CHEMISTRY 202 Hour Exam III. Dr. D. DeCoste T.A (60 pts.) 31 (10 pts.) 32 (20 pts.) 33 (30 pts.) Total (120 pts)

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1 CHEMISTRY 202 Hour Exam III December 4, 2014 Dr. D. DeCoste Name Signature T.A. This exam contains 33 questions on 12 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 (60 pts.) 31 (10 pts.) 32 (20 pts.) 33 (30 pts.) Total (120 pts) Useful Information: R = J/Kmol k = Ae -Ea/RT k 2 E ln( ) = a 1 [ k1 R T T ] 2

2 Hour Exam III Page No Given that the ionization energy of Si(g) = 780 kj/mol and that the electron affinity of Si(g) = 134 kj/mol, determine the ionization energy of Si (g). a) 780 kj/mol b) 134 kj/mol c) 646 kj/mol d) 646 kj/mol e) The ionization energy of Si cannot be determined with these data. 2. Which of the following correctly ranks the following from largest to smallest ionic radius? a) F -, O 2-, Na +, Mg 2+ b) Mg 2+, Na +, F -, O 2- c) Mg 2+, Na +, O 2-, F - d) Na +, Mg 2+, F -, O 2- e) O 2-, F -, Na +, Mg For an isoelectronic series, the species with the highest ionization energy is a) the species with the largest positive charge. b) the species with the smallest positive charge. c) the species with the largest negative charge. d) the species that is neutral (no charge). e) All ionization energies are the same since they are part of an isoelectronic series Consider the following orderings to answer the following questions. I. Li < B < F II. Mg 2+ < Mg + < Mg III. Cl < Br < I IV. Cl - < Ar < K + 4. Which of these give(s) a correct trend in size (radius)? a) II, III b) I, II c) IV d) III e) I, III, IV 5. Which of these give(s) a correct trend in ionization energy? a) I, II, III b) III, IV c) III d) I, IV e) II, III Which of the following has a Lewis structure most like that of the carbonate ion? a) the nitrate ion b) the sulfate ion c) carbon dioxide d) ozone (O 3 ) e) nitrogen dioxide

3 Hour Exam III Page No Which of the following is a linear molecule in which the central atom has three lone electron pairs? a) MgF 2 b) SF 2 c) KrF 2 d) OF 2 e) BeH 2 8. Which of the following molecules is nonpolar even though it has polar bonds? a) BF 3 b) NH 3 c) ClF 3 d) N 2 e) Cl 2 O 9. How many of the following molecules can correctly be described as bent? O 3 H 2 S SO 2 CO 2 XeCl 2 a) 1 b) 2 c) 3 d) 4 e) Which of the following names correctly describes the shape of SF 4? a) tetrahedral b) square planar c) trigonal bipyramid d) see-saw e) square pyramid 11. Consider a compound with the formula MN x, where M is the central atom and x = 2-6. How many of the following statements are false? (Note: could be true = false). I. If the formula is MN 4, the geometry of the molecule is tetrahedral. II. If the geometry is the same as the shape, the molecule is non-polar. III. If M is from the second row of the periodic table, x cannot equal 5 or 6. IV. If the formula is MN 2, the shape of the molecule is linear. a) 0 b) 1 c) 2 d) 3 e) Which of the following correctly labels the molecules as polar or non-polar? SO 2 BH 3 CO 2 a) polar polar polar b) non-polar non-polar non-polar c) polar non-polar polar d) polar non-polar non-polar e) non-polar polar polar 13. Identify the strongest intermolecular force for the given molecule C 2 H 6 NH 3 CH 3 OCH 3 a) LDF H-bond H-bond b) dipole-dipole LDF dipole-dipole c) H-bond H-bond dipole-dipole d) LDF dipole-dipole dipole-dipole e) LDF H-bond dipole-dipole

4 Hour Exam III Page No Consider 1.0 mol samples of the following gases at STP (0 C, 1 atm): CO, He, SO 2, and N 2. Arrange the gases from most ideal behavior to least ideal behavior. most ideal least ideal a) N 2 He SO 2 CO b) N 2 He CO SO 2 c) He N 2 SO 2 CO d) He CO N 2 SO 2 e) He N 2 CO SO Consider the following compounds: ethane (C 2 H 6 ) methanol (CH 3 OH) water (H 2 O) formaldehyde (H 2 CO) A mixture of which two would result in ΔH soln most closely equal to zero? a) Ethane and methanol b) Ethane and water c) Methanol and water d) Formaldehyde and water e) Ethane and formaldehyde Recall the iodine-clock demonstration from lecture in which a solution containing the iodate ion (IO 3 ) is mixed with a solution containing the metabisulfite ion (S 2 O 5 ) in the presence of a starch indicator. The reaction is known to be first order in [S 2 O 5 ]. The mechanism is complicated, but when [S 2 O 5 ] = 0 M, the solution turns from colorless to deep blue. We will assume the reaction happens quickly enough to use the method of initial rates. You carry out the reaction in two trials (both at 25.0 C) and obtain the following data: Trial [IO 3 ] 0 [S 2 O 5 ] 0 Time until blue I x 10-3 M 8.68 x 10-4 M 28.0 s II x 10-3 M 8.68 x 10-4 M 14.0 s 16. Determine the order for [IO 3 ] in the rate law. a) 0 b) 1 c) 2 d) 3 e) More info. is needed 17. Determine the value of k (units in terms of M and seconds). a) 8.74 x 10-2 b) 15.5 c) 435 d) 6750 e) 1.40 x If you run the reaction with [IO 3 ] 0 = 3.14 x 10-3 M and [S 2 O 5 ] 0 = 8.68 x 10-4 M, determine the time for the solution to turn blue. a) 15.0 s b) 18.3 s c) 20.5 s d) 21.5 s e) 24.2 s 19. You carry out Trial I again, but this time at 41.0 C. The solution turns blue after 18 seconds. Determine the activation energy of the reaction. a) 12.2 kj/mol b) 21.5 kj/mol c) 33.7 kj/mol d) 67.4 kj/mol e) 83.0 kj/mol

5 Hour Exam III Page No Which of the following statements is false? a) Reaction rate generally increases linearly with temperature. b) The magnitude of E for a reaction does not have an effect on reaction rate. c) The frequency factor, A, in the Arrhenius equation is technically temperature dependent, but we use it as though it is temperature independent. d) Catalysts provide reaction pathways with lower activation energies. e) All of the above statements (a-d) are true. 21. Consider the reaction aa Products in which [A] 0 = M. After 10 seconds, [A] = M and after a total of 30 seconds [A] = M. Determine [A] after a total of 45 seconds. a) M b) M c) M d) M e) M 22. How many of the following statements are true? I. The rate law of an elementary step comes directly from the coefficient(s) in the balanced elementary step equation. II. Adding a catalyst increases the amount of product formed from a given amount of reactants. III. In the steady-state approximation we assume the concentration of the intermediate is zero. IV. Changing the temperature changes the activation energy of a reaction. a) 0 b) 1 c) 2 d) 3 e) Recall the elephant s toothpaste demonstration which involved the decomposition of a 30% hydrogen peroxide solution at 25 C. You are studying the same reaction at different conditions (at constant temperature): 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) You record [H 2 O 2 ] over time and obtain the following data: Time (seconds) [H 2 O 2 ] (M) Determine the rate law constant for this reaction (using units of M and seconds). a) 7.50 x 10-3 b) c) d) e)

6 Hour Exam III Page No You are probably familiar with phenolphthalein as an acid-base indicator. In an acidic solution phenolphthalein appears colorless. When enough base is added to turn the solution slightly basic, the solution turns pink. It turns out that the color of a phenolphthalein solution will fade over time in a basic solution. We can monitor the disappearance of the color much as you did in the ferroin lab in Chemistry 203. We will use P 2 to designate the pink form of phenolphthalein and we can represent the reaction as: P 2 (aq) + OH (aq) POH 3 (aq) (pink) (colorless) and the rate law as: rate = k[p 2- ] m [OH] n where rate = - d [P 2- ] dt We run two experiments and collect the following data: Experiment #1: [OH ] = 0.30 M Experiment #2: [OH ] = 0.15 M [P 2- ] M Time (s) [P 2- ] M Time (s) 3.80 x x x x x x x x x x x x x x x x x x Determine rate law for the reaction. a) k[p 2- ] b) k[p 2- ][OH - ] c) k[p 2- ][OH - ] 2 d) k[p 2- ] 2 [OH - ] e) k[oh - ] 25. Determine the value of the rate law constant (using M and seconds) a) 1.23 x 10-6 b) 3.14 x 10-3 c) 6.30 x 10-3 d) e) 0.429

7 Hour Exam III Page No The following mechanism is proposed for the decomposition of A 2 B(g) to A 2 (g) and B 2 (g): 1. A 2 B(g) A 2 (g) + B(g) 2. A 2 B(g) + B(g) A 2 (g) + B 2 (g) 26. 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? 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. a) rate = k[a 2B] [A ] 2 2 b) rate = k[a 2 B] 2 c) rate = k[a 2B] [A ] 2 d) rate = k[a 2 B] 2 [A 2 ] e) rate = k[a 2 B] d [B ] 27. 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) The rate law appears first order in [A 2 B] at any [A 2 B]. b) The rate law appears second order in [A 2 B] at any [A 2 B]. c) The rate law appears first order in [A 2 B] at very low [A 2 B] and second order in [A 2 B] at very high [A 2 B]. d) The rate law appears first order in [A 2 B] at very high [A 2 B] and second order in [A 2 B] at very low [A 2 B]. e) The rate law has an inverse relationship with [A 2 B].

8 Hour Exam III Page No 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 t ½ vs. [A] for reaction type aa products which is second order in A. D 29. A plot of [NOBr] vs. time for a reaction NOBr(g) 2NO(g) + Br 2 (g) which has the following mechanism: 1. NOBr(g) NO(g) + Br(g) (slow) 2. NOBr(g) + Br(g) NO(g) + Br 2 (g) (fast) E 30. A plot of [NOBr] vs. time for a reaction NOBr(g) 2NO(g) + Br 2 (g) which has the following mechanism: 1. 2NOBr(g) NOBr-BrON(g) (slow) 2. NOBr-BrON(g) 2NO(g) + Br 2 (g) (fast) E

9 Hour Exam III Page No For each of the following questions, fully explain your answer, using Lewis structures and VSEPR theory when appropriate. a. The periodic table is an amazing achievement in human intellectual progress. Developed before atomic theory was widely accepted, it helps explain a variety of observations. For example, being in the same column, chlorine, bromine, and iodine are all diatomic in their standard states. It turns out, though that at room conditions, one is a gas, one is a liquid, and one is a solid. Match each element in its standard state with its phase and explain your answer. [5 points] Use Lewis structures to show these are all linear and non-polar. Since they are non-polar, the strongest IMFs are LDFs for each of these. With more electrons, the LDFs are stronger, thus the IMFs are strongest for I 2 and weakest for Cl 2. The stronger the IMF the more energy required to separate the molecules, thus the higher the boiling point or melting point. Thus, I 2 has the highest boiling and melting points and Cl 2 has the lowest boiling and melting points. Thus, Cl 2 is the gas, Br 2 is the liquid, and I 2 is the solid. b. You and your Chemistry 205 lab partner (yes, you made it!) are finishing up the final lab of the semester. You both accidentally and independently synthesize a compound with the formula XeCl 2 F 2. Interestingly, your compound is a gas and your partner s is a liquid, each at lab conditions. Explain how these compounds can have the same formulas yet exist in different phases at the same temperature and pressure. [5 points] Draw Lewis structures and use VSEPR to show that these are square planar. The chlorine atoms can either be 90 or 180 from each other. If 90, the molecule is polar; if 180, the molecule is non-polar. The polar molecule has stronger IMFs, meaning more energy required to separate the molecules, meaning a higher boiling point The polar molecule is the liquid, the non-polar molecule is the gas.

10 Hour Exam III Page No You are taking a vacation and on your flight you happen to sit next to a fellow chemistry major from another university. You talk about what you learned this past semester and you mention you ve just discussed why sodium chloride forms a compound with +1 and -1 charges. The new acquaintance treats you to an eye-roll, laughs and says We learned that in high school. Elements want to be like Noble gases. Since you are polite you stifle your laugh, grab a napkin and pen, get out your copy of Chemical Principles from your carry-on bag (yeah, I know), and get to explaining. Use the Tables below and justify your answers with words and numbers. Be polite and encouraging in your explanations. a. Giving your new friend the benefit of the doubt that want equates to being thermodynamically favorable, how do the data in Table 12.6 contradict the idea that elements want to be like Noble gases? Cite two specific examples with actual numbers from the Tables above in your explanation. Be complete! [4 points] Examples may vary but could include Na and Mg. If an element wanted to be like a Noble gas, it would be thermodynamically favorable for Na to lose an electron and for Mg to lose two electrons. Note that energy is required to remove an electron from Na (495 kj/mol) and a great deal of energy is required to remove two electrons from Mg ( = 2180 kj/mol). Could also say: the amount of energy to remove the second electron from Mg is essentially the same as the energy to remove an electron from Ar; or, the change in IE as we go from Na to Ar does not suddenly jump at Ar (that is, it is not somehow out-of-line stable). b. Your seatmate is a bit less arrogant at this point, but counters with: Yeah, but chlorine wants an electron and it s strong enough to take it from sodium. In the spirit of being encouraging and polite, first explain what is correct about this statement (again, assuming wants means it is thermodynamically likely) and then discuss why it is overall incorrect. Use numbers from the Tables above to support your answer. [4 points] It is correct that the electron affinity for chlorine is negative, meaning that it is an exothermic process, or thermodynamically favorable. However, 495 kj/mol is required to remove an electron from Na and only 348.7kJ/mol are released when Cl takes the electron. Thus, the process is overall endothermic (146.3 kj/mol) and overall unfavorable.

11 Hour Exam III Page No. 10 c. Before this person can say anything back, you say Ah, but I am afraid it s even worse! and you write the following chemical equation on the napkin: Na(s) + ½Cl 2 (g) NaCl(s). Explain all factors that show why the process is even less thermodynamically stable than portrayed in part b. Make sure to discuss entropy and Gibb s free energy (we will treat this as a constant pressure and constant temperature process). [4 points] From part b we see that the overall process is endothermic to remove an electron from Na(g) and to add an electron to Cl(g) [note the phases]. But as we see from the chemical equation, we start with Na(s) and Cl 2 (g). Thus, we also have to include the energy involved is the process of Na(s) Na(g) and in breaking the Cl 2 bond. Both of these will require energy and make the process even less favorable. Overall, then reaction as written about shows a decrease in entropy (of the system) since we are making an ordered solid structure and one of the reactants is a gas. We therefore need a large negative enthalpy to make up for this. d. Now that your seatmate is regretting not attending UIUC, the question is sheepishly asked So, why then does NaCl even form? Explain the term that is missing and how it affects the overall spontaneity of the reaction. [4 points] The term that is missing is lattice energy. This is the energy released when a cation and anion come together to form the ionic solid lattice. This is a very exothermic process that more than makes up for the other steps and makes the overall process quite exothermic. The process needs to be exothermic since S for this process is negative (the ionic solid is very structured). e. This person thanks you for your time, but has a couple of follow up questions. But why then would magnesium lose two electrons? And if it is somehow better to lose two, why doesn t sodium do this? Answer these questions with words and numbers. [4 points] Lattice energy is directly related to the charge of the ions. To form a (+2, 2) structure, for example, would release four times as much energy due to lattice energy as a (+1, 1) structure. Note that while the 2 nd IE of Mg is larger than the first (1445 compared to 735 kj/mol), the difference is made up by lattice energy. Since the 2 nd IE for Na is very much greater that the first for Na (4560 compared to 495 kj/mol), the lattice energy difference would not be enough.

12 Hour Exam III Page No Pure N 2 O was placed in a rigid vessel and allowed to decompose. The following data were collected in an experiment at a constant temperature of 298K. Note that this is total pressure. Assume the rate of the reverse reaction is negligent in this problem. 2N 2 O(g) 2N 2 (g) + O 2 (g) rate = d( P 2 dt N O ) P total (atm) Time (min) atm atm atm atm atm atm a. Determine the rate law for the reaction including the order of the reaction and the value (with units) for the rate constant. Show all work. [6 pts.] P total (atm) = P N2O + P N2 + P O2 = ( x) + 2x + x = x P N2O (atm) Time (min) atm 0 Note that half-lives appear constant (~24 min); first order atm atm rate = k[n 2 O] atm atm Use integrated rate law; get k = min atm b. Determine the partial pressure of N 2 (g) at min. Show all work. [5 pts.] ln(p N2O ) = ( min -1 )(110.0 min) + ln(1.000 atm) P N2O at min = atm Thus, change in pressure for N 2 O = = atm Change in pressure for N 2 O = change in pressure for N 2 ; thus P N2 = atm c. The reaction is carried out again at 303K. The first half-life is noted to be at 4.58 min. Determine the activation energy of the reaction. Show all work. [5 pts.] k = ln(2)/(t 1/2 ) = (ln(2)/4.58min) = min -1. Thus, the rate increases by a factor of ( min -1 / min -1 ) = 5.28 E ln(5.28) = a 1 [ R ] Ea = kj/mol 303

13 Hour Exam III Page No. 12 d. The formula of the reactant, N 2 O, doesn t tell us if the skeletal structure is N-O-N or N-N-O, nor does it tell us if there is a best Lewis structure. Your goal is to find the best Lewis structure for the compound, using formal charge as the primary basis. Define formal charge in your answer. Justify your answer completely. [8 pts.] Formal charge: (# valence electrons in free atom) (# valence electrons assigned to the atom in the molecule) Five possible Lewis structures (note: lone pairs are not included): I II III IV V N=N=O N N O N N O N=O=N N O N Formal charges for I: -1, +1, 0 (N: 5-6 = -1, N: 5-4 = +1, O: 6-6 = 0) Formal charges for II: 0, +1, -1 (N: 5-5 = 0, N: 5-4 = +1, O: 6-7 = -1) Formal charges for III: -2, +1, +1 (N: 5-7 = -2, N: 5-4 = +1, O: 6-5 = +1) Formal charges for IV: -1, +2, -1 (N: 5-6 = -1, O: 6-4 = +2, N: 5-6 = -1) Formal charges for V: 0, +2, -2 (N: 5-5 = 0, O: 6-4 = +2, O: 5-7 = -2) We want to minimize formal charge, so structures I and II are the best. Since oxygen is more electronegative than nitrogen, oxygen should have the negative formal charge (since it has more attraction for a shared electron). So, structure II is better than structure I when taking formal charge into account. e. Using the bond energy below and your answers to parts a, c, and d above, develop a mechanism for the reaction and support your answer. [6 pts.] Bond Bond energy (kj/mol) N=O 607 N O 201 N=N 418 N N 941 The reaction is first order in N 2 O, the structure is closest to N N O (although partially like N=N=O as well), and the activation energy is ~250 kj/mol, which is close to the bond energy of N O (and between that of N O and N=O). Thus, it makes sense that the slow step would be the breaking of the N O bond. A plausible mechanism is: 1. N 2 O(g) N 2 (g) + O(g) (slow) 2. N 2 O(g) + O(g) N 2 (g) + O 2 (g) (fast) This gives the proper rate law and adds up to the overall reaction.