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CHEMSTRY FREE-RESPONSE QUESTIONS (Form B) 1. Answer the following problems about gases. a) The average atomic mass of naturally occurring neon is 20.18 amu. There are two common isotopes of naturally occurring Neon as indicated in the table below. Isotope Ne-20 Ne-22 Mass (amu) 19.99 21.99 (i) Using the information above, calculate the percent abundance of each isotope. Let x represent the natural abundance of Ne-20. 19.99 x 21.99(1 x) 20.18 19.99 x 21.99 21.99 x 20.18 2 x 1.81 x 0.905 Therefore, percent abundances are: Ne-20: 90.5% Ne-22: 1 – 90.5% = 9.5% (ii) Calculate the number of N2-22 atoms in a 12.55 g sample of naturally occurring neon. 12.55 g Ne 1 mol Ne 0.095 mol Ne -22 6.022 1023 Ne-22 atoms 3.6 1022 Ne-22 atoms 20.18 g Ne 1 mol Ne 1 mol Ne - 22 b) A major line in the emission spectrum of neon corresponds to a frequency of 1.34 x 10 14 s1. Calculate the wavelength, in nanometers, of light that corresponds to this line. c c 3.0 10 m/s 6.9 107 m = 690 nm 14 4.34 10 Hz c) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet (UV) radiation, as shown by the equation below, Ozone serves to block harmful ultraviolet radiation that comes from the sun. 8 UV O3(g) → O2(g) + O(g) A molecule of O3(g) absorbs a photon with a frequency of 1.00 x 1015 s-1. (i) How much energy, in joules, does the O3(g) molecule absorb per photon? E h 6.63 1034 J s 1.00 1015 s-1 6.63 1019 J per photon (ii) The minimum energy needed to break and oxygen-oxygen bond in ozone is 387 kJ mol-1. Does a photon with frequency of 1.00 x 1015 s-1 have enough energy to break this bond? Support your answer with a calculation. 6.63 1019 J 6.022 1023 photons 1 kJ 3 399 kJ mol-1 387 kJ mol-1 1 photon 1 mol 10 J Therefore, the bond can be broken. 2. The following questions relate to the synthesis reaction represented by the chemical equation in the box above. (a) Calculate the value of the standard free energy change, ΔSo298, for the reaction. G298 H 298 T S298 264 kJ mol-1 298 K 0.278 kJ mol-1 K 1 181 kJ mol (b) Determine the temperature at which the equilibrium constant Keq, for the reaction is equal to 1 .00 (Assume that ΔHo and ΔSo are independent of temperature). When Keq = 1, then GT RT ln Keq RT ln1 0 Therefore, GT HT T ST 0 -1 HT 264 kJ mol-1 950 K ST 0.278 kJ mol-1 K 1 (c) Calculate the standard enthalpy change, ΔHo, That occurs when 0.256 mol sample of NF3(g) is formed from N2(g) and F2(g) at 1.0 atm and 298K. From the chemical equation, -264 kJ standard enthalpy change occurs when 2 moles of NF3(g) are produced at 1.0 atm and 298 K, so 264 kJ 0.256 mol NF3 ( g ) 33.8 kJ 2.00 mol NF3 ( g ) T The enthalpy change in a chemical reaction is the difference between energy absorbed in breaking bonds in the reactants and energy released by bond formation, in the products. (d) How many bonds are formed when two molecules of NF3 are produced according to the equation in the box above? 1 each molecule of NF3, there are 3 N-F bonds. Therefore, there are 6 N-F bonds when two molecules of NF3 are produced. (e) Use both the information in the box above and the table of average bond enthalpies below to calculate the average enthalpy of the F – F bond. Bond Average Bond Enthalpy NN N-F F-F 946 272 ? In the above chemical reaction, 1 N=N bond and 3 F-F bonds are broken. Meanwhile, 6 N-F bonds are formed. H 298 Ebonds broken Ebonds formed 264 kJ mol-1 BEN N 3BEF F 6 BEN F 946 kJ mol-1 3BEF F 6 272 kJ mol-1 3BEF F 264 946 1632 kJ mol -1 BEF F 141 kJ mol-1 3. Answer the following questions that relate to the analysis of chemical compounds. (a) A compound containing the elements C, H, N and O is analyzed. When a 1.2359 g sample is burned in excess oxygen, 2.241 g of C02(g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H. (i) Determine the mass, in grams, of C in the 1.2359 g sample of the compound. All C in CO2 comes from the C in the sample. Since 2.241 g of CO2 (g) is produced, the moles of C in CO2 is then 2.241 g / 44.01 g/mol = 0.051 mol CO2 is produced. Therefore, there are 0.051 mol of C in CO2. that is, there are 0.051 mol of C in the sample. The mass of C in the sample is 0.051 mol * 12.011 g/mol = 0.6116 g C. 1 mol CO2 1 mol C 12.011 g C 2.241 g CO2 ( g ) 0.6116 g C 44.01 g CO2 1 mol CO 2 1 mol C (ii) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound. Since N is 28.84% in mass, the mass in the sample is 1.2359 g sample 28.84% = 0.3564 g N. (iii) Determine the mass, in grams of O in the original 1.2359 g sample of the compound. Because the compound contains only C, H, N, and O, Mass of O = mass of sample – (Mass of H + C + N) = 1.2359 – (0.0648 + 0.6116 + 0.3564) = 0.2031 g. (iv) Determine the empirical formula of the compound. First converting all masses to moles, 1 mol C 0.6116 g C 0.05092 mol C 12.011 g C 1 mol H 0.0648 g H 0.06429 mol H 1.0079 g H 1 mol N 0.3564 g N 0.02544 mol N 14.007 g N 1 mol O 0.2031 g O 0.01269 mol O 16.00 g O Divide all mol quantities by the smallest number of moles: 0.05092 mol / 0.01269 mol = 4.013 0.06429 mol / 0.01269 mol = 5.066 0.02544 mol / 0.01269 mol = 2.005 0.01269 mol / 0.01269 mol = 1.000 Therefore, the empirical formula is C4H5N2O A different compound, which has the empirical formula CH2Br, has a vapor density of 6.00 g L-1 at 375 K and 0.983 atm. Using these data, determine the following. (i) The molar mass of the compound. Apply the ideal gas law, PV nRT n 0.983 atm 1.00 L PV 0.0319 mol RT 0.0821 L atm mol-1 K 1 375 K So the molar mass of the gas is M = 6.00 g / 0.0319 mol = 188 g/mol. (ii) The molecular formula of the compound. Each CH2Br unit has mass of 12.011 + 2(1.0079) + 79.90 = 93.9 g. And 188 / 93.9 = 2, so there must be two CH2Br units in each molecule. The molecular formula of the compound is C2H4Br2. Answer question 4 below. The section II score weighting for this question is 10 percent. 4. For each of the following three reactions in part (i) write a balanced equation for the reaction and in part (ii) answer the question about the reaction, In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances extremely ionized. Omit formulas for any ions or molecules chat are unchanged by the reaction. You may use the empty space at the bottom of the ‘next page for scratch work but only the equations that are written in the answer boxes provided will be graded, Example: A strip of Magnesium is added to a solution of silver(1) nitrate (i) Balanced equation (ii) Which substance is oxidized in the reaction. (a) A solution of sodium hydroxide is added to a solution of lead(II) nitrate. (i) Balanced equation 2 OH- + Pb2+ Pb(OH)2 (ii) If 1.0 L volumes of 1.0 M solutions of sodium hydroxide and lead(II) nitrate are mixed together, how many moles of the product(s) will be produced? Assume the reaction goes to completion. ______________________________________________________________________ The moles of NaOH is (1.0 L) * (1.0 M) = 1 mol. From the balanced equation, we know that 2 moles of OH- produce 1 mole of Pb(OH)2. Therefore, with 1 mol of NaOH, a total of 0.5 mo of Pb(OH)2 will be produced. ______________________________________________________________________ (b) Excess nitric acid is added to solid calcium carbonate. (i) Balanced equation 2H+ + CaCO3 Ca2+ + H2O + CO2 (ii) Briefly explain why statues made of marble (calcium carbonate) displayed outdoors in urban areas are deteriorating. . The H+ ions in the acid rain react with the marble statues and the soluble compounds of Ca that are formed wash away. ______________________________________________________________________ (c) A solution containing silver(1) ion (un oxidizing agent) is mixed with a solution containing iron(II) ion (a reducing agent). (i) Balanced equation Ag+ + Fe2+ Ag + Fe3+ (ii) If the contents of the reaction mixture described above are filtered, what substance(s), if any, would remain on the filter paper. ______________________________________________________________________ The precipitated solid silver will remain on the filter paper. ______________________________________________________________________ (a) Solid ammonium carbonate decomposes as it is heated. (i) Balanced equation (NH4)2CO3 2NH3 + CO2 + H2O (ii) Predict the algebraic sign ΔSo for the reaction. Explain your reasoning. ______________________________________________________________________ The algebraic sign of ΔSo for the reaction will be positive because one mole of solid (with relatively low entropy) is converted into four moles of gas (with much greater entropy). ______________________________________________________________________ (b) Chlorine gas, an oxidizing agent, is bubbled into a solution of potassium bromide. (i) Balanced equation Cl2 2 Br 2Cl Br2 (ii) What is the oxidation number of chlorine before the reaction occurs. What is the oxidation number of chlorine after the reaction occurs. ___________________________________________________________________ The oxidation number of chlorine is 0 before the reaction since it exists as Cl2 and -1 after the reaction. ______________________________________________________________________ (c) A small piece of sodium is placed in a beaker of distilled water. (i) Balanced equation 2Na + 2H2O H2 + 2Na+ + 2OH- (ii) The reaction is exothermic, and sometimes small flames are observed as the sodium reacts with the water. Identify the product of the reaction that burns to produce the flames. ______________________________________________________________________ It is the H2 gas that burns. ______________________________________________________________________ Answer Question 5 and Question 6. The Section 11 score weighting for these questions is 15 percent each. Your responses to these Questions will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad, diffusive responses. 5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) —> 5Fe3+(aq) + Mn2+(aq) + 4 H2O(l) 5. The mass percent of iron in a soluble iron(II) compound is measured using a titration based on the balanced equation above. (a) What is the oxidation number of manganese in the permanganate ion, MnO4-(aq) The oxidation number of O in MnO4- is -2. The total is -1 So assume Mn as x x + 2 * (-4) = -1 x = +7 The oxidation number of Mn is +7 in the permanganate ion, MnO4-(aq). (b) Identify the reducing agent in the reaction represented above. The oxidation number of Fe changes from +2 to +3 during the reaction. Therefore, it loses an electron. So the reducing agent is Fe2+ (aq). The mass of a sample of the iron(II) compound is carefully measured before the sample is dissolved in distilled water. The resulting solution is acidified with H2SO4(aq) The solution is then titrated with MnO4-(aq) until the end point is reached. (c) Describe the color change that occurs in the flask when the end point of the titration has been reached. Explain why the color of the solution changes at the end point. The solution in the flask changes from colorless to faint purple-pink at the end point of the titration. At the endpoint, there is no Fe2+ (aq) left in the flask to reduce the colored permanganate ion, MnO4-(aq), so when a small amount of permanganate ion is added after the endpoint, the un-reacted permanganate ion present in the solution colors the solution to faint purple/pink. (d) Let the variables g, M, and V be defined as follows: g = the mass, in grams, of the sample of the iron(II) compound M= the molarity of the MnO4-(aq)) used as the titrant V = the volume, in liters, of MnO4-(aq)) added to reach the end point In terms of these variables, the number of moles of MnO4-(aq)) added to reach the end point of the titration is expressed a M x V. Using the variables defined above the molar mass of iron (55.85 g mol-1) and the coefficients in the balanced chemical equation, write the expression for each of the following quantities. (i) The number of moles of iron in the sample The moles of permanganate ion, MnO4-(aq) is M* V. From the balanced equation, we know that 1 mole permanganate ion, MnO4-(aq) reacts with 5 moles of Fe2+. So the mole of Fe2+ in the sample is 5 * M * V. (ii) The mass of iron in the sample, in grams The molar mass of Fe is 55.85 g/mol. So Mass of Fe = 5 * M * V * 55.85 (iii) The mass percent of iron in the compound Mass of Fe 5 M V 55.85 Mass % Fe = 100= 100 Mass of smaple g (e) What effect will adding too much titrant have on the experimentally determined value of the mass percent of iron in the compound? Justify your answer. The experimentally determined mass percent of iron in the compound will be too large. When more titrant is added, V is too large, so expression in (d) (iii) above is too large. 6. Answer the fol1owing questions, which pertain to binary compounds. (a) In the box provided below draw a complete Lewis electron - dot diagram for the IF3 molecule. (b) On the basis of the Lewis electron—dot diagram that you drew (a), predict the molecular geometry of the IF3 molecule. Trigonal planar electronic geometry or T-shaped. (c) In the SO2 molecule, both of the bonds between sulfur and oxygen have the same length. Explain this observation, supporting your explanation by drawing in the box below a Lewis electron—dot diagram (or diagrams) for the SO2, molecule. O == S == O The bonds are the same length because they are both double bonds. (d) On the basis of your Lewis election—dot diagram(s) in part (c), identify the hybridization of the sulfur atom, in the SO2 molecu1e. 2 sp The reaction between SO2(g) and O2(g) to form SO3(g) is represented below 2 SO2(g) + O2(g) ↔ 3 SO3(g) The reaction is exothermic. The reaction is slow a 25oC; however, a catalyst will cause the reaction to proceed faster. (e) Using the axes provided in the next page draw the complete potential—energy diagram for both the catalyzed and uncatalyzed reactions. Clearly label the curve that represents the catalyzed reaction. For the uncatalyzed reaction curve, Ea 0 and H 0 for this exothermic reaction. For the catalyzed curve, Ea < uncatalyed Ea, and the catalyzed reaction begin and end at the same energies as the uncatalyzed curve. (f) Predict how the ratio of the equilibrium pressures PSO2/PSO3, would change when the temperature of the uncatalyzed reaction mixture is increased. Justify your prediction. The ratio PSO2/PSO3 would increase as the temperature increases. Because the reaction is exothermic ( H 0 ), as the temperature is raised, the reaction shifts to the left-hand side. (g) How would the presence of a catalyst affect the change in the ration described in part (f)? Explain. The catalyst would not affect the value of the two equilibrium ratios but would increase the rate of the shifting of the system to the new equilibrium position. The catalyst does this by providing an alternate path with a lower activation energy. 6. Answer each of the following in terms of principles of molecular behavior arid chemical concepts. (a) The structures of glucose, C6H12O6, and cyclohexane, C6H12, are shown below Identify the type(s) of intermolecular attractive forces in (i) pure glucose. hydrogen bonding, or dipole-dipole interaction, (ii) pure cyclohexane London dispersion forces (b) Glucose is soluble in water but cyclohexane is not soluble in water. Explain. Glucose is polar and cyclohexane is nonpolar. Polar solutes are generally soluble in polar solvents such as water. Nonpolar solutes are not soluble in polar solvents. (c) Consider the processes represented below: Process 1: H2O(l) → H2O(g) ΔHo = +44.0 kJ mol-1 Process 2: H2O(l) → H2(g) + 1/2 O2 (g) ΔHo = +286.0 kJ mol-1 (i) For each of the two processes, identify the type(s) of intermolecular or intraamolecular attractive forces that must be overcome for the process to occur. In process 1, hydrogen bonds in liquid water are overcome to produce distinct water molecules in the vapor phase. In process 2, covalent bonds within water molecules must be broken to allow the atoms to recombine into molecular hydrogen and oxygen. (ii) Indicate whether you agree or disagree with the statement in the box below. Support your answer with a short explanation. When water boils, H2O molecules break apart to form hydrogen molecules and oxygen molecules. Disagree. Boiling is simply process 1, in which only intermolecular forces are broken and the water molecules stay intact. No covalent bonds break in this process. (d) Consider the four reaction-energy profile diagrams (i) Identify the two diagrams that could represent catalyzed and an uncatalyzed reaction. Indicate which of the two diagrams represents the catalyzed reaction pathway for the reaction. For the catalyzed curve, Ea < uncatalyed Ea, and the catalyzed reaction begin and end at the same energies as the uncatalyzed curve. Therefore, Diagram 1 represents a catalyzed pathway and diagram 2 represents an uncatalyzed pathway for the same reaction. (ii) Indicate whether you agree or disagree with the statement in the box below. Support your answer with a short explanation. Adding a catalyst to a reaction mixture adds energy that causes the reaction to proceed more quickly. Disagree. A catalyst does not add energy, but provides an alternate reaction pathway with a lower activation enrgy. 2007 AP CHEMISTRY FREE-RESPONSE QUESTIONS (FORM B) Element 1 Element 2 Element 3 Element 4 First Ionization Energy kJ mol-1 1.251 496 738 1,000 Second Ionization Energy kJ mol-1 2,300 4,560 1,450 2,250 Third Ionization Energy kJ mol-1 3,820 6,910 7,730 3,360 6. The table above shows the first three ionization energies for atoms of four elements from the third period of the periodic table. The elements are numbered randomly. Use the information in the table to answer the following questions (a) Which element is most metallic in character? Explain your reasoning. Element 2. It has the lower first-ionization energy. Metallic elements lose electrons when they become ions, and element 2 requires the least amount of energy to remove an electron. (b) Identify element 3. Explain your reasoning. Magnesium Mg. Element 3 has low first and second ionization energies relative to the third ionization energy, indicating that the element has two valence electrons, which is true for Mg. (c) Write the complete electron configuration for an atom of element 3. 2 2 1s 2s 2 p 6 3s 2 (d) What is the expected oxidation state for the most common ion of element 2? +1. The 2nd and 3rd ionization energy is dramatically higher. So the most common oxidation state should be +1. (e) Na What is the chemical symbol for element 2? (f) A neutral atom of which of the four elements has the smallest radius? Element 1 because it has the largest 1st ionization energy. Answer EITHER Question 7 below OR Question 8 printed on page 14. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 15 percent. 7. Answer the following questions about the structures of ions that contain only sulfur arid fluorine. (a) The compounds SF4 arid BF3 react to form and ionic compound according to the following equation. SF4 + BF3 —> SF3 BF4 (i) (ii) Draw a complete Lewis structure for the SF3+ cation in SF3BF4 Identify the type of hybridization exhibited by sulfur in the SF3+ cation. The hybridization is sp3 Identify the geometry o the SF3+ cation that is consistent with the Lewis structure drawn in part (a)(i). Trigonal pyramidal (iii) (iv) Predict whether the F-S-F bond angle in the SF3+ cation is larger than, equal to, or smaller than 109.5°. Justify your answer. The F-S-F bond angle in the SF3+ caton is expected to be slightly smaller than 109.5° because the repulsion between the nonbonding pair of electrons and the S-F bonding pairs electrons squeezes the F-S-F bond angles together slightly. (b) The compounds SF4 and CsF react to form an ionic compound according to the following equation. SF4 + CsF —> CsSF5 (i) Draw a complete Lewis structure for the SF5- anion in CsSF5. (ii) Identify the type of hybridization exhibited by sulfur in the SF5anion. sp3d 2 (iii) Identify the geometry of the SF5- anion that is consistent with the Lewis structure drawn in part (b)(i). Square pyramidal (iv) Identify the oxidation number of sulfur in the compound CsSF5. Cs: +1 F: -1 The total is 0 +1 + x + 5*(-1) = 0 x = +4 so the oxidation number of sulfur is +4 in the compound CsSF5.