CHEMISTRY 202 Hour Exam III December 3, 2015 Dr. D. DeCoste Name Signature T.A. This exam contains 33 questions on 13 numbered pages. Check now to make sure you have a complete exam. You have two hours to complete the exam. Determine the best answer to the first 30 questions and enter these on the special answer sheet. Also, circle your responses in this exam booklet. Show all of your work and provide complete answers to questions 31, 32, and 33. 1-30 (60 pts.) 31 (20 pts.) 32 (20 pts.) 33 (20 pts.) Total (120 pts) Useful Information: R = 8.314 J/Kmol = 0.08206 Latm/molK k = Ae -Ea/RT k 2 E ln( ) = a 1 [ k1 R T 1 1 T ] 2
Hour Exam III Page No. 1 1. Which of the following best completes the following sentence concerning trends on the periodic table? While there can be exceptions, in general a) smaller atoms have larger ionization energies, and smaller electronegativity values. b) smaller atoms have smaller ionization energies, and smaller electronegativity values. c) smaller atoms have smaller ionization energies, and larger electronegativity values. d) smaller atoms have larger ionization energies, and larger electronegativity values. e) There are no general trends among atomic radius, ionization energy, and electronegativity. 2. Given that the first two ionization energy values for Rb(g) = 403 kj/mol and 2633 kj/mol, respectively, determine the electron affinity of Rb(g). a) 403 kj/mol b) 2230 kj/mol c) 2633 kj/mol d) 3036 kj/mol e) The electron affinity of Rb(g) cannot be determined with these data. 3. The following graph plots the first, second, and third ionization energies for Be, B, and C. #1 #2 #3 Which of the following corrects matches the plot to the element? a) #1 = Be #2 = B #3 = C b) #1 = Be #2 = C #3 = B c) #1 = C #2 = B #3 = Be d) #1 = C #2 = Be #3 = B e) #1 = B #2 = C #3 = Be
Hour Exam III Page No. 2 4. Consider a compound formed from a Group 2 element (such as Be or Mg and designated as M below) and a Group 6 element (such as O or S and designated as X below). Which of the following is the most reasonable statement? a) An ionic compound with the ions M 2+ and X 2- is the only stable form of the compound. b) An ionic compound with the ions M + and X - could be thermodynamically favorable, but an ionic compound with the ions M 2+ and X 2- is more thermodynamically favorable. c) The compound between M and X will be in a 1:1 ratio but it is best to think of the bond as polar covalent since atoms do not want to lose electrons (thermodynamically speaking). d) The bond between M and X is polar covalent as explained in c above, but we cannot determine the molecular formula without a molar mass. e) An ionic compound will form but we cannot predict the charges with any degree of accuracy. 5. Consider the following reaction: C 2 H 4 (g) + X 2 (g) CH 2 XCH 2 X(g) ΔH = 549 kj Estimate the C X bond energy given that the C H bond energy is 413 kj/mol, the C C bond energy is 347 kj/mol, the C=C bond energy is 614 kj/mol, and the X X bond energy is 154 kj/mol. Note: you will need to draw Lewis structures for the reactants and products to determine the possible presence of multiple bonds. a) 198 kj b) 352 kj c) 485 kj d) 704 kj e) 970 kj 6. Consider the following molecules/ions: O 3 NO 3 PO 4 3 CH 4 SO 3 How many total Lewis structures (resonance structures) can be drawn for the molecules/ions above that obey the octet rule? a) 5 b) 8 c) 10 d) 12 e) 15 7. Consider the following molecules: SF 4 ICl 5 XeF 4 NF 3 For how many of the molecules above are the geometry (electron-pair arrangement) and shape (molecular structure) the same? a) 0 b) 1 c) 2 d) 3 e) 4
Hour Exam III Page No. 3 8. How many of the following statements is/are false? I. When the difference in electronegativity between two atoms is very large, the bond most likely to form is an ionic bond. II. Covalent bonding results from the sharing of valence electrons between two atoms. III. The valence electrons in a polar bond are found nearer (on the average) to the more electronegative atom in the bond. IV. VSEPR theory states that the central atom in a molecule has the bonded atoms and lone pairs arranged so to minimize electron-electron repulsions. V. If a molecule has polar bonds it is a polar molecule. VI. It is possible for a molecule with polar bonds to have no overall dipole moment. a) 0 b) 1 c) 2 d) 3 e) 4 9. What type(s) of intermolecular force(s) is/are exhibited by propane (C 3 H 8 )? a) hydrogen bonding and London dispersion forces b) hydrogen bonding only c) London dispersion forces only d) dipole-dipole and London dispersion forces e) dipole-dipole only 10. For which of the following mixtures is ΔH soln expected to be the most positive? a) C 6 H 14 and C 7 H 16 b) H 2 O and CH 3 CH 2 OH c) (CH 3 ) 2 CO and H 2 O d) CH 3 CH 2 OH and CH 3 OH e) C 7 H 16 and H 2 O 11. When table salt (solid NaCl) is dissolved in water, the overall process is slightly endothermic (the change in enthalpy is 4 kj/mol). The lattice energy for NaCl is 787 kj/mol. Determine the enthalpy of hydration. a) 783 kj/mol b) 791 kj/mol c) 783 kj/mol d) 791 kj/mol e) 4 12. Consider two pure gaseous substances A and B each made of molecules of approximately the same size. Substance A consists of molecules which are more polar than those of substance B. How many of the following statements is/are true? I. Substance A has a higher vapor pressure than substance B. II. Substance A has a higher boiling point than substance B. III. Substance A is a more ideal gas than substance B IV. The bonds in molecule A must be more polar than the bonds in molecule B. a) 0 b) 1 c) 2 d) 3 e) 4
Hour Exam III Page No. 4 13-15. For each of the following molecules, choose the correct molecular geometry, shape, and polarity. 13. Sulfur tetrafluoride (sulfur is the central atom) GEOMETRY SHAPE POLARITY a) trigonal bipyramid see-saw polar b) tetrahedral tetrahedral polar c) tetrahedral tetrahedral non-polar d) octahedral square planar non-polar e) trigonal bipyramid trigonal pyramid polar 14. Sulfur dioxide (sulfur is the central atom) GEOMETRY SHAPE POLARITY a) trigonal planar bent non-polar b) linear linear non-polar c) tetrahedral bent polar d) linear linear polar e) trigonal planar bent polar 15. Nitrogen triiodide (nitrogen is the central atom) GEOMETRY SHAPE POLARITY a) trigonal planar trigonal planar non-polar b) tetrahedral trigonal planar non-polar c) tetrahedral tetrahedral non-polar d) tetrahedral trigonal pyramid polar e) trigonal planar bent polar ---------------------------------------------------------------------------------------------------------------- 16. Radioactive nuclei decay according to first-order kinetics. If a radioactive sample decays from 1.00 x 10 4 nuclei to 625 nuclei in 10.0 minutes, determine the half-life of this radioactive species. a) 0.667 min b) 1.50 min c) 2.00 min d) 2.50 min e) 5.33 min 17. Consider the reaction type aa products which is zero-order in A. How many of the following statements is/are true? I. [A] remains constant with time. II. The half-life is equal to the rate constant, k. III. The rate of the reaction remains constant over time. IV. An increase in temperature will not affect the rate of the reaction. a) 0 b) 1 c) 2 d) 3 e) 4 18. For a reaction: aa Products, [A] 0 = 4.00 M, and the first two half-lives are 48 and 24 minutes, respectively. Calculate [A] at t= 81 minutes. a) 0.48 M b) 0.62 M c) 1.24 M d) 1.49 M e) 2.81 M
Hour Exam III Page No. 5 19-21. Consider the following data concerning the equation: H 2 O 2 (aq) + 3I (aq) + 2H + (aq) I 3 (aq) + 2H 2 O [H 2 O 2 ] [I ] [H + ] rate I 0.100 M 5.00 x 10-4 M 1.00 x 10-2 M 0.137 M/sec II. 0.100 M 1.00 x 10-3 M 1.00 x 10-2 M 0.268 M/sec III. 0.200 M 1.00 x 10-3 M 1.00 x 10-2 M 0.542 M/sec IV. 0.400 M 1.00 x 10-3 M 2.00 x 10-2 M 1.084 M/sec 19. The rate law for this reaction is a) rate = k[h 2 O 2 ][I ][H + ] b) rate = k[h 2 O 2 ] 2 [I ] 2 [H + ] 2 c) rate = k[i ][H + ] d) rate = k[h 2 O 2 ][H + ] e) rate = k[h 2 O 2 ][I ] 20. The average value for the rate constant (units with M and sec) is a) 108 b) 137 c) 2.71 x 10 3 d) 2.74 x 10 4 e) 3.14 x 10 7 21. Two mechanisms are proposed: I. H 2 O 2 + I H 2 O + OI II. H 2 O 2 + I + H + H 2 O + HOI OI + H + HOI HOI + I + H + I 2 + H 2 O HOI + I + H + I 2 + H 2 O I 2 + I I 3 I 2 + I - I 3 Which of the following describes a potentially correct mechanism? a) Mechanism I with the first step the rate determining step. b) Mechanism I with the second step the rate determining step. c) Mechanism II with the first step rate determining. d) Mechanism II with the second step rate determining. e) Neither mechanism supports the correct rate law. ---------------------------------------------------------------------------------------------------------------- 22. How many of the following are true concerning catalysts? I. The catalyzed reaction has a different rate constant from the uncatalyzed reaction. II. The catalyzed reaction has a different value of ΔH rxn from the uncatalyzed reaction. III. The catalyzed reaction has a different equilibrium constant from the uncatalyzed reaction. IV. The catalyzed reaction has a different activation energy from the uncatalyzed reaction. a) 0 b) 1 c) 2 d) 3 e) 4
Hour Exam III Page No. 6 23-24. Consider the following reaction: The following mechanism is proposed: 2NO(g) + H 2 (g) N 2 O(g) + H 2 O(g) 1. 2NO(g) N 2 O 2 (g) 2. N 2 O 2 (g) + H 2 (g) N 2 O(g) + H 2 O(g) Answer the following questions. Note: the k terms in the rate laws takes into account the rate law constants for the elementary steps and are not necessarily equivalent in the choices below. 23. Assuming the first step is a fast equilibrium step and the second step is rate-determining, which of the following best represents the rate law? a) rate = k[h 2 ] b) rate = k[no][h 2 ] c) rate = k[no][h 2 ] 2 d) rate = k[no] 2 [H 2 ] e) rate = k[no] 2 [H 2 ] 2 d [H ] 24. Using rate = 2, use the steady-state approximation to determine the rate law for the dt proposed mechanism and choose the correct statement below. a) If [H 2 ] is very low, the rate law appears to be the same as the rate law you determined in #23. If [H 2 ] is very high, the rate law appears to be rate = k[no] 2. b) If [H 2 ] is very high, the rate law appears to be the same as the rate law you determined in #23. If [H 2 ] is very low, the rate law appears to be rate = k[no] 2. c) If [NO] is very low, the rate law appears to be the same as the rate law you determined in #23. If [NO] is very high, the rate law appears to be rate = k[h 2 ]. d) If [NO] is very high, the rate law appears to be the same as the rate law you determined in #23. If [NO] is very low, the rate law appears to be rate = k[h 2 ]. e) The rate law appears the same as what you determined in #23 regardless of [NO] or [H 2 ]. ------------------------------------------------------------------------------------------------------------------------ 25. Recall the hydrogen chloride cannon demo from lecture in which a mixture of hydrogen gas and chlorine gas was initiated by a burning magnesium strip: H 2 (g) + Cl 2 (g) 2HCl(g) Suppose the activation energy for the reaction is changed from 239 kj/mol to 155.0 kj/mol at 298K by the introduction of a catalyst. Calculate the ratio of rate(catalyzed):rate(uncatalyzed). Assume the pre-exponential factor, A, is the same for the catalyzed and uncatalyzed reactions. a) 0.967 b) 1.03 c) 1.54 d) 2.30 x 10 7 e) 5.30 x 10 14
Hour Exam III Page No. 7 26. Recall the activated complex demonstration performed in lecture (the one which reacted to form a green transition state, and then turned pink to reveal the original catalyst). Suppose we run two trials, one at 55 C and one at 65 C. The times for the reaction are 85 seconds and 41 seconds, respectively. Determine the activation energy for this reaction. Assume the frequency factor, A, to be constant. a) 261 J/mol b) 663 J/mol c) 48.5 kj/mol d) 67.2 kj/mol e) 98.3 kj/mol 27. A first-order reaction is 42% complete at the end of 17 minutes. Determine the value of the rate constant (units are min -1 ). a) 3.2 x 10-2 b) 5.1 x 10-2 c) 1.8 d) 20 e) 31 --------------------------------------------------------------------------------------------------------------------------- 28-30. Choose the correct graph for the plots described below. A graph can be chosen once, more than once, or not at all. a) b) c) d) e) 28. A plot of [A] vs. time for reaction type aa products which is zero-order in A. [C] 29. A plot of t ½ vs. [A] for reaction type aa products which is first-order in A. [E] 30. t ½ vs k at constant temperature for a reaction type aa products which is first-order in A. [D]
Hour Exam III Page No. 8 31. We discussed in lecture how compounds can have the same chemical formula but different structures, and therefore different properties. Four different compounds exist with the chemical formula C 3 H 9 N. One such compound is isopropyl amine which is used in chemical weapons. Isopropyl amine has the following Lewis structure: a. Draw the three other distinct Lewis structures with the chemical formula C 3 H 9 N. For each of these structures you should minimize formal charge. Can you choose a best Lewis structure based on formal charge? If so, rank the structures. If not, explain why not. [8 pts.] All have 0 formal charge on all atoms; each equally good. b. Of the three compounds in part a, one is a gas at room conditions and the other two are liquids. Label which is the gas and which are liquids, and rank the two liquids in terms of boiling point (which boils at the higher temperature?). Support your answer with a discussion of VSEPR theory, explain the nature of any relevant intermolecular force, and explain the relationship between intermolecular forces and boiling points. [12 pts.] Trimethylamine is a gas and n-propylamine has a higher boiling point than N-methylethylamine. All same size, thus need to look at types of intermolecular forces. All tetrahedral/trigonal planar around the nitrogen, thus all polar (dipole-dipole interactions). Trimethylamine exhibits dipole-dipole interactions but not hydrogen bonding. N-methylethylamine contains an N-H bond which gives rise to hydrogen bonding. And n-propylamine contains two N-H bonds rather than one. H-bonding strong form of dipole-dipole interactions because N-H bond is particularly polar and the hydrogen atom is small. Strong IMF the more attraction between the molecules, thus the more energy required to separate the molecules to form a gas (higher boiling point).
Hour Exam III Page No. 9 32. In the future, you are visiting friends at their 10 th College Reunion (not UIUC). They are looking through an old chemistry notebook when they come across the following chemical equation: H 2 (g) + Cl 2 (g) 2HCl You all notice that there is no phase for HCl. I think it is a solid, states one of your friends. It is made up of H + and Cl ions. Another friend replies, I think it is a gas. The bond between hydrogen and chlorine is covalent not ionic. They are both ready to call off their lifelong friendship when you smile, take your Chemistry 202 textbook from your backpack, and tell them that thermodynamics can solve this problem. So, which friend is correct?; that is, which is the correct chemical equation for the reaction of hydrogen and chlorine gases? Your choices are: H 2 (g) + Cl 2 (g) 2HCl(s) or H 2 (g) + Cl 2 (g) 2HCl(g) Use the data below to provide quantitative support to your answer; that is, provide numbers and calculations along with your discussion. A complete answer will include a discussion of ΔH, ΔS, ΔS surr, and ΔS univ along with calculating any of these values you can while estimating the sign (with explanation) for those you cannot calculate. NOTE: Assume the lattice energy of HCl(s) = 1125 kj/mol
Hour Exam III Page No. 10 32. Provide your answer in the space below. See the previous page for what is required. [20 pts.] ===================================================================== Consider H 2 (g) + Cl 2 (g) 2HCl(g) We need ΔS univ to be positive for the process to be spontaneous. In this case, ΔS 0 because 2 moles of gas are becoming 1 mole of gas. ΔS surr due to enthalpy change of reaction, and we can estimate this with bond energies: Form 2HCl(g): 2(427 kj/mol) Break H 2 (g): +432 kj/mol Break Cl 2 (g): +239 kj/mol Overall: 183 kj Thus ΔH rxn is negative (exothermic), so ΔS surr is positive (heat released to surroundings) Thus, ΔS univ = ΔS surr + ΔS = (positive) + ( 0) = positive; should be spontaneous. ------------------------------------------------------------------------------------------------------------------- Consider H 2 (g) + Cl 2 (g) 2HCl(s) We need ΔS univ to be positive for the process to be spontaneous. In this case, ΔS is negative because 2 moles of gas are becoming 1 mole of solid. Thus, ΔS surr must be positive for the reaction to occur spontaneously. ΔS surr due to enthalpy change of reaction, and we can determine this as follows: H 2 2H (bond energy = +432 kj) 2H 2H + + 2e (2 x ionization energy = 2(~ +1200 kj/mol) = ~ 2400 kj Cl 2 2Cl (bond energy = +239 kj) 2Cl +2e 2Cl (2 x electron affinity = 2(~ 350 kj/mol) = ~ 700 kj 2H + + 2Cl 2HCl(s) (2 x lattice energy) = 2( 1125 kj/mol) = 2250 kj Overall = +121 kj Thus ΔH rxn is positive (endothermic), so ΔS surr is negative (heat released to surroundings) Thus, ΔS univ = ΔS surr + ΔS = (negative) + (negative) = negative; not spontaneous. Thus, reaction is H 2 (g) + Cl 2 (g) 2HCl(g)
Hour Exam III Page No. 11 33. The following data were collected in two studies of the reaction 2H 2 (g) + 2NO(g) N 2 (g) + 2H 2 O(g) EXPERIMENT #1 EXPERIMENT #2 TIME (sec) [NO] [NO] 0 1.00 x 10-2 M 1.00 x 10-2 M 10.0 6.67 x 10-3 M 5.00 x 10-3 M 20.0 5.00 x 10-3 M 3.33 x 10-3 M 30.0 4.00 x 10-3 M 2.50 x 10-3 M 40.0 3.33 x 10-3 M 2.00 x 10-3 M In experiment #1, [H 2 ] 0 = 10.0 M In experiment #2, [H 2 ] 0 = 20.0 M The following three mechanisms are proposed: I H 2 + NO H 2 O + N (slow) N + NO N 2 + O O + H 2 H 2 O II 2NO N 2 O 2 (fast equil.) N 2 O 2 + H 2 N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O III H 2 + 2NO N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O Your overall goal is to evaluate these mechanisms and choose the best mechanism. Provide a complete defense of your answer, which includes an explanation of which mechanism(s) you exclude and why. If you believe two (or even all three) mechanisms are equally good, explain why Along the way answer the questions on the following pages.
Hour Exam III Page No. 12 33 (con t). a. Use the concentration vs. time data to determine the rate law for the reaction. Show all work/explain your thinking (support your answer). Full credit is given for answers not relying on graphs. [4 pts.] rate = k[h 2 ][NO] 2 Data from Expt #2 show that the half-life increases by a factor of two as [NO] changes. Thus the reaction is second order with respect to NO. Between Expt #1 and Expt #2, [H 2 ] was doubled, and the half-life was cut by a factor of 2 (from 20 seconds to 10 seconds). Thus, doubling [H 2 ] doubled the rate. This means that the reaction is first order with respect to H 2. b. Solve for the rate constant (k) for the reaction. Include units. Show all work/explain your thinking (support your answer). [4 pts.] Using t=0 and t =10 from Expt #2 (for example): 1/(5.00 x 10-3 ) = k (10) + 1/(1.00 x 10-2 ); k = 10 = k[h 2 ]; 10 = k[20] k = 0.500 M -2 s -1 Using t=0 and t =30 from Expt #1 (for example): 1/(4.00 x 10-3 ) = k (30) + 1/(1.00 x 10-2 ); k = 5 = k[h 2 ]; 5 = k[10] k = 0.500 M -2 s -1 k = 0.500 M -2 s -1 c. Calculate the concentration of NO(g) in experiment #2 at t=90.0 seconds. Show all work. [2 pts.] 1/[NO] = (10)(90) + 1/(1.00 x 10-2 ) [NO] = 1.00 x 10-3 M
Hour Exam III Page No. 13 33. (con t) d. Choose the best mechanism. Provide a complete defense of your answer, which includes an explanation of which mechanism(s) you exclude and why. If you believe two (or even all three) mechanisms are equally good, explain why. [10 pts.] I H 2 + NO H 2 O + N (slow) N + NO N 2 + O O + H 2 H 2 O Adds up to correct stoichiometry Seems chemically reasonable Gives wrong rate law REJECTED rate = k 1 [NO][H 2 ] II 2NO N 2 O 2 (fast equil.) N 2 O 2 + H 2 N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O Adds up to correct stoichiometry Seems chemically reasonable Gives correct rate law POSSIBLE rate = k 2 [N 2 O 2 ][H 2 ] k 1 [NO] 2 = k -1 [N 2 O 2 ] [N 2 O 2 ] = k 1 /k -1 [NO] 2 rate = k 2 k 1 /k -1 [NO] 2 [H 2 ] III H 2 + 2NO N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O Adds up to correct stoichiometry Seems chemically reasonable Gives correct rate law POSSIBLE rate = k 1 [NO] 2 [H 2 ] Thus, mechanisms II and III are possible. However, mechanism II is a slightly better mechanism because it does not rely on a termolecular elementary step (which is possible, but unlikely).