AP* Bonding & Molecular Structure Free Response Questions page 1

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1 AP* Bonding & Molecular Structure Free Response Questions page 1 Essay Questions 1991 a) two points ΔS will be negative. The system becomes more ordered as two gases orm a solid. b) two points ΔH must be negative. For the reaction to be spontaneous, ΔG must be negative, so ΔH must be more negative than -TΔS is positive. c) two points As T increases, -TΔS increases. Since ΔS is negative, the positive -TΔS term will eventually exceed ΔH (which is negative), making ΔG positive. (In the absence o this, ΔG = ΔH - TΔS and general discussion o the eect o T and ΔS gets 1 point.) d) two points The equilibrium constant is 1. The system is at equilibrium at this temperature with an equal tendancy to go in either direction. OR ΔG = 0 at equilibrium so K = 1 in ΔG = -RT ln K (In the absence o these, ΔG = -RT ln K gets 1 point). The above concludes the AP scoring standards published in The ollowing is simply alternate ways o answering which the AP readers may or may not have given ull credit to. a) The amount o entropy goes down, ΔS is negative. b) ΔG = ΔH - TΔS. I ΔS is negative, then ΔH must also be negative to get a negative ΔG. c) Let us say ΔG is positive when ΔH is positive and ΔS is positive. As T goes up - TΔS becomes more negative until it makes ΔG (which equals ΔH - TΔS) become negative. d) At the temperature when the direction changes, the rate orward = the rate reverse. Since K = k / kr, this equals 1. (1) AP is a registered trademark o the College Board. The College Board was not involved in the production o and does not endorse this product. (2) Test Questions are Copyright by College Entrance Examination Board, Princeton, NJ. All rights reserved. For ace-to-ace teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited. Page 1

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6 AP CHEMISTRY 1999 SCORING GUIDELINES Question 5 (cont.) (a) (i) S is negative ( ) OR S < 0 OR entropy is decreasing. 1 pt 3 moles o gaseous particles are converted to 2 moles o solid/liquid. 1 pt One point earned or correct identiication o ( ) sign o S One point earned or correct explanation (mention o phases is crucial or point) No point earned i incorrect S sign is obtained rom the presumed value o G (ii) H drives the reaction. 1 pt The decrease in entropy ( S < 0) cannot drive the reaction, so the decrease in enthalpy ( H < 0) MUST drive the reaction. OR 1 pt G = H T S ; or a spontaneous reaction G < 0, and a negative value o S positive G. causes a One point earned or identiying H as the principal driving orce or the reaction One point earned or correct justiication Justiication point earned by mentioning the eects o changes in entropy and enthalpy on the spontaneity o the reaction OR by a mathematical argument using the Gibbs-Helmholtz equation and some implication about the comparison between the eects o S and H (iii) Given that G = H T S and S < 0, an increase in temperature causes an increase in the value o G ( G becomes less negative). 1 pt One point earned or the description o the eect o an increase in temperature on S and consequently on G No point earned or an argument based on Le Châtelier s principle (b) (i)the reaction rate depends on the reaction kinetics, which is determined by the value o the activation energy, E act. I the activation energy is large, a reaction that is thermodynamically spontaneous may proceed very slowly (i at all). 1 pt One point earned or linking the rate o the reaction to the activation energy, which may be explained verbally or shown on a reaction proile diagram (ii) The catalyst has no eect on the value o G. 1 pt The catalyst reduces the value o E act, increasing the rate o reaction, but has no eect on H and S, so it cannot aect the thermodynamics o the reaction. the values o 1 pt One point earned or indicating no change in the value o G One point earned or indicating (verbally, or with a reaction-proile diagram) that the catalyst aects the activation energy Copyright 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo are registered trademarks o the College Entrance Examination Board. 10 Page 6

7 AP Chemistry 2000 Scoring Standards Question 6 (10 points) (a) H = 33 ( ) kj = 200 kj 1 pt. Correct set up (work), a numerical result, and a sign, earn this point No math error point deducted or computational mistakes (b) S or this reaction should be small or negligible (near zero or zero) 1 pt. No point earned or statements that S is constant or the same because the number o moles (o gas) is the same on each side o the equation. Point may be earned even i irst point not earned 1 pt. (c) G is negative ( ) at 298 K 1 pt. Point earned only i consistent with answers given in parts (a) and (b) because G = H T( S ), H is negative, and S is near zero, thus it ollows that G will be negative. Equation need not be written explicitly to earn point 1 pt. Notes: Responses that G is positive (+), or not able to be determined, are accepted together with appropriate explanations IF they are ully consistent with incorrect responses given in part (a) and/or part (b). Since question concerns the reaction at 298 K, discussion o what might happen at high or low temperatures does not earn any explanation point. (d) When [O 3 ] is held constant and [NO] is doubled (as in Experiments 1 and 2), 1 pt. the rate also doubles reaction is irst-order in [NO]. When [NO] is held constant and [O 3 ] is doubled (as in Experiments 1 and 3), the rate also doubles reaction is irst-order in [O 3 ]. Rate = k [O 3 ] [NO] (or Rate = k [O 3 ] 1 [NO] 1, or Rate = 10 6 (x) [O 3 ] [NO] ) 1 pt. 1 pt. Notes: First point or correctly determining the order with respect to [NO] and work shown, second point earned or correctly determining the order with respect to [O 3 ] and work shown, and third point earned or correctly writing a rate-law expression that is consistent with answers to analysis o experimental results. Copyright 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark o the College Entrance Examination Board. Page 7

8 AP Chemistry 2000 Scoring Standards Question 6 (continued) (e) Step 1 must be rate-determining, as the rate law or this elementary step is 2 pts. the only one that agrees with the rate law determined experimentally in part (d) above. The ollowing arguments are acceptable, but not required: I Step 2 were the rate-determining step, then the reaction would be second-order with respect to [O 3 ] this is inconsistent with the kinetic data. I Step 3 were the rate-determining step, then the reaction would be second-order with respect to [NO] this is inconsistent with the kinetic data. Notes: Two points are earned or identiying Step 1 as the rate-determining step and giving a valid explanation relating the similarity o the rate law o the slow step to that ound rom the experimental results. Only one point is earned or identiying Step 1 as the rate-determining step but only a minimal or partly incorrect explanation is given (e.g., an explanation based solely on the act that both reactant molecules are only present in Step 1). Two points may be earned or identiying Step 2 or Step 3 as the rate-determining step IF the rate law determined in part (d) contains [O 3 ] 2 or [NO] 2, respectively. Copyright 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark o the College Entrance Examination Board. Page 8

9 AP CHEMISTRY 2003 SCORING GUIDELINES Question 7 7. Answer the ollowing questions that relate to the chemistry o nitrogen. (a) Two nitrogen atoms combine to orm a nitrogen molecule, as represented by the ollowing equation. 2 N(g) N 2 (g) Using the table o average bond energies below, determine the enthalpy change, H, or the reaction. Bond N N N == N N N Average Bond Energy (kj mol 1 ) H = 950 kj The reaction is exothermic because the chemical equation shows the ormation o the N N bond. 1 point or correct sign 1 point or magnitude (b) The reaction between nitrogen and hydrogen to orm ammonia is represented below. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H = 92.2 kj Predict the sign o the standard entropy change, S, or the reaction. Justiy your answer. S is negative. There are ewer moles o product gas (2 mol) compared to reactant gases (4 mol), so the reaction is becoming more ordered. 1 point or correct sign 1 point or indicating ewer moles o products compared to reactants (in the gas phase) (c) The value o G or the reaction represented in part (b) is negative at low temperatures but positive at high temperatures. Explain. G = H T S H and S are negative. At low temperatures, the T S term is smaller than H, and G is negative. At high temperatures, the T S term is higher than H, and G is positive. 1 point each or using G = H T S to explain the sign o G at high and low temperatures. Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 18 Page 9

10 AP CHEMISTRY 2003 SCORING GUIDELINES Question 7 (cont d.) (d) When N 2 (g) and H 2 (g) are placed in a sealed container at a low temperature, no measurable amount o NH 3 is produced. Explain. Even though the reaction is spontaneous at low temperature, the reaction is very slow. The speed o a reaction depends on the raction o colliding molecules with energy that exceeds the activation energy or the reaction. At low temperature, ew reactant particles collide with an energy greater than the activation energy. 1 point or indicating that the requency o collision (or kinetic energy) o molecules is low at low temperature (thus the rate is slow) 1 point or indicating that at low temperature the kinetic energy will likely be too small to exceed the activation energy Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 19 Page 10

11 AP CHEMISTRY 2004 SCORING GUIDELINES (Form B) Question 7 N 2 (g) + 2 H 2 (g) o N 2 H 4 (g) H 298 = kj mol 1 o ; S 298 = 176 J K 1 mol 1 7. Answer the ollowing questions about the reaction represented above using principles o thermodynamics. (a) On the basis o the thermodynamic data given above, compare the sum o the bond strengths o the reactants to the sum o the bond strengths o the product. Justiy your answer. Bond energy (B.E.) o reactants is greater than bond energy o products. Reaction is endothermic, so more energy is required to break bonds o reactants than is given o when new bonds orm in products: H = (B.E.) reactants (B.E.) products > 0 1 point or indicating that reactants have greater bond strength 1 point or correct explanation (b) Does the entropy change o the reaction avor the reactants or the product? Justiy your answer. Entropy change avors reactants. Since there are three moles o reactants in gas phase compared to only one mole o products, there are more possible arrangements o reactant molecules compared to product molecules. 1 point or indicating which, reactants or products, are avored 1 point or explanation (c) For the reaction under the conditions speciied, which is avored, the reactants or the product? Justiy your answer. Reactants are avored because G or reaction is positive. G = H T S, so a positive H and a negative S means G is always positive, independent o temperature. Note: Calculation o G is acceptable with explanation. 1 point or indicating which, reactants or products, are avored 1 point or explanation Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (or AP proessionals) and (or AP students and parents). 14 Page 11

12 AP CHEMISTRY 2004 SCORING GUIDELINES (Form B) Question 7 (cont d.) (d) Explain how to determine the value o the equilibrium constant, K eq, or the reaction. (Do not do any calculations.) Solve ormula G = RT lnk eq or K eq and plug in value o G calculated in part (c), value o temperature (298 K), and value o R (8.31 J mol 1 K 1 ). 1 point or correct mathematical equation and substitution (e) Predict whether the value o K eq or the reaction is greater than 1, equal to 1, or less than 1. Justiy your answer. K eq value is less than 1 or the reaction as written. G = RT lnk eq, and since G is positive, lnk eq will be a negative number which means that K eq is less than one. OR 1 point or the correct prediction with an explanation H > 0 and S < 0, thus G > 0, which means that K eq < 1. Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (or AP proessionals) and (or AP students and parents). 15 Page 12

13 AP* Bonding & Molecular Structure Free Response Questions page 1 Problems (1) AP is a registered trademark o the College Board. The College Board was not involved in the production o and does not endorse this product. (2) Test Questions are Copyright by College Entrance Examination Board, Princeton, NJ. All rights reserved. For ace-to-ace teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited. Page 13

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18 AP CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 3 Total Score 10 points 3. In an experiment, a sample o an unknown, pure gaseous hydrocarbon was analyzed. Results showed that the sample contained g o carbon and g o hydrogen. (a) Determine the empirical ormula o the hydrocarbon. n C = g C 1 mol C = mol C g C n H = g H 1 mol H = mol H g H mol C mol H : mol C : mol H 3 (1 mol C : mol H) = 3 mol C : mol H The empirical ormula is C 3 H 8 1 point or number o moles o carbon and number o moles o hydrogen 1 point or ratio o moles o carbon to moles o hydrogen 1 point or correct ormula (b) The density o the hydrocarbon at 25 C and 1.09 atm is 1.96 g L 1. (i) Calculate the molar mass o the hydrocarbon. PV = nrt grams PV = molar mass RT molar mass = grams RT V P L atm mol K 298 K molar mass = 1.96 g L atm molar mass = 44.0 g mol 1 1 point or correct substitution and 1 point or answer OR 1 point or calculation and 1 point or units 1.96 g L L mol 1 = 43.9 g mol 1 (1 point maximum) (ii) Determine the molecular ormula o the hydrocarbon. Empirical mass n = molar mass Empirical mass or C 3 H 8 is 44 g mol 1 44 g mol 1 n = 44 g mol 1 n = 1, so the molecular ormula is the same as the empirical ormula, C 3 H 8 1 point or reporting correct ormula with veriication Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 8 Page 18

19 AP CHEMISTRY 2003 SCORING GUIDELINES (Form B) Question 3 (cont d.) In another experiment, liquid heptane, C 7 H 16 (l), is completely combusted to produce CO 2 (g) and H 2 O(l), as represented by the ollowing equation. C 7 H 16 (l) + 11 O 2 (g) 7 CO 2 (g) + 8 H 2 O(l) comb The heat o combustion, H, or one mole o C 7 H 16 (l) is kj. (c) Using the inormation in the table below, calculate the value o H or C 7 H 16 (l) in kj mol 1. Compound H (kj mol 1 ) CO 2 (g) H 2 O(l) H = H (products) H (reactants) = 7 H (CO 2 ) + 8 H (H 2 O) [ H (C 7 H 16 ) + 11 H (O 2 )] kj kj = 7( mol mol )+ 8( mol ) [ H (C 7 H 16 ) + 11 (0 kj mol ) ] 4,850 kj 4,850 H kj mol = 2,754 kj mol (C 7 H 16 ) = 191 kj mol kj 2,286 mol H (C 7 H 16 ) (d) A mol sample o C 7 H 16 (l) is combusted in a bomb calorimeter. (i) Calculate the amount o heat released to the calorimeter. 1 point or correct coeicients 1 point or the correct substitution into H equation q released = mol C 7 H kj 1 mol C 7 H 16 = 52.4 kj o heat released 1 point or the amount o heat released (ii) Given that the total heat capacity o the calorimeter is kj C 1, calculate the temperature change o the calorimeter. Q = C p T 52.4 kj = kj C 1 T T = 52.4 kj kj C 1 = 5.65 C T = ( 52.4 kj)/(9.273 kj ) C 1 = C 1 point or the correct change in temperature (Must be consistent with answer in part (d)(i)) Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 9 Page 19

20 AP CHEMISTRY 2004 SCORING GUIDELINES Question 2 2 Fe(s) O 2 (g) Fe 2 O 3 (s) H = 824 kj mol 1 Iron reacts with oxygen to produce iron(iii) oxide, as represented by the equation above. A 75.0 g sample o Fe(s) is mixed with 11.5 L o O 2 (g) at 2.66 atm and 298 K. (a) Calculate the number o moles o each o the ollowing beore the reaction begins. (i) Fe(s) n Fe = 75.0 g Fe 1molFe g Fe = 1.34 mol Fe 1 point or number o moles o Fe(s) (ii) O 2 (g) PV = nrt PV n O2 = RT = 2.66 atm 11.5 L L atm K mol K 1 point or number o moles o O 2 (g) n O2 = 1.25 mol O 2 (b) Identiy the limiting reactant when the mixture is heated to produce Fe 2 O 3 (s). Support your answer with calculations. 1.5 mol O2 n O2 reacting = 1.34 mol Fe 2 mol Fe = 1.01 mol O 2 There is 1.25 mol O 2 initially, so there is an excess o O 2, and Fe is the limiting reactant. 1 point or identiying limiting reactant OR 2 mol Fe n Fe reacting = 1.25 mol O mol O2 = 1.67 mol Fe There is 1.34 mol Fe initially, so there is not enough Fe, and Fe is the limiting reactant. 1 point or supporting calculation Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (or AP proessionals) and (or AP students and parents). 4 Page 20

21 AP CHEMISTRY 2004 SCORING GUIDELINES Question 2 (cont d.) (c) Calculate the number o moles o Fe 2 O 3 (s) produced when the reaction proceeds to completion. 1molFe2O3 n Fe2 O = 1.34 mol Fe 3 2molFe = mol Fe 2 O 3 1 point or number o moles o Fe 2 O 3 produced (d) The standard ree energy o ormation, (i) Calculate the standard entropy o ormation, your answer. G, o Fe 2 O 3 (s) is 740. kj mol 1 at 298 K. S, o Fe 2 O 3 (s) at 298 K. Include units with G = H T S 740. kj mol 1 = 824 kj mol 1 (298 K) +84 kj mol 1 = (298 K) S = + 84 kj mol 298 K 1 S = 0.28 kj mol 1 K 1 S 1 point or calculation o 1 point or correct units S (ii) Which is more responsible or the spontaneity o the ormation reaction at 298 K, the standard enthalpy o ormation, H, or the standard entropy o ormation, S? Justiy your answer. H is the more important actor. The reaction is exothermic, which avors spontaneity. S is negative, which means the system becomes more ordered as the reaction proceeds. Greater order will not increase the spontaneity o the reaction. 1 point or indicating that H is responsible and or an explanation that addresses the signs o H and S Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (or AP proessionals) and (or AP students and parents). 5 Page 21

22 AP CHEMISTRY 2004 SCORING GUIDELINES Question 2 (cont d.) The reaction represented below also produces iron(iii) oxide. The value o per mole o Fe 2 O 3 (s) ormed. H or the reaction is 280. kj 2 FeO(s) O 2 (g) Fe 2 O 3 (s) (e) Calculate the standard enthalpy o ormation, H, o FeO(s). H rxn = Σ H (products) Σ H (reactants) H rxn = H Fe 2 O 3 (s) [2 H o FeO(s) + 1 H 2 O 2 (g)] 280. kj mol 1 = 824 kj mol 1 [2 H FeO(s) (0)] +544 kj mol 1 = 2 H FeO(s) 272 kj mol 1 = H FeO(s) 1 point or correct stoichiometry 1 point or correct calculation Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (or AP proessionals) and (or AP students and parents). 6 Page 22

23 AP CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 3 3. Answer the ollowing questions about the thermodynamics o the reactions represented below. Reaction X: Reaction Y: 1 2 I 2 (s) Cl 2 (g) ICl(g) H = 18 kj mol 1, S 298 = 78 J K 1 mol I 2 (s) Br 2 (l) IBr(g) H = 41 kj mol 1, S 298 = 124 J K 1 mol 1 (a) Is reaction X, represented above, spontaneous under standard conditions? Justiy your answer with a calculation. G = H T S = (18 kj mol 1 ) ( 298 K)(0.078 kj mol 1 K 1 ) = 5 kj mol 1 Reaction is spontaneous because G < 0. One point is earned or the correct value o G. One point is earned or a correct justiication o spontaneity. (b) Calculate the value o the equilibrium constant, K eq, or reaction X at 25 C. G = RT ln K eq ln K eq = (-5 kj mol )(10 J kj ) ln K eq = (8.31 J mol K )(298 K) K eq = e = ( ) = 8 DG RT = One point is earned or the correct answer. (c) What eect will an increase in temperature have on the equilibrium constant or reaction X? Explain your answer. G = RT ln K = H T S ln K eq eq = DH DS - + RT R Since H is positive, an increase in T will cause H /RT to become a smaller negative number, thereore K eq will increase. OR The reaction is endothermic ( H = +18 kj mol 1 ); an increase in temperature shits the reaction to avor more products relative to the reactants, resulting in an increase in the value o K eq. One point is earned or the correct choice with a correct explanation The College Board. All rights reserved. Visit apcentral.collegeboard.com (or AP proessionals) and (or students and parents). 6 Page 23

24 AP CHEMISTRY 2006 SCORING GUIDELINES (Form B) Question 3 (continued) (d) Explain why the standard entropy change is greater or reaction Y than or reaction X. Both reaction X and reaction Y have solid iodine as a reactant, but the second reactant in reaction X is chlorine gas whereas the second reactant in reaction Y is liquid bromine. Liquids have lower entropies than gases, thus in reaction Y the reactants are more ordered (and have lower entropies) than in reaction X. The products o both reaction X and reaction Y have about the same disorder, so the change in entropy rom reactants to products is greater in reaction Y than in reaction X. One point is earned or a correct explanation. (e) Above what temperature will the value o the equilibrium constant or reaction Y be greater than 1.0? Justiy your answer with calculations. G = H T S K eq = 1 when G = 0 T S = H T = H S = 41 kj mol kj mol K = 330 K So when T > 330 K, G < 0 kj mol 1 K eq > 1.0 One point is earned or G = 0. One point is earned or the correct temperature. () For the vaporization o solid iodine, I 2 (s) I 2 (g), the value o DH 298 is 62 kj mol 1. Using this inormation, calculate the value o DH 298 or the reaction represented below. I 2 (g) + Cl 2 (g) 2 ICl(g) I 2 (s) + Cl 2 (g) 2 ICl(g) DH 298 = 2 18 kj mol 1 I 2 (g) I 2 (s) DH 298 = 62 kj mol 1 I 2 (g) + Cl 2 (g) 2 ICl(g) DH 298 = 26 kj mol 1 One point is earned or DH 298 o either the irst or second equation. One point is earned or the correct sum o the DH 298 values The College Board. All rights reserved. Visit apcentral.collegeboard.com (or AP proessionals) and (or students and parents). 7 Page 24