CHEMISTRY 0 Hour Exam III (Multiple Choice Section) December 7, 07 Dr. D. DeCoste Name Signature T.A. This exam contains 0 questions on 4 numbered pages. Chec now to mae sure you have a complete exam. Determine the best answer to these questions and enter these on the special answer sheet. Also, circle your responses in this exam boolet. Each multiple choice question is worth 4 points (total of 80 points). The free response section is worth a total of 80 points. Useful Information: PV = nrt R = 8.34 J/Kmol = 0.0806 Latm/molK = Ae Ea/RT E ln( ) = a [ R T T ]
Hour Exam III (Multiple Choice Section) Page No.,. The following is a seletal structure of the amino acid histidine (as seen in lecture and the textboo!). Complete the Lewis structure such that all atoms have a formal charge of zero, and answer the following questions.. How many multiple bonds are present in the completed Lewis structure?. Which of the following best describes the shape around the carbon atom labeled and the nitrogen atom labeled? Carbon atom (#) Nitrogen atom (#) a) tetrahedral tetrahedral b) trigonal pyramid trigonal planar c) trigonal planar trigonal pyramid d) trigonal pyramid trigonal pyramid e) trigonal planar tetrahedral --------------------------------------------------------------------------------------------------------------------- 3. How many molecular geometries can result in a shape described as bent? 4. How many of the following have polar bonds but are overall nonpolar molecules? ICl 5 NH 3 CCl 4 N SeO CO SF 4 XeCl a) b) 3 c) 5 d) 6 e) 8 5. Octane (C 8 H 8 ) has a higher boiling point than methanol (CH 3 OH). How many of the following is/are true? The higher boiling point is because there is more hydrogen bonding in octane than in methanol. Octane is also expected to have a higher vapor pressure than methanol. The higher boiling point is because the net dipole moment is greater for octane than methanol. The shape around the carbon atoms in octane contributes to its greater polarity, thus its higher boiling point. The higher boiling point for octane has to do with its larger size than methanol.
Hour Exam III (Multiple Choice Section) Page No. 6. An ion has the formula [XO F]. The shape around the central atom, X, is T-shape. Which of the following could be the identity of X? a) Ne b) Br c) Cl d) At least two of these (a-c) could be the identity of X. e) None of these (a-c) could be the identity of X. 7. Use the following data (all in J/mol) to determine the second electron affinity (EA) value for the sulfur atom. The standard states for sodium and sulfur are Na(s) and S(s), respectively. ΔH f Lattice energy st IE for metal st EA for non-metal ΔH sub for metal ΔH sub for non-metal Na S 365 03 495 00. 09 77 a) 88 J/mol b) 553 J/mol c) 79 J/mol d) 57 J/mol e) More data are needed. 8. Consider the reaction as shown and determine the initial rate (in Ms ) for Experiment 4. (CH 3 ) 3 CBr(aq) + OH (aq) (CH 3 ) 3 COH(aq) + Br (aq) [(CH 3 ) 3 CBr] 0 [OH ] 0 Initial Rate. 0.0 M 0.0 M.0 0 4 Ms. 0.0 M 0.0 M.0 0 4 Ms 3. 0.0 M 0.0 M.0 0 4 Ms 4. 0.30 M 0.0 M a).7 0 3 b) 9.0 0 4 c) 4.5 0 4 d) 3.0 0 4 e).0 0 4 ------------------------------------------------------------------------------------------------------------------------ 9, 0. Consider the generic reaction aa + bb + cc dd + ee where the rate law is d[a] = [A] [B][C] An experiment is carried out in which [B] 0 =.000 M, [C] 0 = 3.000 M, and [A] =.000 x 0 4 M. After 3.000 minutes, [A] = 3.40 x 0 5 M. 9. Determine the value of the rate constant, (units using M and minutes), for the reaction. a).87 0 5 b) 6.435 0 3 c) 0.386 d) 4 e) 78 0. Determine [A] after 4.00 minutes. a) 7.73 0 5 M b) 4.06 0 5 M c) 3.704 0 5 M d) 8.933 0 6 M e) 4.49 0 7 M
Hour Exam III (Multiple Choice Section) Page No. 3. For the reaction NOCl(g) NO(g) + Cl (g), successive half-lives are observed to be 0.0 min, 0.0 min, and 40.0 min. If [NOCl] 0 = 0.00 M, determine [NOCl] at 80.0 minutes. a) 0.096 M b) 0.05 M c) 0.0 M d) 0.0093 M e) 3.9 x 0 4 M. The rate law for 4PH 3 (g) P 4 (g) + 6H (g) is second order in PH 3 (g). When [PH 3 ] 0 = 0.00 M, the reaction is 0.0% complete in 40.0 minutes. Determine the first half-life of the reaction. a) 0.0 min b) 00. min c) 4 min d) 37 min e) 60. min -------------------------------------------------------------------------------------------------------------------- 3 7. Consider the reaction N O 5 (g) 4NO (g) + O (g). At a given temperature you run this reaction and obtain the following data: time (minutes) [N O 5 ] (M) 0.4 x 0 0 6.6 x 0 3 30 4.83 x 0 3 50.58 x 0 3 70.38 x 0 3 3. Determine [O ] after the reaction has been running for 0.0 minutes. a).67 x 0 3 M b) 3.34 x 0 3 M c) 4.53 x 0 3 M d) 6.68 x 0 3 M e) 9.06 x 0 3 M You run the reaction again twice. Once at 98 K and determine that =. x 0 3, and another at 38K and determine that = 0.0. ( values in units of molarity and minutes). 4. Determine the activation energy for this reaction. a) 04 J b) 8 J c) J d) 859 J e) 0,500 J 5. Determine the temperature at which the first reaction (for #3) was run. a) 78 K b) 98 K c) 38 K d) 38 K e) 338 K 6. Determine the first half-life if you were to run this at 348K ([N O 5 ] 0 =.4 x 0 M). a) 45.8 seconds b).7 min c) 5.0 min d) 9.4 min e) 9. min 7. From these data, we can tell that the decomposition of N O 5 (g) must be a) exothermic. b) endothermic. c) spontaneous at high temperatures but not low temperatures. d) spontaneous at low temperatures but not high temperatures. e) We cannot tell any of the above (a-d) from these data.
Hour Exam III (Multiple Choice Section) Page No. 4 8. Consider the following two generic reactions: d[a] aa bb rate = = =.30 x 0 M/min at 5 C cc dd d[c] rate = = [C] In a particular experiment, A and C are placed in separate containers (at 5 C) under conditions such that [A] 0 = 0.5 M and [C] 0 = 0.50 M. After the reactions had progressed for 5.00 min, it was found that [A] = [C]. Determine the value of (units of Lmol min ) at 5 C. a).30 x 0 b) 4.90 x 0 c) 0.599 d) 9.0 e) 39. ------------------------------------------------------------------------------------------------------------------------------ 9-0. Most enzyme reactions are found to obey the following mechanism in which the enzyme, E, binds to the substrate, S, to form a complex E S. This complex subsequently decomposes:. E + S E S. E S Products + E d[products] 9. Determine the rate law for the reaction assuming a steady state for E S; rate =. a) rate = [E][S] [E][S] b) rate = - + [S] c) rate = [E][S] d) rate = e) rate = [E][S] - + [E][S] 0. Which of the following plots best represents reaction rate as a function of [S]? a) b) c) d) e)