trends of atomic properties. They must also understand the difference between peri odic trends and the explanation for what is responsible for such trends. " Some students need help in distinguishing between the first-ionization energy and electron affinity. " Students need to practice the reasoning sequence employed in understanding molecular structure: complete a Lewis electron-dot diagram, use this to determine the electron-pair orientation and thus the molecular shape, and then use the shape to find the central atom hybridization and molecular polarity. The role of symmetry in determining polarity needs to be stressed. Question 6 W hat was the intent of this question? This question explored the importance of intermolecular interactions in phase changes and dissolution. To earn full credit, a student had to identify the relevant forces involved in each process. In part (a) students had to explain that pyridine’s ability to hydrogen bond with water distinguishes its aqueous solubility from that of benzene. In part (b) students had to recognize that while ethanol and dimethyl ether (structural isomers) have similar dispersion forces, the hydrogen bonding between ethanol molecules leads to a higher boiling point. Part (c) required st udents to contrast the melting points of a network covalent solid (in which strong cova lent bonds are broken in the melting transition) and a molecular solid (in which only relatively weak intermolecular attractio ns must be overcome). In part (d) students had to recognize that the London/dispersion interactions between Cl 2 molecules must be greater than the total intermolecular forces between HCl molecules, and then attribute the di fference to the larger number of electrons in the Cl 2 molecules. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 8
How well did students perform on this question? Students did poorly on this question. The mean score was only 1.86 out of 8 possible points. Both the median and modal scores were 1. Answers reveal ed a widespread misunderstanding concerning the differences between the interactions between molecu les and the bonds that hold atoms together. W hat were com m on student errors or om issions? In all parts of the question, many students showed uncertainty about the distinctions between molecules and atoms, and between intermolecular forces and covalent (intramolecular) bonds. Many responses contained the ambiguous phrase “the forces holding the molecules together” or similar constructions, and it was difficult for Readers to determine whether the st udents intended to refer to inter- or intramolecular forces. Part (a): Many students said that pyridine molecules “d issociated” or “came apart” when they dissolved in water, while the benzene bonds were so strong that th ey could not come apart. Students often attributed the solubility of pyridine to the solubility of amm onium or nitrate compounds. Appropriate discussions of the nature of the interaction between pyridine and water were rare; the adag e that “like dissolves like” was used by a vast majority of the students, but it did not by itself earn credit because it was not an explanation or a discussion of the interaction between either of the solutes and water. Part (b): The fundamental error made in a pluralit y (if not a majority) of the responses was that the covalent bonds within dimethyl ether and ethanol must be broken for the material to boil. A very common answer indicated that the C–C bond in ethanol was stronger than the C–O bonds in dimethyl ether, so that more energy was needed to break apart ethanol. A variation on this theme was to say that ethanol’s oxygen was easier (or harder) to remove than the (protected, less-conspicuous, or less-exposed) oxygen in the center of dimethyl ether. Students frequently identified the (covalent) O–H bond in ethanol as a hydrogen bond, and they cited the ease of breaking the O–H bond in ethanol as the reason for the difference in boiling points (indeed, students often identified any covalent bond to hydrogen as a hydrogen bond). Students often referred to ethanol’s hydrogen bonding as “the strongest bond,” stronger than any of dimethyl ether’s covalent bonds. Dimethyl ether was almost uniformly (and in correctly) classified as nonpolar, and many students attempted incorrect explanations of boiling point differences based on th e apparent linearity of dimethyl ether or the length of the hydrocarbon chain in ethanol. Part (c): A very common error based the differen ce in melting points on the covalent bond orders in Lewis electron-dot diagrams for SO 2 and SiO 2 molecules. Comparison of covalent and ionic bond strengths was another common approa ch. Differences in electronegativities among the three elements (S, Si, O) were frequently cited, as were their relativ e positions on the periodic table. The properties of elemental S, Si, and O were also repeatedly invoked. Many responses displayed the misunderstanding that “n etwork covalent bonds are stronger than regular covalent bonds,” rather than compari ng the network covalent bond strength in SiO 2 to the strength of the intermolecular forces between SO 2 molecules. Students often classified SiO 2 as an ionic compound, and many responses referred to Si as a transition metal. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 9
Part (d): It was apparent that many students ar e fundamentally confused about the difference between intramolecular Cl–Cl and H–Cl bonds and the intermolecular interactions between Cl 2 and HCl molecules. Responses often attributed double bonds to Cl 2 (and HCl), and comparisons between the triple (or quadruple) bond in Cl 2 and the double bond in HCl were common. The Cl 2 molecule was often said to be polar, while HCl was nonpolar. Students often attributed some property (or properties) to Cl 2 because “it is diatomic.” Students ofte n cited periodic trends or positions of H and Cl on the periodic table. Students interchangeably identified Cl 2 and HCl as ionic, polar, or nonpolar and as having ionic bonding, covalent bonding, hydrogen bonding, dipole-dipole, ion-dipole, and dispersion forces. Any selection of these forces, in any combination or order, could be found as students tried to justify the difference in boiling points. Responses frequently tried to invoke the dissociation of HCl as a strong acid to explain its low boiling point. Based on your experience of student responses at the AP Reading, what m essage would you like to send to teachers that m ight help them to im prove the perform ance of their students on the exam ? Students have clearly been exposed to the material co vered in this question. Wh ile they used the correct vocabulary, they generally used it incorrectly. The correct phrase “network covalent bonds” appeared in answers to part (c), for example, but the subsequent prose showed that students did not know what it meant. The bonds between atoms in molecules must be distinguished from the interactions that keep the molecules attracted to each other. The forces within a molecule are different from the forces between them. Phrases like “the intermolecular forces within the molecule” illuminate a major misunderstanding that must be addressed. The phrase “the forces holding molecules together” (and similar constructions) is ambiguous and should be avoided in favor of clear language, such as “forces between molecules” and “forces within molecules.” © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 10
A P ® C hem istry 2008 Scorin g G uidelin es The College Board: Connecting Students to College Success The College Board is a not-for-profit memb ership association whose mission is to connect students to college success and opportunity. Founded in 1900, the associat ion is composed of more than 5,400 sch ools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, a nd 3,500 colleges through major programs and se rvices 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. © 2008 The College Board. All rights reserv ed. College Board, AP Central, Advanced Placement Program, AP, SAT, and the acorn logo are re gistered trademarks of 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 tr ademarks of their respective owners. Permission to use copyrighted College Bo ard 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 online home for AP teachers: apcentral.collegeboard.com.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 1 C( s) + CO 2(g) →← 2 CO( g) Solid carbon and carbon dioxide gas at 1,160 K were pl aced in a rigid 2.00 L cont ainer, and the reaction represented above occurred. As the reaction proceeded, th e total pressure in the container was monitored. When equilibrium was reached, there was still some C( s) remaining in the container. Results are recorded in the table below. Time (hours) Total Pressure of Gases in Container at 1,160 K (atm) 0.0 5.00 2.0 6.26 4.0 7.09 6.0 7.75 8.0 8.37 10.0 8.37 (a) Write the expression for the equilibrium constant, Kp, for the reaction. Kp = 2 2 CO CO ()P P One point is earned for the correct expression. (b) Calculate the number of moles of CO 2(g) initially placed in the container. (Assume that the volume of the solid carbon is negligible.) (5.00 atm)(2.00 L) 0.105 mol L atm (0.0821 )(1,160 K) mol K PV n RT !! ! One point is earned for the correct setup. One point is earned for the correct answer. (c) For the reaction mixture at equilibrium at 1,160 K , the partial pressure of the CO 2(g) is 1.63 atm. Calculate (i) the partial pressure of CO( g) , and 2 CO CO total P P P += 2 CO CO total P P P =− = 8.37 atm − 1.63 atm = 6.74 atm One point is earned for the correct answer supported by a correct method .
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 1 (continued) (ii) the value of the equilibrium constant, Kp. Kp = 2 2 CO CO ()P P = 2 (6.74 atm) 1.63 atm = 27.9 One point is earned for a correct setup that is consistent with part (a). One point is earned for the correct answer according to the setup. (d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to th e final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of th e solid catalyst is negligible.) The total pressure of the gases at equilibrium with a catalyst present would be equal to the total pressure of the gases without a catalyst. Although a catalyst w ould cause the system to reach the same equilibrium state more quickly, it would not affect the extent of the reaction, which is determined by the value of the equilibrium constant, Kp. One point is earned for the correct answer with justification . In another experiment involving the same reac tion, a rigid 2.00 L container initially contains 10.0 g of C( s) , plus CO( g) and CO 2(g) , each at a partial pressure of 2.00 atm at 1,160 K. (e) Predict whether the partial pressure of CO 2(g) will increase, decrease, or remain the same as this system approaches equilibrium. Justify your prediction with a calculation. Q = 2 2 CO CO ()P P = 2 (2.00 atm) 2.00 atm = 2.00 < Kp ( = 27.9) , therefore 2 COP will decrease as the system approaches equilibrium. One point is earned for a correct calculation involving Q or ICE calculation. One point is earned for a correct conclusion based on the calculation.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 2 Answer the following questions relating to gravimetric analysis. In the first of two experiments, a student is assigned th e task of determining the number of moles of water in one mole of MgCl 2⋅nH2O . The student collects the data shown in the following table. Mass of empty container 22.347 g Initial mass of sample and container 25.825 g Mass of sample and container after first heating 23.982 g Mass of sample and container after second heating 23.976 g Mass of sample and container after third heating 23.977 g (a) Explain why the student can correctly conclude th at the hydrate was heated a sufficient number of times in the experiment. No additional mass was lost during the third heating, indicating that all the water of hydration had been driven off. One point is earned for the correct explanation. (b) Use the data above to (i) calculate the total number of moles of water lost when the sample was heated, and mass of H2O lost = 25.825 − 23.977 = 1.848 g OR 25.825 − 23.976 = 1.849 g 1.848 g H 2O × 2 2 1 mol H O 18.02 g H O = 0.1026 mol H 2O One point is earned for calculating the correct number of moles of water. (ii) determine the form ula of the hydrated compound. mass of anhydrous MgCl 2 = 23.977 − 22.347 = 1.630 g 1.630 g MgCl 2 × 2 2 1 mol MgCl 95.20 MgCl g = 0.01712 mol MgCl 2 2 2 0.1026 mol H O 0.01712 mol MgCl = 5.993 ≈ 6 mol H 2O per mol MgCl 2 ! formula is MgCl 2!6H 2O One point is earned for calculating the correct number of moles of anhydrous MgCl 2 . One point is earned for writing the correct formula (with supporting calculations).
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 2 (continued) (c) A different student heats the hydrat e in an uncovered crucible, and some of the solid spatters out of the crucible. This spattering will have what effect on the calculated mass of the water lost by the hydrate? Justify your answer. The calculated mass (or moles) of water lost by the hydrate will be too large because the mass of the solid that was lost will be assumed to be water when it actually included some ! MgCl 2!as well. One point is earned for the correct answer w ith justification. In the second experiment, a student is given 2.94 g of a mixtu re containing anhydrous MgCl 2 and KNO 3 . To determine the percentage by mass of MgCl 2 in the mixture, the student uses excess AgNO 3(aq ) to precipitate the chloride ion as AgCl( s) . (d) Starting with the 2.94 g sample of the mixture dissolved in water, briefly describe the steps necessary to quantitatively determine the mass of the AgCl precipitate. Add excess AgNO 3 . - Separate the AgCl precipitate (by filtration). - Wash the precipitate and dry the precipitate completely. - Determine the mass of AgCl by difference. Two points are earned for all three major steps : filtering the mixture, drying the precipitate, and determining the mass by difference. One point is earned for any two steps. (e) The student determines the mass of the AgCl precipitate to be 5.48 g. On the basis of this information, calculate each of the following. (i) The number of moles of MgCl 2 in the original mixture 5.48 g AgCl × 1 mol AgCl 143.32 A gCl g = 0.0382 mol AgCl 0.0382 mol AgCl × 1 mol Cl 1 mol A gCl × 2 1 mol MgCl 2 mol Cl = 0.0191 mol MgCl 2 One point is earned for calculating the number of moles of AgCl. One point is earned for conversion to moles of MgCl 2. (ii) The percent by mass of MgCl 2 in the original mixture 0.0191 mol MgCl 2 × 2 2 95.20 MgCl g 1 mol MgCl = 1.82 g MgCl 2 2 MgCl 1.82 g 2.94 g sample × 100% = 61.9% MgCl 2 by mass One point is earned for calculating the correct percentage.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 3 Answer the following questions related to ch emical reactions invol ving nitrogen monoxide, NO( g) . The reaction between solid copper and nitric acid to form copper(II) ion, nitrogen monoxide gas, and water is represented by the following equation. 3 Cu( s) + 2 NO 3−(aq ) + 8 H +(aq ) → 3 Cu 2+(aq ) + 2 NO( g) + 4 H 2O( l) E° = + 0.62 V (a) Using the information above and in the table below, calculate the standard reduction potential, E°, for the reduction of NO 3− in acidic solution. Half-Reaction Standard Reduction Potential, E° Cu 2+(aq ) + 2 e− → Cu( s) + 0.34 V NO 3−(aq ) + 4 H +(aq ) + 3 e− → NO( g) + 2 H 2O( l) ? rxnE! = 3 NOE " ! − 2+CuE! = 3 NOE " ! − 0.34 V = 0.62 V ! 3 NOE " ! = 0.62 V + 0.34 V = 0.96 V One point is earned for the correct calculation of the standard reduction potential. (b) Calculate the value of the standard free energy change, ΔG°, for the overall reaction between solid copper and nitric acid. ΔG° = −nFE° = −(6)(96,500 C mol −1)(0.62 V) = − 360,000 J mol −1 = − 360 kJ mol −1 One point is earned for the correct value of n, the number of moles of electrons. One point is earned for calculating the correct value of ΔG°, with correct sign and consistent units. (c) Predict whether the value of the standard entropy change, ΔS°, for the overall reaction is greater than 0, less than 0, or equal to 0. Justify your prediction. ΔS° > 0. Even though there is a loss of 7 moles of ions in solution, the value of ΔS° for the overall reaction will be greater than zero because two moles of NO gas will be produced (there are no gaseous reactants). One point is earned for the correct answer with a justification that is based on the gaseous state of one of the products.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 3 (continued) Nitrogen monoxide gas, a product of the reaction a bove, can react with oxygen to produce nitrogen dioxide gas, as represented below. 2 NO( g) + O 2(g) → 2 NO 2(g) A rate study of the reaction yielded the data recorded in the table below. Experiment Initial Concentration of NO (mol L −1) Initial Concentration of O 2 (mol L −1) Initial Rate of Formation of NO 2 (mol L −1 s−1) 1 0.0200 0.0300 8.52 × 10 −2 2 0.0200 0.0900 2.56 × 10 −1 3 0.0600 0.0300 7.67 × 10 −1 (d) Determine the order of the reacti on with respect to each of the follo wing reactants. Give details of your reasoning, clearly explaini ng or showing how you arri ved at your answers. (i) NO Comparing experiments 1 and 3, the tripling of the initial concentration of NO while the initial concentration of oxygen remained constant at 0.0300 mol L −1 resulted in a nine-fold increase in the initial rate of formation of NO 2. Since 9 = 3 2, the reaction is second order with respect to NO. One point is earned for the correct answer with justification. (ii) O 2 Comparing experiments 1 and 2, the tripling of the initial concentration of O2 while the initial concentration of NO !remained constant at 0.0200 mol L −1 resulted in a tripling in the initial rate of formation of NO 2. Since 3 = 3 1, the reaction is first order with respect to O2. One point is earned for the correct answer with justification. (e) Write the expression for the rate law for the re action as determined from the experimental data. rate = k[NO] 2[O 2] One point is earned for the correct expression for the rate law.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 3 (continued) (f) Determine the value of the rate constant for the reaction, clearly indicating the units. Because the coefficient for NO 2 in the balanced equation is 2, the rate of the reaction is defined as 1 2 the rate of the appearance of NO 2 . From part (e) above, k = 2 2 reaction rate [NO] [O ] = () 2 2 2 1 rate of formation of NO 2 [ NO] [O ] "#$%&' Substituting data from experiment 1, k = () () 2 1 1 1 2 1 1 8.52 10 mol L s 2 (0.0200 mol L ) (0.0300 mol L ) − − − −− × = 3.55 × 10 3 L 2 mol −2s−1 One point is earned for calculating the correct value of the rate constant. One point is earned for including the correct units. Note: a rate constant value of 7.10 × 10 3 L 2 mol −2s−1 earns the point if the rate of reaction is assumed to be the same as the rate of formation of NO 2.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 4 (a) Aqueous sodium hydroxide is added to a saturated solution of aluminum hydroxide, forming a complex ion. (i) Balanced equation: Al(OH) 3 + OH − → [Al(OH) 4]− Al(OH) 3 + 3 OH − → [Al(OH) 6]3− Al 3+ + 4 OH − → [Al(OH) 4]− Al 3+ + 6 OH − → [Al(OH) 6]3− One point is earned for the correct reactants. Two points are earned for a correct product. One point is earned for balancing the equation. (ii) If the resulting mixture is acidified, would the concentration of the complex ion increase, decrease, or remain the same? Explain. The [Al(OH !4] − will decrease because … (If equilibrium exists), the H+ added would react with the OH − in solution, reducing the [OH −] and shifting the equilibrium toward the reactants, thus reducing the concentration of the complex ion. OR (If the reaction has gone to completion), the H+ added would react with the [Al(OH) 4]−, thus reducing the concentration. [Al(OH) 4]− + H + → Al(OH) 3 + H 2O One point is earned for a correct answer with an explanation.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 4 (continued) (b) Hydrogen chloride gas is oxidized by oxygen gas. (i) Balanced equation 4 HCl + O 2 → 2 H 2O + 2 Cl 2 Some other acceptable equations and products: 4 HCl + 3 O 2 → 2 H 2O + 4 ClO 4 HCl + 5 O 2 → 2 H 2O + 4 ClO 2 4 HCl + 7 O 2 → 2 H 2O + 4 ClO 3 2 HCl + O 2 → 2 HClO HCl + O 2 → HClO 2 2 HCl + 3 O 2 → 2 HClO 3 HCl + 2 O 2 → HClO 4 One point is earned for the correct reactants. Two points are earned for the correct products. One point is earned for balancing the equation. (ii) If three moles of hydrogen chloride gas and three moles of oxygen gas react as completely as possible, which reactant, if any, is pr esent in excess? Justify your answer. O 2!would be in excess because of the stoichiometry of the reaction; 4 moles of !HCl !are consumed for 1 mole of O2.!!(It takes only 0.75 mole of !O2!to react with 3 moles of !HCl , leaving an excess of 2.25 moles of !O2.) For other acceptable equations and products, the excess reactant must be based on the stoichiometry of the reaction given by the student. One point is earned for a correct answer that is based on the balanced chemical equation and that has an appropriate justification.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 4 (continued) (c) Solid potassium oxide is added to water. (i) Balanced equation: K2O + H 2O → 2 K + + 2 OH − One point is earned for the correct reactants. Two points are earned for the correct products. One point is earned for balancing the equation. (ii) If a few drops of phenolphthalein are added to th e resulting solution, what w ould be observed? Explain. The solution would turn pink because the production of ! OH −!makes the solution basic. In basic solutions, phenolphthalein turns pink. One point is earned for the correct answer with an explanation.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 5 Using principles of atomic and mo lecular structure and the informati on in the table below, answer the following questions about atomic fluorine, oxygen, and xenon, as we ll as some of their compounds. Atom First Ionization Energy (kJ mol −1) F 1,681.0 O 1,313.9 Xe ? (a) Write the equation for the ionization of atomic fluorine that requires 1,681.0 kJ mol "1. F(g) → F +(g) + e− One point is earned for the correct equation. (Phase designations are not required.) (b) Account for the fact that the firs t ionization energy of atomic fluorine is greater than that of atomic oxygen. (You must dis cuss both atoms in your response.) In both cases the electron removed is from the same energy level (2 p), but fluorine has a greater effective nuclear charge due to one more proton in its nucleus (the electrons are held more tightly and thus take more energy to remove). One point is earned fo r recognizing that the effective nuclear charge of !F!is greater than that of !O. (c) Predict whether the first ionization energy of atomic xenon is greater than, less than, or equal to the first ionization energy of atomic fl uorine. Justify your prediction. The first ionization energy of !Xe !should be less than the first ionization energy of !F . To ionize the !F!atom, an electron is removed from a 2 p orbital. To ionize the !Xe !atom, an electron must be removed from a 5 p orbital. The 5 p is a higher energy level and is farther from the nucleus than 2 p, hence it takes less energy to remove an electron from !Xe. One point is earned for a prediction based on size and/or energy level.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 5 (continued) (d) Xenon can react with oxygen and fluorine to form compounds such as XeO 3 and XeF 4. In the boxes provided, draw the complete Lewis electron-dot diagram for each of the molecules represented below. XeO 3 XeF 4 One point is earned for each correct Lewis electron-dot diagram. Omission of lone pairs of electrons on the O or F atoms results in a one-time, 1-point deduction. (e) On the basis of the Lewis electron-dot diagrams you drew for pa rt (d), predict the following: (i) The geometric shape of the XeO 3 molecule Trigonal pyramidal One point is earned for a shape that is consistent with the Lewis electron-dot diagram. (ii) The hybridization of the valence orbitals of xenon in XeF 4 sp 3d2 One point is earned for the hybridization consistent with the Lewis electron-dot diagram. (f) Predict whether the XeO 3 molecule is polar or nonpol ar. Justify your prediction. The !XeO 3!molecule would be polar because it contains three polar !Xe – O !bonds that are asymmetrically arranged around the central Xe atom (i.e., the bond dipoles do not cancel but add to a net molecular dipole with the Xe atom at the positive end). One point is earned for the answer that is consistent with the shape indicated in part (e)(i). One point is earned for an explanation correctly related to the shape in part (e)(i).
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 6 (a) Structures of the pyridine molecule and the benzen e molecule are shown below. Pyridine is soluble in water, whereas benzene is not soluble in water. Account for the difference in solubility. You must discuss both of the substances in your answer. Pyridine is polar (and cap able of forming hydrogen bonds with water), while the nonpolar benzene is not capable of forming hydrogen bonds. Pyridine will dissolve in water because of the strong hydrogen bonds (or dipole-dipole intermolecular interactions) that exist between the lone pair of electrons on pyridine’s nitrogen atom and the solvent water molecules. No such strong intermolecular interaction can exist between benzene and water, so benzene is insoluble in water. One point is earned for identifying a relevant structural difference between pyridine and benzene. One point is earned for indicating that pyridine is soluble in water because pyridine can form strong dipole-di pole interactions (or hydrogen bonds) with water, while benzene cannot. (b) Structures of the dimethyl ether molecule and the ethanol molecule are shown below. The normal boiling point of dimethyl ether is 250 K, whereas the normal boiling point of ethanol is 351 K. Account for the difference in boiling points. You must discuss both of the substances in your answer.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 6 (continued) The intermolecular forces of attraction among molecules of dimethyl ether consist of London (dispersion) forces and weak dipole-dipole interactions. In addition to London forces and dipole-dipole interactions that are comparable in strength to those in dimethyl ether, ethanol can form hydrogen bonds between the H of one molecule and the O of a nearby ethanol molecule. Hydrogen bonds are particularly strong intermolecular forces, so they require more energy to overcome during the boiling process. As a result, a higher temperature is needed to boil ethanol than is needed to boil dimethyl ether. One point is earned for recognizing that ethanol molecules can form intermolecular hydrogen bonds, whereas dimethyl ether molecules do not form intermolecular hydrogen bonds. One point is earned for recognizing that, compared to the energy required to overcome the weaker intermolecular forces in liquid dimethyl ether, more energy is required to overcome the stronger hydrogen bonds in liquid ethanol, leading to a higher boiling point. (c) SO 2 melts at 201 K, whereas SiO 2 melts at 1,883 K. Account for the difference in melting points. You must discuss both of the substances in your answer. In the solid phase, SO 2 consists of discrete molecules with dipole-dipole and L ondon (dispersion) forces among the molecules. These forces are relatively weak and are easily overcome at a relatively low temperature, consistent with the low melting point of SO 2. In solid !SiO 2, a network of !Si!and !O!atoms, linked by strong covalent bonds, exists . These covalent bonds are much stronger than typical intermolecular interactions, so very high temperatures are needed to overcome the covalent bonds in !SiO 2. This is consistent with the very high melting point for !SiO 2. One point is earned for recognizing that SO 2 is a molecular solid with only weak dipole-dipole and London forces among SO 2 molecules. One point is earned for recognizing that !SiO 2!is a covalent network solid, and that strong covalent bonds must be broken for !SiO 2!to melt.
A P® C H EM ISTR Y 2008 SC O R IN G G U ID ELIN ES © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Question 6 (continued) (d) The normal boiling point of Cl 2(l) (238 K) is higher than the normal boiling point of HCl( l) (188 K). Account for the difference in normal boiling points base d on the types of intermolecular forces in the substances. You must discuss both of the substances in your answer. The intermolecular forces in liquid !Cl 2!are London (dispersion) forces, whereas the intermolecular forces in liquid !HCl !consist of London forces and dipole-dipole interactions. Since the boiling point of !Cl 2!is higher than the boiling point of !HCl , the London forces among !Cl 2!molecules must be greater than the L ondon and dipole-dipole forces among !HCl !molecules. The greater strength of the London forces between !Cl 2! molecules occurs because !Cl 2!has more electrons than !HCl , and the strength of the London interaction is proportional to the total number of electrons. One point is earned for recognizing that the London forces among Cl 2 molecules must be larger than the intermolecular forces (London and dipole-dipole) among HCl molecules. One point is earned for recognizing that the strength of the London forces among molecules is proportional to the total number of electrons in each molecule.
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