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key_exam_22 - {N 65 ‘ fl 1 Last Name First Name Lab Sec...

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Unformatted text preview: {N 65 ‘/ fl 1 Last Name First Name Lab Sec. # ; TA: ; Lab day/time: Andreas Toupadakis, Ph.D. Spring 2007 CHEMISTRY 2C Section A EXAM 2 - . Multi le Choice Instruct/ans. . curcle one) CLOSED BOOK EXAM! No books, notes, or additional scrap paper are permitted. All information required is contained on the exam. Place all work in the space provided. If you require additional space, use the back of the exam. A scientific calculator may be used (if it is a programmable calculator, its memory must be cleared before the exam). This exam has 8 pages total. (1) Read each question carefully. (2) For Part I there is some partial credit. There is no partial credit given for part II. There is partial credit for part III. (3) The last page contains a periodic table and some useful information. You may remove it for easy access. (4) If you finish early, RECHECK YOUR ANSWERS! U.C. Davis is an Honor Institution Possible Points _l' Earned Points #1—12 (3 points each) / 36 # 13—18 (7 points each) / 42 #19 (14 points) / 14 _# 20 (8 points) _L_ / 08 Total Score (100) l / 100 Name Exam 2 (Page 2 of 9) Part I: Concepts (3 points each) Some partial credit is available 1. While PCI5 exists, NH5 does not exist. Why? The nitrogen atom does not have any d orbitals available for bonding, the phosphorus atom does. Consider the fab/e be/ow and answer the fo/low/ng two questions (2 and 3): ©ATZOO7SPR Element Ionization Energy Standard Oxidation Melting Point (kJ/mole) Potential (V) (°C) Li l 520 l 3.05 180 | Na l 495 l— 2.71 98 l 2. According to their ionization energies, Na is oxidized easier than Li. According to their standard oxidation potentials, Li is oxidized easier than Na. Explain this apparent disparity For a particular element, each oxidation process (electron transfer I e. electron removal) L takes place under different conditions. The values 520 and 495 (ionization energy) refer to electron transfer In the gas phase, [Energy + M(g) —> M (g) + e ]. In contrast, the values 3 .05 and 2.71 (standard oxidation potential) refer to electron transfer In water, [M(s) —> M (aq) + e + E]. In the gas phase, the endothermic electron transfer involves the energy of one step. In water, the electron transfer involves the energy of the same step as in the gas phase but in addition to that it involves the energy of two additional steps. One is the endothermic sublimation step and the other is the exothermic hydration step. [Energy + M(s) ——> M(g)] then [Energy + M(g) —> M (g) + e ]and finally [M '(g) + water —> M‘(aq) + energy]. Even though the required energy for the sublimation of Li is higher than that for No, because ‘of its much smaller size (higher charge density) the released hydration energy for Li is much 1 higher than that for Na. Thus In water, the oxidation of Li is more thermodynamically favored than the oxidation of Na. 3. Even though according to their standard oxidation potentials, Li is oxidized easier than Na, the reaction of sodium with water is more vigorous than the reaction of lithium with water. Explain why. {L “Easier" relates to energy/thermodynamics. “Vigorous" relates to speed/kinetics._ Na has a much lower melting point tthan Li and during the exothermic reaction it melts i. becoming a very large number of _t_____iny s_p_____heres thus acquiring a much larger surface {L area thus reacting faster. 4. Give the Lewis structure of potassium superoxide and the oxidation number of each of its two oxygen atoms (show which oxygen atom has what oxidation number). I< L970? I I —l O Name Exam 2 (Page 3 of 9) 6. 7. All metal aX/des of group 2 show basic properties. Only beryllium oxide shows acidic properties. Please answer the following two questions (5 and 6 ): ©AT2007SPR Why is only beryllium oxide amphoteric among all the other oxides in group 2? Because of the much smaller size of the beryllium atom or ion relatively to the other elements in the group. It has larger positive charge density thus attracting and bond to hydroxide ions. Demonstrate the acidic properties of beryllium oxide by providing a balanced chemical equation. BeO(s) + 20H'(aq) + H20(l) ——> Be(OH)ii2'(aq) Give the structure of the following three molecules: O l o/IP\O lo I no.0 ’9‘ o , icy/kc 0‘7 \ O O Tetraphosphorus Decoxide Name Exam 2 (Page 4 of 9) 8. If increasing The concenTraTion of A in a chemical reacTion causes no increase in The raTe of The reacTion, Then we may say The reacTion raTe is firsT order in [A]. a) True b) False 9. If The half-life of a reacTanT is independenT of HS iniTial concenTraTion, The reacTion order is O. a) True b) False 10. Energy of acTivaTion has no effecT on reacTion raTe. a) True b) False 11. Adding a caTalysT lowers The acTivaTion energy of a reacTion. a) True b) False 12. A heTerogeneous caTalysT is a caTalysT ThaT is in Two phases of maTTer. a) True b) False Name 13. 14. 15. Exam 2 (Page 5 of 9) Part II: Short Calculations (7 points each) No partial creclit (right or wrong) The rate of formation of NH3 in the reaction N2(g) + 3H2(g) —+ 2NH3(g) was reported as 1.2 mmolL'ls'1 under a certain set of conditions. What is the rate of consumption of hydrogen? a) 3.6 mmolL'ls'1 b) 2.4 mmolL‘ls'1 c) 1.2 mmolL'ls‘1 d) 1.8 mrnoIL'ls'1 e) 0.8 mmolL'ls‘1 A reaction has a rate law of the form k[A]2[B]. What are the units of the rate constant k if the reaction rate is measured in mol L'1 3'1? a) mol'1 L2 s‘1 b) mol'2 L2 s'1 c) mol'2 L s"1 d) mol'2 L2 s e) mol'2 L3 s"1 A certain first-order reaction has a aaIf-life of 20.0 minutes. How much time is required for this reaction to be 75% complete? a) 20. min b) 40. min c) 15. min d) 30. min e) 5 min Name 16. 17. 18. Exam 2 (Page 6 of 9) The rate constants of the forward and reverse elementary reactions for the dimerization of proflavin, an antibac'erial agent that inhibits the biosynthesis of DNA by intercalating between adjacent base pairs, were found to be 8.1x108 mol'1 L 5'1 (second order) and 2.0x106 3'1 (first order), respectively. The equilibrium constant for the dimerization is: a) 2.5x10‘3 b) 4.0x102 c) 4.0x103 d) 5.0x102 e) 2.0x10‘2 The conversion of cyclopropene to propane is a first-order reaction. Therefore the pre—exponential factor, A (an Arrhenius parameter) has units: ©AT2007SPR a) Lmols‘1 b) Lmol'ls‘2 c) mol'ls'1 d) Lmol'ls‘1 e) s‘1 Iodine atoms combine to form molecular iodine in the gas phase: I(g)+I(g) —> I2(9) This reaction follows second-order kinetics and has the high rate constant 7.0x109 M'1~ s‘1 at 23°C. If the initial concentration of I(g) was 0.086 M, the concentration after 2.0 min is: a) 1.2x10‘12 M b) 1.7x10'18 M c) 8.4x1045 M d) 1.2x1022 M e) 7.4x104 M Name Exam 2 (Page 7 of 9) ParT III: ParTial crediT may be given Please show all work for calculations 19. (14 poinTs) The reacTion of niTric oxide wiTh hydrogen aT 1280 °C is: 2NO(g) + 2H2(g) —» N2(g) + 2H20(g) From The following daTa collecTed aT This TemperaTure: a) DeTermine The raTe law expression for The reacTion. b) DeTermine The value of The raTe consTanT. c) DeTermine The raTe of The react on when [N0] = 12.0x10'3 M and [H2] = 6.0x10‘3 M. l EXP [NO] (M) [F2] (M) l7 IniTial Rafe (M/s) ‘ l 1 5.0x10'3 2.0x10'3 1.3x10'5 l l 2 10.0x10’3 2.0x10'3 5.0x10'5 I l 3 | 10.0x10'3 4.0x10'3 10.0x10‘5 a) RaTe law expression is: x r y QQTC: lK£NOj LHI] - .3 x .3 \) kClo-OXTO ml (ZOHO M ‘j Qalcizl coxm'SM/é sq- Qfif (‘i\ ‘lfia x Io'g M IS l< (C0 MODE/Mlx (2.0 x :6le ‘ A»; , _ ._ , ‘ 11 ,L , (T m X~1 L<(“Jamal/Ml)((Li.oj§lc§2AM;J o.__. «w... a... ., onlbi'l lO.O'Xl6§Nl§ ., 1 'j k(lO.OXlD‘;M\X (1.0 \UQ’KN) “J ._—————-—-——-———-——- ’ quem‘, go “0”;ng - ’L ‘rlAeve/l’ove r~> Rafe = k [N012 [H2] b) RaTe consTanT is: Yxow 419.1: M Yale , {fox lo“g NI; k = 2.5x102M'2 -s'1 TNGT‘ [Hz] («0.0 x (O‘BMlLU‘O XIOJ‘AH c) RaTe of reacTion is: 1 ~ "‘3 ’L Qale:(’lsx\o/M1.<l(l1.owo Ml~ _ —4 ‘ (60 X ‘0’ 3M) /'\_/) Rafe — 2.2)(10 M/s Name Exam 2 (Page 8 of 9) 20. (8 points) Consider the decomposition of hydrogen peroxide at T°C: 2 H202 (0CD —* ZHZOU) + 02 (9) Derive the equation which relates the reaction rate based on the rate of oxygen production as a function of oxygen pressure change AP, instead oxygen concentration change, A[Oz]. Qfliié: VAEO’LI (3t PtEO1—SQT Ag. (3?: [)[OAX QT AEO’lj : #9., RT Quite: BRA... : 9.3.}. QT DJ: 0% QT Q'QCJC‘ROW Qote: Ef— i. at QT Name Exam 2 (Page 9 of 9) Some useful equations and constants: PLEASE NOTE: Important values and equations required for calculations are given with the respective problem. The following may or may not be of any use. Constants: R = 8.3145 J / mol K NA = 6.022 x1023 1atm = 760 torr R = 0.0821 L atm / mol K h = 6.626 x10‘34 J 5 1nm :10-9 m 1 L = 1000 cm3 Equations: Slope : Ay/Ax TK = Toc+ 273.15 T0}: — 32 = 1.8 Toc ha AC°=AH°—TAS° AG=AG°+RTan AG°=—RT1nKeq k= Ae R’ m 2m 1n£=_Ea[i_i] k2 R T T rule :k[A]° 2k [A], =—kt+[A]0 Q = [31,10 ””6 2 HA] lnlA], = —kt+1n[A]0 ‘% = 29kg m’e=k[/‘1]2 szt+ 1 t. = 1 [A], [A]0 3 k[A]0 Periodlic Table l l l I .0079»! J 90121“ 12 28 5 7: '79 no u l § 195.078 [96 M55 200- ill ”2 ( (3'71) {173'} 73 76 Ta Re Os ”09‘1“? "UK-U 1862117 19023 I922” “)5 106 ._ . ills ' 3 : _ (BM: 1265) Ni 85934 46 d Pt I l0 269: ...
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