Hooperslab.weebly.com



Equilibrium - Free Response Questions from 2011-20182011 AP Chemistry Free Response – Form B1.Answer the following questions about the solubility and reactions of the ionic compounds M(OH)2 and MCO3, where M represents an unidentified metal.a.) Identify the charge of the M ion in the ionic compounds above.b.) At 25 oC, a saturated solution of M(OH)2 has an [OH-] = 1.4x10-5 Mi.) Write the solubility product expression for M(OH)2.ii.) Calculate the value of the solubility product constant, Ksp, for M(OH)2 at 25 oC.c.) For the metal carbonate, MCO3, the value of the solubility product constant, Ksp, is 7.4 x 10-14 at 25 oC. On the basis of this information and your results in part b), which compound, M(OH)2 or MCO3, has the greater molar solubility in water at 25 oC? Justify your answer with a calculation.d.) MCO3 decomposes at high temperatures, as shown by the reaction represented below.MCO3(s) ? MO(s) + CO2(g)A sample of MCO3 is placed in a previously evacuated container, heated to 423 K, and allowed to come to equilibrium. Some solid MCO3 remains in the container. The value of KP for the reaction at 423 K is 0.0012.i.) Write the equilibrium constant expression for KP of the reaction.ii.) Determine the pressure, in atm, of CO2(g) in the container at equilibrium at 423 K.iii.) Indicate whether the value of ΔG for the reaction at 423 K is positive, negative, or zero. Justify your answer.Rubric 2011 Form B Question 1 (10 points) Answer the following questions about the solubility and reactions of the ionic compounds M(OH)2 and MCO3, where M represents an unidentified metal. (a) Identify the charge of the M ion in the ionic compounds above. + 1 point is earned for the correct charge. b.) At 25 oC, a saturated solution of M(OH)2 has an [OH-] = 1.4x10-5 M(i) Write the solubility-product constant expression for M(OH)2 . Ksp = [M2+] [OH?]2 1 point is earned for the correct expression. (ii) Calculate the value of the solubility-product constant, Ksp , for M(OH)2 at 25°C. [M2+] = ? [OH?]= ? (1.4 × 10?5 M) = 7.0 × 10?6 1 point is earned for the correct relationship between [M2+] and [OH?]. Ksp = [M2+] [OH?]2 = (7.0 × 10?6)(1.4 × 10?5)2 = 1.4 × 10?15 1 point is earned for the correct value. (c) For the metal carbonate, MCO3, the value of the solubility-product constant, Ksp, is 7.4 × 10?14 at 25°C. On the basis of this information and your results in part (b), which compound, M(OH)2 or MCO3, has the greater molar solubility in water at 25°C? Justify your answer with a calculation. For M(OH)2 : [M2+] and molar solubility = 7.0 × 10?6 M For MCO3: Ksp = 7.4 × 10?14 = [M2+][CO32?][M2+] and molar solubility = 2.7 × 10?7 M Because 7.0 × 10?6 M > 2.7 × 10?7 M, M(OH)2 has the greater molar solubility. 1 point is earned for the molar solubility of MCO3.1 point is earned for an answer consistent with the calculated molar solubility. 2012 B FRQ # 33.CaSO4·2H2O(s) CaSO4(s) + 2 H2O(g)The hydrate CaSO4·2H2O(s) can be heated to form the anhydrous salt, CaSO4(s), as shown by the reaction represented above.a.Using the data in the table below, calculate the value of ΔGo, in kJ / molrxn, for the reaction at 298 K.SubstanceΔGfo at 298 K (kJ / mol)CaSO4·2H2O(s)-1795.70CaSO4(s)-1320.30H2O(g)-228.59b.Given that the value of ΔHo for the reaction at 298 K is +105 kJ / molrxn, calculate the value of ΔSo for the reaction at 298 K. Include units with your answer. A sample of CaSO4·2H2O(s) is placed in a cylinder with a movable piston as shown in the diagram below. The air above the solid is at 1.00 atm and is initially dry (partial pressure of H2O(g) = 0 atm.) c. Write the expression for the equilibrium constant, Kp, for the reaction. d. Given that the equilibrium constant, Kp is 6.4 x 10-4 at 298 K, determine the partial pressure, in atm, of water vapor in the cylinder at equilibrium at 298 K. e. If the volume of the system is reduced to one-half of its original volume and the system is allowed to reestablish equilibrium at 298 K, what will be the pressure, in atm, of the water vapor at the new volume? Justify your answer.In the laboratory, the hydrate CaSO4·2H2O(s) can be heated in a crucible to completely drive off the water of hydration to form the anhydrous salt, CaSO4(s) f. A 2.49 g sample of pure CaSO4·2H2O(s) is heated several times until the mass is constant. Calculate the mass, in grams, of the solid that remains after the dehydration reaction is complete.Rubric 2012 BFRQ #3 2013 #1 Free Response1.Answer the following questions about the solubility of some fluoride salts of alkaline earth metals.a. A student prepares 100. mL of a saturated solution of MgF2 by adding 0.50 g of solid MgF2 to 100. mL of distilled water at 25 oC and stirring until no more solid dissolves. (Assume that the volume of the undissolved MgF2 is neglibibly small.) The saturated solution is analyzed, and it is determined that [F-] in the solution is 2.4 x 10-3 M.i. Write the chemical equation for the dissolving of solid MgF2 in water.ii. Calculate the number of moles of MgF2 that dissolved.iii. Determine the value of the solubility product constant, Ksp, for MgF2 at 25C.b. A beaker contains 500. mL of a solution in which both Ca2+(aq) and Ba2+(aq) are present at a concentration of 0.10 M at 25oC. A student intends to separate the ions by adding 0.20 M NaF solution one drop at a time from a buret. At 25 oC the value of Ksp for CaF2 is 3.5 x 10-11; the value of Ksp for BaF2 is 1.8 x 10-6.i. Which salt will precipitate first, CaF2 or BaF2? Justify your answer.For parts b ii) and b iii) below, assume that the addition of the NaF solution does not significantly affect the total volume of the liquid in the beaker.ii. Calculate the minimum concentration of F-(aq) necessary to initiate precipitation of the salt selected in part b i.iii. Calculate the minimum volume of 0.20 M NaF that must be added to the beaker to initiate precipitation of the salt selected in part b i.c. There are several ways to dissolve salts that have limited solubility. Describe one procedure to redissolve the precipitate formed in part b.Rubric2013 Form B Free Response1. NH4Cl(s) NH3(g) + HCl(g)When solid ammonium chloride is heated, it decomposes as represented above. The value of Kp for the reaction is 0.0792 at 575 K. A 10.0 g sample of solid ammonium chloride is placed in a rigid, evacuated 3.0 L container that is sealed and heated to 575 K. The system comes to equilibrium with some solid NH4Cl remaining in the container.a. Write the expression for the equilibrium constant for the reaction in terms of partial pressures (i.e. Kp).b. Calculate the partial pressure of NH3(g), in atm, at equilibrium at 575 K.c. A small amount of NH3(g) is injected into the equilibrium mixture in the 3.0 L container at 575 K.i. As the new equilibrium is being established at 575 K, does the amount of NH4Cl(s) in the container increase, decrease, or remain the same? Justify your answer.ii. After the new equilibrium is established at 575 K, is the value of Kp greater than, less than, or equal to the value before the NH3(g) was injected into the container? Justify your answer.d. When the temperature of the container is lowered to 500 K, the number of moles of NH3(g) in the container decreases. On the basis of this observation, is the decomposition of NH4Cl(s) endothermic or exothermic? Justify your answer.Rubric2014 AP Chemistry Form AMass of KI tablet0.425 gMass of thoroughly dried filter paper1.462 gMass of filter paper + precipitate after first drying1.775 gMass of filter paper + precipitate after second drying1.699 gMass of filter paper + precipitate after third drying1.698 g1. A student is given the task of determining the I- content of tablets that contain KI and an inert, water-soluble sugar as a filler. A tablet is dissolved in 50.0 mL of distilled water, and an excess of 0.20 M Pb(NO3)2(aq) is added to the solution. A yellow precipitate forms, which is then filtered, washed, and dried. The data from the experiment are shown in the table above.a. For the chemical reaction that occurs when the precipitate forms,i. write a balanced, net-ionic equation for the reaction, andii. explain why the reaction is best represented by a net ionic equation.b. Explain the purpose of drying and weighing the filter paper with the precipitate three times.c. In the filtrate solution, is [K+] greater than, less than, or equal to [NO3-]? Justify your answer.d. Calculate the number of moles of precipitate that is produced in the experiment.e. Calculate the mass percent of I- in the tablet.f. In another trial, the student dissolves a tablet in 55.0 mL of water instead of 50.0 mL of water. Predict whether the experimentally determined mass percent of I- will be greater than, less than, or equal to the amount calculated in part (e). Justify your answer.g. A student in another lab also wants to determine the I- content of a KI tablet but does not have access to Pb(NO3)2. However, the student does have access to 0.20 M AgNO3, which reacts with I-(aq) to produce AgI(s). The value of Ksp for AgI is 8.5 x 10-17.i. Will the substitution of AgNO3 for Pb(NO3)2 result in the precipitation of the I- ion from solution? Justify your answer.ii. The student only has access to one KI tablet and a balance that can measure to the nearest 0.01 g. Will the student be able to determine the mass of AgI produced to three significant figures? Justify your answer.2015 #44. Answer the following questions about the solubility of Ca(OH)2 (Ksp = 1.3 x 10-6).a. Write a balanced chemical equation for the dissolution of Ca(OH)2(s) is 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. Represent water molecules as Rubric2. Answer the following questions about Fe and Al compounds.a. Fe2O3(s) and Al2O3(s) have similar chemical properties; some similarities are due to the oxides having similar lattice energies. Give two reasons why the lattice energies of the oxides are similar.Use the following reactions that involve Fe and Al compounds to answer parts b and c.In distilled waterReaction 1:Fe2O3(s) + 3 H2O(l) 2 Fe(OH)3(s)Reaction 2:Al2O3(s) + 3 H2O(l) 2 Al(OH)3(s)In baseReaction 3:Fe(OH)3(s) + NaOH(aq) no reactionReaction 4:Al(OH)3(s) + NaOH(aq) NaAl(OH)4(aq)In acidReaction 5:Fe(OH)3(s) + 3 HCl(aq) FeCl3(aq) + 3 H2O(l)Reaction 6:Al(OH)3(s) + 3 HCl(aq) AlCl3(aq) + 3 H2O(l)Reaction 7:NaAl(OH)4(aq) + HCl(aq) Al(OH)3(s) + NaCl(aq) + H2O(l)When heatedReaction 8:2 Fe(OH)3(s) heat Fe2O3(s) + 3 H2O(g)Reaction 9:2 Al(OH)3(s) heat Al2O3(s) + 3 H2O(g)CompoundKspFe(OH)34 x 10-38Al(OH)31 x 10-33b. The Ksp values of Fe(OH)3 and Al(OH)3 are given in the table above. A 1.0 g sample of powdered Fe2O3(s) and a 1.0 g sample of powdered Al2O3(s) are mixed together and placed in a 1.0 L of distilled water.i. Which ion, Fe3+(aq) or Al3+(aq), will be present in the higher concentration? Justify your answer with respect to the Ksp values provided.ii. Write a balanced chemical equation for the dissolution reaction that results in the production of the ion that you identified in part i.c. Students are asked to develop a plan for separating Al2O3(s) from a mixture of powdered Fe2O3(s) and powdered Al2O3(s) using chemical reactions and laboratory techniques.i. One student proposes that Al2O3(s) can be separated from the mixture by adding water to the mixture and then filtering. Explain why this approach is not reasonable.ii. A second student organizes a plan using a table. The first two steps have already been entered in the table, as shown below. Complete the plan by listing the additional steps that are needed to recover the Al2O3(s). List the steps in the correct order and refer to the appropriate reaction by number, if applicable.StepDescriptionReactions1Add NaOH(aq) to convert Al2O3(s) to Al(OH)3(s) and then to NaAl(OH)4(aq)2 and 42Filter out the solid Fe(OH)3 from the mixture and save the filtrate.345iii. The second student recovers 5.5 g of Al2O3(s) from a 10.0 g sample of the mixture. Calculate the percent of Al by mass in the mixture of the two powdered oxides. (The molar mass of Al2O3 is 101.96 g/mol and the molar mass of Fe2O3 is 159.70 g/mol.)Rubric2015 B 5. Answer the following questions about two isomers, methyl methanoate and ethanoic acid. The molecular formula of the compounds is C2H4O2.a. Complete the Lewis electron dot diagram of methyl methanoate in the box below. Show all valence electrons.A student puts 0.020 mol of methyl methanoate into a previously evacuated rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid instead of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic acid is due to the following reversible reactionCH3COOH(g) + CH3COOH(g) (CH3COOH)2(g)b. Assume that when equilibrium has been reached, 50. percent of the ethanoic acid molecules have reacted.i. Calculate the total pressure in the vessel at equilibrium at 450 K.ii. Calculate the value of the equilibrium constant, KP, for the reaction at 450 K.Rubric20166.Ba2+(aq) + EDTA4-(aq) Ba(EDTA)2-(aq)K = 7.7 x 107The polyatomic ion C10H12N2O84- 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 1.00 L. After equilibrium has been reestablished, is the number of moles of Ba2+(aq) 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 is diluted? Justify your answer.Rubric2016 International has been left out here because it is used as the mock exam.20172. 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 acidIsocyanic acidb. Using the Lewis electron dot diagrams of fulminic acid and isocyanic acid shown in the boxes above and the table of average bond enthapies below, determine the value of ΔHo for the reaction of HCNO(g) to form HNCO(g).BondEnthalpyBond (kJ / mol)BondEnthalpy (kJ / mol)BondEnthalpy (kJ / mol)N – O201C = N615H – C413C = O745C N891H – N391c. A student claims that ΔSo 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.Rubric2017 6. Answer the following questions about Mg(OH)2. At 25 oC, the value of the solubility product constant, Ksp, for Mg(OH)2(s) is 1.8 x 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 oC.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.ReactionLattice Energy (kJ / mol)Mg(OH)2(s) Mg2+(g) + 2 OH-(g)2900Sr(OH)2(s) Sr2+(g) + 2 OH-(g)2300Rubric2017 International2.H2O2(aq) + OCl-(aq) H2O(l) + Cl-(aq) + O2(g)A student investigates the reaction between H2O2(aq) and NaOCl(aq), which is represented by the net ionic equation shown above.a. Is the reaction represented above a redox reaction? Justify your answer.To better understand the reaction, the student looks up thermodynamic data for the reaction. For the reaction represented above, the value of ΔG298o is -197 kJ / molrxn and the value of ΔS298o is 144 J / (K molrxn).b. Calculate the value of ΔH298o for the reaction in kJ / molrxnc. Does the temperature inside the flask increase, decrease, or remain the same as the reaction proceeds? Justify your answer.d. Calculate the value of the equilibrium constant, K, for the reaction at 298 K.Rubric2017 International6.Answer the following questions about the solubility of AgCl(s). The value of Ksp of AgCl(s) is 1.8 x 10-10.a.Calculate the value of [Ag+] in a saturated solution of AgCl in distilled water.b.The concentration of Cl-(aq) in seawater is 0.54 M.i.Calculate the molar solubility of AgCl(s) in seawater.ii.Explain why AgCl(s) is less soluble in seawater than in distilled water.Rubric2.2 NO(g) + O2(g) 2 NO2(g)A student investigates the reaction 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)ΔH298oΔS298oΔG298o-40.4 kJ / molrxn-138.5 J / (K molrxn)0.87 kJ / molrxnb.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 P(NO) and P(NO2) in the vessel are initially 1.0 atm, will P(N2O3) at equilibrium be equal to 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.Rubric2017 International2.Answer the following questions relating to HCl, CH3Cl and CH3Br.a.HCl(g) can be prepared by the reaction of concentrated H2SO4(aq) with NaCl(s), as represented by the following equationH2SO4(aq) + 2 NaCl(s) 2 HCl(g) + Na2SO4(aq)i.A student claims that the reaction is a redox reaction. Is the student correct? Justify your answer.ii.Calculate the mass, in grams, of NaCl(s) needed to react with excess H2SO4(aq) to produce 3.00 g of HCl(g). Assume that the reaction goes to completion.HCl(g) can react with methanol vapor, CH3OH(g), to produce CH3Cl(g) as represented by the following equation.CH3OH(g) + HCl(g) CH3Cl(g) + H2O(g)Kp = 4.7 x 103 at 400 Kb.CH3OH(g) and HCl(g) are combined in a 10.00 L sealed reaction vessel and allowed to reach equilibrium at 400 K. The initial partial pressure of CH3OH(g) in a vessel is 0.250 atm and that of HCl(g) is 0.600 atm.i.Does the total pressure in the vessel increase, decrease, or remain the same as the equilibrium is approached? Justify your answer in terms of the reaction stoichiometry.ii.Considering the value of Kp, calculate the final partial pressure of HCl(g) after the system inside the vessel reaches equilibrium at 400 K.iii.The student claims that the final partial pressure of CH3OH(g) at equilibrium is very small but not exactly zero. Do you agree or disagree with the student’s claim? Justify your answer.Rubric ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download