AP Chem ALL FRQs

a) (i) 35.7 torr x 1 atm/760 torr = 0.0470 atm Pethene = Ptotal - Pwater = 0.822 atm - 0.0470 atm = 0.775 atm n = PV/RT = (0.775 atm)(0.0854 L)/(0.08206 L atm mol-1 K-1)(305 K) = 0.00264 mol (ii) 0.200 g C2H5OH x 1 mol C2H5OH/46.1 g C2H5OH x 1 mol C2H4/ 1mol C2H5OH = 0.00434 mol C2H4 produced b) % yield = actual yield/maximum possible yield x 100 = 0.00264 mol/0.00434 mol x 100 = 60.8% c) Yes, the data support the student's claim. △G° = △H° -T△S° = 45.5 kJ/molrxn - (298 K)(0.126 kJ/(K∙molrxn)) = 8.0 kJ/molrxn Because △G° > 0, the value of Kp = e -△G° /RT) < 1.00 d) Diagram should include all bonding pairs plus two nonbonding pairs on the O atom. (A line may be used to represent an electron pair.) e) The bond angle is approximately 109° f) Ethene is only slightly soluble in water because the weak dipole/induced dipole intermolecular attractions between nonpolar ethene molecules and polar water molecules are weaker than the hydrogen bonds between water molecules. Ethanol molecules are soluble in water because they are polar and form hydrogen bonds with water molecules as they dissolve.

2015 FRQ #2 Ethene, C2H4(g) (molar mass 28.1 g/mol), may be prepared by the dehydration of ethanol, C2H5OH(g) (molar mass 46.1 g/mol), using a solid catalyst. A setup for the lab synthesis is shown in the diagram above. The equation for the dehydration reaction is given below. A student added a 0.200 g sample of C2H5OH(l) to a test tube using the setup shown above. The student heated the test tube gently with a Bunsen burner until all of the C H OH(l) 2 5 evaporated and gas generation stopped. When the reaction stopped, the volume of collected gas was 0.0854 L at 0.822 atm and 305 K. (The vapor pressure of water at 305 K is 35.7 torr.) (a) Calculate the number of moles of C2H4(g) (i) that are actually produced in the experiment and measured in the gas collection tube and (ii) that would be produced if the dehydration reaction went to completion. (b) Calculate the percent yield of C2H4(g) in the experiment. Because the dehydration reaction is not observed to occur at 298 K, the student claims that the reaction has an equilibrium constant less than 1.00 at 298 K. (c) Do the thermodynamic data for the reaction support the student's claim? Justify your answer, including a calculation of DG298 D for the reaction. The Lewis electron-dot diagram for C2H4 is shown below in the box on the left. In the box on the right, complete the Lewis electron-dot diagram for C2H5OH by drawing in all of the electron pairs. (e) What is the approximate value of the C-O-H bond angle in the ethanol molecule? (f) During the dehydration experiment, C2H4(g) and unreacted C2H5OH(g) passed through the tube into the water. The C2H4 was quantitatively collected as a gas, but the unreacted C2H5OH was not. Explain this observation in terms of the intermolecular forces between water and each of the two gases.

a) (i) Ecell = 0.34 V- (-1.31 V) = 1.65 V (ii) (The arrow should point to the left.) b) (i) The mass increases. (ii) Oxygen gas from the air reacts with Zn(s) in the cell, producing ZnO(s), which has more mass than the original Zn(s). c (i) The cell potential will be lower. (ii) 02(g), a reactant in the cell reaction, will be at a lower partial pressure at the higher elevation; thus the reaction has a greater value of Q (closer to K). Deviations in partial pressure that take the cell closer to equilibrium will decrease the magnitude of the cell potential. d) For Na, 1.0 g Na x 1.0 mol Na/22.99 g Na x 1.0 mol e-/1.0 mol Na = 0.043 mol e- For Ca, 1.0 g Ca x 1.0 mol Ca/40.08 g Ca x 2.0 mol e-/1.0 mol Ca = 0.050 mol e- The cell with the Ca anode would transfer more electrons. e) (i) 1s2 2s22p6 3s2 3p6 4s2 3d10 or [Ar] 4s2 3d10 (ii) 4s sublevel

2015 FRQ #1 Metal-air cells are a relatively new type of portable energy source consisting of a metal anode, an alkaline electrolyte paste that contains water, and a porous cathode membrane that lets in oxygen from the air. A schematic of the cell is shown above. Reduction potentials for the cathode and three possible metal anodes are given in the table below. (a) Early forms of metal-air cells used zinc as the anode. Zinc oxide is produced as the cell operates according to the overall equation below. 2 Zn(s) + O2(g) → 2 ZnO(s) (i) Using the data in the table above, calculate the cell potential for the zinc-air cell. (ii) The electrolyte paste contains OH− ions. On the diagram of the cell above, draw an arrow to indicate the direction of migration of OH− ions through the electrolyte as the cell operates. (b) A fresh zinc-air cell is weighed on an analytical balance before being placed in a hearing aid for use. (i) As the cell operates, does the mass of the cell increase, decrease, or remain the same? (ii) Justify your answer to part (b)(i) in terms of the equation for the overall cell reaction. (c) The zinc-air cell is taken to the top of a mountain where the air pressure is lower. (i) Will the cell potential be higher, lower, or the same as the cell potential at the lower elevation? (ii) Justify your answer to part (c)(i) based on the equation for the overall cell reaction and the information above. d) Metal-air cells need to be lightweight for many applications. In order to transfer more electrons with a smaller mass, Na and Ca are investigated as potential anodes. A 1.0 g anode of which of these metals would transfer more electrons, assuming that the anode is totally consumed during the lifetime of a cell? Justify your answer with calculations. (e) The only common oxide of zinc has the formula ZnO. (i) Write the electron configuration for a Zn atom in the ground state. (ii) From which sublevel are electrons removed when a Zn atom in the ground state is oxidized?

a) H+ + C6H7O2- ⇄ HC6H7O2 b) 1.25 mol HCl/1000 mL = x mol HCl/29.95 mL x = 0.0374 mol HCl 0.0374 mol C6H7O2-/45.0 mL = x mol C6H7O2-/1000 mL ⇒ 0.832 M c) Thymol blue; it has a pl<a close to the pH at the equivalence point, so it will change color near the equivalence point. d) pH = pKa = —log (1.7 x 10-5) = 4.77 e) [The pH curve should have the correct shape.] f) For sorbic acid Ka = [H+][C6H7O2-]/[HC6H7O2] thus [C6H7O2-]/{HC6H7O2] = Ka/[H+] = 1.7 x 10-5/10-3.37 ≈ 0.04 ⇒ [HC6H702] > [C6H702-] OR The concentrations of HC6H7O2 and C6H7O2- are equal at the half-equivalence point. A pH of 3.37 is lower than that at the half-equivalence point, so the protonated form, HC6H7O2, has a higher concentration in the soft drink.

2015 FRQ #3 Potassium sorbate, KC6H7O2 (molar mass 150. g/mol) is commonly added to diet soft drinks as a preservative. A stock solution of KC6H7O2(aq) of known concentration must be prepared. A student titrates 45.00 mL of the stock solution with 1.25 M HCl(aq) using both an indicator and a pH meter. The value of Ka forsorbic acid, HC6H7O2, is 1.7 × 10−5 (a) Write the net-ionic equation for the reaction between KC6H7O2(aq) and HCl(aq). (b) A total of 29.95 mL of 1.25 M HCl(aq) is required to reach the equivalence point. Calculate [KC6H7O2] in the stock solution. (c) The pH at the equivalence point of the titration is measured to be 2.54 Which of the following indicators would be the best choice for determining the end point of the titration? Justify your answer. (d) Calculate the pH at the half-equivalence point. (e) The initial pH and the equivalence point are plotted on the graph below. Accurately sketch the titration curve on the graph below. Mark the position of the half-equivalence point on the curve with an X. (f) The pH of the soft drink is 3.37 after the addition of the KC6H7O2(aq). Which species, HC6H7O2 or C6H7O2 − , has a higher concentration in the soft drink? Justify your answer.

a) Ca(OH)2 ⇄ Ca2+ + 2 OH- b) Ksp = [Ca2+] [OH-]2 1.3 x 10-6 = (0.10 + x) (2x)2 ≈ (0.10) 4x2 [assuming x << 0.10] 1.3 x 10-5 = 4x2 x = 0.0018 M Molar solubility of Ca(OH)2 = 0.0018M c) [The diagram should show the oxygen side of the water molecules oriented closer to the Ca2+ ion.]

2015 FRQ #4 Answer the following questions about the solubility of 6 Ca(OH) Ksp= 1.3 x 10-6 (a) Write a balanced chemical equation for the dissolution of Ca(OH)2(s) in pure water. (b) Calculate the molar solubility of Ca(OH)2 in 0.10 M Ca(NO3)2 . (c) In the box below, complete a particle representation diagram that includes four water molecules with proper orientation around the Ca2+ ion.

a) First order b) ["Increasing the concentration of the food coloring" should be circled.] If the initial concentration of blue food coloring is increased, then more time is required (regardless of the reaction order indicated in part (a)) for the bleach to oxidize the additional blue food coloring. c) The spectrophotometer should be set to a different wavelength.

2015 FRQ #5 Na2C37H34N2S3O9 + OCl− → products blue colorless Blue food coloring can be oxidized by household bleach (which contains OCl− ) to form colorless products, as represented by the equation above. A student used a spectrophotometer set at a wavelength of 635 nm to study the absorbance of the food coloring over time during the bleaching process. In the study, bleach is present in large excess so that the concentration of OCl− is essentially constant throughout the reaction. The student used data from the study to generate the graphs below. (a) Based on the graphs above, what is the order of the reaction with respect to the blue food coloring? (b) The reaction is known to be first order with respect to bleach. In a second experiment, the student prepares solutions of food coloring and bleach with concentrations that differ from those used in the first experiment. When the solutions are combined, the student observes that the reaction mixture reaches an absorbance near zero too rapidly. In order to correct the problem, the student proposes the following three possible modifications to the experiment. • Increasing the temperature • Increasing the concentration of the food coloring • Increasing the concentration of the bleach Circle the one proposed modification above that could correct the problem, and explain how that modification increases the time for the reaction mixture to reach an absorbance near zero (c) In another experiment, a student wishes to study the oxidation of red food coloring with bleach. How would the student need to modify the original experimental procedure to determine the order of the reaction with respect to the red food coloring?

a) Lil and KI. Lil has a small cation and a large anion and KI has a large cation and the same large anion. The melting point of Lil (with its smaller cation) is lower than that of KI. OR Lil and LiF. Lil has a small cation and a large anion and LiF has the same small cation and a small anion. The melting point of Lil (with its larger anion) is lower than that of LiF. OR Lil and NaF. Lil has a small cation and a large anion and NaF has a relatively small cation and a small anion. The melting point of Lil (with its larger anion) is lower than that of NaF. b) Either LiF or NaF is acceptable. F- + H20 ⇄ HF + OH

2015 FRQ #6 . A student learns that ionic compounds have significant covalent character when a cation has a polarizing effect on a large anion. As a result, the student hypothesizes that salts composed of small cations and large anions should have relatively low melting points. (a) Select two compounds from the table and explain how the data support the student's hypothesis. (b) Identify a compound from the table that can be dissolved in water to produce a basic solution. Write the net ionic equation for the reaction that occurs to cause the solution to be basic.

a) (i) q = mc△T = (110.0 g)(4.18 J/g∙℃))(35.6℃ - 15.0℃) = 9,470 J = 9.47 kJ (ii) 10.0 g LiCl x 1 mol LiCl/42.39 g LiCl = 0.236 mol LiCl -9.47 kJ/0.236 mol LiCl = -40.1 kJ/molrxn b) 1s2 2s2 2p6 c) The valence electrons in the Na ion are in a higher principal energy level than the valence electrons in the Li+ ion. Electrons in higher based on occupied principal energy levels. principal energy levels are, on average, farther from the nucleus. d) LiCl. Because the Li+ ion is smaller than the Na+ ion, the Coulombic attractions between ions in LiCl are stronger than in NaCl. This results in a greater lattice enthalpy. e) Cl ↘↖ Li+ See diagram above. f) There are interactions between Li+ ions and polar water molecules and between Cl- ions and polar water molecules. These are ion-dipole interactions.

2016 FRQ #1 A student investigates the enthalpy of solution, ΔHsoln , for two alkali metal halides, LiCl and NaCl. In addition to the salts, the student has access to a calorimeter, a balance with a precision of ±0.1 g, and a thermometer with a precision of ±0.1°C. (a) To measure ΔHsoln for LiCl, the student adds 100.0 g of water initially at 15.0°C to a calorimeter and adds 10.0 g of LiCl(s), stirring to dissolve. After the LiCl dissolves completely, the maximum temperature reached by the solution is 35.6°C. (i) Calculate the magnitude of the heat absorbed by the solution during the dissolution process, assuming that the specific heat capacity of the solution is 4.18 J/(g·°C). Include units with your answer. (ii) Determine the value of ΔHsoln for LiCl in kJ/molrxn . To explain why ΔHsoln for NaCl is different than that for LiCl, the student investigates factors that affect ΔHsoln and finds that ionic radius and lattice enthalpy (which can be defined as the ΔH associated with the separation of a solid crystal into gaseous ions) contribute to the process. The student consults references and collects the data shown in the table below. b) Write the complete electron configuration for the Na+ ion in the ground state. (c) Using principles of atomic structure, explain why the Na+ ion is larger than the Li+ ion. (d) Which salt, LiCl or NaCl, has the greater lattice enthalpy? Justify your answer. (e) Below is a representation of a portion of a crystal of + ( Li or Cl− ) LiCl. Identify the ions in the representation by writing the appropriate formulas in the boxes below. (f) The lattice enthalpy of LiCl is positive, indicating that it takes energy to break the ions apart in LiCl. However, the dissolution of LiCl in water is an exothermic process. Identify all particle-particle interactions that contribute significantly to the dissolution process being exothermic. For each interaction, include the particles that interact and the specific type of intermolecular force between those particles.

a) It is an acid-base reaction. The weak acid HC2H3O2 reacts with the weak base HCO3- with HC2H3O2 donating a proton. OR It is an acid-base reaction. No solid precipitates, so it is not a precipitation reaction. None of the oxidation numbers change, so it is not a redox reaction. b) 2.24 g NaHCO3 x 1 mol NaHCO3/84.0 g = 0.0267 mol NaHCO3 60.0 mL x 0.875 mol HC2H3O2/1000 mL = 0.0525 mol HC2H3O2 The NaHC03(s) and HC2H302(aq) react in a 1: 1 ratio, so the limiting reactant is NaHC03(s). c) As the reaction proceeds, both reactants are consumed and their concentrations decrease. Collisions between reactant particles become less likely as their concentations decrease, thus the reaction rate slows. d) (i) Entropy only should be circled. (ii) △G° = △H° - T△S° Reactions are thermodynamically favorable when △G° is negative. Since the reaction is endothermic (the flask gets cooler, △H° is positive), the reaction is not driven by enthalpy, because enthalpy does not help make △G° negative. Because there are no gases in the reactants and one of the products is a gas, △S° must be positive, which helps make △G° negative. e) See diagram above f) C2H3O2- + H30+ → HC2H3O2 + H2O OR C2H3O2- + H+ → HC2H302

2016 FRQ #2 A student designs an experiment to study the reaction between NaHCO3 and HC2H3O2 . The reaction is represented by the equation above. The student places 2.24 g of NaHCO3 in a flask and adds 60.0 mL of 0.875 M HC2H3O2 . The student observes the formation of bubbles and that the flask gets cooler as the reaction proceeds. (a) Identify the reaction represented above as an acid-base reaction, precipitation reaction, or redox reaction. Justify your answer. (b) Based on the information above, identify the limiting reactant. Justify your answer with calculations. (c) The student observes that the bubbling is rapid at the beginning of the reaction and gradually slows as the reaction continues. Explain this change in the reaction rate in terms of the collisions between reactant particles. (d) In thermodynamic terms, a reaction can be driven by enthalpy, entropy, or both. (i) Considering that the flask gets cooler as the reaction proceeds, what drives the chemical reaction between NaHCO3(s) and HC2H3O2(aq) ? Answer by drawing a circle around one of the choices below. Enthalpy only Entropy only Both enthalpy and entropy (ii) Justify your selection in part (d)(i) in terms of ΔG°. (e) The HCO3 − ion has three carbon-to-oxygen bonds. Two of the carbon-to-oxygen bonds have the same length and the third carbon-to-oxygen bond is longer than the other two. The hydrogen atom is bonded to one of the oxygen atoms. In the box below, draw a Lewis electron-dot diagram (or diagrams) for the HCO3 − ion that is (are) consistent with the given information. (f) A student prepares a solution containing equimolar amounts of HC2H3O2 and NaC2H3O2 . The pH of the solution is measured to be 4.7. The student adds two drops of 3.0 M HNO3(aq) and stirs the sample, observing that the pH remains at 4.7. Write a balanced, net-ionic equation for the reaction between HNO3(aq) and the chemical species in the sample that is responsible for the pH remaining at 4.7.

a) 127.570 - 126.549 - 1.021 g I2 1.021 g I2 x 1 mol I2/253.80 g I2 = 0.004023 mol I2 b) Number of moles of I2 = number of moles of M 1.284 g MI2 - 1.021 g I2 = 0.263 g M Molar mass of M = 0.263 g M/0.004023 mol M = 65.4 g/mol c) The student could dissolve the compound in water or melt the compound and see if the solution/melt conducts electricity. Ifthe solution/melt conducts electricity, mobile ions capable of carrying charge must be present, thus the compound is likely to be Ionic. OR The student could heat the compound until it melts or boils. If the melting/boiling point is very high, then the compound is likely to be ionic. d) Both Br2 and I2 molecules are nonpolar molecules, therefore the only possible intermolecular forces are London dispersion forces. The London dispersion forces are stronger in 12 because it is larger in size with more electrons and/or a more polarizable electron cloud. The stronger London dispersion forces in 12 result in a higher melting point, which makes 12 a solid at room temperature. e) [Na2S203(aq) should be circled.] The reaction between S2032 (aq) and 12(s) will be thermodynamically favorable because E° for the reaction is positive (E° = 0.54 - 0.08 = +0.46 V), from which it follows that △G° is negative because △G° = -nFE° f) I2 + 2S2032- → 2 I- + S4062-

2016 FRQ #3 M + I2 → MI2 To determine the molar mass of an unknown metal, M, a student reacts iodine with an excess of the metal to form the water-soluble compound MI2 , as represented by the equation above. The reaction proceeds until all of the I2 is consumed. The MI2(aq) solution is quantitatively collected and heated to remove the water, and the product is dried and weighed to constant mass. The experimental steps are represented below, followed by a data table. (a) Given that the metal M is in excess, calculate the number of moles of I2 that reacted. (b) Calculate the molar mass of the unknown metal M. The student hypothesizes that the compound formed in the synthesis reaction is ionic. (c) Propose an experimental test the student could perform that could be used to support the hypothesis. Explain how the results of the test would support the hypothesis if the substance was ionic. The student hypothesizes that Br2 will react with metal M more vigorously than I2 did because Br2 is a liquid at room temperature. (d) Explain why I2 is a solid at room temperature whereas Br2 is a liquid. Your explanation should clearly reference the types and relative strengths of the intermolecular forces present in each substance. While cleaning up after the experiment, the student wishes to dispose of the unused solid I2 in a responsible manner. The student decides to convert the solid I2 to I −(aq) anion. The student has access to three solutions, H2O2(aq), Na2S2O3(aq), and Na2S4O6(aq), and the standard reduction table shown below. (e) Which solution should the student add to I2(s) to reduce it to I−(aq)? Circle your answer below. Justify your answer, including a calculation of E° for the overall reaction. H2O2(aq) Na2S2O3(aq) Na2S4O6(aq) (f) Write the balanced net-ionic equation for the reaction between I2 and the solution you selected in part (e)

a) Ka = [C6H5O-][H3O+]/[C6H5OH] 1.12 x 10-10 = x2/(0.75 - x) Assume that x << 0.75 x2 = 8.4 x 10-11 x = √8.4 x 10-11 x = 9.2 x 10-6 M pH = -log[H+] = -log (9.2 x 10-6) = 5.04 b) Numbers 10 through 14 should be circled. When pH > pKa , the deprotonated form will predominate. pKa = —log(l.12 x 10-10) = 9.95, therefore at pH 10 and above, [C6H5O-] > [C6H5OH]

2016 FRQ #4 C6H5OH(aq) + H2O(l) ←→ C6H5O−(aq) + H3O+(aq) Ka = 1.12 × 10−10 4. Phenol is a weak acid that partially dissociates in water according to the equation above. (a) What is the pH of a 0.75 M C6H5OH(aq) solution? (b) For a certain reaction involving C6H5OH(aq) to proceed at a significant rate, the phenol must be primarily in its deprotonated form, C6H5O−(aq). In order to ensure that the C6H5OH(aq) is deprotonated, the reaction must be conducted in a buffered solution. On the number scale below, circle each pH for which more than 50 percent of the phenol molecules are in the deprotonated form (C6H5O−(aq)). Justify your answer.

a) For trial 1, n/v = 0.020 mol/L (or assume the volume of the vessel is 1.0 L; the number of moles of C4H6 in the vessel would then be 0.020 mol). PV = nRT P = nRT/V = (0.020 mol)(0.08206 L atm mol-1 K-1)(625 K)/1.0 L = 1.0 atm b) Second order (because the plot of 1/[C4H6] is a straight line). c) From the second-order rate law (differential form): rate = k[C4H6]2 ⇒ k = rate/([C4H6])2 = 0.0010 mol/(L ∙ s)/(0.020 mol/L)2 = 2.5 l/(mol ∙ s) OR 1/[C4H6]t = 2kt + 1/[C4H6]0 The coefficient oft is equal to 2k because of the reaction stoichiometry. The slope of the line in the plot of 1/[C4H6] versus time is 2k. Thus slope = 5.0 L/(mol •s) = 2k, therefore k = 2.5 L/(mol• s). Note: Students who choose the second method of determining k but omit the factor of 2, thereby getting an answer of 5.0 L/(mol• s), still earn the point.

2016 FRQ #5 2 C4H6(g) → C8H12(g) At high temperatures the compound C4H6 (1,3-butadiene) reacts according to the equation above. The rate of the reaction was studied at 625 K in a rigid reaction vessel. Two different trials, each with a different starting concentration, were carried out. The data were plotted in three different ways, as shown below. (a) For trial 1, calculate the initial pressure, in atm, in the vessel at 625 K. Assume that initially all the gas present in the vessel is C4H6 . (b) Use the data plotted in the graphs to determine the order of the reaction with respect to C4H6 . (c) The initial rate of the reaction in trial 1 is 0.0010 mol/(L⋅s). Calculate the rate constant, k, for the reaction at 625 K.

a) Based on the K value, the reaction goes essentially to completion. Ba2+(aq) is the limiting reactant. The concentration of Ba2+ when the solutions are first mixed but before any reaction takes place is 0.20 M/2 = 0.10 M. Thus the equilibrium concentration of Ba(EDTA)2 (aq) is 0.10M. b) The number of moles of Ba2+(aq) increases because the percent dissociation of Ba(EDTA)2-(aq) increases as the solution is diluted. OR A mathematical justification such as the following: The dilution from 100.0 mL to I .00 L reduces the concentrations of all species to one tenth oftheir original values. Immediately after the dilution, the reaction quotient, Q, can be determined as shown below. Q = ⅟10[Ba(EDTA)2-]/⅟10[Ba2+] x ⅟10[EDTA4-] = 10 K Because Q > K, the net reaction will produce more reactants to move toward equilibrium, so the number of moles of Ba2+(aq) will be greater than the number in the original solution.

2016 FRQ #6 Ba2+(aq) + EDTA4−(aq) R Ba(EDTA)2−(aq) K = 7.7 × 107 The polyatomic ion C10H12N2O8 4− is commonly abbreviated as EDTA4−. The ion can form complexes with metal ions in aqueous solutions. A complex of EDTA4− with Ba2+ ion forms according to the equation above. A 50.0 mL volume of a solution that has an EDTA4−(aq) concentration of 0.30 M is mixed with 50.0 mL of 0.20 M Ba(NO3) 2 to produce 100.0 mL of solution (a) Considering the value of K for the reaction, determine the concentration of Ba(EDTA)2−(aq) in the 100.0 mL of solution. Justify your answer. (b) The solution is diluted with distilled water to a total volume of 2+ Ba (aq) 1.00 L. After equilibrium has been reestablished, is the number of moles of present in the solution greater than, less than, or equal to the number of moles of Ba2+(aq) present in the original solution before it was diluted? Justify your answer.

a) 37.30 mL - 5.65 mL- 31.65 mL b) At the equivalence point, moles OH- added — moles of H+ consumed. Because HA is monoprotic: (0.110 M)(0.03165 L) x 1 mol HA/a mol NaOH x 1/0.0250 L = 0.139 M OR moles of H= consumed = MaVa MaVa = MbVb Therefore, Ma = MbVb/Va = (0.110 M)(0.03165 L)/0.0250 L = 0.139 M c) The error would increase the calculated acid concentration. A volume of NaOH(aq) larger than the actual volume needed to reach the equivalence point, would lead to a calculation of moles of base that would be greater than the moles of acid actually present in the solution. The assumption that the moles of acid are the same as the moles of base would lead to a calculated concentration of acid that would be higher than the actual concentration.

2016 FRQ #7 A student is given a 25.0 mL sample of a solution of an unknown monoprotic acid and asked to determine the concentration of the acid by titration. The student uses a standardized solution of 0.110 M NaOH(aq), a buret, a flask, an appropriate indicator, and other laboratory equipment necessary for the titration. (a) The images below show the buret before the titration begins (below left) and at the end point (below right). What should the student record as the volume of NaOH(aq) delivered to the flask? (b) Based on the given information and your answer to part (a), determine the value of the concentration of the acid that should be recorded in the student's lab report. (c) In a second trial, the student accidentally added more NaOH(aq) to the flask than was needed to reach the end point, and then recorded the final volume. Would this error increase, decrease, or have no effect on the calculated acid concentration for the second trial? Justify your answer.

(a) (i) n = PV/RT = 0.40 atm x 25.0 L/0.08206 (L∙atm)/(mol∙K) x 393 K = 0.31 mol Cl2(g) (ii) 0.31 mol Cl2 x 1 mol CS2/3 mol Cl2 x 76.13 g CS2/1 mol CS2 = 7.9 g CS2 b) (i) At the higher temperature the particles have a greater average kinetic energy than at the lower temperature. Thus there are more collisions with sufficient energy to overcome the activation energy. (ii) 1 point is earned for a curve that has a peak above and to the left of the given curve. 1 point is earned for a curve that is below the given curve in the region beyond the activation energy. c) (i) CL - S - S - CL See correct diagram above. (ii) Any value between 104° and 110° d) (i) Dipole-dipole forces, London dispersion forces (ii) The intermolecular forces among CC14 molecules must be stronger than those among HCI molecules because the CC14 condenses at a higher temperature than HCI.

2017 FRQ #1 CS2(g) + 3 Cl2(g) → CCl4(g) + S2Cl2(g) Carbon tetrachloride, CCl4(g), can be synthesized according to the reaction represented above. A chemist runs the reaction at a constant temperature of 120°C in a rigid 25.0 L container. (a) Chlorine gas, Cl2(g), is initially present in the container at a pressure of 0.40 atm. (i) How many moles of Cl2(g) are in the container? (ii) How many grams of carbon disulfide, CS2(g), are needed to react completely with the Cl2(g) ? (b) At 30°C the reaction is thermodynamically favorable, but no reaction is observed to occur. However, at 120°C, the reaction occurs at an observable rate. (i) Explain how the higher temperature affects the collisions between the reactant molecules so that the reaction occurs at an observable rate at 120°C. (ii) The graph below shows a distribution for the collision energies of reactant molecules at 120°C. Draw a second curve on the graph that shows the distribution for the collision energies of reactant molecules at 30°C. (c) S2Cl2 is a product of the reaction. (i) In the box below, complete the Lewis electron-dot diagram for the S2Cl2 molecule by drawing in all of the electron pairs. (ii) What is the approximate value of the Cl−S−S bond angle in the S2Cl2 molecule that you drew in part (c)(i) ? (If the two Cl−S−S bond angles are not equal, include both angles.) (d) CCl4(g) can also be produced by reacting CHCl3(g) with Cl2(g) at 400°C, as represented by the equation below. CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) At the completion of the reaction a chemist successfully separates the CCl (g) 4 from the HCl(g) by cooling the mixture to 70°C, at which temperature the CCl4(g) condenses while the HCl(g) remains in the gaseous state. (i) Identify all types of intermolecular forces present in HCl(l). (ii) What can be inferred about the relative strengths of the intermolecular forces in CCl4(l) and HCl(l) ? Justify your answer in terms of the information above.

a) In the diagram on the left, the C atom has a formal charge of zero and the O atom has a formal charge of —1. In the diagram on the right, the C atom has a formal charge of —1 and the O atom has a formal charge of zero. The diagram on the left is the better representation because it puts the negative formal charge on oxygen, which is more electronegative than carbon. Fulminic acid can convert to isocyanic acid according to the equation below. HCNO(g) fulminic acid ⇄ HNCO(g) isocyanic acid fulminic acid H - C ≡ N - O isocyanic acid H - N = C = O b) Compound HCNO HNCO Bond Enthalpies (kJ/mol) HCNO 413 + 891 + 201 HNCO 391 + 615 +745 Total Bond Enthalpy (kJ/mol) HCNO 1505 HNCO 1751 △H° = ∑(enthalpies of bonds broken) - ∑(enthalpies of bonds formed) =1505 kJ/mol - 1751 kJ/mol =-246 kJ/molrxn c) The change from fulminic acid to isocyanic acid is a rearrangement of atoms with no change in phase or number of molecules. d) Isocyanic acid (HNCO) will be present in higher concentration. △G° is essentially equal to △H° because △S° is essentially zero, so △G° ≈ -246 kJ/molrxn, indicating the forward reaction is thermodynamically favorable. Since △G° is negative, K> 1 (△G° = -RT 1n K), resulting in a higher concentration of product than reactant at equilibrium. e) (i) From inspecting the data table or the graph, it is evident that the decomposition reaction has a constant half-life, which indicates that the reaction is a first- order reaction. (ii) k = 0.693/t½ = 0.693/10. h = 0.069 h-1 OR k = In[A]0 - In[A]t/t = In(0.1000) - In(0.500)/10. h = 0.069 h-1 f) Perform the experiment at a different concentration of OH-(aq) and measure how the concentration of CO(NH2)2 changes over time. (Other variables, such as temperature, should be held constant.)

2017 FRQ #2 Answer the following questions about the isomers fulminic acid and isocyanic acid. Two possible Lewis electron-dot diagrams for fulminic acid, HCNO, are shown below (a) Explain why the diagram on the left is the better representation for the bonding in fulminic acid. Justify your choice based on formal charges. Fulminic acid can convert to isocyanic acid according to the equation below. HCNO(g)<-> HNCO(g) fulminic acid isocyanic acid (b) Using the Lewis electron-dot diagrams of fulminic acid and isocyanic acid shown in the boxes above and the table of average bond enthalpies below, determine the value of 'H° for the reaction of HCNO(g) to form HNCO(g). (c) A student claims that ΔS° for the reaction is close to zero. Explain why the student's claim is accurate. (d) Which species, fulminic acid (HCNO) or isocyanic acid (HNCO), is present in higher concentration at equilibrium at 298 K? Justify your answer in terms of thermodynamic favorability and the equilibrium constant. The ammonium salt of isocyanic acid is a product of the decomposition of urea, CO(NH2)2 , represented below. CO(NH2)2(aq) R NH4 +(aq) + OCN−(aq) A student studying the decomposition reaction runs the reaction at 90°C. The student collects data on the concentration of urea as a function of time, as shown by the data table and the graph below. (e) The student proposes that the rate law is rate = k[CO(NH2)2]. (i) Explain how the data support the student's proposed rate law. (ii) Using the proposed rate law and the student's results, determine the value of the rate constant, k. Include units with your answer. (f) The student learns that the decomposition reaction was run in a solution with a pH of 13. Briefly describe an experiment, including the initial conditions that you would change and the data you would gather, to determine whether the rate of the reaction depends on the concentration of OH−(aq).

a) Kp = (Pno)2/(Pn2)(Po2) b) N2(g) + O2(g) ⇄ 2 NO(g) Initial 6.01 1.61 0 Change -x -x +2x Equilibrium 6.01-x 1.61-x 0.122 2x = 0.122 atm ⇒ x = 0.0610 atm Kp = (0.122)2/(5.95)(1.55) = 0.00161 c) (i) NaOH will neutralize some of the HN02 to produce N02-. The resulting solution contains a mixture of a weak acid and its conjugate base, which is a buffer solution. HNO2 + OH- → NO2- + H2O (ii) The student should add 50.0 mL of 0.100 MNaOH(aq). When half of the HN02 is converted to the conjugate base, [HN02] = [N02-], therefore the buffer has a pH equal to pKa . OR pH = PKa + log[NO2-]/[HNO2], thus pH = pKa when [HNO2] = [NO2-] d) The buffer made by the second student is more resistant to changes in pH because it contains a higher concentration of HN02 and N02- to react with added H+ or OH ions. e) The pH of the solution is less than 3.40. If [HN02] = [N02-], pH = pKa, and the pH of the solution would be 3.40. Since [HN02] > [N02-], as represented in the diagram, the solution has a pH less than 3.40. OR pH = pKa + log [NO2-]/[HNO2] ⇒ pH = 3.40 + log 5/10 ⇒ pH = 3.10

2017 FRQ #3 N2(g) + O2(g) ←→ 2 NO(g) At high temperatures, N2(g) and O2(g) can react to produce nitrogen monoxide, NO(g), as represented by the equation above. (a) Write the expression for the equilibrium constant, Kp , for the forward reaction. (b) A student injects N2(g) and O2(g) into a previously evacuated, rigid vessel and raises the temperature of the vessel to 2000°C. At this temperature the initial partial pressures of N2(g) and O2(g) are 6.01 atm and 1.61 atm, respectively. The system is allowed to reach equilibrium. The partial pressure of NO(g) at equilibrium is 0.122 atm. Calculate the value of Kp . Nitrogen monoxide, NO(g), can undergo further reactions to produce acids such as HNO2 , a weak acid with a Ka of 4.0 × 10−4 and a pKa of 3.40. (c) A student is asked to make a buffer solution with a pH of 3.40 by using 0.100 M HNO2(aq) and 0.100 M NaOH(aq). (i) Explain why the addition of 0.100 M NaOH(aq) to 0.100 M HNO2(aq) can result in the formation of a buffer solution. Include the net ionic equation for the reaction that occurs when the student adds the NaOH(aq) to the HNO2(aq). (ii) Determine the volume, in mL, of 0.100 M NaOH(aq) the student should add to 100. mL of 0.100 M HNO2(aq) to make a buffer solution with a pH of 3.40. Justify your answer. (d) A second student makes a buffer by dissolving 0.100 mol of NaNO2(s) in 100. mL of 1.00 M HNO2(aq). Which is more resistant to changes in pH when a strong acid or a strong base is added, the buffer made by the second student or the buffer made by the first student in part (c) ? Justify your answer. (e) A new buffer is made using HNO2(aq) as one of the ingredients. A particulate representation of a small representative portion of the buffer solution is shown below. (Cations and water molecules are not shown.) Is the pH of the buffer represented in the diagram greater than, less than, or equal to 3.40 ? Justify your answer.

a) Dye C is the least polar because it moved the farthest. Nonpolar dyes are more strongly attracted to the nonpolar solvent. AND/OR Nonpolar dyes are least strongly retained by the polar paper. b) Dye A is present in the unknown sample. The unknown sample moves to a position that is midway between the origin and the solvent front, and so does dye A. OR Dye A has a retention factor (RI) that is close to 0.50 on the chromatogram with the three dyes, and the unknown also has a retention factor close to 0.50.

2017 FRQ #4 A student investigates various dyes using paper chromatography. The student has samples of three pure dyes, labeled A, B, and C, and an unknown sample that contains one of the three dyes. The student prepares the chromatography chambers shown above on the left by putting a drop of each dye at the indicated position on the chromatography paper (a polar material) and standing the paper in a nonpolar solvent. The developed chromatograms are shown above on the right. (a) Which dye (A, B, or C) is the least polar? Justify your answer in terms of the interactions between the dyes and the solvent or between the dyes and the paper. (b) Which dye is present in the unknown sample? Justify your answer.

a) q = mc△°T = (125.00 g)(4.18 J/g∙℃)) 51.1℃- 22.0℃) = 15,200 J = 15.2 kJ b) 1 mol C3H7OH x 60.09 g C3H7OH/1 mol C3H7OH x 15.2 kJ/0.55 g C3H7OH = 1661 kJ = 1.7 x 103 kJ c) The final temperature measured by the second student would be less than that measured by the first student because: the actual mass of C3H70H(l) combusted will be less than 0.55 g OR combustion of the contaminated sample will also require vaporization of the water in the sample.

2017 FRQ #5 A student performs an experiment to determine the enthalpy of combustion of 2-propanol, C3H7OH(l), which combusts in oxygen according to the equation above. The student heats a sample of water by burning some of the C3H7OH(l) that is in an alcohol burner, as represented below. The alcohol burner uses a wick to draw liquid up into the flame. The mass of C3H7OH(l) combusted is determined by weighing the alcohol burner before and after combustion. (a) Calculate the magnitude of the heat energy, in kJ, absorbed by the water. (Assume that the energy released from the combustion is completely transferred to the water.) (b) Based on the experimental data, if one mole of C3H7OH(l) is combusted, how much heat, in kJ, is released? Report your answer with the correct number of significant figures. (c) A second student performs the experiment using the same mass of water at the same initial temperature. However, the student uses an alcohol burner containing C3H7OH(l) that is contaminated with water, which is miscible with C3H7OH(l). The difference in mass of the alcohol burner before and after the combustion in this experiment is also 0.55 g. Would the final temperature of the water in the beaker heated by the alcohol burner in this experiment be greater than, less than, or equal to the final temperature of the water in the beaker in the first student's experiment? Justify your answer.

a) 1.8 x 10-11 = [Mg2+][OH-]2 = (x)(2x)2 = 4x3 x = ∛1.8 x 10-11/4 =m 1.65 x 10-4 M = [Mg2+] = [Mg(OH)2] 0.100 L x 1.65 x 10-4 mol/1 L x 58.32 g Mg(OH)2/1 mol Mg(OH)2 = 9.6 x 10-4 g Mg(OH)2 b) The Sr2+ ion is larger than the Mg2+ ion because it has additional occupied energy levels (or shells). Coulomb's law states that the force of attraction between cation and anion is inversely proportional to the square of the distance between them. Since the distance between Mg2+ and OH- is shorter than the distance between Sr2+ and OH , the attractive forces in Mg(OH)2 are stronger and, therefore, its lattice energy is greater.

2017 FRQ #6 Answer the following questions about Mg(OH)2 . At 25°C, the value of the solubility product constant, Ksp, for Mg(OH)2(s) is 1.8 × 10−11 . (a) Calculate the number of grams of Mg(OH)2 (molar mass 58.32 g/mol) that is dissolved in 100. mL of a saturated solution of Mg(OH)2 at 25°C. (b) The energy required to separate the ions in the Mg(OH)2 crystal lattice into individual Mg2+(g) and OH−(g) ions, as represented in the table below, is known as the lattice energy of Mg(OH)2(s). As shown in the table, the lattice energy of Sr(OH)2(s) is less than the lattice energy of Mg(OH)2(s). Explain why in terms of periodic properties and Coulomb's law.

a) (i) E° = 1.33 - 0.70 0.63 V (ii) E° = -1.84+ 1.77 = -0.07 V b) (i) The student should use the dichromate ion for the titration because, for the reaction, the value of E° is positive, which means that the reaction is thermodynamically favorable. OR △G° = -nFE° and n, F, and E° are all positive numbers, therefore △G° < 0, which means that the reaction is thermodynamically favorable. (ii) △G° = -nFE° = -6(96,485 C/mol)(0.63 J/C)(1 kJ/1000 J) = -360 kJ/molrxn

2017 FRQ #7 A student wants to determine the concentration of H2O2 in a solution of H2O2(aq). The student can use one of two titrants, either dichromate ion, Cr2O7 2-(aq), or cobalt(II) ion, Co2+(aq). The balanced chemical equations for the two titration reactions are shown below. The half-reactions and the E° values for the systems related to the titrations above are given in the following table. (a) Use the information in the table to calculate the following. (i) E° for the reaction between Cr2O7 2-(aq) and H2O2(aq) at 298 K (ii) E° for the reaction between Co2+(aq) and H2O2(aq) at 298 K (b) Based on the calculated values of E°, the student must choose the titrant for which the titration reaction is thermodynamically favorable at 298 K. (i) Which titrant should the student choose? Explain your reasoning. (ii) Calculate the value of ΔG°, in kJ/molrxn, for the reaction between the chosen titrant and H2O2(aq).

a) +1 b) 100.00 mL x 0.500 mol Na2S2O3/1000 mL x 158.10 g Na2S2O3/1 mol NaS2O3 = 7.90 g Na2S2O3 c) NaOCl is the limiting reactant because all of the substances started with the same amount, so the highest coefficient in the reactants will show which is depleted first. d) From the graph the final temperature is 32.5°C. △T = Tf - Ti = 32.5°C - 20.0°C = 12.5°C e) q = mc△T = (15.21 g)(3.94 J/g∙°C))(12.5°C) = 749 J ii) nNaOCl= 5.0 mL x 0.5 mol NaOCl/1000 mL = 0.00250 mol NaOCl nRxn= 0.00250 mol NaOCl x 1 mol rxn/ 4 mol NaOCl = 0.000625 mol rxn H= -0.749 kJ/ 0.000625 mol= 1.20 x 10^3 kJ/molrxn f) By doubling the volumes, the number of moles of the reactants are doubled, which doubles the amount of energy produced. Therefore the amount of heat per mole will remain the same. OR In the second experiment, △H°rxn = 2mc△T/2n = mc△T/n = △H°rxn Thus the magnitude is the same as calculated in the first experiment. g) S2032-(aq) + 4 OCI-(aq) + 2 OH (aq) ---> 2 S042-(aq) + 4 Cl-(aq) + H20(l)

2018 FRQ #1 Na 2S2O3(aq) + 4 NaOCl(aq) + 2 NaOH(aq) → 2 Na2SO4(aq) + 4 NaCl(aq) + H2O(l) A student performs an experiment to determine the value of the enthalpy change, change in H rxn , for the oxidation-reduction reaction represented by the balanced equation above. (a) Determine the oxidation number of Cl in NaOCl. (b) Calculate the number of grams of Na2S2O3 needed to prepare 0.500 M Na2S2O3 100.00 mL of (aq). In the experiment, the student uses the solutions shown in the table below. (c) Using the balanced equation for the oxidation-reduction reaction and the information in the table above, determine which reactant is the limiting reactant. Justify your answer. 20.0°C, are combined in an insulated calorimeter. The temperature of the reaction mixture is monitored, as shown in the graph below. (d) According to the graph, what is the temperature change of the reaction mixture? (e) The mass of the reaction mixture inside the calorimeter is 15.21 g. (i) Calculate the magnitude of the heat energy, in joules, that is released during the reaction. Assume that the specific heat of the reaction mixture is 3.94 J/(g·°C) and that the heat absorbed by the calorimeter is negligible. (ii) Using the balanced equation for the oxidation-reduction reaction and your answer to part (c), calculate D the value of the enthalpy change of the reaction, change Hrxn , in kJ/molrxn . Include the appropriate algebraic sign with your answer The student repeats the experiment, but this time doubling the volume of each of the reactants, as shown in the table below. (f) The magnitude of the enthalpy change, DHD rxn , in kJ/molrxn , calculated from the results of the second experiment is the same as the result calculated in part (e)(ii). Explain this result. (g) Write the balanced net ionic equation for the given reaction.

a) See student example above. (8 molecules of NO and 2 molecules of 02) b) (i) △G° = -RT1nK k = e-△G°/RT k = e - 870 J/mol/(8.314 J mol-1 K-1)(298 K) k = 0.70 (ii) No, the pressure will not equal 1.0 atm. PN2O3 would only equal 1.0 atm if the reaction goes to completion. OR The value of K indicates that a substantial amount of reactants will be present at equilibrium. c) Disagree. Because the reaction is exothermic, increasing the temperature and a correct justification. of the reaction will favor the formation of the reactants (according to Le Chatelier's principle). d) (i) See sample response above. (Line segments can be used to represent electron pairs.) (ii) sp2 e) (i) 20. mL KOH x 0.100 mol HOH/1000 mL KOH = 0.20020 mol KOH added ⇒ 0.0020 mol HNO2 in 100. ml of solution because the stoichiometry of the neutralization reaction is I to 1. 0.0020 mol HNO2/0.100 L = 0.020 M HNO2 (ii) The value of pKa is about 3.4. f) N02-(aq) The titration is past the half-equivalence point; therefore, there will be more conjugate base present than acid.

2018 FRQ #2 2 NO(g) + O2(g) → 2 NO2(g) A student investigates the reactions of nitrogen oxides. One of the reactions in the investigation requires an equimolar mixture of NO(g) and NO2(g), which the student produces by using the reaction represented above. (a) The particle-level representation of the equimolar mixture of NO(g) and NO2(g) in the flask at the completion of the reaction between NO(g) and O2(g) is shown below in the box on the right. In the box below on the left, draw the particle-level representation of the reactant mixture of NO(g) and O2(g) that would yield the product mixture shown in the box on the right. In your drawing, represent oxygen atoms and nitrogen atoms as indicated below The student reads in a reference text that NO(g) and NO2(g) will react as represented by the equation below. Thermodynamic data for the reaction are given in the table below the equation. NO(g) + NO2(g) <-> N2O3(g) (b) The student begins with an equimolar mixture of NO(g) and NO2(g) in a rigid reaction vessel and the mixture reaches equilibrium at 298 K. (i) Calculate the value of the equilibrium constant, K, for the reaction at 298 K. (ii) If both PNO and PNO2 in the vessel are initially 1.0 atm, will PN O2 at equilibrium be equal to 3 1.0 atm? Justify your answer (c) The student hypothesizes that increasing the temperature will increase the amount of N2O3(g) in the equilibrium mixture. Indicate whether you agree or disagree with the hypothesis. Justify your answer. N2O3(g) reacts with water to form nitrous acid, HNO2 (aq), a compound involved in the production of acid rain. The reaction is represented below. N2O3(g) + H2O(l) → 2 HNO2(aq) (d) The skeletal structure of the HNO2 molecule is shown in the box below. (i) Complete the Lewis electron-dot diagram of the HNO2 molecule in the box below, including any lone pairs of electrons. (ii) Based on your completed diagram above, identify the hybridization of the nitrogen atom in the HNO2 molecule. To produce an aqueous solution of HNO2 , the student bubbles N2O3(g) into distilled water. Assume that the reaction goes to completion and that HNO2 is the only species produced. To determine the concentration of HNO2(aq) in the resulting solution, the student titrates a 100. mL sample of the solution with 0.100 M KOH(aq). The neutralization reaction is represented below. HNO 2(aq) + OH−(aq) → NO2 −(aq) + H2O(l) The following titration curve shows the change in pH of the solution during the titration. (e) Use the titration curve and the information above to (i) determine the initial concentration of the HNO2(aq) solution (ii) estimate the value of pKa for HNO2(aq) (f) During the titration, after a volume of 15 mL of 0.100 M KOH(aq) has been added, which species, HNO2(aq) or NO2 −(aq), is present at a higher concentration in the solution? Justify your answer

a) 1s2 2s2 3s2 3/76 3d6 OR [Ar] 3d6 b) Both ions have the same nuclear charge; however, the greater number of electrons in the outermost shell of Fe2+ results in greater electron-electron repulsion within that shell, leading to a larger radius. c) Coulomb's law: F ∝ q1q2/r2 (need not be explicitly stated) In comparison to the Fe2+ ion, the Fe3+ ion has a higher charge. OR The smaller size of Fe3+ allows it to get closer to a water molecule. d) Fe2+(aq) → Fe3+(aq) + e- e) 17.48 mL x 0.0350 mol KMnO4/1000 mL = 0.000612 mol KMnO4 0.000612 mol KMnO4 x 5 mol Fe2+/1 mol KMnO4 = 0.003059 mol Fe2+ 0.003059 mol Fe2+/0.0100 L = 0.306 M Fe2+ f) The volumetric flask is designed to contain only 25.00 mL precisely. g) 7.531 g Fe2O3 x 1 mol Fe2O3/159.70 g Fe2O3 = 0.04716 mol Fe2O3 0.04716 mol Fe2O3 x 2 mol Fe/1 mol Fe2O3 = 0.09431 mol Fe h) 0.09431 mol Fe x 55.85 g Fe/ 1 mol = 5.267 g Fe 5.267 g Fe/6.724 g sample x 100 = 78.33% (i) The calculated mass percent of Fe would be lower than the actual mass percent of Fe. A sample that contains any Feo (rather than Fe203) will have a higher actual mass percent of Fe than a completely oxidized sample would have. Therefore, when the moles of Fe are calculated (assuming all the mass of the sample is Fe203) the calculated number of moles of Fe, and hence the calculated mass percent of Fe, will be lower.

2018 FRQ #3 Answer the following questions relating to Fe and its ions, Fe2+ and Fe3+. ( a) Write the ground-state electron configuration of the Fe2+ ion. (b) The radii of the ions are given in the table above. Using principles of atomic structure, explain why the radius of the Fe2+ ion is larger than the radius of the Fe3+ ion. (c) Fe3+ ions interact more strongly with water molecules in aqueous solution than Fe2+ ions do. Give one reason for this stronger interaction, and justify your answer using Coulomb's law. A student obtains a solution that contains an unknown concentration of Fe2+(aq). To determine the concentration of Fe2+(aq) in the solution, the student titrates a sample of the solution with MnO4 (aq) − , which converts Fe2+(aq) to Fe3+(aq), as represented by the following equation. 5 Fe2+(aq) + MnO4 −(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) (d) Write the balanced equation for the half-reaction for the oxidation of Fe2+(aq) to Fe3+(aq). (e) The student titrates a 10.0 mL sample of the Fe2+(aq) solution. Calculate the value of [Fe2+] in the solution if it takes 17.48 mL of added 0.0350 M KMnO4(aq) to reach the equivalence point of the titration To deliver the 10.0 mL sample of the Fe2+(aq) solution in part (e), the student has the choice of using one of the pieces of glassware listed below. • 25 mL buret • • 25 mL graduated cylinder • 25 mL beaker 25 mL volumetric flask (f) Explain why the 25 mL volumetric flask would be a poor choice to use for delivering the required volume of the Fe2+(aq) solution. In a separate experiment, the student is given a sample of powdered Fe(s) that contains an inert impurity. The student uses a procedure to oxidize the Fe(s) in the sample to Fe2O3(s). The student collects the following data during the experiment. (g) Calculate the number of moles of Fe in the Fe2O3(s) produced. (h) Calculate the percent by mass of Fe in the original sample of powdered Fe(s) with the inert impurity. (i) If the oxidation of the Fe(s) in the original sample was incomplete so that some of the 7.531 g of product was FeO(s) instead of Fe2O3(s), would the calculated mass percent of Fe(s) in the original sample be higher, lower, or the same as the actual mass percent of Fe(s)? Justify your answer.

a) CS2 has only London dispersion forces, while COS has London dispersion forces and dipole-dipole forces. The London dispersion forces in CS2 are stronger than the combination of London dispersion forces and dipole-dipole forces in COS. b) 10.0 gCS2 x 1 mol CS2/76.13 g CS2 = 0.131 mol CS2 P = nRT/V = (0.131 mol)(0.08206 L atm mol-1K-1)(325 K)/5.0 L = 0.70 atm

2018 FRQ #4 The table above gives the molecular structures and boiling points for the compounds CS2 and COS. (a) In terms of the types and relative strengths of all the intermolecular forces in each compound, explain why the boiling point of CS2(l) is higher than that of COS(l). (b) A 10.0 g sample of CS2(l) is put in an evacuated 5.0 L rigid container. The container is sealed and heated to 325 K, at which temperature all of the CS2(l) has vaporized. What is the pressure in the container once all of the CS2(l) has vaporized?

a) HF is a weak acid and is only partially ionized. This fact is consistent with Figure 1, which shows that one out of eight (—13%) HF molecules is ionized (to form one H30+ and one F-). OR Figure 2 cannot represent HF because it represents 100% ionization of the acid. b) Assume [H30+] = [F-] in HF(aq) [h3O+]/0.0350 M = 0.130 ⇒ [H3O+] = 0.00455 M HF(aq) + H2O(l) ⇄ F-(aq) + H3O+(aq) I 0.0350 0 ~0 C -0.00455 +0.00455 +0.00455 E 0.0304 0.00455 0.00455 Ka = [H3O+][F-]/[HF] = (0.00455)2/(0.0304) = 6.81 x 10-4 c) The percent ionization of HF in the solution would increase. Doubling the volume of the solution decreases the initial concentration of each species by one-half; therefore, Q = (½[H3O+]i)(½[F-]i)/½[HF]i = ½Ka ⇒ Q<Ka∙ Consequently the equilibrium position will shift toward the products and increase the percent ionization. OR New volume = twice original volume, thus new [HF]i = 0.035/2 = 0.0175 M Ka = [H3O+][F-]/[HF] = 6.81 x 10-4 (value from part (b)) Let [H3o+][F-]/[HF] = [F-]= x Then 6.81 x 10-4 = (x)(x)/(0.0175 - x) ≈ x2/(0.0175) ⇒ x ≈ 0.00345 M Percent ionization = 0.00345 M/0.0175 M x 100 = 20% 20.% > 13.0%; therefore, the percent ionization increases.

2018 FRQ #5 HF(aq) + H2O(l) <-> F-(aq) + H3O+(aq) The ionization of HF(aq) in water is represented by the equation above. In a 0.0350 M HF(aq) solution, the percent ionization of HF is 13.0 percent. (a) Two particulate representations of the ionization of HF molecules in the 0.0350 M HF(aq) solution are shown below in Figure 1 and Figure 2. Water molecules are not shown. Explain why the representation of the ionization of HF molecules in water in Figure 1 is more accurate than the representation in Figure 2. (The key below identifies the particles in the representations.) (b) Use the percent ionization data above to calculate the value of Ka for HF. (c) If 50.0 mL of distilled water is added to 50.0 mL of 0.035 M HF(aq), will the percent ionization of HF(aq) in the solution increase, decrease, or remain the same? Justify your answer with an explanation or calculation.

a) The salt bridge is missing. The salt bridge allows for the migration of ions to maintain charge balance in each half-cell. b) (i) E°cell = E°red(cathode) - E°red(anode) +1.54 = +0.80 v — X x = +0.80 v - (+1.54 V) = -0.74 v (ii) 3 Ag+(aq) + Cr(s) 3 Ag(s) + Cr3+(aq) (iii) △G° = -nFE° = -(3 mol e-/1 molrxn)(96,485 c/mol e-)(1.54 j/c) = -4.46 x 105 J/molrxn

2018 FRQ #6 A student sets up a galvanic cell at 298 K that has an electrode of Ag(s) immersed in a 1.0 M solution of Ag (aq) and an electrode of Cr(s) immersed in a 1.0 M solution of Cr3+(aq), as shown in the diagram above. (a) The student measures the voltage of the cell shown above and discovers that it is zero. Identify the missing component of the cell, and explain its importance for obtaining a nonzero voltage. (b) The student adds the missing component to the cell and measures Ecell D to be +1.54 V. As the cell operates, Ag+ ions are reduced. Use this information and the information in the table above to do the following. (i) Calculate the value of for the half-reaction Cr 3+(aq) + 3 e− → Cr(s). (ii) Write the balanced net-ionic equation for the overall reaction that occurs as the cell operates. (iii) Calculate the value of ΔG° for the overall cell reaction in J/molrxn.

a) The element is nitrogen, N. b) k = 0.693/t½ = 0.693/10. min = 0.069 min-1 c) 64 → 32 → 16 → 8 → 4 → 2 → 1 6 half-lives are required. 6 x 10. min = 60. min 1n[A]t - 1n[A]0 = -kt t = 1n(l) - 1n(64)/-0.069 min-1 = 60. min

2018 FRQ #7 . The complete photoelectron spectrum of an element is represented above. (a) Identify the element. A radioactive isotope of the element decays with a half-life of 10. minutes. (b) Calculate the value of the rate constant, k, for the radioactive decay. Include units with your answer. (c) If 64 atoms of the radioactive isotope are originally present in a sample, what is the expected amount of time that will pass until only one atom of the isotope remains? Show how you arrived at your answer.

a) sp2 b) A dashed line should connect a hydrogen atom in water to a nitrogen or oxygen atom in urea or an oxygen atom in water to a hydrogen atom in urea. One possible correct response is shown above. c) 5.39 g H2NCONH2 x 1mol/60.6g = 0.0897 mol 0.0897 mol/0.00500 L= 17.9M d) The increased solubility at the higher temperature implies that the dissolution of urea is endothermic. If a saturated solution of urea is heated, then the equilibrium system is stressed. The stress is counteracted by the endothermic dissolution of more urea. e) mass of urea, mass of water, initial temperature of water, final temperature of solution f) △S°soln =S°(H2NCONH2(aq)) - S°(H2NCONH2(s)) 70.1 J/(mol∙K) = S°(H2NCONH2(aq)) - 104.6 J/(mol∙K) S°(H2NCONH2(aq)) = 174.7 J/(mol∙K) g) Urea molecules in solution have a greater number of possible arrangements than in solid urea. This increased number of arrangements corresponds to a positive △S°soln• h) Thermodynamic favorability for a process at standard conditions is determined by the sign of △G°, with △G° = △H° — T△S°. Since △S° is positive, the T△S° term makes the value of △G° smaller and thus makes the dissolution more thermodynamically favorable.

2019 FRQ #1 1. The compound urea, H2NCONH2 , is widely used in chemical fertilizers. The complete Lewis electron-dot diagram for the urea molecule is shown above. (a) Identify the hybridization of the valence orbitals of the carbon atom in the urea molecule. ( b) Urea has a high solubility in water, due in part to its ability to form hydrogen bonds. A urea molecule and four water molecules are represented in the box below. Draw ONE dashed line (----) to indicate a possible location of a hydrogen bond between a water molecule and the urea molecule. H2NCONH2(s) ←→ H2NCONH2(aq) The dissolution of urea is represented by the equation above. A student determines that 5.39 grams of H2NCONH2 (molar mass 60.06 g/mol) can dissolve in water to make 5.00 mL of a saturated solution at 20.°C. (c) Calculate the concentration of urea, in mol/L, in the saturated solution at 20.°C. (d) The student also determines that the concentration of urea in a saturated solution at 25°C is 19.8 M. Based on this information, is the dissolution of urea endothermic or exothermic? Justify your answer in terms of Le Chatelier's principle. (e) The equipment shown above is provided so that the student can determine the value of the molar heat of solution for urea. Knowing that the specific heat of the solution is 4.18 J/(g⋅°C), list the specific measurements that are required to be made during the experiment. (f) The entropy change for the dissolution of urea, change in enthalpy, is 70.1 J/(mol⋅K) at 25°C. Using the information in the table above, calculate the absolute molar entropy, S°, of aqueous urea. (g) Using particle-level reasoning, explain why the change in enthalpy soln is positive for the dissolution of urea in water. (h) The student claims that ΔS° for the process contributes to the thermodynamic favorability of the dissolution of urea at 25°C. Use the thermodynamic information above to support the student's claim.

a) 12 has the longest bond length because the radius of the I atom is greater than the radii of the other halogen atoms. Thus, the distance between the nuclei of atoms in 12 is greater than it is in smaller halogens. b) 2 Br- + Cl2 → Br2 + 2Cl- E° = E° (reduced species) - E° (oxidized species) = 1.36 V - 1.07 V = +0.29 V. Because E° for the reaction has a positive value, the reaction is thermodynamically favorable. c) The only intermolecular attractions in Br2(l) are London forces, while those in BrCl(l) include both London forces and dipole-dipole forces. However, due to the greater polarizability of the electron cloud of Br2 compared to that of BrCl, the London forces in Br2(l) are stronger than the combined intermolecular forces in BrCl(l). Thus, the boiling point of Br2(l) is greater than that of BrCl(l). d) P = nRT/V = (0.100 mol)(0.08206 L atm mol-1 K-1)(298 K)/2.00 L = 1.22 atm e) Keq = [Br2][Cl2]/[BrCl]2 or Keq = Pbr2Pcl2/(PBrCl)2 f) 2BrCl(g) ⇄ Br2(g) + Cl2(g) I 1.22 0 0 C -2x +x +x E 0.71 0.26 0.26 PBrCl decomposed = (0.42)(1.22 atm) = 0.51 atm 2x = 0.51 atm ⇒ x=0.26 atm Keq = (0.26)(0.26)/(0.71)2 = 0.13 g) △H° = ∑(bond energies)broken - ∑(bond energies)formed 1.6kJ/mol = 2(Br-Cl bond energy) - (193 kJ/mol +243 kJ/mol) (436 + 1.6) kJ/mol = 2(Br—Cl bond energy) Br—Cl bond energy = 219 kJ/mol

2019 FRQ #2 Answer the following questions relating to the chemistry of the halogens. ( a) The molecular formulas of diatomic bromine, chlorine, fluorine, and iodine are written below. Circle the formula of the molecule that has the longest bond length. Justify your choice in terms of atomic structure. Br2 Cl2 F2 I2 A chemistry teacher wants to prepare Br2 . The teacher has access to the following three reagents: NaBr(aq), Cl2(g), and I2(s). (b) Using the data in the table above, write the balanced equation for the thermodynamically favorable reaction that will produce Br2 when the teacher combines two of the reagents. Justify that the reaction is thermodynamically favorable by calculating the value of E° for the reaction. Br2 and Cl2 can react to form the compound BrCl. (c) The boiling point of Br2 is 332 K, whereas the boiling point of BrCl is 278 K. Explain this difference in boiling point in terms of all the intermolecular forces present between molecules of each substance. The compound BrCl can decompose into Br2 and Cl2 , as represented by the balanced chemical equation below 2 BrCl(g) <--> Br2(g) + Cl2(g) ΔH° = 1.6 kJ/molrxn A 0.100 mole sample of pure BrCl(g) is placed in a previously evacuated, rigid 2.00 L container at 298 K. Eventually the system reaches equilibrium according to the equation above. (d) Calculate the pressure in the container before equilibrium is established. (e) Write the expression for the equilibrium constant, Keq , for the decomposition of BrCl. After the system has reached equilibrium, 42 percent of the original BrCl sample has decomposed. (f) Determine the value of Keq for the decomposition reaction of BrCl at 298 K. (g) Calculate the bond energy of the Br-Cl bond, in kJ/mol, using ΔH° for the reaction (1.6 kJ/molrxn) and the information in the following table.

a) Ca2+(aq) + CO3-(aq) → CaCO3(s) b) The drawing shows one Ca2+ ion. c) 0.93g CaCO3 x 1 mol CaCO3/100.1 g = 0.0093 mol CaCO3 0.0093 mol CaCO3 x 1 mol NaCo3/1 mol CaCo3 = 0.0093 mol Na2Co3 d) Disagree. The presence of water in the solid will cause the measured mass of the precipitate to be greater than the actual mass of CaCO3. As a result, the calculated number of moles of CaCO3 and moles of Na2CO3 will be greater than the actual moles present. Therefore the calculated concentration of Na2C03(aq) will be too high. e) The liquid conducts electricity because ions (Na+(aq), Ca2+(aq), and N03-(aq)) are present in the solution. f) Determine the pH of the solution using a pH meter. g) First determine [OH ] using POH = 14 - pH, then [OH ] = 10-POH Then, use the Kb expression and an ICE table (see example below) to determine [C032-] and [HC03-] at equilibrium. The initial concentration of C032- , Ci , is equal to the sum of the equilibrium concentrations of CO32- and HCO3-. CO32-(aq) + H2O(l) ⇄ HCO3-(aq) + OH-(aq) I ci --- ⇄ 0 0 C -x --- ⇄ +x +x E ci-x --- ⇄ x x Kb = (x)(x)/ci-x ⇒ ci = (x)(x)/Kb + x g) Less than. The small value of Kb, 2.1 x 10-4, indicates that the reactants are favored. h) No, the Na2C03 solution is not suitable. The pKa of HC03- is 10.32. Buffers are effective when the required pH is approximately equal to the pKa of the weak acid. An acid with a pKa of 10.32 is not appropriate to prepare a buffer with a pH of 6.

2019 FRQ #3 A student is given 50.0 mL of a solution of Na2CO3 of unknown concentration. To determine the concentration of the solution, the student mixes the solution with excess 1.0 M Ca(NO3)2(aq), causing a precipitate to form. The balanced equation for the reaction is shown below. Na2CO3(aq) + Ca(NO3)2(aq) → 2 NaNO3(aq) + CaCO3(s) (a) Write the net ionic equation for the reaction that occurs when the solutions of Na2CO3 and Ca(NO3)2 are mixed. (b) The diagram below is incomplete. Draw in the species needed to accurately represent the major ionic species remaining in the solution after the reaction has been completed. The student filters and dries the precipitate of CaCO3 (molar mass 100.1 g/mol) and records the data in the table below. (c) Determine the number of moles of Na2CO3 in the original 50.0 mL of solution. (d) The student realizes that the precipitate was not completely dried and claims that as a result, the calculated Na2CO3 molarity is too low. Do you agree with the student's claim? Justify your answer. (e) After the precipitate forms and is filtered, the liquid that passed through the filter is tested to see if it can conduct electricity. What would be observed? Justify your answer. The student decides to determine the molarity of the same Na2CO3 solution using a second method. When Na2CO3 is dissolved in water, CO3 2−(aq) hydrolyzes to form HCO3 −(aq), as shown by the following equation. CO3 2−(aq) + H2O(l) <-> HCO3 −(aq) + OH−(aq) Kb = [HCO3-][OH-] / [CO3^2- ] = 2.1 × 10−4 (f) The student decides to first determine [OH−] in the solution, then use that result to calculate the initial concentration of CO3 2−(aq). (i) Identify a laboratory method (not titration) that the student could use to collect data to determine [OH−] in the solution. (ii) Explain how the student could use the measured value in part (f)(i) to calculate the initial concentration of CO3 2−(aq). (Do not do any numerical calculations.) (g) In the original Na2CO3 solution at equilibrium, is the concentration of HCO3 −(aq) greater than, less than, or equal to the concentration of CO3 2−(aq) ? Justify your answer. (h) The student needs to make a CO3 2−/HCO3 − buffer. Is the Na2CO3 solution suitable for making a buffer with a pH of 6? Explain why or why not

a) The average speed of the molecules increases as temperature increases. b) P1/T1 ⇒ 0.70 atm/299 k = P2/425 K ⇒ P2=0.99 atm c) Faster-moving gas particles collide more frequently with the walls of the container, thus increasing the pressure. OR Faster-moving gas particles collide more forcefully with the walls of the container, thus increasing the pressure. d) The attractive forces between C02 molecules result in a pressure that is lower than that predicted by the ideal gas law.

2019 FRQ #4 . A student is doing experiments with CO2(g). Originally, a sample of the gas is in a rigid container at 299 K and 0.70 atm. The student increases the temperature of the CO2(g) in the container to 425 K. (a) Describe the effect of raising the temperature on the motion of the CO2(g) molecules. (b) Calculate the pressure of the CO2(g) in the container at 425 K. (c) In terms of kinetic molecular theory, briefly explain why the pressure of the CO2(g) in the container changes as it is heated to 425 K. (d) The student measures the actual pressure of the CO2(g) in the container at 425 K and observes that it is less than the pressure predicted by the ideal gas law. Explain this observation.

a) 1s2 2s2 2p6 3s2 3p6 4s2 or [Ar] 4s2 element Ca b) Energy (E) required = 0.980 x 10-18 J E = hv = hc/ʎ ⇄ ʎ = hc/E ʎ = (6.626 x 10-34 Js)(2.998 x 108 ms-1)/0.980 x 10-18J ʎ = 2.03 x 10-7 m

2019 FRQ #5 The complete photoelectron spectrum of an element in its ground state is represented below. (a) Based on the spectrum, (i) write the ground-state electron configuration of the element, and (ii) identify the element. (b) Calculate the wavelength, in meters, of electromagnetic radiation needed to remove an electron from the valence shell of an atom of the element.

a) The linear graph of 1/[NO2] vs. time indicates a second-order reaction. b) rate = k[NO2]2 c) Mechanism 1 Step 1: NO2(g) + NO2(g) → NO(g) + NO3(g) Step 2: NO3(g) → NO(g) + O2(g) Yes. Step 1 is slow, therefore it is the rate-determining step of this mechanism. The rate law of this elementary reaction is rate = k[NO2][NO2] = k[NO2]2 which is consistent with the second-order rate law in part (b). Mechanism 2 Step 1: NO2(g) + NO2(g) ⇄ N2O4(g) Step 2: N2O4(g) → 2NO(g) + O2(g) Yes. Step 2 is slow; therefore, it is the rate- determining step of this mechanism. The rate law of this elementary reaction is rate = k[N2O4]. Because N2O4 is an intermediate, it cannot appear in the rate law of the overall reaction. Because Keq = [N2O4]/[NO2]2 in step 1, [N204] = Keq[NO2]2. Then, substituting Keq[NO2]2 for [N2O4] in the rate law of step 2 gives rate = (k Keq)[NO2]2, which is consistent with the rate law in part (b).

2019 FRQ #6 Nitrogen dioxide, NO2(g), is produced as a by-product of the combustion of fossil fuels in internal combustion engines. At elevated temperatures NO2(g) decomposes according to the equation below. 2 NO2(g) → 2 NO(g) + O2(g) The concentration of a sample of NO2(g) is monitored as it decomposes and is recorded on the graph directly below. The two graphs that follow it are derived from the original data. (a) Explain how the graphs indicate that the reaction is second order. (b) Write the rate law for the decomposition of NO2(g). (c) Consider two possible mechanisms for the decomposition reaction. (i) Is the rate law described by mechanism I shown below consistent with the rate law you wrote in part (b) ? Justify your answer. (ii) Is the rate law described by mechanism II shown below consistent with the rate law you wrote in part (b) ? Justify your answer.

a) MnO4- is reduced to Mn2+ as the oxidation number of Mn changes from +7 to +2, indicating a gain of 5 electrons. b) 29.55 mL- 3.35 mL = 26.20 mL c) (0.02620 mol/L) = 0.000616 mol d) No. The 0.00143 M titrant solution is so diluted that the volume of titrant needed to reach the end point would be much greater than the 50 mL capacity of the buret.

2019 FRQ #7 6 H+(aq) + 2 MnO4 −(aq) + 5 H2C2O4(aq) → 10 CO2(g) + 8 H2O(l) + 2 Mn2+(aq) A student dissolved a 0.139 g sample of oxalic acid, H2C2O4 , in water in an Erlenmeyer flask. Then the student titrated the H2C2O4 solution in the flask with a solution of KMnO4 , which has a dark purple color. The balanced chemical equation for the reaction that occurred during the titration is shown above. (a) Identify the species that was reduced in the titration reaction. Justify your answer in terms of oxidation numbers. (b) The student used a 50.0 mL buret to add the KMnO4(aq) to the H2C2O4(aq) until a faint lavender color was observed in the flask, an indication that the end point of the titration had been reached. The initial and final volume readings of the solution in the buret are shown below. Write down the initial reading and the final reading and use them to determine the volume of KMnO4(aq) that was added during the titration. (c) Given that the concentration of KMnO4(aq) was 0.0235 M, calculate the number of moles of MnO4 − ions that completely reacted with the H2C2O4 . \(d) The student proposes to perform another titration using a 0.139 g sample of H2C2O4 , but this time using 0.00143 M KMnO4(aq) in the buret. Would this titrant concentration be a reasonable choice to use if the student followed the same procedure and used the same equipment as before? Justify your response.

a) Ka= [H3O+] [HCOO-] / [HCOOH] b) ICE table; [H3O ] = x, then 1.8 x 10 ^-4 = x^2/ (0.25 ) ; x=0.0067M pH= -log[H3O+]= -log(0.0067)=2.17 c) *double bond from C to above O d) i) H2NNH2 + HCOOH <-> H2NNH3+ +HCOO- ii) Acidic. The Ka of H2NNH3 + is greater than the Kb of HCOO−, so the production of H3O+(aq) occurs to a greater extent than the production of OH−(aq). e) Yes. The oxidation number of hydrogen changes from +1 in HCOOH to zero in H2. OR • Yes. The oxidation number of carbon changes from +2 in HCOOH to +4 in CO2 f) 24 atm total × 1 atm CO2 / 2 atm of product = 12 atm CO2; then PV=nRT = 2.1 mol CO2 g) It would remain the same. In a catalyzed reaction the net amount of catalyst is constant.

2021 FRQ #1 HCOOH(aq) + H2O(l) <--> H3O (aq ) + HCOO (aq); Ka= 1.8 x 10^-4 (a) Write the expression for the equilibrium constant, Ka, for the reaction. (b) Calculate the pH of a 0.25 M solution of HCOOH. (c) In the box below, complete the Lewis electron-dot diagram for HCOOH. Show all bonding and nonbonding valence electrons. H2NNH2 (aq ) + H2O(l) <-> H2NNH3 (aq ) + OH (aq) Kb= 1.3 x 10^-6 d) In aqueous solution, the compound H2NNH2 reacts according to the equation above. A 50.0 mL sample of M H2NNH2 0.25 aq)( is combined with a 50.0 mL sample of 0.25 M HCOOH( ) aq . (i) Write the balanced net ionic equation for the reaction that occurs when H2NNH2 is combined with HCOOH. (ii) Is the resulting solution acidic, basic, or neutral? Justify your answer. When a catalyst is added to a solution of HCOOH(aq), the reaction represented by the following equation occurs. HCOOH(aq )→ H2(g ) + CO2 (e) Is the reaction a redox reaction? Justify your answer (f) The reaction occurs inarigid 4.3 L vessel at 25°C, and the total pressure is monitored, as shown in the graph above. The vessel originally did not contain any gas. Calculate the number of moles of CO2( )g produced in the reaction. (Assume that the amount of CO2(g) dissolved in the solution is negligible.) (g) After the reaction has proceeded for several minutes, does the amount of catalyst increase, decrease, or remain the same? Justify your answer

a) i) 14 protons and 14 neutrons ii) 1s2 2s2 2p6 3s2 3p2 OR [Ne] 3s2 3p b) SiH4 is composed of molecules, for which the only intermolecular forces are London dispersion forces. SiO2 is a network covalent compound with covalent bonds between silicon and oxygen atoms. London dispersion forces are much weaker than covalent bonds, so SiH4 boils at a much lower temperature than SiO2. c) SiH4 (g) -> Si(s) + 2 H2 (g) d) The H2(g) molecules are more highly dispersed than the Si(s) atoms and, therefore, have a higher absolute molar entropy. Silicon is a solid; therefore, its atoms are in fixed positions, are less dispersed, and have a lower absolute molar entropy e) S= (18 + 2(131))-205= +75 J/(molrxn * K) f) High temperature is required for the reactant particles to have sufficient thermal energy to overcome the activation energy of the reaction. g) * a little to the right of 1 and up 2 levels h) The valence electrons of a Ge atom occupy a higher shell (n=4) than those of a Si atom (n=3), so the average distance between the nucleus and the valence electrons is greater in Ge than in Si. This greater separation results in weaker Coulombic attractions between the Ge nucleus and its valence electrons, making them less tightly bound and, therefore, easier to remove compared to those in Si. i) E= hv = h(c/wavelength) = 6.626 x 10 ^-34 J *s (2.998 x 10^8 m s-1/ 4.00 x 10 ^-7 m) = 4.97 x 10 ^-19

2021 FRQ #2 Answer the following questions about the element Si and some of its compounds. (a) The mass spectrum of a pure sample of Si is shown below. (i) How many protons and how many neutrons are in the nucleus of an atom of the most abundant isotope of Si ? (ii) Write the ground-state electron configuration of Si. Two compounds that contain Si are SiO2 and SiH4. (b) At 161 K, SiH4 boils but SiO2 remains as a solid. Using principles of interparticle forces, explain the difference in boiling points. At high temperatures, SiH4 decomposes to form solid silicon and hydrogen gas. (c) Write a balanced equation for the reaction. A table of absolute entropies of some substances is given below (d) Explain why the absolute molar entropy of Si(s) is less than that of H2(g). (e) Calculate the value, in J/(mol*K), of ∆S° for the reaction. (f) The reaction is thermodynamically favorable at all temperatures. Explain why the reaction occurs only at high temperatures. (g) A partial photoelectron spectrum of pure Si is shown below. On the spectrum, draw the missing peak that corresponds to the electrons in the 3p sublevel. (h) Using principles of atomic structure, explain why the first ionization energy of Ge is lower than that of Si (i) A single photon with a wavelength of 4.00 10−7 m is absorbed by the Si sample. Calculate the energy of the photon in joules.

a) q=mcT (15.0 g)(0.72 J/(g C))(39.7 C 22.0 C)=190 J b) qsys= -qsurr; -190 J x 1 kJ/1000 J x 1 mol rxn/ -1650 KJ= 0.00012 mol rxn; 0.00012 mol x 4 mol Fe/1mol x 55.85 g Fe/ 1 mole Fe= 0.027 g Fe c) Greater than. A greater mass of iron provides a greater number of moles of reaction, which would transfer a greater quantity of thermal energy to the same mass of sand and therefore lead to a greater maximum temperature.

2021 FRQ #4 4 Fe(s) + 3O2 (g) -> 2Fe2O3 (s) H= -1650 KJ/mol 4. A student investigates a reaction used in hand warmers, represented above. The student mixes Fe(s) with a catalyst and sand in a small open container. The student measures the temperature of the mixture as the reaction proceeds. The data are given in the following table. (a) The mixture (Fe(s), catalyst, and sand) has a total mass of 15.0 g and a specific heat capacity of 0.72 J/(g°C). Calculate the amount of heat absorbed by the mixture from 0 minutes to 4 minutes (b) Calculate the mass of Fe(s), in grams, that reacted to generate the amount of heat calculated in part (a). (c) In a second experiment, the student uses twice the mass of iron as that calculated in part (b) but the same mass of sand as in the first experiment. Would the maximum temperature reached in the second experiment be greater than, less than, or equal to the maximum temperature in the first experiment? Justify your answer.

a) goes towards Mg (l) under power source b) No, because 2.0 V is less than 3.73 V, which is the minimum voltage needed for electrolysis to occur. Ecell = -2.37 V + (-1.36 V)= -3.73 V c) 2.00 g Mg x 1 mol Mg/24.30 g Mg x 2 mol e- / 1 mol Mg= 0.165 mol e- 0.165 mol e- x 96,485 C/ 1 mol e- x 1s/5.00 C= 3,180 s

2021 FRQ #5 Molten MgCl2 can be decomposed into its elements if a sufficient voltage is applied using inert electrodes. The products of the reaction are liquid Mg (at the cathode) and Cl2 gas (at the anode). A simplified representation of the cell is shown above. The reduction half-reactions related to the overall reaction in the cell are given in the table. (a) Draw an arrow on the diagram to show the direction of electron flow through the external circuit as the cell operates. (b) Would an applied voltage of 2.0 V be sufficient for the reaction to occur? Support your claim with a calculation as part of your answer. (c) If the current in the cell is kept at a constant 5.00 amps, how many seconds does it take to produce 2.00 g of Mg(l) at the cathode?

a) Ionic solids do not have free-moving ions that are required to carry an electric current. Therefore, there is no conduction of electricity b) CaSO4. The greater electrical conductivity of the CaSO4 solution relative to the PbSO4 solution implies a higher concentration of ions, which comes from the dissolution (dissociation) of CaSO4 to a greater extent. c)The drawing shows solid PbSO4 at the bottom of the beaker (similar to the solid shown for CaSO4) and fewer dissociated Pb2+ and SO4 2− ions in the solution d) The additional precipitate is CaSO4 that forms in response to the increased [SO4 2−] in solution. According to Le Chatelier's principle (Q > Ksp), the introduction of SO4 2− as a common ion shifts the equilibrium towards the formation of more CaSO4(s).

2021 FRQ #6 A student is studying the properties of CaSO4 and PbSO4. The student has samples of both compounds, which are white powders. (a) The student tests the electrical conductivity of each solid and observes that neither solid conducts electricity. Describe the structures of the solids that account for their inability to conduct electricity The student places excess CaSO4(s) in a beaker containing 100 mL of water and places excess PbSO4(s) in another beaker containing 100 mL of water. The student stirs the contents of the beakers and then measures the electrical conductivity of the solution in each beaker. The student observes that the conductivity of the solution in the beaker containing the CaSO4(s) is higher than the conductivity of the solution in the beaker containing the PbSO4(s). (b) Which compound is more soluble in water, CaSO4(s) or PbSO4(s) ? Justify your answer based on the results of the conductivity test. The left side of the diagram below shows a particulate representation of the contents of the beaker containing the CaSO4(s) from the solution conductivity experiment. (c) Draw a particulate representation of PbSO4(s) and the ions dissolved in the solution in the beaker on the right in the diagram. Draw the particles to look like those shown to the right of the beaker. Draw an appropriate number of dissolved ions relative to the number of dissolved ions in the beaker on the left. (d) The student attempts to increase the solubility of CaSO4 (s) by adding 10.0 mL of 2 M H2SO4 to the beaker, and observes that additional precipitate forms in the beaker. Explain this observation

a) Ba^2+ (aq) + SO4^2- (aq) -> BaSO4 (s) b) 1.136 g - 0.764 g = 0.372 g BaSO4 0.372 g x 1 mol/233.39 g = 0.00159 mol c) 0.00159 mol BaSO4 x 1 mol CuSO4 / 1 mol BaSO4 = 0.00159 mol CuSO4 ; 0.00159 mol CuSO4/ 0.05000 L = 0.0318 M CuSO4 d) VOOM VOOM; (0.0500 M) (50.00mL) / (0.1000 M)= 25.0 mL e) First, measure out the correct volume of 0.1000 M CuSO4 solution with a 25.0 mL volumetric pipet (graduated cylinder or buret is acceptable). Transfer the 25.0 mL of solution to a 50.00 mL volumetric flask and dilute the solution with water up to the 50.00 mL mark. f) between 0.032 M and 0.038 M g) The concentration will be less than that determined in part (f). The additional water will decrease the concentration of CuSO4 in the cuvette. Therefore, there will be a decrease in absorbance (according to the Beer-Lambert law). This dilution results in a lower estimated concentration of CuSO4

2021 FRQ# 3 A student is given the task of determining the molar concentration of a CuSO4 solution using two different procedures, precipitation and spectrophotometry. For the precipitation experiment, the student adds 20.0 mL of 0.200 M Ba(NO3)2 to 50.0 mL of the CuSO4 (aq) . The reaction goes to completion and a white precipitate forms. The student filters the precipitate and dries it overnight. The data are given in the following table. (a) Write a balanced net ionic equation for the precipitation reaction. (b) Calculate the number of moles of precipitate formed. (c) Calculate the molarity of the original CuSO4 solution. For the spectrophotometry experiment, the student first makes a standard curve. The student uses a 0.1000 M solution of CuSO4 (aq) to make three more solutions of known concentration (0.0500 M, 0.0300 M, and 0.0100 M) in 50.00 mL volumetric flasks. (d) Calculate the volume of 0.1000 M CuSO4(aq) needed to make 50.00 mL of 0.0500 M CuSO4(aq). (e) Briefly describe the procedure the student should follow to make 50.00 mL of 0.0500 M CuSO4(aq) using 0.1000 M CuSO4aq, a 50.00 mL volumetric flask, and other standard laboratory equipment. Assume that all appropriate safety precautions will be taken. The standard curve is given below. (f) The absorbance of the CuSO4 solution of unknown concentration is 0.219. Determine the molarity of the solution. (g) A second student performs the same experiment. There are a few drops of water in the cuvette before the second student adds the CuSO4 aq solution of unknown concentration. Will this result in a CuSO4 aq concentration for the unknown that is greater than, less than, or equal to the concentration determined in part (f) ? Justify your answer.

a) 0.325 mol O2 x 32.00 g O2/ 1 mol O2= 10.4 g O2 ; 10.4 g/7.95 L = 1.31 g/L b) No, the density of the gas remains constant because P, R, and T remain constant AND the mass and volume of O2 decrease proportionately. c) As the gas cools, the average kinetic energy (speed) of the O2 molecules decreases. The molecules rebound with less energy when they collide with each other and the walls of the container. The spacing between particles decreases, causing the volume occupied by the gas to decrease d) The ideal gas law assumes that gas particles do not experience interparticle attractions. As a real gas cools further, the intermolecular forces have greater effect as the average speed of the molecules decreases, resulting in inelastic collisions. To maintain a gas pressure of 1.00 atm, the volume must decrease to accommodate more collisions with less energy

FRQ #7 A student investigates gas behavior using a rigid cylinder with a movable piston of negligible mass, as shown in the diagram above. The cylinder contains 0.325 mol of O2(g). (a) The cylinder has a volume of 7.95 L at 25°C and 1.00 atm. Calculate the density of the O2(g), in g/L, under these conditions. (b) Attempting to change the density of the O2(g), the student opens the valve on the side of the cylinder, pushes down on the piston to release some of the gas, and closes the valve again. The temperature of the gas remains constant at 25°C. Will this action change the density of the gas remaining in the cylinder? Justify your answer. (c) The student tries to change the density of the O2(g) by cooling the cylinder to −55°C, which causes the volume of the gas to decrease. Using principles of kinetic molecular theory, explain why the volume of the O2(g) decreases when the temperature decreases to −55°C. (d) ) The student further cools the cylinder to −180°C and observes that the measured volume of the O2(g) is substantially smaller than the volume that is calculated using the ideal gas law. Assume all equipment is functioning properly. Explain why the measured volume of the O2(g) is smaller than the calculated volume. (The boiling point of O2( )l is −183°C.)