# AP Chemistry 1999 Scoring Guidelines

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1 AP Chemistry 1999 Scoring Guidelines The materials included in these files are intended for non-commercial use by AP teachers for course and exam preparation; permission for any other use must be sought from the Advanced Placement Program. Teachers may reproduce them, in whole or in part, in limited quantities, for face-to-face teaching purposes but may not mass distribute the materials, electronically or 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 for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, 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 opportunity. ounded in 1900, the association is composed of more than 3,900 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents,,000 high schools, and 3,500 colleges, through major 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.

2 9 Points AP CHEMISTRY Question 1 ne point deduction for mathematical error (maximum once per question) ne point deduction for error in significant figures* (maximum once per question) *number of significant figures must be correct within +/- one digit. (except for ph: +/- two digits) (a) + [NH 4 ][H = [NH ] 3 ] K ph = 3.5 (b) [H ] = { or } ph = [H + ] = ( ) 5 5 (c) K b = = (or ) pts Note: 1 st point for [NH 4 + ] = [H ] = ; nd point for correct answer (d) % ionization = 100 % = 3.11% (or ) (e) NH 3 + H + NH 4 + (i) mol NH 3 = M L = mol = mol H + needed vol HCl solution = mol = L = 30.0 ml M (ii) mol H + added = mol M L = mol H + added = mol NH + 4 produced [NH mol 4 ] = = M = [NH 3 ] L Note: Point earned for mol, or M [NH 3 ] or [NH + 4 ], or statement halfway to equivalence point.

3 Question 1 (cont.) + [NH ][H [NH ] ] K b = = = [H ] ph = ph = (= ) (= 4.759) (= 9.41) (iii) 10.0 ml past equivalence point L M = mol H + in 60.0 ml mol [ H ] = = M L ph = log ( ) =.700 ne point deduction for mathematical error (maximum once per question) ne point deduction for error in significant figures* (maximum once per question) *number of significant figures must be correct within +/ one digit (except for ph: +/ two digits) 3

4 Question 9 points ne point deduction for mathematical error (maximum once per question) ne point deduction for error in significant figures* (maximum once per question) *number of significant figures must be correct within +/- one digit. (a) (i) nm/sec m/sec 14 1 (, = ) = sec 9 c ν = = or λ 495 nm m (ii) E = hν = ( J sec)( sec 1 ) = J (iii) ( J)( mol 1 )( kj/j) = 4 kj/mol Note: No units required if answers are numerically same as above. No penalty if answers are correct with different units and units are explicitly indicated (e.g., for part (ii), kj is acceptable) (b) (i) Energy is emitted. The n = 6 state is at a higher energy than the n = state. Going from a high energy state to a low energy state means that energy must be emitted. Note: The key idea is that the energy of the n = 6 state is higher (more excited) than the lower (less excited) n = state. The argument that e is farther away at the n = 6 level is not accepted. (ii) E J 0 6 = J 19 = = J, E E = E 6 E = J ( = 6 19 J) = J J R, E = ( ) J = (0.) J = J 6 Note: Point earned for determining the energy of transition. Negative energies acceptable. 4

5 Question (cont.) E = hc λ λ = hc E R, = h E ν J 14 1 = sec 34 = Jsec Note: Point earned for writing or using hc E =, or for calculating the frequency, ν λ 8 1 c m sec 1 nm λ = = ν sec 10 m = 411 nm R, ( J sec)( nm sec ) λ = = 411 nm J Note: Point earned for correct wavelength; λ = 4.11 x 10 7 m accepted. Negative wavelength not accepted. (iii) The positive charge holding the electron is greater for He +, which has a + nucleus, than for H with its 1+ nucleus. The stronger attraction means that it requires more energy for the electron to move to higher energy levels. Therefore, transitions from high energy states to lower states will be more energetic for He + than for H. Note: ther arguments accepted, such as, E is proportional to Z. Since Z = for He + and Z = 1 for H, all energy levels in He + are raised (by a factor of 4). ther accepted answers must refer to the increased charge on the He + nucleus, and NT the mass. ne point deduction for mathematical error (maximum once per question) ne point deduction for error in significant figures* (maximum once per question) *number of significant figures must be correct within +/ one digit 5

6 9 points: AP CHEMISTRY Question 3 ne point deduction for mathematical error (maximum once per question) ne point deduction for error in significant figures* (maximum once per question) *number of significant figures must be correct within +/- one digit. (a) (b) Rate of Br (g) loss occurs at ½ the rate of NBr(g) formation. Rate of Br (g) loss = sec M = M sec 1 (or mol L 1 sec 1 ) Note: No penalty for missing units; ignore + or signs Comparing experiments 1 and, [N] remains constant, [Br ] doubles, and rate doubles; therefore, rate [Br ] 1 reaction is first-order with respect to [Br ] x 1 = x k[n] [Br ] = x k[n] [Br ] k[0.0160] k[0.030] x x [0.040] [0.0060] x 1 = 4 = = x = reaction is second-order with respect to [N] 4 pts. Note: ne point earned for a proper set-up, comparing experiments and 3 (as is shown here) or experiments 1 and 3. Second point earned for solving the ratios correctly and determining that the exponent =. Also, credit can be earned for a non-mathematical approach (e.g., one point for describing the change in [Br ] and subsequent effect on rate, another point for describing the change in [N] and subsequent effect on rate). (c) (i) Rate = k[n] [Br ] Note: Point earned for an expression that is not inconsistent with the answer in part (b) (ii) k = 1 Rate M sec = 3 [N] [Br ] (0.0160) (0.010) M = 105 M sec 1 (or 105 L mol sec 1 ) pts. (Using rate of Br (g) loss = M sec 1 k = 5.7 M sec 1 is also correct.) 6

7 Question 3 (cont.) Note: ne point for solving for the value of the rate constant consistent with the (c). rate-law expression found in part (b) or stated in part (c); one point for the correct units consistent with the rate-law expression found in part (b) or stated in part (d) No, it is not consistent with the given experimental observations. This mechanism gives a reaction that is first-order in [N], and first-order in [Br ], as those are the two reactants in the rate-determining step. Kinetic data show the reaction is second-order in [N] (and first-order in [Br ]), so this cannot be the mechanism. Note: ne point earned for No [or for Yes if rate = k[n][br ] in part (b)]. ne point earned for justifying why this mechanism is inconsistent with the observed rate-law [or consistent with rate law stated earlier in response]. ne point deduction for mathematical error (maximum once per question) ne point deduction for error in significant figures* (maximum once per question) *number of significant figures must be correct within +/ one digit 7

8 Question 4 15 points: Students choose five of the eight reactions. nly the answers in the boxes are graded (unless clearly marked otherwise). Each correct answer earns 3 points, 1 point for reactants and points for products. All products must be correct to earn both product points. Equations do not need to be balanced and phases need not be indicated. Any spectator 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. A fully molecular equation (when it should be ionic) earns a maximum of one point. Ion charges must be correct. (a) Ca + H Ca(H) No penalty for the set of products { Ca +, H, and Ca(H) } (b) NH 4 N 3 N + + H R NH 4 N 3 N + H Two points earned for NH 4 N N + H No penalty for other oxides of nitrogen (e.g., N, N, N 3, N 4 but not N 5 ) (c) Br + I Br + I (d) PbC 3 + H + + HS 4 (or S 4 ) PbS 4 + C + H (or HC 3 ) No reactant point earned for H S 4 No product point earned for H C 3 (e) e 3 + Al Al 3 + e No penalty for the set of products { e, e, and Al 3 } (f) CH 3 NH + H CH 3 NH H Two points earned for MeNH + H MeNH H (g) C + Na Na C 3 (h) Ba + + Cr 4 BaCr 4 8

9 8 Points AP CHEMISTRY Question 5 (a) PV = nrt AND m n =, R molar mass = M mrt DRT, R M = PV P (b) temperature, atmospheric pressure, volume of the gas, and mass of gas (mass of pressurized container before and after releasing the gas) 3 pts Note: 1 point earned for any two of the above, points earned for any three of them. The mass of the gas is acceptable as a measurement for the 1 st or nd point. Extraneous measurements (e.g., density, volume of liquid, etc.) are ignored. To earn 3 rd point, mass of pressurized container before and after releasing the gas, or change in mass of container must be indicated. (c) to equalize internal pressure with room pressure (atmospheric pressure), or the pressure(s) will be the same. (64 58) g (d) % error = 100 % 58 g (or %, or 6 58 ) Note: No points earned for generic response (e.g., (expt. theor.) 100 ), theor. or for %. No penalty if 100% is absent or if value (10%) 64 is not calculated. (e) Pressure will be larger, therefore number of moles will be larger mass molar mass =, therefore calculated molar mass will be smaller moles R, mrt M = (or PV DRT M = ), and the denominator, PV, will be too large. pts P Therefore, the value of the molar mass ( = mrt or PV DRT ) will be too small. P R, The pressure is larger, or the number of moles is larger, or since P total = (P unknown P water ) we know that P total > P unknown. Note: If 8 points: m n = is missing in part (a) but present in part (e), 1 point is earned for part (a). M 9

10 Question 5 (cont.) (a) (i) S is negative ( ) R S < 0 R entropy is decreasing. 3 moles of gaseous particles are converted to moles of solid/liquid. ne point earned for correct identification of ( ) sign of S ne point earned for correct explanation (mention of phases is crucial for point) No point earned if incorrect S sign is obtained from the presumed value of G (ii) H drives the reaction. The decrease in entropy ( S < 0) cannot drive the reaction, so the decrease in enthalpy ( H < 0) MUST drive the reaction. R G = H T S ; for a spontaneous reaction G < 0, and a negative value of S positive G. causes a ne point earned for identifying H as the principal driving force for the reaction ne point earned for correct justification Justification point earned by mentioning the effects of changes in entropy and enthalpy on the spontaneity of the reaction R by a mathematical argument using the Gibbs-Helmholtz equation and some implication about the comparison between the effects of S and H (iii) Given that G = H T S and S < 0, an increase in temperature causes an increase in the value of G ( G becomes less negative). ne point earned for the description of the effect of an increase in temperature on S and consequently on G No point earned for an argument based on Le Châtelier s principle (b) (i)the reaction rate depends on the reaction kinetics, which is determined by the value of the activation energy, E act. If the activation energy is large, a reaction that is thermodynamically spontaneous may proceed very slowly (if at all). ne point earned for linking the rate of the reaction to the activation energy, which may be explained verbally or shown on a reaction profile diagram (ii) The catalyst has no effect on the value of G. The catalyst reduces the value of E act, increasing the rate of reaction, but has no effect on H and S, so it cannot affect the thermodynamics of the reaction. the values of ne point earned for indicating no change in the value of G ne point earned for indicating (verbally, or with a reaction-profile diagram) that the catalyst affects the activation energy 10

11 8 points: AP CHEMISTRY Question 7 a. 1 point C H 5 H (lask #3) 1 point Ethanol, a nonelectrolyte, does not break up or dissociate in solution. ne point earned for identifying C H 5 H ne point earned for the correct explanation. Explanation point earned for a description of a nonelectrolyte (e.g., something that does not break up or does not dissociate.) No point earned for describing C H 5 H as organic, or as the compound that contains the most hydrogens. b. 1 point MgCl (lask #) 1 point The freezing-point depression is proportional to the concentration of solute particles. All solutes are at the same concentration, but the van't Hoff factor (i) is largest for MgCl. ne point earned for identifying MgCl. ne point earned for the correct explanation. c. 1 point 11

12 Question 7 (cont.) C H 5 H (lask #3) 1 point The lowering of vapor pressure of water is directly proportional to the concentration of solute particles in solution. C H 5 H is the only nonelectrolyte, so it will have the fewest solute particles in solution. ne point earned for identifying C H 5 H. ne point earned for the correct explanation. d. 1 point Na (lask #1) 1 point The ion, generated upon dissolution of Na, is a weak base. It is the only solution with ph > 7. ne point earned for identifying Na. ne point earned for the correct explanation. 1

13 Question 8 8 points (a) (i) - pts C C ne point earned for each Lewis electron-dot structure Indication of lone pairs of electrons are required on each structure Resonance forms of C 3 are not required (ii) In C, the C interactions are double bonds, R, in C 3 the C interactions are resonance forms (or figures below.) C C C The carbon-oxygen bond length is greater in the resonance forms than in the double bonds. 1 st point earned for indicating double bonds are present in C R resonance occurs in C 3 nd point earned for BTH of the above AND indicating the relative lengths of the bond types (b) (i) pts C S.. ne point earned for each Lewis electron-dot structure Lone pairs of electrons are required on each structure 13

14 Question 8 (cont.) (ii) C 4 has a tetrahedral geometry, so the bond dipoles cancel, leading to a nonpolar molecule. With five pairs of electrons around the central S atom, S 4 exhibits a trigonal bipyramidal electronic geometry, with the lone pair of electrons. In this configuration, the bond dipoles do not cancel, and the molecule is polar. ne point earned for each molecule for proper geometry and explanation C. S 14

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