02/05/09 Last 4 Digits of USC ID: Dr. Jessica Parr
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1 Chemistry 05 B First Letter of PLEASE PRINT YOUR NAME IN BLOCK LETTERS Exam last Name Name: 02/05/09 Last 4 Digits of USC ID: Dr. Jessica Parr Lab TA s Name: Question Points Score Grader Total 00 Please Sign Below: I certify that I have observed all the rules of Academic Integrity while taking this examination. Signature: Instructions:. You must show work to receive credit. 2. If necessary, please continue your solutions on the back of the preceding page (facing you). 3. YOU MUST use black or blue ink. (No pencil, no whiteout, no erasable ink.) 4. There are 0 problems on 2 pages. Please count them before you begin. A periodic table and some useful equations can be found on the last page. 5. Good luck!! =)
2 . (2 pt) Fill in the blanks in the following sentences. a. For a rate mechanism to be appropriate the rate law from the must match the experimentally determined rate law. b. The molecularity of the following elementary step is : 2 A (g) B (g) + 2 C (g) c. The is the minimum amount of energy that must be overcome to produce a chemical reaction. d. A(n) catalyst is in the same state as the reactants. e. A(n) is produced in one elementary step and consumed in a subsequent step. f. In order for molecules to react they must overcome some minimum amount of energy as well as have the correct with respect to each other. 2
3 2. (9 pt) Circle the correct answer for the following multiple choice questions. a. Determine the rate law for the following reaction based on the initial rate data provided. 2 H 2 (g) + 2 NO (g) 2 H 2 O (g) + N 2 (g) Experiment [H 2 ] o (M) [NO] o (M) Initial Rate (M/s) i. rate = k[h 2 ][NO] ii. rate = k[h 2 ] 2 [NO] iii. rate = k[h 2 ][NO] 2 iv. rate = k[h 2 ] 2 [NO] 2 v. cannot be determined b. A reaction was found to have a rate law of: rate = k[clo 2 - ][OH - ]. A rate of M/s was found when the initial concentration of ClO 2 - was M and that of OH - was M. What is the value of the rate constant? i x 0-5 M - s - ii. 3.8 M - s - iii x 0-5 s - iv. 3.8 s - v x 0-5 M/s v. 3.8 M/s c. A reaction is found to be second order in NO 2 and zero order in CO. If the rate constant for the reaction is 2.08 x 0-4 M - s -, the initial concentration of NO 2 is M and the initial concentration of CO is 0.43 M, what is the initial rate under these conditions? i x 0-5 M/s ii x 0-5 M/s iii..09 x 0-4 M/s iv x 0-5 M/s v x 0-5 M/s 3
4 3. (9 pt) The rate of the reaction: O (g) + NO 2 (g) NO (g) + O 2 (g) was studied at a certain temperature. In a set of experiments, NO 2 was in large excess at a concentration of.0 x 0 3 molecules/cm 3 and the following plots were found: [O] (molecules/cm^3) 6.00E E E E E+09.00E E E+09 y = -.55E+x E ln[o] y = -00.5x Time (s) Time (s) /[O] (cm^3/molecules) 4.50E E E E E E-09.50E-09.00E E E E E-09 y =.23E-07x E Time (s) a. What is the order of the reaction with respect to O? b. The reaction is known to be first order with respect to NO 2. Determine the overall rate law and the value of the rate constant (with appropriate units). 4
5 4. (2 pt) The rate law for the reaction: 2 NOBr (g) 2 NO (g) + Br 2 (g) at some temperature is k[nobr] 2, where k is M - s -. a. What is the half-life of this reaction if the initial concentration is M? b. What percentage of NOBr remains after 9.00 s if the initial concentration is.50 M? 5
6 5. (8 pt) The activation energy for a particular reaction is 78.9 kj/mol. As the temperature is increased from 32.0 o C to a higher temperature, the rate constant increases by a factor of Calculate the higher temperature. 6
7 6. (2 pt) Determine whether the following statements are true or false. If the statement is false correct the part in bold to make it true. a. Equilibrium is reached in chemical reactions when the rates of the forward and the reverse reactions are equal to each other. True False b. The equilibrium constant for a particular reaction is 3.5 x 0 8. In this reaction the equilibrium lies to the left. True False c. If a system was initially at equilibrium and the concentration of one of the products was doubled, the equilibrium constant would also double. True False d. Pure solids are never included in the equilibrium expression. True False 7
8 7. (9 pt) Circle the correct answer for the following multiple-choice questions. a. For the following reaction, 2.00 moles of A and 3.00 moles of B are placed in a 6.00 L container: A (g) + 2 B (g) C (g). At equilibrium, the concentration of A is M. What is the concentration of B at equilibrium? i M ii M iii M iv M v. cannot be determined b. Consider the reaction C (s) + CO 2 (g) 2 CO (g). At 273 K the K p value is What is the P CO at equilibrium if the P CO2 is 0.0 atm at this temperature? i..4 atm ii. 2.0 atm iii. 4. atm iv. 6.7 atm v. 250 atm c. If the equilibrium constant for A (g) + B (g) C (g) is 0.23, then the equilibrium constant for 3 C (g) 3 A (g) + 3 B (g) is: i..86 x 0-3 ii iii iv. 8.3 v
9 8. (8 pt) At 650 K, the value of the equilibrium constant, K, for the ammonia synthesis reaction (given below) is.22. If a vessel contains a reaction mixture in which [N 2 ] = 0.00 M, [H 2 ] = M, and [NH 3 ] = M, will more ammonia form? (Circle Yes or No below, AND show work to support your answer.) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Circle the correct answer: YES NO 9
10 9. (9 pt) At a certain temperature, K = 9. x 0-4 for the reaction: FeSCN 2+ (aq) Fe 3+ (aq) + SCN - (aq) Calculate the concentrations of Fe 3+, SCN -, and FeSCN 2+ in a solution that is initially 2.0 M FeSCN 2+. [FeSCN 2+ ] [Fe 3+ ] [SCN - ] 0
11 0. (2 pt) Consider the decomposition of the compound C 5 H 6 O 3 as follows: C 5 H 6 O 3 (g) C 2 H 6 (g) + 3 CO (g) When a 5.63 g sample of pure C 5 H 6 O 3 (g) was sealed into an otherwise empty 2.50 L flask and heated to 200. o C, the pressure in the flask gradually rose to.63 atm and remained at that value. a. What is the initial pressure of C 5 H 6 O 3 in the flask? b. What are the pressures of C 5 H 6 O 3, C 2 H 6 and CO in the flask at equilibrium? P C5H6O3 P C2H6 P CO c. What is K p for this reaction?
12 2 H He Li Be B C N O F Ne 20.8 Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc (99) 44 Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po (209) 85 At (20) 86 Rn (222) 87 Fr (223) 88 Ra Ac Rf (26) 05 Db (262) 06 Sg (263) 07 Bh (262) 08 Hs (265) 09 Mt (268) Lanthanides 58 Ce Pr Nd Pm (45) 62 Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 75 Actinides 90 Th Pa U Np Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (25) 99 Es (252) 00 Fm (257) 0 Md (258) 02 No (259) 03 Lr (26) R = L*atm/K*mol = 8.34 J/K*mol PV = nrt k = A exp(-e a /RT) = 2 2 ln T T R E k k a ] 0 [ ] [ A kt A + = 0 ] ln[ ] ln[ A kt A + = ] 0 [ ] [ A kt A + =
Last 4 Digits of USC ID:
Chemistry 05 B Practice Exam Dr. Jessica Parr First Letter of last Name PLEASE PRINT YOUR NAME IN BLOCK LETTERS Name: Last 4 Digits of USC ID: Lab TA s Name: Question Points Score Grader 8 2 4 3 9 4 0
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