Name: YOU MUST: Put your name and student ID on the bubble sheet correctly. Put all your answers on the bubble sheet. Please sign the statement on the last page of the exam. Please make sure your exam has 6 pages total. Please keep your eyes on your own paper and your answers covered. Use the exam as scratch paper. We will not grade anything on the exam itself. Turn in both the exam and bubble sheet when you are done. Good luck! Questions 1-10 compare two quantities. For the following pairs, select A if the item in column A is larger, B if the item in column B is larger, C if they are equal, or D if you cannot tell from the information provided Which is greater? Column A Column B 1. 8.31 J/(mole K) vs. 0.0821 (L atm)/(mole K) 2. 1 vs. The order of the 14 C radioactive decay reaction 3. The rate of an exothermic reaction vs. The rate of an exothermic reaction at at T= 200K at T = 400K 4. K c for an exothermic reaction vs. K c for an exothermic reaction at at T= 200K at T = 400K 5. 1 vs. K c for a reactant-favored reaction 6. Q for a reaction vs. K c for the same reaction 7. Q for the reaction X (g) Y (g) vs. K c for this reaction at time t = 0 if 0.1 moles X is added at time t = 0 8. K c for a product-favored reaction vs. K c for the reverse of this reaction 9. K c at 298K for a gaseous reaction vs. K p at 298K for the same reaction with Δn = -2 10. The overall order of the reaction vs. 2 X + Y Z 1
11. The following item is excerpted from The Daily Collegian on October 20 th, 2005: [A UMass student] failed a breath blood-alcohol test 8 hours later If a student records a blood alcohol level of 0.0167M (or 0.1% by volume), what was the student s alcohol level 8 hours before this measurement if the zero order rate constant for the oxidation of ethanol in the liver is 0.00425 M/hour? A. 0.0162M ( = 0.097% by volume) B. 0.0173M ( = 0.103% by volume) C. 0.0254M ( = 0.152% by volume) D. 0.0434M ( = 0.260% by volume) E. 0.0507M ( = 0.303% by volume) 12. Which is the correct form of the balanced chemical equation that corresponds to the following equilibrium constant expression? [NH K c = 3 ] [N 2 ] 1/2 [H 2 ] 3/2 A. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) B. 2 NH 3 (g) N 2 (g) + 3 H 2 (g) C. NH 3 (g) 1/2 N 2 (g) + 3/2 H 2 (g) D. 1/2 N 2 (g) + 3/2 H 2 (g) NH 3 (g) E. 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) The next two problems refer to the following data, collected at 904 C for the reaction: 2 NO(g) + 2 H 2 (g) N 2 (g) + 2 H 2 O(g) Experiment [NO] [H 2 ] rate of appearance of N 2 1 0.420 M 0.122 M 1.36 x 10-4 M/sec 2 0.140 M 0.122 M 1.51 x 10-5 M/sec 3 0.140 M 0.244 M 3.02 x 10-5 M/sec 13. From the above data, the rate equation for this reaction is: A. rate = k[no][h 2 ] B. rate = k[no] 2 [H 2 ] C. rate = k[no][h 2 ] 2 D. rate = k[no] 2 [H 2 ] 2 E. None of the above 14. Using the data above, calculate the rate constant k for the reaction. A. 0.00631 M -2 sec -1 B. 0.0150 M -2 sec -1 C. 0.0218 M -3 sec -1 D. 0.00265 M -2 sec -1 E. None of the above 2
15. What is the activation energy of a reaction if the rate constant k drops from 5.42 x 10-5 sec -1 to 1.15 x 10-5 sec -1 when the temperature is reduced from 500K to 400K? A. 0.254 kj/mole B. 1.29 kj/mole C. 11.2 kj/mole D. 25.8 kj/mole E. 78.3 kj/mole The next three questions refer to the following equilibria: Reaction 1: Ca(OH) 2 (s) Ca 2+ (aq) + 2 OH - (aq) K c = 6.5 10-6 Reaction 2: H 2 O(liq) H + (aq) + OH - (aq) K c = 1.0 10-14 16. What is the correct form of the equilibrium constant expression for Reaction 1? A. K c = [Ca 2+ ][OH - ] 2 / [Ca(OH) 2 ] B. K c = [Ca(OH) 2 ] / [Ca 2+ ][OH - ] 2 C. K c = [Ca 2+ ][OH - ] 2 D. K c = [Ca 2+ ][OH - ] E. None of these is correct 17. Which (if any) of the reactions is product favored? A. Reaction 1 B. Reaction 2 C. Both Reactions 1 and 2 D. Neither Reaction 1 nor 2 18. From the data above, determine the equilibrium constant for the following reaction Ca 2+ (aq) + 2 H 2 O(liq) Ca(OH) 2 (s) + 2 H + (aq) A. K c = 6.5 10-34 B. K c = 1.5 10-23 C. K c = 6.5 10-20 D. K c = 1.5 10-9 E. K c = 3.1 10-9 3
19. The reaction for the formation of ammonia from nitrogen and hydrogen is: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ΔH = -92.2 kj/mole Which of the following will drive the equilibrium system to the right? A. Adding NH 3 (g) B. Removing N 2 (g) C. Decreasing the temperature D. Increasing the temperature E. Adding a catalyst The next three questions refer to the following information: The splitting of the sugar lactose into glucose and galactose occurs via a first order process with a first order rate constant of 0.0367 sec -1. We initially have 0.200 moles of lactose in a 1 L container and we want to reduce that amount to 0.0200 moles. Lactose glucose + galactose 20. What is the concentration of lactose after 60 seconds? A. 0.0200 M B. 0.0221 M C. 0.181 M D. 0.593 M E. 1.81 M 21. How long will it take to get to 0.100 moles of lactose in the above reaction? A. 0.0530 sec B. 0.254 sec C. 2.72 sec D. 18.9 sec E. 136 sec 22. Which of the following plots would you expect to be linear for the above lactose reaction? A. [Lactose] vs. time B. ln[lactose] vs. time C. 1/[Lactose] vs. time D. [Lactose] vs. temperature E. ln[lactose] vs. temperature 4
23. An empty 4.00 L flask is filled with 0.75 mol SO 3, 2.50 mol SO 2, and 1.30 mol O 2, and then the container is allowed to reach equilibrium. Predict the effect on the concentrations of SO 3 as the system approaches equilibrium. At the temperature of the reaction, K c = 2. 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) A. [SO 3 ] will decrease because Q > K. B. [SO 3 ] will decrease because Q < K. C. [SO 3 ] will increase because Q < K. D. [SO 3 ] will increase because Q > K. E. [SO 3 ] will remain the same because Q = K. 24. At 1000 K, the equilibrium constant, K c, for the following reaction is 1.10 10-7. 2 H 2 S(g) 2 H 2 (g) + S 2 (g) A reaction vessel at 1000 K initially contains 1.00 M of H 2 S. The reaction then proceeds to equilibrium, where the concentration of H 2 is recorded as 6.02 x 10-3 M. What are the equilibrium concentrations of H 2 S and S 2? A. [H 2 S] = 0.994 M and [S 2 ] = 3.01 x 10-3 M B. [H 2 S] = 0.994 M and [S 2 ] = 6.02 x 10-3 M C. [H 2 S] = 0.997 M and [S 2 ] = 3.01 x 10-3 M D. [H 2 S] = 0.997 M and [S 2 ] = 6.02 x 10-3 M E. None of the above 25. The decomposition of HI gas HI(g) ½ H 2 (g) + ½ I 2 (g) is second order with respect to HI, with a rate constant of 30 M -1 min -1 at 716K. How long does it take for the concentration of HI to drop from 0.100 M to 0.0100 M at 716K? A. 0.0030 min B. 0.077 min C. 0.33 min D. 3.0 min E. 2700 min Please sign the following statement at the completion of the exam: I did not cheat on this exam. (name) (signature) 5
PERIODIC TABLE OF THE ELEMENTS 1A 2A 3B 4B 5B 6B 7B 8B 8B 8B 1B 2B 3A 4A 5A 6A 7A 8A 1 H 1.008 3 Li 6.939 11 Na 22.99 19 K 39.10 37 Rb 85.47 55 Cs 132.9 87 Fr (223) 4 Be 9.012 12 Mg 24.31 20 Ca 40.08 38 Sr 87.62 56 Ba 137.3 88 Ra 226.0 21 Sc 44.96 39 Y 88.91 57 La 138.9 89 Ac 227.0 22 Ti 47.90 40 Zr 91.22 72 Hf 178.5 104 Unq (261) USEFUL INFORMATION: ln(a/a 0 ) = -kt 1/A-1/A 0 = kt A 0 -A = kt 23 V 50.94 41 Nb 92.91 73 Ta 181.0 105 Unp (262) 24 Cr 52.00 42 Mo 95.94 74 W 183.8 106 Unh (263) 25 Mn 54.94 43 Tc (99) 75 Re 186.2 107 Uns (262) 26 Fe 55.85 44 Ru 101.1 76 Os 190.2 108 Uno (265) 27 Co 58.93 45 Rh 102.9 77 Ir 192.2 109 Une (266) 28 Ni 58.71 46 Pd 106.4 78 Pt 195.1 29 Cu 63.55 47 Ag 107.9 79 Au 197.0 30 Zn 65.39 48 Cd 112.4 80 Hg 200.6 1 atm = 760 mm Hg 5 B 10.81 13 Al 26.98 31 Ga 69.72 49 In 114.8 81 Tl 204.4 6 C 12.01 14 Si 28.09 32 Ge 72.61 50 Sn 118.7 82 Pb 207.2 Room Temperature = 25 C = 298K T ( C) + 273 = T (K) k = Ae -Ea/RT R = 8.31 J/(mol K) = 0.0821 (L atm)/(mol K) ln(k 2 /k 1 ) = (E a /R)(1/T 1-1/T 2 ) t 1/2 = 0.693/k ΔG = -RT ln K K p = K c (RT) Δn 7 N 14.01 15 P 30.97 33 As 74.92 51 Sb 121.8 83 Bi 209.0 8 O 16.00 16 S 32.07 34 Se 78.96 52 Te 127.6 84 Po (209) 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9 85 At (210) 2 He 4.003 10 Ne 20.18 18 Ar 39.95 36 Kr 83.80 54 Xe 131.3 86 Rn (222) 6