Sample Responses from the Chemistry Practice Exam

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1 Sample Responses from the Chemistry Practice Exam Sample Questions Scoring Guidelines Commentaries on the Responses Effective Fall 2013

2 About the College Board The College Board is a mission-driven not-for-profit organization that connects students to college success and opportunity. Founded in 1900, the College Board was created to expand access to higher education. Today, the membership association is made up of over 6,000 of the world s leading educational institutions and is dedicated to promoting excellence and equity in education. Each year, the College Board helps more than seven million students prepare for a successful transition to college through programs and services in college readiness and college success including the SAT and the Advanced Placement Program. The organization also serves the education community through research and advocacy on behalf of students, educators and schools. For further information, visit College Board, Advanced Placement Program, AP and the acorn logo are registered trademarks of the College Board. All other products and services may be trademarks of their respective owners. Visit the College Board on the Web: B

3 Contents Preface...E Question Information for Free-Response Question Scoring Guidelines for Free-Response Question Commentaries and Scores...15 Question Information for Free-Response Question Scoring Guidelines for Free-Response Question Commentaries and Scores...28 Question Information for Free-Response Question Scoring Guidelines for Free-Response Question Commentaries and Scores Question Information for Free-Response Question Scoring Guidelines for Free-Response Question Commentaries and Scores Question Information for Free-Response Question Scoring Guidelines for Free-Response Question Commentaries and Scores...57 C

4 Question Information for Free-Response Question Scoring Guidelines for Free-Response Question Commentaries and Scores Question Information for Free-Response Question Scoring Guidelines for Free-Response Question Commentaries and Scores...77 D

5 Preface Preface This publication is designed to help teachers and students understand and prepare for the revised AP Chemistry Exam. The publication includes sample free-response questions, scoring guidelines, student responses at various levels of achievement, and reader commentaries. Collectively, these materials accurately reflect the design, composition, and rigor of the revised exam. The sample questions are those that appear on the AP Chemistry Practice Exam, and the responses were collected during a field test of the exam. The students gave permission to have their work reproduced at the time of the field test, and the responses were read and scored by AP Chemistry Readers in Following each free-response question, its scoring guideline, and three student samples, you will find a commentary about each sample. Commentaries include the score that each response would have earned, as well as a brief rationale to support the score. E

6 Free-Response Question 1 CHEMISTRY Section II 7 Constructed-Response Questions (Time 90 minutes) YOU MAY USE YOUR CALCULATOR FOR SECTION II Directions: Questions 1, 2, and 3 are long constructed-response questions that should require about 20 minutes each to answer. Questions 4, 5, 6, and 7 are short constructed-response questions that should require about seven minutes each to answer. Read each question carefully and write your response in the space provided following each question. Your responses to these questions will be scored on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Specific answers are preferable to broad, diffuse responses. For calculations, clearly show the method used and the steps involved in arriving at your answers. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. 1. A student performs an experiment in which the conductivity of a solution of Ba(OH) 2 is monitored as the solution is titrated with 0.10 M H 2 SO 4. The original volume of the Ba(OH) 2 solution is 25.0 ml. A precipitate of BaSO 4 (K sp = ) formed during the titration. The data collected from the experiment are plotted in the graph above. (a) As the first 30.0 ml of 0.10 M H 2 SO 4 are added to the Ba(OH) 2 solution, two types of chemical reactions occur simultaneously. On the lines provided below, write the balanced net-ionic equations for (i) the neutralization reaction and (ii) the precipitation reaction. (i) Equation for neutralization reaction: (ii) Equation for precipitation reaction: GO ON TO THE NEXT PAGE. 1

7 (b) The conductivity of the Ba(OH) 2 solution decreases as the volume of added 0.10 M H 2 SO 4 changes from 0.0 ml to 30.0 ml. (i) Identify the chemical species that enable the solution to conduct electricity as the first 30.0 ml of 0.10 M H 2 SO 4 are added. (ii) On the basis of the equations you wrote in part (a), explain why the conductivity decreases. (c) Using the information in the graph, calculate the molarity of the original Ba(OH) 2 solution. (d) Calculate the concentration of Ba 2+ (aq) in the solution at the equivalence point (after exactly 30.0 ml of 0.10 M H 2 SO 4 are added). (e) The concentration of Ba 2+ (aq) in the solution decreases as the volume of added 0.10 M H 2 SO 4 increases from 30.0 ml to 31.0 ml. Explain. GO ON TO THE NEXT PAGE. 2

8 Information for Free-Response Question 1 Timing Essential Knowledge/ Big Ideas Science Practices Learning Objectives The student should spend approximately 20 minutes on this question. 1.E.2 Conservation of atoms makes it possible to compute the masses of substances involved in physical and chemical processes. Chemical processes result in the formation of new substances, and the amount of these depends on the number and the types and masses of elements in the reactants, as well as the efficiency of the transformation. Big Idea 3: Changes in matter involve the rearrangement and/or reorganization of atoms and/or the transfer of electrons. 3.A.1 A chemical change may be represented by a molecular, ionic, or net ionic equation. 6.C.3 The solubility of a substance can be understood in terms of chemical equilibrium. 1.5 The student can re-express key elements of natural phenomena across multiple representations in the domain. 4.2 The student can design a plan for collecting data to answer a particular scientific question. 5.1 The student can analyze data to identify patterns or relationships. 7.1 The student can connect phenomena and models across spatial and temporal scales The student can design, and/or interpret data from, an experiment that uses gravimetric analysis to determine the concentration of an analyte in a solution. 3.1 Students can translate among macroscopic observations of change, chemical equations, and particle views. 3.2 The student can translate an observed chemical change into a balanced chemical equation and justify the choice of equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, ph) that influence the solubility. 3

9 Scoring Guidelines for Free-Response Question 1 Question 1 (10 Points) Conductivity Volume of 0.10 M H 2 SO 4 Added (ml) A student performs an experiment in which the conductivity of a solution of Ba(OH) 2 is monitored as the solution is titrated with 0.10 M H 2 SO 4. The original volume of the Ba(OH) 2 solution is 25.0 ml. A precipitate of BaSO 4 (Ksp = ) formed during the titration. The data collected from the experiment are plotted in the graph above. (a) As the first 30.0 ml of 0.10 M H 2 SO 4 are added to the Ba(OH) 2 solution, two types of chemical reactions occur simultaneously. On the lines provided below, write the balanced net-ionic equations for (i) the neutralization reaction and (ii) the precipitation reaction. (i) Equation for neutralization reaction: (ii) Equation for precipitation reaction: Ba 2+ (aq) + SO 4 2 (aq) BaSO 4 (s) H + (aq) + OH (aq) H 2 O(l) 1 point is earned for each correct product. 1 point is earned for the correct reactants with atoms and charges balanced in both reactions. (b) The conductivity of the Ba(OH) 2 solution decreases as the volume of added 0.10 M H 2 SO 4 changes from 0.0 ml to 30.0 ml. (i) Identify the chemical species that enable the solution to conduct electricity as the first 30.0 ml of 0.10 M H 2 SO 4 are added. Ba 2+ (aq) and/or OH (aq) 1 point is earned for either ion. (ii) On the basis of the equations you wrote in part (a), explain why the conductivity decreases. As the titration approaches the equivalence point, Ba 2+ (aq) ions are removed from solution by the precipitation reaction, and OH (aq) ions are removed from solution by the neutralization reaction. 1 point is earned for each correct explanation. Note: response must refer to both reactions for full credit. 4

10 (c) Using the information in the graph, calculate the molarity of the original Ba(OH) 2 solution. moles Ba(OH) 2 = moles H 2 SO 4 (at equivalence point) moles H 2 SO 4 = mol L = mol 10. L [Ba(OH) 2 ] = mol Ba(OH) 2 volume of original solution = mol L = 0.12 M 1 point is earned for the correct determination of the number of moles of titrant added at the equivalence point (can be implicit). 1 point is earned for the correct calculation of the original concentration of Ba(OH) 2 (aq). (d) Calculate the concentration of Ba 2+ (aq) in the solution at the equivalence point (after exactly 30.0 ml of 0.10 M H 2 SO 4 are added). K sp = [Ba 2+ ] [SO 4 2 ] = [Ba 2+ ] = [SO 4 2 ] [Ba 2+ ] = = M 1 point is earned for the correct calculation based on K sp. (e) The concentration of Ba 2+ (aq) in the solution decreases as the volume of added 0.10 M H 2 SO 4 increases from 30.0 ml to 31.0 ml. Explain. Because of the common ion effect, adding sulfate ions to an equilibrium reaction involving sulfate ions will cause the reaction to consume the added ions as a new equilibrium is established. Consequently, more BaSO 4 (s) is formed, causing the Ba 2+ (aq) concentration to decrease. 1 point is earned for a correct explanation, which must use an equilibrium argument (for example, citing the common ion effect or Le Chatelier s principle) rather than a stoichiometric argument. 5

11 Sample: 1A 6

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14 Sample: 1B 9

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17 Sample: 1C 12

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20 2013 Practice Exam Scoring Commentary Note: Student samples are quoted verbatim and may contain grammatical errors. Free-Response Question 1 Commentary Overview This question provided an opportunity for students to quantitatively and qualitatively describe the chemical processes that occur during the course of a conductivity titration. Part (a) provided an opportunity to illustrate the chemical reactions (neutralization and precipitation) that occurred during the titration using net ionic equations. Part (b) required students to identify any species that would cause the solution to conduct electricity during the titration as well as discuss why the conductivity drops as the titration proceeded. Parts (c) and (d) requested that students use data from the titration coupled with knowledge of stoichiometry and solubility equilibrium to determine the initial concentration of the Ba(OH) 2 solution (part [c]) and the concentration of barium cation at the equivalence point (part [d]). Finally, part (e) required students to recognize that the solubility equilibrium persists past the equivalence point, resulting in a decrease in the concentration of the barium cation as the addition of sulfate anion shifts the solubility equilibrium toward solid barium sulfate. Sample: 1A Score: 9 In part (a), the response earned all 3 points for the correct net ionic equations for both the precipitation and neutralization reactions. In part (b), the response earned a total of 3 points. Part (i) earned 1 point for listing Ba 2+ and OH as the species that conduct electricity in the solution. Part (ii) earned 2 points for recognition that both of these ions were being consumed during the reactions, causing their concentrations to decrease. In part (c), 2 points were earned for the correct calculation of 0.12 M for the initial concentration of the Ba(OH) 2 solution. In part (d), 0 points were earned. Although there was recognition of solubility equilibrium at the equivalence point, an incorrect attempt was made to calculate the equilibrium concentration of the sulfate anion via dilution of the two solutions. This yielded an incorrect answer for the concentration of the barium cation at equilibrium. In part (e), 1 point was earned for a clear discussion of the reduction of the concentration of Ba 2+ at equilibrium via the common ion effect, because when more SO 4 2 are added from H 2 SO 4, it shifts the reaction to the left, resulting in a decrease in the Ba 2+ ion concentration at equilibrium. 15

21 Sample: 1B Score: 6 In part (a), 3 points were earned for the correct net ionic equations for both reactions. Although the reactions were not in the requested location on the exam, each was clearly labeled, in order, on the following page. In part (b), a total of 1 point was earned. In part (i), 1 point was earned for the correct identification of the barium cation as a conductive species. Part (ii) earned 0 points because no clear discussion was given of the decrease in ion concentration in solution as the reaction proceeded. In part (c), 1 point was earned for the correct calculation of the moles of Ba(OH) 2 in the original solution. The incorrect calculation of the initial molarity of the solution did not earn the second point. In part (d), 1 point was earned for the recognition of the solubility equilibrium condition at the equivalence point of the titration followed by the correct calculation of [Ba 2+ ] from the K sp data. In part (e), 0 points were earned because there was no recognition that the solubility equilibrium of the barium sulfate is maintained past the equivalence point. This response revealed considerable misconceptions about solubility equilibrium, since it was implied that the barium and sulfate ions would still react stoichiometrically after the equivalence point had been reached. Sample: 1C Score: 3 In part (a), 1 point was earned for the recognition that H 2 O was the product of the neutralization reaction. Because the other reaction was incorrect, additional points were not earned. In part (b), 0 points were earned. There was no explicit mention of any ions as the agents of electrical conduction in the solution, or that ion concentration decreased as the reaction proceeded. In part (c), 2 points were earned. The response correctly calculated the initial concentration of the barium hydroxide solution. Note that the work in this response was clearly shown at both the top of the paper as well as in part (b). In part (d), 0 points were earned. The concentration of barium cation at the equivalence point was incorrectly calculated, and no work was shown. In part (e), 0 points were earned. No explicit mention was made of an equilibrium condition past the equivalence point. In addition, the response implied that ions were still reacting stoichiometrically, which was a common misconception as revealed in responses to this question. 16

22 Free-Response Question 2 2. A student is given the task of determining the enthalpy of reaction for the reaction between HCl(aq) and NaOH(aq). The following materials are available M HCl(aq) 1.00 M NaOH(aq) distilled water 2.00 M HCl(aq) 2.00 M NaOH(aq) goggles insulated cups with covers gloves lab coat thermometer (± 0.1 C) stirring rod The student may select from the glassware listed in the table below. Glassware Items Precision 250 ml Erlenmeyer flasks ± 25 ml 100 ml beakers ± 10 ml 100 ml graduated cylinders ± 0.1 ml (a) The student selects two 100 ml beakers, uses them to measure 50 ml each of 1.00 M HCl(aq) solution and 1.00 M NaOH(aq) solution, and measures an initial temperature of 24.5 C for each solution. Then the student pours the two solutions into an insulated cup, stirs the mixture, covers the cup, and records a maximum temperature of 29.9 C. (i) Is the experimental design sufficient to determine the enthalpy of reaction to a precision of two significant figures? Justify your answer. (ii) List two specific changes to the experiment that will allow the student to determine the enthalpy of reaction to a precision of three significant figures. Explain. (b) A second student is given two solutions, 75.0 ml of 1.00 M HCl and 75.0 ml of 1.00 M NaOH, each at 25.0 C. The student pours the solutions into an insulated cup, stirs the mixture, covers the cup, and records the maximum temperature of the mixture. (i) The student calculates the amount of heat evolved in the experiment to be 4.1 kj. Calculate the student s experimental value for the enthalpy of reaction, in kj/mol rxn. (ii) The student assumes that the thermometer and the calorimeter do not absorb energy during the reaction. Does this assumption result in a calculated value of the enthalpy of reaction that is higher than, lower than, or the same as it would have been had the heat capacities of the thermometer and calorimeter been taken into account? Justify your answer. (iii) One assumption in interpreting the results of the experiment is that the reaction between HCl(aq) and NaOH(aq) goes to completion. Justify the validity of this assumption in terms of the equilibrium constant for the reaction. (c) A third student calculates a value for the enthalpy of reaction that is significantly higher than the accepted value. (i) Identify a specific error in procedure made by the student that will result in a calculated value for the enthalpy of reaction that is higher than the accepted value. (Vague statements like human error or incorrect calculations will not earn credit.) (ii) Explain how the error that you identified in part (c)(i) leads to a calculated value for the enthalpy of reaction that is higher than the accepted value. GO ON TO THE NEXT PAGE. 17

23 Information for Free-Response Question 2 Timing Essential Knowledge Science Practices Learning Objectives The student should spend approximately 20 minutes on this question. 5.B.3 Chemical systems undergo three main processes that change their energy: heating/cooling, phase transitions, and chemical reactions. 5.B.4 Calorimetry is an experimental technique that is used to measure the change in energy of a chemical system. 6.A.2 The current state of a system undergoing a reversible reaction can be characterized by the extent to which reactants have been converted to products. The relative quantities of reaction components are quantitatively described by the reaction quotient, Q. 6.A.4 The magnitude of the equilibrium constant, K, can be used to determine whether the equilibrium lies toward the reactant side or the product side. 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. 2.3 The student can estimate numerically quantities that describe natural phenomena. 4.2 The student can design a plan for collecting data to answer a particular scientific question. 5.1 The student can analyze data to identify patterns or relationships. 5.6 The student is able to use calculations or estimations to relate energy changes associated with heating/cooling a substance to the heat capacity, relate energy changes associated with a phase transition to the enthalpy of fusion/vaporization, relate energy changes associated with a chemical reaction to the enthalpy of the reaction, and relate energy changes to PΔV work. 5.7 The student is able to design and/or interpret the results of an experiment in which calorimetry is used to determine the change in enthalpy of a chemical process (heating/cooling, phase transition, or chemical reaction) at constant pressure. 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g., reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K. 6.7 The student is able, for a reversible reaction that has a large or small K, to determine which chemical species will have very large versus very small concentrations at equilibrium. 18

24 Scoring Guidelines for Free-Response Question 2 Question 2 (10 Points) A student is given the task of determining the enthalpy of reaction for the reaction between HCl(aq) and NaOH(aq). The following materials are available M HCl(aq) 1.00 M NaOH(aq) distilled water 2.00 M HCl(aq) 2.00 M NaOH(aq) goggles insulated cups with covers gloves lab coat thermometer (± 0.1 C) stirring rod The student may select from the glassware listed in the table below. Glassware Items Precision 250 ml Erlenmeyer flasks ± 25 ml 100 ml beakers ± 10 ml 100 ml graduated cylinders ± 0.1 ml (a) The student selects two 100 ml beakers, uses them to measure 50 ml each of 1.00 M HCl(aq) solution and 1.00 M NaOH(aq) solution, and measures an initial temperature of 24.5 C for each solution. Then the student pours the two solutions into an insulated cup, stirs the mixture, covers the cup, and records a maximum temperature of 29.9 C. (i) Is the experimental design sufficient to determine the enthalpy of reaction to a precision of two significant figures? Justify your answer. No. The use of the beakers to measure 50 ml ±10 ml of solutions limits the precision of the volume measurements and of the calculations to ±20% or 1 significant figure. 1 point is earned for the correct answer with the correct explanation. (ii) List two specific changes to the experiment that will allow the student to determine the enthalpy of reaction to a precision of three significant figures. Explain. Use graduated cylinders to measure the volumes of acid and base allowing a volume precision of ±0.1 ml or 3 significant figures for a volume of 50.0 ml. AND Use the 2.00 M HCl and 2.00 M NaOH solutions (instead of 1.00 M) to get a larger T, thereby improving the relative precision in T to ±1%. 1 point is earned for the change of glassware to graduated cylinders with a proper explanation. 1 point is earned for using the 2.00 M solutions for improved relative precision in temperature. Note: doubling the volumes will not increase ΔT or significantly improve volume precision. (b) A second student is given two solutions, 75.0 ml of 1.00 M HCl and 75.0 ml of 1.00 M NaOH, each at 25.0 C. The student pours the solutions into an insulated cup, stirs the mixture, covers the cup, and records the maximum temperature of the mixture. (i) The student calculates the amount of heat evolved in the experiment to be 4.1 kj. Calculate the student s experimental value for the enthalpy of reaction, in kj/mol rxn. 19

25 75.0 ml H = 1.00 mol HCl (or NaOH) 1000 ml 4. 1 k J = 55 kj/mol mol of reactants rxn = mol HCl (or NaOH) 1 point is earned for the correct calculation of moles of reactants. 1 point is earned for the correct substitution and answer. (ii) The student assumes that the thermometer and the calorimeter do not absorb energy during the reaction. Does this assumption result in a calculated value of the enthalpy of reaction that is higher than, lower than, or the same as it would have been had the heat capacities of the thermometer and calorimeter been taken into account? Justify your answer. The calculated value of the enthalpy of reaction will be lower (smaller or less negative) than it would have been had the thermometer and calorimeter been taken into account. The thermometer and calorimeter will absorb some of the heat of reaction. This lost heat is ignored in the original calculation of H rxn, making it smaller in magnitude (less negative). OR The actual heat capacity of the system is the sum of the heat capacities of the water, thermometer, and calorimeter. The assumed heat capacity of the system (water only) is less than the actual value, resulting in a lower (less negative) calculated value of H rxn. 1 point is earned for the correct prediction. 1 point is earned for an acceptable justification. (iii) One assumption in interpreting the results of the experiment is that the reaction between HCl(aq) and NaOH(aq) goes to completion. Justify the validity of this assumption in terms of the equilibrium constant for the reaction. H + + OH H 2 O, the reaction between HCl(aq) and NaOH(aq), is the reverse of H 2 O H + + OH, the autoionization of water (for which K = K w = ). Thus the value of K for the neutralization reaction is the reciprocal of K w, or , a very large number. Thus the neutralization reaction goes virtually to completion. 1 point is earned for the correct justification. 20

26 (c) A third student calculates a value for the enthalpy of reaction that is significantly higher than the accepted value. (i) Identify a specific error in procedure made by the student that will result in a calculated value for the enthalpy of reaction that is higher than the accepted value. (Vague statements like human error or incorrect calculations will not earn credit.) The student read the thermometer incorrectly in such a way to result in a calculated value of T that was too high (either read T i too low or read T f too high). OR The student mistakenly used 2.00 M acid and 2.00 M base, thinking they were both 1.00 M. 1 point is earned for an acceptable procedural error that results in a higher calculated value. (ii) Explain how the error that you identified in part (c)(i) leads to a calculated value for the enthalpy of reaction that is higher than the accepted value. The calculation of the molar enthalpy of reaction may be expressed as mass soln c T Molar H rxn =. nrxn If there is a measurement error that results in a T that is too high, the magnitude (i.e., the absolute value) of the calculated molar enthalpy of reaction will be too high. 1 point is earned for an explanation that is consistent with the stated procedural error. 21

27 Sample: 2A* * Due to a question and scoring guideline change after the piloting of this exam, all of the responses to Question 2 represent a combination of student/teacher work. 22

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29 Sample: 2B 24

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31 Sample: 2C 26

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33 Free-Response Question 2 Commentary Overview This question provided an opportunity for students to perform an in-depth examination of an enthalpy of neutralization experiment. Part (a) required students to examine the design of the experiment, determining whether the design was sufficient to allow for the enthalpy of neutralization to be calculated to two significant figures (i), and proposing, with explanation, changes to the design that would guarantee a calculated enthalpy of three significant figures (ii). Part (b) requested students to calculate the enthalpy of reaction (part i) and examine the experimental and theoretical assumptions made in the experiment (parts ii and iii). Finally, in part (c), students were required to identify a specific error that caused the calculated heat of reaction to be higher than that of the literature value. In addition, students needed to explain why this error led to an artificially inflated value of ΔH for the reaction. Sample: 2A Score: 10 In part (a), the response earned 3 points. In (i), 1 point was earned since it was clearly stated that beakers have a precision of only one significant figure. Thus, using them for volume measurement will result in an answer of one significant figure for the enthalpy of reaction. In (ii), 2 points were earned for the suggestion to use graduated cylinders with a precision of three significant figures, as well as the suggestion to use 2.00 M reactant solutions. The latter suggestion would increase the value of ΔT to three significant figures. Each of these changes would improve the precision of the experiment to the desired level. In part (b), the maximum 5 points were earned. In (i), the value of -54 kj/mol rxn was clearly calculated from the data in the problem. In (ii) it was correctly stated that ignoring energy loss to the calorimeter and the thermometer would result in a ΔH that is too low, because the lost energy could not be added back into the calculation. Finally, part (iii) earned 1 point because the equilibrium constant of 1 x for the reaction of HCl and NaOH is clearly stated, resulting in a reaction that is essentially a limiting reactant problem. In part (c), 2 points were earned. The correct error identified is the student believing 2.00 M solutions were being used, while in reality the concentration of each solution is just 1.00 M. Such reasoning would result in a calculated ΔH that was too large. The correct rationale for why this error would lead to a higher calculated value for enthalpy of reaction is that dividing the larger q, calculated from q = mcδt, by the value of the moles of reactant that was too low (because the student thought the molarity of the solution was 1.00 M instead of 2.00 M) would result in a calculated value of ΔH that was artificially high by a factor of two. Sample: 2B Score: 6 In part (a), the response earned 2 points. In (i), 1 point was earned for noting that the precision of the beakers is only accurate to one significant figure. In (ii), 1 point was awarded for selecting graduated cylinders to measure the volumes of reactants which are good to the tenths place. Although the response also suggests using 2.00 M solutions, there is no explanation as to why these solutions would increase precision to three significant figures. Thus, the final point in the problem was not earned. 28

34 In part (b), a total of 4 points were earned. In (i), 1 point was earned for the correct calculation of moles of reactants. However, although the magnitude of the ΔH was correct, the sign on the answer implies that the reaction is endothermic. Thus, the second point was not earned. In (ii), both points were earned because the response correctly predicts that the calculated enthalpy will be lower, as indicated by the explanation that they are missing out on the energy lost to the calorimeter and thermometer. In (iii), 1 point was earned for recognizing that the equilibrium constant for the neutralization reaction was extremely large. In part (c), 0 points were earned, since an improper measurement in the volumes of the solutions would not result in a ΔH of reaction that was too large. Sample: 2C Score: 4 In part (a), the response earned 2 points. In (i), it was correctly noted that beakers only provide one significant figure in a volume measurement. Thus, the enthalpy of reaction could not be determined to two significant figures. In (ii), the suggestion of using the more precise graduated cylinder earned 1 point. Although using a more precise thermometer could also improve the precision of the enthalpy measurement to three significant figures, this answer did not receive credit because more precise thermometers were not available in the list of equipment. In part (b), 2 points were earned. In (i), the correct calculation of ΔH, with proper sign, earned 2 points. In (ii), the response incorrectly predicts that the enthalpy of reaction will be higher. Thus, 0 points were earned. In (iii), it is mentioned that HCl and NaOH will react strongly. However, the response incorrectly states that the equilibrium constant for the reaction will be small, as opposed to the correct value of 1 x 10 14, which would result in the reaction essentially going to completion. In part (c), 0 points were earned. The error in mixing that was discussed would not cause a ΔH that was significantly higher than the accepted value. 29

35 Free-Response Question 3 SO 2 Cl 2 (g) Æ SO 2 (g) + Cl 2 (g) 3. A 4.32 g sample of liquid SO 2 Cl 2 is placed in a rigid, evacuated 1.50 L reaction vessel. As the container is heated to 400. K, the sample vaporizes completely and starts to decompose according to the equation above. The decomposition reaction is endothermic. (a) If no decomposition occurred, what would be the pressure, in atm, of the SO 2 Cl 2 (g) in the vessel at 400. K? (b) When the system has reached equilibrium at 400. K, the total pressure in the container is 1.26 atm. Calculate the partial pressures, in atm, of SO 2 Cl 2 (g), SO 2 (g), and Cl 2 (g) in the container at 400. K. (c) For the decomposition reaction at 400. K, (i) write the equilibrium-constant expression for K p for the reaction, and (ii) calculate the value of the equilibrium constant, K p. (d) The temperature of the equilibrium mixture is increased to 425 K. Will the value of K p increase, decrease, or remain the same? Justify your prediction. (e) In another experiment, the original partial pressures of SO 2 Cl 2 (g), SO 2 (g), and Cl 2 (g) are 1.0 atm each at 400. K. Predict whether the amount of SO 2 Cl 2 (g) in the container will increase, decrease, or remain the same. Justify your prediction. GO ON TO THE NEXT PAGE. 30

36 Information for Free-Response Question 3 Timing Essential Knowledge Science Practices Learning Objectives The student should spend approximately 20 minutes on this question. 2.A.2 The gaseous state can be effectively modeled with a mathematical equation relating various macroscopic properties. A gas has neither a definite volume nor a definite shape; because the effects of attractive forces are minimal, we usually assume that the particles move independently. 6.A.3 When a system is at equilibrium, all macroscopic variables, such as concentrations, partial pressures, and temperature, do not change over time. Equilibrium results from an equality between the rates of the forward and reverse reactions, at which point Q = K. 6.B.1 Systems at equilibrium respond to disturbances by partially countering the effect of the disturbance (Le Chatelier s principle). 1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. 2.3 The student can estimate numerically quantities that describe natural phenomena. 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 2.6 The student can apply mathematical relationships or estimation to determine macroscopic variables for ideal gases. 6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants as equilibrium is approached. 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K. 6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction. 6.8 The student is able to use Le Chatelier s principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium. 31

37 Scoring Guidelines for Free-Response Question 3 Question 3 (10 Points) SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) A 4.32 g sample of liquid SO 2 Cl 2 is placed in a rigid, evacuated 1.50 L reaction vessel. As the container is heated to 400. K, the sample vaporizes completely and starts to decompose according to the equation above. The decomposition reaction is endothermic. (a) If no decomposition occurred, what would be the pressure, in atm, of the SO 2 Cl 2 (g) in the vessel at 400. K? Assuming no decomposition, m g molesso Cl = = = mol 2 2 M g/mol p SO 2 Cl 2 = nrt V = atm ( mol)( L. atm / mol. K)(400. K) = 1.50 L 1 point is earned for the correct calculation of moles of SO 2 Cl 2 (may be implicit). 1 point is earned for the correct calculation of the pressure. (b) When the system has reached equilibrium at 400. K, the total pressure in the container is 1.26 atm. Calculate the partial pressures, in atm, of SO 2 Cl 2 (g), SO 2 (g), and Cl 2 (g) in the container at 400. K. Pressures at equilibrium at 400. K: SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) Total x x x x p total = x = 1.26 atm x = p SO2 = p Cl2 = 0.56 atm = x = 0.14 atm p SO Cl point is earned for the correct setup. 1 point is earned for the correct calculation of pressures. (c) For the decomposition reaction at 400. K, (i) write the equilibrium-constant expression for K p for the reaction, and K p = p p p. SO2 Cl2 SO2Cl2 1 point is earned for the correct K p expression. Note: the pressure subscripts must be specific (i.e., SO 2, Cl 2, and SO 2 Cl 2 NOT, e.g., A, B, C, and D). (ii) calculate the value of the equilibrium constant, K p. K p = ( )( ) = point is earned for the correct calculation of K p that is consistent with the K p expression stated in part (c)(i) and with the partial pressures calculated in part (b). (d) The temperature of the equilibrium mixture is increased to 425 K. Will the value of K p increase, decrease, or remain the same? Justify your prediction. 32

38 At a higher temperature, K p will increase. According to Le Chatelier s principle, raising the temperature of an endothermic reaction at equilibrium adds a thermal stress that increases the value of K p and produces more products. 1 point is earned for the correct prediction. 1 point is earned for a proper justification in terms of Le Chatelier s principle. (e) In another experiment, the original partial pressures of SO 2 Cl 2 (g), SO 2 (g), and Cl 2 (g) are 1.0 atm each at 400. K. Predict whether the amount of SO 2 Cl 2 (g) in the container will increase, decrease, or remain the same. Justify your prediction. The amount of SO 2 Cl 2 in the container will decrease. Initially Q p = 1.0 < 2.2. = K p, thus the reaction will consume SO 2 Cl 2 as it proceeds in the forward direction to reestablish equilibrium. 1 point is earned for the correct prediction. 1 point is earned for an acceptable justification. Note: the justification must consider the relative values of Q p and K p. 33

39 Sample: 3A 34

40 35

41 Sample: 3B 36

42 37

43 Sample: 3C 38

44 39

45 Free-Response Question 3 Commentary Overview This question provided an opportunity for students to quantitatively analyze a gas phase decomposition reaction in chemical equilibrium. Part (a) required students to calculate the initial pressure of the reaction system before any reaction occurred. Part (b) required students to calculate the equilibrium pressures of all reactants and products given the total equilibrium pressure of the reaction system. Part (c) required students to write the equilibrium constant expression K p for the decomposition reaction, and to determine the value of K p using the equilibrium pressures determined in part (b). Part (d) requested that students predict the effect of a temperature increase on the value of K p, with proper justification utilizing Le Châtelier s principle. Finally, in part (e), students were asked to predict what would happen to the amount of reactant in a subsequent experiment as the reaction system approached equilibrium via a comparison of the reaction quotient Q to the equilibrium constant K. Sample: 3A Score: 10 In part (a), the response earned the maximum 2 points for correct use of the ideal gas law to calculate the initial pressure of the reaction system before any decomposition occurred. In part (b), 2 points were earned for correctly using a RICE chart and the stoichiometry of the reaction to determine the pressures of all reactants and products at equilibrium, given that the total pressure was 1.26 atm. In part (c), the maximum 2 points were awarded for the correct expression for the equilibrium constant K p, and for the proper substitution of the values obtained in part (b) to obtain a value of 2.2. In part (d), 2 points were earned. The first point was earned for the correct prediction that the value of K p will increase because the reaction is endothermic. Proper use of Le Châtelier s principle to justify this prediction earned the second point. In part (e), 2 points were earned for the correct calculation of Q = 1 and for the excellent justification that because Q is less than K, the reaction must shift right to make more products. Therefore the amount of SO 2 Cl 2 in the container will decrease. Sample: 3B Score: 7 In part (a), 2 points were earned for the correct calculation of the initial pressure of SO 2 Cl 2 gas from the ideal gas law before any reaction occurred. In part (b), 0 points were earned. The response correctly indicated that both products would have the same pressure at equilibrium. However, the incorrect setup of an equation that related the pressures of each reaction species to the total pressure of 1.26 atm prevented any points from being earned. In part (c), 1 point was earned. The use of square brackets, [ ], in the K p expression implied that concentrations, as opposed to pressures, should be used to calculate K p. Consequently, no points were awarded in part (i). Proper substitution and calculation of the written equilibrium constant utilizing the pressure values determined in part (b) earned 1 point. 40

46 In part (d), 2 points were earned. The first point was earned for the correct prediction that the value of K p would increase as the temperature of the reaction system increased. The use of Le Châtelier s argument, in which energy could be considered a reactant of the endothermic reaction, causing K p to increase as the temperature increased earned the second point. In part (e), 2 points were earned. The response correctly predicted that the amount of SO 2 Cl 2 increased when equilibrium was reached, which was correct based upon previous calculations. Because the response had to use the value of 0.11 calculated in part (c), the second point was earned by engaging in a correct discussion that the value of Q (mislabeled as K p at that moment ) must decrease from 1.0 to 0.11, causing the concentration of the reactants to increase. Sample: 3C Score: 4 In part (a), 1 point was earned. Although the number of moles of SO 2 Cl 2 before reaction was correctly calculated for 1 point, the incorrect value of R used in the ideal gas law resulted in an incorrect initial pressure for the gas. Thus, the second point was not earned. In part (b), 0 points were earned. The pressures at equilibrium were calculated incorrectly, utilizing a ratio of molar masses as opposed to using stoichiometry and the total pressure of the reaction system at equilibrium. In part (c), the maximum 2 points were earned. One point was earned in (i) for the K p expression, because the brackets around the pressures were interpreted to imply multiplication of the pressures inside of them, and not a concentration of a pressure. Correct substitution and calculation of K p from the pressures of part (b) earned the second point. In part (d), 0 points were earned for the incorrect prediction because the constant is not dependent on the temperature, thus it will remain the same. In part (e), 1 point was earned. The response correctly predicts that for the K p calculated in part (c), the pressure of SO 2 Cl 2 would increase as the reaction system reached equilibrium. However, the second point was not earned because of the incorrect explanation that does not relate the reaction quotient to the equilibrium constant. 41

47 Free-Response Question 4 2 NO 2 (g) + F 2 (g) 2 NO 2 F(g) 4. It is proposed that the reaction represented above proceeds via the mechanism represented by the two elementary steps shown below. Step I: NO 2 + F 2 NO 2 F + F (slow) Step II: NO 2 + F Æ NO 2 F (fast reversible) (a) Step I of the proposed mechanism involves the collision between NO 2 and F 2 molecules. This step is slow even though such collisions occur very frequently in a mixture of NO 2 (g) and F 2 (g). Consider a specific collision between a molecule of NO 2 and a molecule of F 2. (i) One factor that affects whether the collision will result in a reaction is the magnitude of the collision energy. Explain. (ii) Identify and explain one other factor that affects whether the collision will result in a reaction. (b) Consider the following potential rate laws for the reaction. Circle the rate law below that is consistent with the mechanism proposed above. Explain the reasoning behind your choice in terms of the details of the elementary steps of the mechanism. rate = k[no 2 ] 2 [F 2 ] rate = k[no 2 ][F 2 ] GO ON TO THE NEXT PAGE. 42

48 Information for Free-Response Question 4 Timing Essential Knowledge/ Enduring Understanding Science Practices Learning Objectives The student should spend approximately 7 8 minutes on this question. 4.B.1 Elementary reactions can be unimolecular or involve collisions between two or more molecules. 4.B.2 Not all collisions are successful. To get over the activation energy barrier, the colliding species need sufficient energy. Also, the orientations of the reactant molecules during the collision must allow for the rearrangement of reactant bonds to form product bonds. 4.C Many reactions proceed via a series of elementary reactions. 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. 6.5 The student can evaluate alternative scientific explanations. 7.1 The student can connect phenomena and models across spatial and temporal scales. 4.4 The student is able to connect the rate law for an elementary reaction to the frequency and success of molecular collisions, including connecting the frequency and success to the order and rate constant, respectively. 4.5 The student is able to explain the difference between collisions that convert reactants to products and those that do not in terms of energy distributions and molecular orientation. 4.7 The student is able to evaluate alternative explanations, as expressed by reaction mechanisms, to determine which are consistent with data regarding the overall rate of a reaction, and data that can be used to infer the presence of a reaction intermediate. 43

49 Scoring Guidelines for Free-Response Question 4 Question 4 (4 Points) 2 NO 2 (g) + F 2 (g) 2 NO 2 F(g) It is proposed that the reaction represented above proceeds via the mechanism represented by the two elementary steps shown below. Step I: NO 2 + F 2 NO 2 F + F (slow) Step II: NO 2 + F NO 2 F (fast reversible) (a) Step I of the proposed mechanism involves the collision between NO 2 and F 2 molecules. This step is slow even though such collisions occur very frequently in a mixture of NO 2 (g) and F 2 (g). Consider a specific collision between a molecule of NO 2 and a molecule of F 2. (i) One factor that affects whether the collision will result in a reaction is the magnitude of the collision energy. Explain. Successful molecular collisions must have sufficient energy in order to result in reaction. Only collisions with sufficient energy to overcome the activation energy barrier, E a, will be able to reach the transition state and begin to break the F F bond. 1 point is earned for a correct explanation that makes reference to the activation energy of the reaction. (ii) Identify and explain one other factor that affects whether the collision will result in a reaction. For a collision to be successful, the molecules must have the correct orientation. Only collisions with the correct orientation will be able to begin to form an N F bond and begin to break an F F bond as the transition state is approached (that is, the molecules must contact each other at very specific locations on their surfaces for the transition state to be accessible). 1 point is earned for identifying the relative orientation of the colliding molecules. 1 point is earned for an explanation that makes reference to specific parts (atoms or bonds) of the reacting molecules. (b) Consider the following potential rate laws for the reaction. Circle the rate law below that is consistent with the mechanism proposed above. Explain the reasoning behind your choice in terms of the details of the elementary steps of the mechanism. rate = k[no 2 ] 2 [F 2 ] rate = k[no 2 ][F 2 ] The rate law that is consistent with the mechanism is the one on the right above (rate = k[no 2 ][F 2 ]). Step I is the slower step and the ratedetermining step in the mechanism. Since Step I is an elementary reaction, its rate law is given by the stoichiometry of the reacting molecules, rate Step I = k 1 [NO 2 ][F 2 ]. 1 point is earned for identifying the correct rate law with a proper explanation. The explanation must correlate the overall rate law with the rate law derived from the stoichiometry of the slow step in the mechanism. Note: a statement relating the coefficients of the reactants in Step I to the exponents in the rate law indicates a correct understanding. 44

50 Sample: 4A* * Due to a question and scoring guideline change after the piloting of this exam, all of the responses to Question 4 represent a combination of student/teacher work. 45

51 Sample: 4B 46

52 Sample: 4C 47

53 Free-Response Question 4 Commentary Overview This question provided an opportunity for students to demonstrate an understanding of reaction mechanisms. Part (a) required students to explain the molecular factors that result in a successful collision. Part (b) required students to correctly identify the rate law for the reaction mechanism given in the problem, using details of the elementary reaction steps to justify their choice. Sample: 4A Score: 4 In part (a), the response earned the maximum 3 points. One point was earned in (i) for correctly describing the need for a successful collision to have enough energy to overcome the activation energy barrier. The remaining points were earned in (ii) as orientation of collision was identified as the second factor necessary for a successful collision. The description of the need for the collision to hit the right way so that the collision starts to break the bond in F 2 and make a bond between N and F provided an excellent explanation of what proper orientation means. In part (b), 1 point was earned because the response selected the proper rate law and provided an explanation that clearly connected the rate law of the slow step of the mechanism to the overall rate law for the reaction. Sample: 4B Score: 3 In part (a), 2 points were earned. In (i), 1 point was earned for a discussion that the molecules need to have enough energy to overcome the activation energy. In (ii), 1 point was earned for identifying orientation as the second factor. However, the explanation of what proper orientation means was not in sufficient depth to earn the second point. In part (b), 1 point was earned for correctly identifying the rate = k[no 2 ][F 2 ] in the response, and for the explanation that stated that the rate law of a reaction is determined by the rate law of slowest step of the mechanism. Sample: 4C Score: 1 In part (a), the response earned 1 point. In (i) the response correctly described the need of a collision to have enough energy to overcome the activation energy barrier. However, the identification of volume in (ii) as the second factor of a successful collision failed to earn additional points. In part (b), 0 points were earned because an incorrect rate law was identified. In addition, the response stated the common misconception that the rate law is determined from the overall stoichiometry of the reaction, as opposed to the stoichiometry of the rate-determining step. 48

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