AP Chemistry 000 Scing Guidelines The materials included in these files are intended f non-commercial use by AP teachers f course and exam preparation; permission f any other use must be sought from the Advanced Placement Program. Teachers may reproduce them, in whole in part, in limited quantities, f face-to-face teaching purposes but may not mass distribute the materials, electronically otherwise. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here. This permission does not apply to any third-party copyrights contained herein. These materials were produced by Educational Testing Service (ETS), which develops and administers the examinations of the Advanced Placement Program f the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opptunity, and their programs, services, and employment policies are guided by that principle. The College Board is a national nonprofit membership association dedicated to preparing, inspiring, and connecting students to college and opptunity. Founded in 1900, the association is composed of me than 3,900 schools, colleges, universities, and other educational ganizations. Each year, the College Board serves over three million students and their parents,,000 high schools, and 3,500 colleges, through maj programs and services in college admission, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT, the PSAT/NMSQT, the Advanced Placement Program (AP ), and Pacesetter. The College Board is committed to the principles of equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. Copyright 001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acn logo are registered trademarks of the College Entrance Examination Board.
AP Chemistry 000 Scing Standards Question 1 (10 points) (a) K c [H [H ] [S ] S] K c (H (H ) (S S) ) No point earned f a K p expression if an exponent is increctly used (b) (i) [H ] [S ] (.0)(3.7 10 1.5 L mol) 5.95 10 M (ii) [H S] 0.10 mol (.0)(3.7 10 1.5 L mol).05 10 M pts. [H S] 0.0800 M 0.0595 M 0.005 M Notes: One point is earned f getting the crect number of moles of H S ; second point is earned f dividing by 1.5 L and getting a consistent answer. Although not crect, one point may be earned f calculating the initial [H S] (see below) and using that value as the equilibrium H S concentration. 3. 40 g 341. g mol [H S] 0. 0798M 15. L (c) K c 3 7 10 (5.95 10 ). 15. (.05 10 ) (5.95 10 ) (.05 10 (0.098) ) 0.50 pts. Notes: One point is earned f crectly using the molarity of the S (dividing the number of moles by the volume). The second point is earned by crectly using the numbers generated in (b) in the expression shown in (a) and getting the appropriate answer. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 1 (continued) (d) P nrt V ( 3.7 10 mol)(0.081 L atm mol 1 K)(483 K) 1.18 atm 1.5 L P [S ]RT (0.0976 mol L )(0.081 L atm mol 1 K ) (483 K) 1.18 atm pts. Note: The first point is earned f crectly setting up either of these expressions. The second point is earned f the crect answer ( an answer consistent with the numbers used). Also, combining Ptotal V ntotal RT with P S X P S total, where is the mole fraction of S, can earn the points. XS (e) K c 1 K c 1 0.50.00 pts. Notes: One point is earned f recognizing that K c must be an inverse related to K c. A second point is earned f recognizing that the square root of K c is involved. Only one point can be earned f the expression below, f crectly calculating just the inverse of K c just K c. K c [H S] [H ][S 1 ] Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question (10 points) (a) (i) Zn(s) Zn + (aq) + e (ii) Co + (aq) + Zn(s) Co(s) + Zn + (aq) (iii) 0.76 V + ( 0.8 V) 0.48 V Note: phase designations not required in part (i) part (ii) (b) (i) G nfe (96,500)(0.55V) pts. 1.1 10 5 J 1.1 10 kj First point earned f n (consistent use of n 4 also accepted) Second point earned f negative sign, crect number ( ± 1 sig. figs.), and appropriate units (kj J kj/mole J/mole) (ii) G RT ln(k) 1.1 10 5 J [8.31 J mol K ][98 K][ln (K)] K.0 10 19 (full credit also f crect use of ne log K ) 0.059 (iii) O + H O H O 0.55 V O + 4 H + + 4 e H O 1.3 V O + 4 H + + 4 e H O 0.68 V pts. O + H + + e H O 0.68 V (not required) Two points earned f crect voltage with suppting numbers (chemical equations not necessary) One point earned f crect chemical equations with increct voltage 100 C 1 mol e 1 mol Cu 63.55 g Cu ( c) 3,600 sec 1 sec 96,500 C mol e 1 mol Cu 119 g Cu pts. Two points earned f crect answer (3 ± 1 sig. figs.) One point earned f any two of these steps: (amp)(sec) coulombs coulombs mol e mol e mol Cu mol Cu g Cu Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 3 (10 points) (a) Molar Mass [9.01 + (1.011) + 4(16.00) + 3(16.00 + (1.0079))] g mol (97.034 + 54.0474) g mol 151.0814 g mol (1.011) % Carbon 100 15.90% 151.0814 No point earned f 0.1590. 3.1g BeCO4 3 HO (b) (i) 0.014 mol BeCO4 3 HO 151.0814 g mol 1mol BeC 4 0.014 mol BeC O 4 3 H O O 0.014 mol BeC O 4 1mol BeC O 3 H O 4 0.014 mol BeC O 4 97.034 g mol.06 g BeC O 4 97.034 g mol % BeC O 4 100% 64.3% 151.0814 g mol Mass BeC O 4 3.1g 0.643.06 g No point earned f an answer larger than iginal mass (3.1g) (ii) 0.014 mol BeC O 4 3 H O 3 mol HO 1mol BeC O 3H O 4 0.0637 mol H O V nrt P (0. 063mol) (0.081Latm mol K ) ((0 + 73) K).67 L HO (735/ 760) atm 3.1 g BeC O 4 3 H O.06 g BeC O 4 1.15 g H O pts. 1.15 g H 18. 0 g mol O HO 0.0639 mol H O use ideal gas law as above Note: One point is earned f determining the amount of H O(g) fmed and one point is earned f the proper use of the ideal gas law with a consistent answer. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 3 (continued) (c) (i) The reducing agent is C O 4 Also accepted: C O 4, BeC O 4, C O 4, 5 C O 4 (ii) n n MnO C O 4 4 (0.0150 mol L 5 n MnO 4 ) (0.01780 5 (.67 10 4 ) L).67 10 6.68 10 4 4 mol MnO mol C O 4 4 Notes: One point is earned f determining determining n MnO 4 n. An increct value of C O 4 MnO 4 and one point is earned f n can lead to an increct, but internally consistent, value f n. A point may be earned if n is C O 4 MnO 4 crectly calculated (i.e., reflects crect stoichiometry of the balanced equation) from an increct value of n. C O 4 (iii) Total n 100. ml n 0.0 ml CO4 MnO4 5 (6.68 10 4 mol C O 4 ) 3.34 10 3 mol C O 4 (iv) mol BeC O 4 mol C O 4 3.34 10 3 mol C O 4 mass BeC O 4 (3.34 10 3 mol BeC O 4 )(97.034 g mol ) 0.3385 g 0.3385 g BeC O4 % BeC O 4 100% 93.9% 0.345 g sample Note: The point is earned if an increct value f moles of BeC O 4 is used crectly in the calculation. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 4 (15 points) Students choose five of the eight reactions. Only the answers in the boxes are graded (unless clearly marked otherwise). Each crect answer earns 3 points; 1 point f reactants and points f products. Reactants must be completely crect to earn the reactant point. All products must be crect to earn both product points; one of the two products crect earns 1 product point. Equations need not be balanced n phases indicated (and no penalty if present but increct). Any spectat ions on the reactant side nullify the 1 possible reactant point, but if they appear again on the product side, there is no product-point penalty. Ion charges must be crect. (a) Ca + H O Ca(OH) + H ( Ca + + OH + H ) 1 product point f Ca + + H, Ca + + OH CaO + H (b) C 4 H 9 OH + O CO + H O C 4 H 10 O structural fmula also accepted f butanol No penalty f CO + CO + H O as products 1 product point f CO + H O (c) NH 3 + Ni + [Ni(NH 3 ) x ] +, where x any integer from 1 through 6 F full credit, charges must be crect (square brackets not required) 1 product point earned if only product err is charge (d) Cu + + S CuS 1 product point earned if reactants crect but response shows increct stoichiometry f a copper sulfide product (e.g., Cu S, CuS, Cu 5 S 9 ) product points earned if reactant charges increct but the product is consistent (e) Sn + + Ag + Sn 4+ + Ag To earn product point, Sn must be in 4+ state Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 4 (continued) (f) H + + HCO 3 H O + CO H 3 O + + HCO 3 H O + CO If the only product is H CO 3, one product point is earned HBr written as undissociated and Br included as a product loses the reactant point only, but if HBr as a reactant produces Br as a product, a product point is lost (g) SrO + H O Sr + + OH ( Sr(OH) ) Only 1 product point f Sr(OH) + H (extra product) (h) CO + Fe O 3 CO + Fe ( CO + FeO ) Iron products that contain reduced iron and oxygen are also accepted Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 5 (10 points) (a) 0 Pure Solvent 0 Solution pts. Temperature ( o C) 15 10 5 Temperature ( o C) 15 10 5 0 Time 0 Time Notes: One point is earned f each crect graph. The first graph should show a line that drops to 10 C, holds steady at 10 C, and then falls steadily to 0 C. There must be a discernable plateau at 10 C to earn this point. The second graph should show a line that drops to below 10 C, levels off ( slants down a bit), and then falls me sharply to 0 C. (b) (i) Measure mass of solute, mass of solvent, mass of solution (two of three must be shown) Measure the T fp ( the freezing point of the solution) Volume of solution (without density), molality, number of moles do not earn points (ii) Given: T ik f m ( T K f m) pts. m (mol solute)/(kg solvent) moles g/(molar mass) Combine to get: molar mass (i)(k f )(g solute)/( T)(kg solvent) Notes: One point is earned f any two equations, and two points are earned f all three equations. Solute and solvent must be clearly identified in the equations. (iii) the difference in the vertical position of the hizontal ptions of the graphs is equal to T fp, the change in freezing point due to the addition of the solute. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 5 (continued) (c) The molar mass is too small. If some of the solvent evapates, then the (kg solvent) term used in the equation in (b) (ii) is larger than the actual value. If the (kg solvent) term used is too large, then the value calculated f the molar mass will be too small. If some of the solvent evapates, then the concentration (molality) of the solute will be greater than we think it is. Me moles of solute results in a smaller molar mass ( since T ik f m, then the T obs would be greater than it should be). Since the molar mass of the unknown solute is inversely proptional to T, an erroneously high value f T implies an erroneously low value f the molar mass (calculated molar mass would be too small). (d) % err % err (16 g mol 6 g mol 1 10 g mol 10 g mol 0 g mol 100% ) 100% Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 6 (10 points) (a) H 33 ( 90. + 143 ) kj 00 kj Crect set up (wk), a numerical result, and a sign, earn this point No math err point deducted f computational mistakes (b) S f this reaction should be small negligible (near zero zero) No point earned f statements that S is constant the same because the number of moles (of gas) is the same on each side of the equation. Point may be earned even if first point not earned (c) G is negative ( ) at 98 K Point earned only if consistent with answers given in parts (a) and (b) because G H T( S ), H is negative, and S is near zero, thus it follows that G will be negative. Equation need not be written explicitly to earn point Notes: Responses that G is positive (+), not able to be determined, are accepted together with appropriate explanations IF they are fully consistent with increct responses given in part (a) and/ part (b). Since question concerns the reaction at 98 K, discussion of what might happen at high low temperatures does not earn any explanation point. (d) When [O 3 ] is held constant and [NO] is doubled (as in Experiments 1 and ), the rate also doubles reaction is first-der in [NO]. When [NO] is held constant and [O 3 ] is doubled (as in Experiments 1 and 3), the rate also doubles reaction is first-der in [O 3 ]. Rate k [O 3 ] [NO] ( Rate k [O 3 ] 1 [NO] 1, Rate 10 6 (x) [O 3 ] [NO] ) Notes: First point f crectly determining the der with respect to [NO] and wk shown, second point earned f crectly determining the der with respect to [O 3 ] and wk shown, and third point earned f crectly writing a rate-law expression that is consistent with answers to analysis of experimental results. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 6 (continued) (e) Step 1 must be rate-determining, as the rate law f this elementary step is pts. the only one that agrees with the rate law determined experimentally in part (d) above. The following arguments are acceptable, but not required: If Step were the rate-determining step, then the reaction would be second-der with respect to [O 3 ] this is inconsistent with the kinetic data. If Step 3 were the rate-determining step, then the reaction would be second-der with respect to [NO] this is inconsistent with the kinetic data. Notes: Two points are earned f identifying Step 1 as the rate-determining step and giving a valid explanation relating the similarity of the rate law of the slow step to that found from the experimental results. Only one point is earned f identifying Step 1 as the rate-determining step but only a minimal partly increct explanation is given (e.g., an explanation based solely on the fact that both reactant molecules are only present in Step 1). Two points may be earned f identifying Step Step 3 as the rate-determining step IF the rate law determined in part (d) contains [O 3 ] [NO], respectively. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 7 (8 points) (a) The isotopes have the same number (34) of protons, but a different number of neutrons. No comment about the number of electrons is necessary (b) 1s s p 6 3s 3p 6 4s 3d 10 4p 4 1s s p 6 3s 3p 6 3d 10 4s 4p 4 No point is earned f [Ar] 4s 3d 10 4p 4, because the question specifically asks f a complete electron configuration. Since there are three different 4p bitals, there must be two unpaired electrons. Notes: The second part should have some explanation of Hund s rule, and may include a diagram. The second point can still be earned even if the first point is not IF the electron configuration is increct, but the answer f the second part is consistent with the electron configuration given in the first part. (c) (i) The ionized electrons in both Se and Br are in the same energy level, but Br has me protons than Se, so the attraction to the nucleus is greater. Note: There should be two arguments in an acceptable answer -- the electrons removed are from the same (4p) bital and Br has me protons (a greater nuclear charge) than Se. (ii) The electron removed from a Te atom is in a 5p bital, while the electron removed from an Se atom is in a 4p bital. The 5p bital is at a higher energy than the 4p bital, thus the removal of an electron in a 5p bital requires less energy. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 7 (continued) (d) F F F Se F F Se F F F Notes: One point earned f a crect Lewis diagram and a sketch. The Lewis diagram and the molecular structure may be combined into one sketch if both aspects (electron pairs and structure) are crect. Dots, lines, a mixture of both can be used in the Lewis diagram. The lone pair of electrons need not be shown in the sketch -- just the atomic positions. No credit earned f just a verbal description of molecular geometry ( see-saw, saw-hse, something distted ), because the question clearly asks the student to sketch the molecular structure. The SeF 4 molecule is polar, because the polarities induced by the bonds and the lone pair of electrons do not cancel. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 8 (8 points) (a) NH 3 (aq) + H + (aq) NH + 4 (aq) NH 3 (aq) + H 3 O + (aq) NH + 4 (aq) + H O(l) Note: phase designations not required to earn point (b) Sketch of Titration Curve: 3 pts. 1.00 11.00 10.00 9.00 8.00 ph 7.00 6.00 5.00 4.00 3.00.00 1.00 0.00 0.00 5.00 10.00 15.00 0.00 5.00 30.00 35.00 40.00 Volume of 0.0 M HCl Added 1 st pt. initial ph must be > 7 (calculated ph 11) nd pt. equivalence point occurs at 15.0 ml ± 1 ml of HCl added (equivalence point must be detectable from the shape of the curve a mark on the curve) 3 rd pt. ph at equivalence point must be < 7 (calculated ph 5). Note: a maximum of 1 point earned f any of the following: - a line without an equivalence point - a random line that goes from high ph to low ph - an upward line with increasing ph (equivalence point MUST be at 15.0 ml) Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP Chemistry 000 Scing Standards Question 8 (continued) (c) Methyl Red would be the best choice of indicat, because the pk a f Methyl Red is closest to the ph at the equivalence point. Notes: explanation must agree with equivalence point on graph alternative explanation that titration involves strong acid and weak base (with product an acidic salt) earns the point d) The resulting solution is basic. K b f NH 3 (1.8 10 5 ) and K a f NH + 4 (5.6 10 0 ) indicate that NH 3 is a stronger base than NH + 4 is an acid [OH ] K b 1.8 10 5 because of the equimolar and equivolume amounts of ammonium and ammonia cancellation in the buffer ph calculation. Thus 0.05 poh 5 and ph 9 (i.e., recognition of buffer, so that log 0 0.05 poh pk b 5 ph 14 poh 9) Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved.