AP CHEMISTRY 2009 SCORING GUIDELINES

2009 SCORING GUIDELINES Question 1 (10 points) Answer the following questions that relate to the chemistry of halogen oxoacids. (a) Use the information in the table below to answer part (a)(i). Acid HOCl K a at 298 K 2.9 10 8 HOBr 2.4 10 9 (i) Which of the two acids is stronger, HOCl or HOBr? Justify your answer in terms of K a. HOCl is the stronger acid because its K a value is greater than the K a value of HOBr. correct answer with justification. (ii) Draw a complete Lewis electron-dot diagram for the acid that you identified in part (a)(i). One point is earned for a correct diagram. (iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a stronger acid or a weaker acid than the acid that you identified in part (a)(i). Justify your prediction in terms of chemical bonding. HOI is a weaker acid than HOCl because the O H bond in HOI is stronger than the O H bond in HOCl. The lower electronegativity (electron-drawing ability) of I compared with that of Cl results in an electron density that is higher (hence a bond that is stronger) between the H and O atoms in HOI compared with the electron density between the H and O atoms in HOCl. OR The conjugate base OCl is more stable than OI because Cl, being more electronegative, is better able to accommodate the negative charge. One point is earned for predicting that HOI is a weaker acid than HOCl and stating that iodine has a lower electronegativity than chlorine and EITHER OR stating that this results in a stronger O H bond in HOI stating that this decreases the stability of the OI ion in solution. 2009 The College Board. All rights reserved.

2009 SCORING GUIDELINES Question 1 (continued) (b) Write the equation for the reaction that occurs between hypochlorous acid and water. HOCl + H 2 O OCl + H 3 O + OR HOCl OCl + H + correct equation. (c) A 1.2 M NaOCl solution is prepared by dissolving solid NaOCl in distilled water at 298 K. The hydrolysis reaction OCl (aq) + H 2 O(l) HOCl(aq) + OH (aq) occurs. (i) Write the equilibrium-constant expression for the hydrolysis reaction that occurs between OCl (aq) and H 2 O(l). K b = [HOCl][OH ] [OCl ] correct expression. (ii) Calculate the value of the equilibrium constant at 298 K for the hydrolysis reaction. K b = K K w a = 1.0 10 2.9 10 14 8 = 3.4 10 7 correct value with supporting work. (iii) Calculate the value of [OH ] in the 1.2 M NaOCl solution at 298 K. [OCl ] [HOCl] [OH ] initial value 1.2 0 0 change x x x equilibrium value 1.2 x x x One point is earned for the correct setup. K hyd = 3.4 10 7 = [OH ][HOCl] [OCl ] = ( x)( x) (1.2 x) 2 x 1.2 One point is earned for the correct answer with supporting calculations. (1.2)(3.4 10 7 ) = x 2 x = [OH ] = 6.4 10 4 M 2009 The College Board. All rights reserved.

2009 SCORING GUIDELINES Question 1 (continued) (d) A buffer solution is prepared by dissolving some solid NaOCl in a solution of HOCl at 298 K. The ph of the buffer solution is determined to be 6.48. (i) Calculate the value of [H 3 O + ] in the buffer solution. [H + ] = 10 6.48 = 3.3 10 7 M correct value. (ii) Indicate which of HOCl(aq) or OCl (aq) is present at the higher concentration in the buffer solution. Support your answer with a calculation. [H + ] = 3.3 10 7 M and K a for HOCl = 2.9 10 8 K a = 2.9 10 8 = [H ][OCl ] [HOCl] 7 (3.3 10 )[OCl ] [HOCl] correct answer with supporting buffer calculations. [OCl ] [HOCl] = 2.9 10 3.3 10 8 7 = 0.088 [HOCl] > [OCl ] 2009 The College Board. All rights reserved.

2009 SCORING GUIDELINES Question 5 (8 points) Reaction Equation H 298 S 298 G 298 X C(s) + H 2 O(g) CO(g) + H 2 (g) +131 kj mol 1 +134 J mol 1 K 1 +91 kj mol 1 Y CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) +41 kj mol 1 +42 J mol 1 K 1 +29 kj mol 1 Z 2 CO(g) C(s) + CO 2 (g)??? Answer the following questions using the information related to reactions X, Y, and Z in the table above. (a) For reaction X, write the expression for the equilibrium constant, K p. pco ph2 K p = p H2O correct expression. (b) For reaction X, will the equilibrium constant, K p, increase, decrease, or remain the same if the temperature rises above 298 K? Justify your answer. K p will increase. If the temperature is increased for an endothermic reaction ( H 298 = +131 kj mol 1 ), then by Le Chatelier s principle the reaction will shift toward products, thereby absorbing energy. With greater concentrations of products at equilibrium, the value of K p will increase. OR Because G = RT ln K p = H 298 T S 298, H then ln K p = 298 S + 298. RT R An increase in T for a positive H 298 results in an increase in ln K p and thus an increase in K p. correct answer with appropriate justification. 2009 The College Board. All rights reserved.

2009 SCORING GUIDELINES Question 5 (continued) (c) For reaction Y at 298 K, is the value of K p greater than 1, less than 1, or equal to 1? Justify your answer. K p for reaction Y is less than 1. For reaction Y, G 298 = +29 kj mol 1, a positive number. Because G = RT ln K and G is positive, then ln K p must be negative. This is true when K p is less than 1. OR A positive G results in a nonspontaneous reaction under standard conditions. This favors reactants over products as equilibrium is approached starting from standard conditions, resulting in a K p less than 1. correct answer with appropriate justification. (d) For reaction Y at 298 K, which is larger: the total bond energy of the reactants or the total bond energy of the products? Explain. The total bond energy of the reactants is larger. Reaction Y is endothermic ( H 298 = + 41 kj mol 1 > 0), so there is a net input of energy as the reaction occurs. Thus, the total energy required to break the bonds in the reactants must be greater than the total energy released when the bonds are formed in the products. correct answer with appropriate explanation. (e) Is the following statement true or false? Justify your answer. On the basis of the data in the table, it can be predicted that reaction Y will occur more rapidly than reaction X will occur. The statement is false. Thermodynamic data for an overall reaction have no bearing on how slowly or rapidly the reaction occurs. correct answer with appropriate justification. 2009 The College Board. All rights reserved.

(f) Consider reaction Z at 298 K. AP CHEMISTRY 2009 SCORING GUIDELINES Question 5 (continued) (i) Is S for the reaction positive, negative, or zero? Justify your answer. S for reaction Z is negative. In reaction Z, two moles of gas with relatively high entropy are converted into one mole of solid and one mole of gas, a net loss of one mole of gas and thus a net loss in entropy. OR Reaction Z can be obtained by reversing reactions X and Y and adding them together. Thus S for reaction Z is the sum of two negative numbers and must itself be negative. correct answer with an appropriate justification. (ii) Determine the value of H for the reaction. Add the values of the negatives of 131 kj mol 1 + (41 kj mol 1 ) = 172 kj mol H 298 for reactions X and Y : One point is earned for the correct answer. (iii) A sealed glass reaction vessel contains only CO(g) and a small amount of C(s). If a reaction occurs and the temperature is held constant at 298 K, will the pressure in the reaction vessel increase, decrease, or remain the same over time? Explain. The pressure in the flask decreases. The reaction would proceed to the right, forming more C(s) and CO 2 (g). Because two moles of CO(g) would be consumed for every mole of CO 2 (g) that is produced, the total number of moles of gas in the flask would decrease, thereby causing the pressure in the flask to decrease. correct answer with an appropriate explanation. 2009 The College Board. All rights reserved.

2009 SCORING GUIDELINES (Form B) Question 5 (9 points) Answer the following questions about nitrogen, hydrogen, and ammonia. (a) In the boxes below, draw the complete Lewis electron-dot diagrams for N 2 and NH 3. The correct structures are shown in the boxes above. Two points are earned for the correct Lewis electron-dot diagrams (1 point each). (b) Calculate the standard free-energy change, G, that occurs when 12.0 g of H 2 (g) reacts with excess N 2 (g) at 298 K according to the reaction represented below. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G 298 = 34 kj mol 1 1 mol H2 12.0 g H 2 2.0 g H 1 mol reaction 3 mol H2 2 34 kj 1 mol reaction = 68 kj One point is earned for the correct stoichiometry. One point is earned for the correct answer. (c) Given that H 298 for the reaction is 92.2 kj mol 1, which is larger, the total bond dissociation energy of the reactants or the total bond dissociation energy of the products? Explain. H 298 = (bond energy of the reactants) (bond energy of the products) Based on the equation above, for H 298 to be negative, the total bond energy of the products must be larger than the total bond energy of the reactants. OR More energy is released as product bonds are formed than is absorbed as reactant bonds are broken. One point is earned for the correct answer with the correct equation and explanation. 2009 The College Board. All rights reserved.

2009 SCORING GUIDELINES (Form B) Question 5 (continued) (d) The value of the standard entropy change, S 298, for the reaction is 199 J mol 1 K 1. Explain why the value of S 298 is negative. All of the reactants and products in the reaction are in the gas phase, so the sign of the entropy change will depend on the number of moles of particles in the reactants and products. There are more moles of reactants (four) compared with moles of products (two), so there is a greater number of microstates in the reactants than in the products. Therefore the entropy decreases as the reaction proceeds (fewer possible microstates), and the sign of the entropy change is negative. One point is earned for the correct explanation. (e) Assume that H and S for the reaction are independent of temperature. (i) Explain why there is a temperature above 298 K at which the algebraic sign of the value of G changes. G = H TS As the temperature increases TS will at some point exceed H. Because both H and S are negative, the sign of G will then change from negative to positive. correct explanation. (ii) Theoretically, the best yields of ammonia should be achieved at low temperatures and high pressures. Explain. Low temperatures: The reaction is exothermic. By Le Chatelier s principle, decreasing the temperature drives the reaction to the right to produce more heat energy, and thus more ammonia is produced. High pressures: For this reaction, higher pressure is achieved by decreasing the volume of the container. As pressure increases, the reaction equilibrium shifts in the direction that reduces the total number of particles (by Le Chatelier s principle). In this case, the product has fewer moles of particles than the reactants; thus product would be favored. Higher pressure therefore results in an increase in the amount of ammonia. One point is earned for explaining increased yield at low temperatures. One point is earned for explaining increased yield at high pressures. 2009 The College Board. All rights reserved.

2009 SCORING GUIDELINES (Form B) Question 6 (9 points) Answer the following questions about electrochemical cells. It is observed that when silver metal is placed in aqueous thallium(i) fluoride, TlF, no reaction occurs. When the switch is closed in the cell represented above, the voltage reading is +1.14 V. (a) Write the reduction half-reaction that occurs in the cell. Ag + + e Ag correct equation. (b) Write the equation for the overall reaction that occurs in the cell. Tl + Ag + Tl + + Ag correct equation. (c) Identify the anode in the cell. Justify your answer. The anode is where oxidation occurs. In the overall reaction Tl is oxidized to Tl +, so the anode is the Tl electrode in the left cell. correct answer with justification. (d) On the diagram above, use an arrow to clearly indicate the direction of electron flow as the cell operates. The arrow should show electron flow in the direction from the Tl electrode through the wire to the Ag electrode. One point is earned for a correct arrow. 2009 The College Board. All rights reserved.

2009 SCORING GUIDELINES (Form B) Question 6 (continued) (e) Calculate the value of the standard reduction potential for the Tl + /Tl half-reaction. E cell = E red E ox +1.14 V = +0.80V E ox E ox = 0.34 V correct setup. correct answer. The standard reduction potential, E, of the reaction Pt 2+ + 2 e Pt is 1.20 V. (f) Assume that electrodes of pure Pt, Ag, and Ni are available as well as 1.00 M solutions of their salts. Three different electrochemical cells can be constructed using these materials. Identify the two metals that when used to make an electrochemical cell would produce the cell with the largest voltage. Explain how you arrived at your answer. E(V) Ni 2+ + 2 e Ni 0.25 Ag + + e Ag 0.80 Pt 2+ + 2 e Pt 1.20 E cell = E red E ox The two metals that yield the largest E cell are those with the biggest difference in E, namely, Pt and Ni (see E cell calculation below). E cell = +1.20 (0.25) = +1.45 V correct answer with justification. (g) Predict whether Pt metal will react when it is placed in 1.00 M AgNO 3 (aq). Justify your answer. When Pt metal is added to 1.00 M AgNO 3, the only redox reaction that could occur would be for Pt to become oxidized as Ag + is reduced. Ecell = Ered Eox Because occur. Ecell = +0.80 V (+1.20 V) = 0.40 V for that reaction is negative, no reaction will One point is earned for comparing E values. One point is earned for the correct interpretation. 2009 The College Board. All rights reserved.

2008 SCORING GUIDELINES Question 1 C(s) + CO 2 (g) 2 CO(g) Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the reaction represented above occurred. As the reaction proceeded, the total pressure in the container was monitored. When equilibrium was reached, there was still some C(s) remaining in the container. Results are recorded in the table below. Time (hours) Total Pressure of Gases in Container at 1,160 K (atm) 0.0 5.00 2.0 6.26 4.0 7.09 6.0 7.75 8.0 8.37 10.0 8.37 (a) Write the expression for the equilibrium constant, K p, for the reaction. K p = ( P ) P CO CO2 2 correct expression. (b) Calculate the number of moles of CO 2 (g) initially placed in the container. (Assume that the volume of the solid carbon is negligible.) n PV (5.00 atm)(2.00 L) 0.105 mol RT L atm (0.0821 )(1,160 K) mol K correct setup. correct answer. (c) For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO 2 (g) is 1.63 atm. Calculate (i) the partial pressure of CO(g), and P + P = P CO CO total 2 P = P P = 8.37 atm 1.63 atm = 6.74 atm CO total CO 2 correct answer supported by a correct method.

(ii) the value of the equilibrium constant, K p. AP CHEMISTRY 2008 SCORING GUIDELINES Question 1 (continued) K p = ( PCO) P CO2 2 = 2 (6.74 atm) 1.63 atm = 27.9 One point is earned for a correct setup that is consistent with part (a). correct answer according to the setup. (d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid catalyst is negligible.) The total pressure of the gases at equilibrium with a catalyst present would be equal to the total pressure of the gases without a catalyst. Although a catalyst would cause the system to reach the same equilibrium state more quickly, it would not affect the extent of the reaction, which is determined by the value of the equilibrium constant, K p. correct answer with justification. In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g of C(s), plus CO(g) and CO 2 (g), each at a partial pressure of 2.00 atm at 1,160 K. (e) Predict whether the partial pressure of CO 2 (g) will increase, decrease, or remain the same as this system approaches equilibrium. Justify your prediction with a calculation. Q = therefore ( PCO) P CO2 equilibrium. 2 PCO 2 = 2 (2.00 atm) 2.00 atm = 2.00 < K p ( = 27.9), will decrease as the system approaches One point is earned for a correct calculation involving Q or ICE calculation. One point is earned for a correct conclusion based on the calculation.

2008 SCORING GUIDELINES Question 6 (a) Structures of the pyridine molecule and the benzene molecule are shown below. Pyridine is soluble in water, whereas benzene is not soluble in water. Account for the difference in solubility. You must discuss both of the substances in your answer. Pyridine is polar (and capable of forming hydrogen bonds with water), while the nonpolar benzene is not capable of forming hydrogen bonds. Pyridine will dissolve in water because of the strong hydrogen bonds (or dipole-dipole intermolecular interactions) that exist between the lone pair of electrons on pyridine s nitrogen atom and the solvent water molecules. No such strong intermolecular interaction can exist between benzene and water, so benzene is insoluble in water. One point is earned for identifying a relevant structural difference between pyridine and benzene. One point is earned for indicating that pyridine is soluble in water because pyridine can form strong dipole-dipole interactions (or hydrogen bonds) with water, while benzene cannot. (b) Structures of the dimethyl ether molecule and the ethanol molecule are shown below. The normal boiling point of dimethyl ether is 250 K, whereas the normal boiling point of ethanol is 351 K. Account for the difference in boiling points. You must discuss both of the substances in your answer.

2008 SCORING GUIDELINES Question 6 (continued) The intermolecular forces of attraction among molecules of dimethyl ether consist of London (dispersion) forces and weak dipole-dipole interactions. In addition to London forces and dipole-dipole interactions that are comparable in strength to those in dimethyl ether, ethanol can form hydrogen bonds between the H of one molecule and the O of a nearby ethanol molecule. Hydrogen bonds are particularly strong intermolecular forces, so they require more energy to overcome during the boiling process. As a result, a higher temperature is needed to boil ethanol than is needed to boil dimethyl ether. One point is earned for recognizing that ethanol molecules can form intermolecular hydrogen bonds, whereas dimethyl ether molecules do not form intermolecular hydrogen bonds. One point is earned for recognizing that, compared to the energy required to overcome the weaker intermolecular forces in liquid dimethyl ether, more energy is required to overcome the stronger hydrogen bonds in liquid ethanol, leading to a higher boiling point. (c) SO 2 melts at 201 K, whereas SiO 2 melts at 1,883 K. Account for the difference in melting points. You must discuss both of the substances in your answer. In the solid phase, SO 2 consists of discrete molecules with dipole-dipole and London (dispersion) forces among the molecules. These forces are relatively weak and are easily overcome at a relatively low temperature, consistent with the low melting point of SO 2. In solid SiO 2, a network of Si and O atoms, linked by strong covalent bonds, exists. These covalent bonds are much stronger than typical intermolecular interactions, so very high temperatures are needed to overcome the covalent bonds in SiO 2. This is consistent with the very high melting point for SiO 2. One point is earned for recognizing that SO 2 is a molecular solid with only weak dipole-dipole and London forces among SO 2 molecules. One point is earned for recognizing that SiO 2 is a covalent network solid, and that strong covalent bonds must be broken for SiO 2 to melt.

2008 SCORING GUIDELINES Question 6 (continued) (d) The normal boiling point of Cl 2 (l) (238 K) is higher than the normal boiling point of HCl(l) (188 K). Account for the difference in normal boiling points based on the types of intermolecular forces in the substances. You must discuss both of the substances in your answer. The intermolecular forces in liquid Cl 2 are London (dispersion) forces, whereas the intermolecular forces in liquid HCl consist of London forces and dipole-dipole interactions. Since the boiling point of Cl 2 is higher than the boiling point of HCl, the London forces among Cl 2 molecules must be greater than the London and dipole-dipole forces among HCl molecules. The greater strength of the London forces between Cl 2 molecules occurs because Cl 2 has more electrons than HCl, and the strength of the London interaction is proportional to the total number of electrons. One point is earned for recognizing that the London forces among Cl 2 molecules must be larger than the intermolecular forces (London and dipole-dipole) among HCl molecules. One point is earned for recognizing that the strength of the London forces among molecules is proportional to the total number of electrons in each molecule.

2008 SCORING GUIDELINES (Form B) Question 1 Answer the following questions regarding the decomposition of arsenic pentafluoride, AsF 5 (g). (a) A 55.8 g sample of AsF 5 (g) is introduced into an evacuated 10.5 L container at 105 C. (i) What is the initial molar concentration of AsF 5 (g) in the container? mol AsF 5 = 55.8 g AsF 5 [AsF 5 ] i = 0.328 mol AsF 5 10.5 L 1 mol AsF5 169.9 g AsF = 0.0313 M 5 = 0.328 mol One point is earned for the correct molar mass. One point is earned for the correct concentration. (ii) What is the initial pressure, in atmospheres, of the AsF 5 (g) in the container? PV = nrt P = 1 1 0.328 mol 0.0821 L atm mol K 378 K 10.5 L = 0.969 atm correct substitution. correct pressure. At 105 C, AsF 5 (g) decomposes into AsF 3 (g) and F 2 (g) according to the following chemical equation. AsF 5 (g) AsF 3 (g) + F 2 (g) (b) In terms of molar concentrations, write the equilibrium-constant expression for the decomposition of AsF 5 (g). K = [AsF 3] [F 2] [AsF ] 5 correct equation. (c) When equilibrium is established, 27.7 percent of the original number of moles of AsF 5 (g) has decomposed. (i) Calculate the molar concentration of AsF 5 (g) at equilibrium. 100.0% 27.7% = 72.3% [AsF 5 ] = 0.723 0.0313 M = 0.0226 M correct concentration.

2008 SCORING GUIDELINES (Form B) Question 1 (continued) (ii) Using molar concentrations, calculate the value of the equilibrium constant, K eq, at 105 C. [AsF 3 ] = [F 2 ] = 0.277 [AsF 5 ] i = 0.277 0.0313 M = 0.00867 M One point is earned for setting [AsF 3 ] = [F 2 ]. Note: the point is not earned if the student indicates that [AsF 3 ] = [F 2 ] = [AsF 5 ]. K eq = [AsF 3] [F 2] [AsF ] 5 = [0.00867] [0.00867] [0.0226] = 0.00333 correct calculation of [AsF 3 ] and [F 2 ]. correct calculation of K eq. (d) Calculate the mole fraction of F 2 (g) in the container at equilibrium. mol AsF 5 = 0.0226 M 10.5 L = 0.237 mol mol F 2 = mol AsF 3 = 0.00867 M 10.5 L = 0.0910 mol mol fraction F 2 = = OR mol fraction F 2 = mol F2 mol F + mol AsF + mol AsF 2 3 5 0.0910 0.0910 + 0.0910 + 0.237 = 0.217 0.00864 0.00864 + 0.00864 + 0.0226 = 0.217 correct calculation of the mole fraction of F 2 (g).

2008 SCORING GUIDELINES (Form B) Question 5 The identity of an unknown solid is to be determined. The compound is one of the seven salts in the following table. Al(NO 3 ) 3. 9H2 O BaCl 2. 2H2 O CaCO 3 CuSO 4. 5H2 O NaCl BaSO 4 Ni(NO 3 ) 2. 6H2 O Use the results of the following observations or laboratory tests to explain how each compound in the table may be eliminated or confirmed. The tests are done in sequence from (a) through (e). (a) The unknown compound is white. In the table below, cross out the two compounds that can be eliminated using this observation. Be sure to cross out these same two compounds in the tables in parts (b), (c), and (d). Al(NO 3 ) 3. 9H2 O BaCl 2. 2H2 O CaCO 3 CuSO 4. 5H2 O NaCl BaSO 4 Ni(NO 3 ) 2. 6H2 O One point is earned for each correctly crossed-out compound. (b) When the unknown compound is added to water, it dissolves readily. In the table below, cross out the two compounds that can be eliminated using this test. Be sure to cross out these same two compounds in the tables in parts (c) and (d). Al(NO 3 ) 3. 9H2 O BaCl 2. 2H2 O CaCO 3 CuSO 4. 5H2 O NaCl BaSO 4 Ni(NO 3 ) 2. 6H2 O One point is earned for each additional correctly crossed-out compound. (c) When AgNO 3 (aq) is added to an aqueous solution of the unknown compound, a white precipitate forms. In the table below, cross out each compound that can be eliminated using this test. Be sure to cross out the same compound(s) in the table in part (d). Al(NO 3 ) 3. 9H2 O BaCl 2. 2H2 O CaCO 3 CuSO 4. 5H2 O NaCl BaSO 4 Ni(NO 3 ) 2. 6H2 O One point is earned for crossing out Al(NO 3 ) 3. 9H2 O or for crossing out Ni(NO 3 ) 2. 6H2 O if it had not been crossed out earlier.

2008 SCORING GUIDELINES (Form B) Question 5 (continued) (d) When the unknown compound is carefully heated, it loses mass. In the table below, cross out each compound that can be eliminated using this test. Al(NO 3 ) 3. 9H2 O BaCl 2. 2H2 O CaCO 3 CuSO 4. 5H2 O NaCl BaSO 4 Ni(NO 3 ) 2. 6H2 O One point is earned for crossing out NaCl or for crossing out either CaCO 3 or BaSO 4 if they had not been crossed out earlier. (e) Describe a test that can be used to confirm the identity of the unknown compound identified in part (d). Limit your confirmation test to a reaction between an aqueous solution of the unknown compound and an aqueous solution of one of the other soluble salts listed in the tables above. Describe the expected results of the test; include the formula(s) of any product(s). Mix an aqueous solution of BaCl 2. 2H2 O with an aqueous solution of CuSO 4. 5H2 O. The BaSO 4 will precipitate. One point is earned for describing a precipitation reaction between the compound left in part (d) and another compound given in the problem. One point is earned for a correct identification of a precipitate that would form upon the mixing of the chosen solutions.

2008 SCORING GUIDELINES (Form B) Question 6 Use principles of thermodynamics to answer the following questions. (a) The gas N 2 O 4 decomposes to form the gas NO 2 according to the equation below. (i) Predict the sign of H for the reaction. Justify your answer. Bonds are broken when NO 2 molecules form from N 2 O 4 molecules. Energy must be absorbed to break bonds, so the reaction is endothermic and the sign of H is positive. correct sign and a correct explanation. (ii) Predict the sign of S for the reaction. Justify your answer. There are two gaseous product molecules for each gaseous reactant molecule, so the product has more entropy than the reactant. The entropy increases as the reaction proceeds, so the sign of S is positive. correct sign and a correct explanation. (b) One of the diagrams below best represents the relationship between G and temperature for the reaction given in part (a). Assume that H and S are independent of temperature. Draw a circle around the correct graph. Explain why you chose that graph in terms of the relationship G = H T S. The leftmost graph should be circled. S is positive, so as T increases, T S becomes a larger positive number. At higher temperatures, you are subtracting larger positive numbers from H to get G, so G decreases with increasing temperature. correct graph selection. explanation.

2008 SCORING GUIDELINES (Form B) Question 6 (continued) (c) A reaction mixture of N 2 O 4 and NO 2 is at equilibrium. Heat is added to the mixture while the mixture is maintained at constant pressure. (i) Explain why the concentration of N 2 O 4 decreases. The reaction is endothermic. For endothermic reactions, increasing the temperature drives the reaction to the right. This increases the equilibrium concentration of NO 2 and decreases the equilibrium concentration of N 2 O 4. correct explanation. (ii) The value of K eq at 25 C is 5.0 10 3. Will the value of K eq at 100 C be greater than, less than, or equal to this value? Because the reaction is endothermic, at higher temperatures the reaction goes further to the right. This means that the value of K eq at 100 C will be greater than the value of K eq at 25 C. correct choice. (No explanation required.) (d) Using the value of K eq at 25 C given in part (c)(ii), predict whether the value of H is expected to be greater than, less than, or equal to the value of T S. Explain. K eq at 25 C is less than 1, hence G must be positive. And in order for G to be positive, H must be greater than T S. correct prediction. explanation.