# AP Chemistry 2003 Scoring Guidelines Form B

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2 Question 1 Total Score 10 points 2 HI(g) H 2 (g) + I 2 (g) 1. After a 1.0 mole sample of HI(g) is placed into an evacuated 1.0 L container at 700. K, the reaction represented above occurs. The concentration of HI(g) as a function of time is shown below. (a) Write the expression for the equilibrium constant, K c, for the reaction. K c = [H 2 ][I 2 ] [HI] 2 1 point for correct expression (b) What is [HI] at equilibrium? From the graph, [HI] eq is 0.80 M 1 point for equilibrium [HI] 2

3 Question 1 (cont d.) (c) Determine the equilibrium concentrations of H 2 (g) and I 2 (g). 2 HI(g) H 2 (g) + I 2 (g) I 1.0 M 0 0 C 0.20 M M M E 0.80 M 0.10 M 0.10 M [I 2 ] = [H 2 ] = 0.10 M 1 point for stoichiometric relationship between HI reacting and H 2 (g) or I 2 (g) forming 1 point for [H 2 ] eq and [I 2 ] eq (d) On the graph above, make a sketch that shows how the concentration of H 2 (g) changes as a function of time. From the graph, [H 2 ] eq is 0.10 M The curve should have the following characteristics: - start at 0 M; - increase to 0.1 M; - reach equilibrium at the same time [HI] reaches equilibrium 1 point for any two characteristics 2 points for all three characteristics (e) Calculate the value of the following equilibrium constants for the reaction at 700. K. (i) K c K c = [H 2 ][I 2 ] [HI] 2 = [0.10][0.10] [0.80] 2 = point for correct substitution (must agree with parts (b) and (c)) (ii) K p K p = K c = The number of moles on the product side is equal to the number of moles on the reactant side K p = K c (RT) n n = 2 2 = 0 K p = K c (RT) 0 K p = K c 1 point for K p = K c (with verification) 3

4 Question 1 (cont d.) (f) At 1,000 K, the value of K c for the reaction is In an experiment, 0.75 mole of HI(g), 0.10 mole of H 2 (g), and 0.50 mol of I 2 (g) are placed in a 1.0 L container and allowed to reach equilibrium at 1,000 K. Determine whether the equilibrium concentration of HI(g) will be greater than, equal to, or less than the initial concentration of HI(g). Justify your answer. Q = [H 2 ][I 2 ] [HI] 2 = [0.10][0.50] [0.75] 2 = K c = Q > K c To establish equilibrium, the numerator must decrease and the denominator must increase. Therefore, [HI] will increase. 1 point for calculating Q and comparing to K c 1 point for predicting correct change in [HI] 4

5 Question 2 Total Score 10 points 2. Answer the following questions that relate to chemical reactions. (a) Iron(III) oxide can be reduced with carbon monoxide according to the following equation. Fe 2 O 3 (s) + 3 CO(g) 2 Fe(s) + 3 CO 2 (g) A 16.2 L sample of CO(g) at 1.50 atm and 200. C is combined with g of Fe 2 O 3 (s). (i) How many moles of CO(g) are available for the reaction? PV = nrt n CO = PV RT = (1.50 atm) (16.2 L) L atm mol K (473 K) = mol CO 1 point for correct substitution 1 point for answer (ii) What is the limiting reactant for the reaction? Justify your answer with calculations. n Fe2 O = g Fe 3 2 O 3 1 mol Fe 2 O g Fe 2 O = mol Fe 2 O 3 3 n CO required = mol Fe 2 O 3 3 mol CO 1 mol Fe 2 O = mol 3 CO required to completely react with mol Fe 2 O mol CO are available, so CO is in excess and Fe 2 O is 3 limiting. OR n Fe2 O 3 required = mol CO 1 mol Fe 2 O 3 3 mol CO = mol mol Fe 2 O 3 corresponds to 33.4 g Fe 2 O 3 (the amount of Fe 2 O 3 required to completely react with mol CO) 1 point for moles of CO or Fe 2 O 3 required 1 point for correct conclusion NOTE: Answer must be consistent with moles of CO calculated in part (a) mol of Fe 2 O 3 is available, so there is not enough Fe 2 O 3 to completely react with mol CO. Therefore, Fe 2 O 3 is the limiting reactant. 5

6 Question 2 (cont d.) (iii) How many moles of Fe(s) are formed in the reaction? n Fe = mol Fe 2 O 3 2 mol Fe 1 mol Fe 2 O = mol Fe produced 3 If mol CO is used as the limiting reactant, then the number of moles of Fe formed is: n Fe = mol CO 2 mol Fe = mol Fe 3 mol CO 1 point for correct number of moles of Fe formed Note: Values must be consistent with answer in part (a) (ii). (b) In a reaction vessel, mol of Ba(NO 3 ) 2 (s) and mol of H 3 PO 4 (aq) are combined with deionized water to a final volume of 2.00 L. The reaction represented below occurs. 3 Ba(NO 3 ) 2 (aq) + 2 H 3 PO 4 (aq) Ba 3 (PO 4 ) 2 (s) + 6 HNO 3 (aq) (i) Calculate the mass of Ba 3 (PO 4 ) 2 (s) formed. n H3 PO 4 = mol Ba(NO 3 ) 2 2 mol H 3 PO 4 3 mol Ba(NO 3 ) = 0.400mol H 3 PO 4 2 required to completely react with mol Ba(NO 3 ) 2. There is mol H 3 PO 4 available. Therefore, H 3 PO 4 is the limiting reactant. 1 point for determining the limiting reactant mass Ba3 (PO 4 ) 2 = mol H 3 PO 4 1 mol Ba 3 (PO 4 ) 2 2 mol H 3 PO g Ba 3 (PO 4 ) 2 = 90.3 g Ba 1 mol Ba 3 (PO 4 ) 2 3 (PO 4 ) 2 If Ba(NO 3 ) 2 is used as the limiting reactant: mass Ba3 (PO 4 ) 2 = mol Ba(NO 3 ) 2 1 mol Ba 3 (PO 4 ) 2 3 mol Ba(NO 3 ) g Ba 3 (PO 4 ) 2 1 mol Ba 3 (PO 4 ) = 120. g Ba 2 3 (PO 4 ) 2 1 point for determining the correct mass of Ba 3 (PO 4 ) 2 6

7 (ii) Calculate the ph of the resulting solution. Question 2 (cont d.) n HNO3 = mol H 3 PO 4 6 mol HNO 3 2 mol H 3 PO = mol HNO 3 4 [HNO 3 ] = mol HNO L = 0.45 M Since all the H 3 PO 4 has reacted, the only acid in the solution is HNO 3. Since HNO 3 is a strong acid it completely dissociates. ph = log[h + ] = log(0.45) = 0.35 If Ba(NO 3 ) 2 is used as the limiting reactant n HNO3 = mol Ba(NO 3 ) 2 6 mol HNO 3 3 mol Ba(NO = 1.2 mol HNO 3 ) 3 2 [HNO 3 ] = 1.2 mol HNO L = 0.60 M ph = log[h + ] = log(0.60) = point for number of moles of H + 1 point for calculation of [H + ] and ph Must be consistent with part (b) (i) (iii) What is the concentration, in mol L 1, of the nitrate ion, NO 3 (aq), after the reaction reaches completion? The final concentration of NO 3 must be the same as the initial concentration. n NO3 = mol Ba(NO 3 ) 2 2 mol NO 3 1 mol Ba(NO 3 ) = 1.2 mol NO 3 2 [NO 3 ] = 1.2 mol NO 3 = 0.60 M NO 2.0 L 3 1 point for concentration of NO 3 7

8 Question 3 Total Score 10 points 3. In an experiment, a sample of an unknown, pure gaseous hydrocarbon was analyzed. Results showed that the sample contained g of carbon and g of hydrogen. (a) Determine the empirical formula of 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 formula is C 3 H 8 1 point for number of moles of carbon and number of moles of hydrogen 1 point for ratio of moles of carbon to moles of hydrogen 1 point for correct formula (b) The density of the hydrocarbon at 25 C and 1.09 atm is 1.96 g L 1. (i) Calculate the molar mass of 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 for correct substitution and 1 point for answer OR 1 point for calculation and 1 point for units 1.96 g L L mol 1 = 43.9 g mol 1 (1 point maximum) (ii) Determine the molecular formula of the hydrocarbon. Empirical mass n = molar mass Empirical mass for C 3 H 8 is 44 g mol 1 44 g mol 1 n = 44 g mol 1 n = 1, so the molecular 1 point for reporting correct formula with verification formula is the same as the empirical formula, C 3 H 8 8

9 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 following equation. C 7 H 16 (l) + 11 O 2 (g) 7 CO 2 (g) + 8 H 2 O(l) comb The heat of combustion, H, for one mole of C 7 H 16 (l) is kj. (c) Using the information in the table below, calculate the value of H for C 7 H 16 (l) in kj mol 1. f Compound H f (kj mol 1 ) CO 2 (g) H 2 O(l) f H = H f (products) H f (reactants) = 7 H f (CO 2 ) + 8 H f (H 2 O) [ H f (C 7 H 16 ) + 11 H f (O 2 )] 4,850 kj mol kj 4,850 mol H f kj kj = 7( )+ 8( mol kj 2,286 = 2,754 kj mol (C 7 H 16 ) = 191 kj mol mol ) [ H mol H (C 7 H 16 ) f f (C 7 H 16 ) + 11 (0 kj mol ) ] (d) A mol sample of C 7 H 16 (l) is combusted in a bomb calorimeter. (i) Calculate the amount of heat released to the calorimeter. 1 point for correct coefficients 1 point for the correct substitution into H f equation q released = mol C 7 H kj 1 mol C 7 H 16 = 52.4 kj of heat released 1 point for the amount of heat released (ii) Given that the total heat capacity of the calorimeter is kj C 1, calculate the temperature change of 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 for the correct change in temperature (Must be consistent with answer in part (d)(i)) 9

10 Question 4 Total Score 15 points 4. (a) Hot hydrogen gas is passed over heated copper(ii) oxide solid. H 2 + CuO Cu + H 2 O Note: Cu 2 O is an acceptable product (b) Solid sodium hydride is added to water. NaH + H 2 O Na + + OH + H 2 (c) Propanone is burned in air. Note: CO is an acceptable product C 3 H 6 O + O 2 CO 2 + H 2 O (d) A solution of lead(ii) nitrate is added to a solution of potassium sulfate. Pb 2+ + SO 4 2 PbSO 4 (e) Ammonia gas is mixed with hydrogen chloride gas. HCl + NH 3 NH 4 Cl Note: 1 product point for NH Cl ; 2 points (total) for NH 3 + H + NH 4 + (f) Sulfur trioxide gas is bubbled into water. SO 3 + H 2 O H + + HSO 4 (g) Excess concentrated potassium hydroxide solution is added to a solution of nickel(ii) chloride. OH + Ni 2+ Ni(OH) 4 2 Note: Ni(OH) 2, Ni(OH) 3, Ni(OH) 5 3, and Ni(OH) 6 4 are acceptable products (h) Solid sodium acetate is added to 1.0 M hydrobromic acid. NaC 2 H 3 O 2 + H + HC 2 H 3 O 2 + Na + 10

11 Question 5 Total Score 10 points 5. Oxalic acid, H 2 C 2 O 4, is a primary standard used to determine the concentration of potassium permanganate, KMnO 4, in solution. The equation for the reaction is as follows. 2 KMnO 4 (aq) + 5 H 2 C 2 O 4 (aq) + 3 H 2 SO 4 2 MnSO 4 (aq) + 10 CO 2 (g) + 8 H 2 O(l) + K 2 SO 4 (aq) A student dissolves a sample of oxalic acid in a flask with 30 ml of water and 2.00 ml of 3.00 M H 2 SO 4. The KMnO 4 solution of unknown concentration is in a 25.0 ml buret. In the titration, the KMnO 4 solution is added to the solution containing the oxalic acid. (a) What chemical species is being oxidized in the reaction? H 2 C 2 O 4 is the substance being oxidized. The half-reaction is: H 2 C 2 O 4 (aq) 2 CO 2 (g) + 2 H + (aq) + 2 e OR The oxidation state of carbon changes from +3 to +4 in CO 2. 1 point for identifying H 2 C 2 O 4 or C 2 O 4 2 as species oxidized (b) What substance indicates the observable endpoint of the titration? Describe the observation that shows the endpoint has been reached. In the reaction, the purple KMnO 4 solution in the buret is added to the colorless solution in the flask. KMnO 4 reacts with H 2 C 2 O 4 upon addition, so the purple KMnO 4 color disappears as it is added to the solution in the flask that contains unreacted H 2 C 2 O 4. As soon as all the H 2 C 2 O 4 has reacted (endpoint), the KMnO 4 is in excess and the solution in the flask will turn pink (pink is the color produced when the more concentrated purple KMnO 4 solution in the buret is diluted in the solution in the flask). 1 point for identifying KMnO 4 as reacting species that indicates the endpoint 1 point for indicating color change is from colorless to pink at the endpoint (c) What data must be collected in the titration in order to determine the molar concentration of the unknown KMnO 4 solution? The mass of oxalic acid, the initial volume of the KMnO 4 solution in the buret, and the final volume of the KMnO 4 solution in the buret 1 point for the mass of oxalic acid 1 point for the initial and final volume or for saying the change in volume of KMnO 4 11

12 Question 5 (cont d.) (d) Without doing any calculations, explain how to determine the molarity of the unknown KMnO 4 solution. Determine the moles of oxalic acid (by dividing the mass of oxalic acid measured/weighed, by the molar mass of oxalic acid). Use the stoichiometric ratio of the amount (2 mol) of KMnO 4 to amount (5 mol) of H 2 C 2 O 4 from the balanced chemical equation to convert from amount (in moles) of H 2 C 2 O 4 to amount (in moles) of KMnO 4. Divide the amount (in moles) of KMnO 4 by the volume, expressed in liters, of KMnO 4 needed to reach the endpoint. 1 point for determining moles of H 2 C 2 O 4 1 point for using correct stoichiometric factor 1 point for dividing moles of KMnO 4 by liters of KMnO 4 solution (e) How would the calculated concentration of the KMnO 4 solution be affected if 40 ml of water was added to the oxalic acid initially instead of 30 ml? Explain your reasoning. There would be no effect on the concentration of the KMnO 4 solution. We are only interested in the moles of oxalic acid. Since it is a solid, the moles of oxalic acid are calculated from the mass of oxalic acid. The volume of water used to dissolve the oxalic acid is independent of the moles of oxalic acid. 1 point for effect 1 point for explanation 12

13 Question 6 Total Score 9 points 6. Answer the following questions about electrochemistry. (a) Several different electrochemical cells can be constructed using the materials shown below. Write the balanced net-ionic equation for the reaction that occurs in the cell that would have the greatest positive value of Ecell. Al(s) Al 3+ (aq) + 3 e Cu 2+ (aq) + 2 e Cu(s) 2 Al(s) + 3 Cu 2+ (aq) 2 Al 3+ (aq) + 3 Cu(s) 1 point for selection of correct two redox couples 1 point for correctly balanced net ionic equation (b) Calculate the standard cell potential, E cell, for the reaction written in part (a). Al 3+ (aq) + 3 e Al(s) Cu 2+ (aq) + 2 e Cu(s) Ecell = Ecathode Eanode E = 1.66 V E = V = V ( 1.66 V) = V 1 point for correct cell E (Must be consistent with part (a)) 13

14 Question 6 (cont d.) (c) A cell is constructed based on the reaction in part (a) above. Label the metal used for the anode on the cell shown in the figure below. The metal is aluminum solid. 1 point for correct metal (Must be consistent with part (a)) (d) Of the compounds NaOH, CuS, and NaNO 3, which one is appropriate to use in a salt bridge? Briefly explain your answer, and for each of the other compounds, include a reason why it is not appropriate. NaOH is not appropriate. The anion, OH, would migrate towards the anode. The OH would react with the Al 3+ ion in solution. CuS is not appropriate. It is insoluble in water, so no ions would be available to migrate to the anode and cathode compartment to balance the charge. NaNO 3 is appropriate. It is soluble in water, and neither the cation nor the anion will react with the ions in the anode or cathode compartment. 1 point for correctly indicating whether each compound is appropriate, along with an explanation (3 points total) 14

15 Question 6 (cont d.) (e) Another standard cell is based on the following reaction. Zn + Pb 2+ Zn 2+ + Pb If the concentration of Zn 2+ is decreased from 1.0 M to 0.25 M, what effect does this have on the cell potential? Justify your answer. E cell = Ecell ln [Zn 2+ ] [Pb 2+ ] If [Zn 2+ ] is reduced, then the ratio [Zn 2+ ] [Pb 2+ < 1, therefore ] ln [Zn 2+ ] [Pb 2+ < 0. Thus E ] cell increases (becomes more positive). 1 point for correctly indicating how E cell is affected 1 point for explanation in terms of Nernst equation and Q 15

17 Question 7 (cont d.) (c) Compounds containing Kr have been synthesized, but there are no known compounds that contain He. Helium has a filled shell (the first shell), so does not tend to lose or gain electrons. Therefore, helium does not react. Krypton, while having filled 4s and 4p sublevels, has empty 4d and 4f sublevels. These empty orbitals affect the reactivity of Kr. 1 point for filled shell for He 1 point for indicating presence of empty d orbitals in Kr Note: Also acceptable is a comparison of the ionization energies of helium, and krypton and then the justification for krypton being more reactive. (d) The first ionization energy of Be is 900 kj mol 1, but the first ionization energy of B is 800 kj mol 1. The electron configuration for Be is 1s 2 2s 2, whereas the electron configuration for B is 1s 2 2s 2 2p 1. The first electron removed in boron is in a 2p subshell, which is higher in energy than the 2s subshell, from which the first electron is removed in beryllium. The higher in energy the subshell containing the electron to be removed (ionized), the lower the ionization energy. 1 point for indicating the difference in the subshell where the first electron is removed for each element 1 point for associating higher energy sublevel with lower ionization energy 17

18 Question 8 Total Score 8 points 8. The decay of the radioisotope I-131 was studied in a laboratory. I-131 is known to decay by beta ( 0 1 e ) emission. (a) Write a balanced nuclear equation for the decay of I-131. Note: β for I Xe e 1 point for correct equation 0 1 e is acceptable (b) What is the source of the beta particle emitted from the nucleus? A neutron spontaneously decays to an electron and a proton. 1 point for identifying a neutron as the source of the beta emission The radioactivity of a sample of I-131 was measured. The data collected are plotted on the graph below. (c) Determine the half-life, t 1/2, of I-131 using the graph above. The half-life is 8 days. That is the time required for the disintegration rate to fall from 16,000 to one-half its initial value, 8, point for half-life 18

19 Question 8 (cont d.) (d) The data can be used to show that the decay of I-131 is a first-order reaction, as indicated on the graph below. (i) Label the vertical axis of the graph above. The label on the y-axis should be ln or log one of the following: disintegrations or moles or atoms or [I-131] or disintegration rate. 1 point for correct label on y-axis (ii) What are the units of the rate constant, k, for the decay reaction? From the graph, the units on the rate constant are days 1 (Units of time 1 is acceptable) 1 point for correct units (iii) Explain how the half-life of I-131 can be calculated using the slope of the line plotted on the graph. The slope of the line is k. The slope is negative, so k is a positive number. The half-life can then be calculated using the relationship t 1/2 = k 1 point for indicating slope is k 1 point for half-life equation (d) Compare the value of the half-life of I-131 at 25 C to its value at 50 C. The half-life will be the same at the different temperatures. The half-life of a nuclear decay process is independent of temperature. 1 point 19

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