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Practice Exam 3 2009

# Practice Exam 3 2009 - CHM 2046 Exam 3 Bubble in A B or C...

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CHM 2046 Exam 3 April 13, 2006 Bubble in A , B or C as the test form code at the top right of your answer sheet. See the end for useful information. Come on, now. I hear you're feeling down. Well I can ease your pain, get you on your feet again. Relax. I need some information first. Just the basic facts, can you show me where it hurts? --------- Comfortably Numb VERSION 0 In this version, the correct answer for each question is answer a. 1. Calculate ΔGº at 298 K for the reaction 3 CH 3 OH (g) + 2 N 2 (g) → 4 NH 3 (g) + 3 CO (g) if you are given the information below: N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) ΔGº = -32.0 kJ CH 3 OH (g) → 2 H 2 (g) + CO (g) ΔGº = +24.7 kJ a. +10.1 kJ b. -7.3 kJ c. -56.7 kJ d. +14.6 kJ e. cannot be determined 2. Calculate ΔG at 298 K for the reaction below: A (g) + B (g) → C (g) ΔGº = -280 kJ if the gas pressures (p) are p(A) = 2.0 atm, p(B) = 30 atm, and p(C) = 0.40 atm. 3. What is the equilibrium constant of the reaction below at 298 K Xe (g) + F 2 (g) XeF 2 (g) given that ΔHº = -292 kJ, and ΔSº = -40.3 J/K? 4. A hot, molten mixture of NaCl (l), PbBr 2 (l) and CuF 2 (l) is placed in an electrolysis cell and electrolyzed. The products are: a. Cu (l) and Br 2 (g) b. Na (l) and Cl 2 (g) c. Pb (l) and Br 2 (g) d. Na (l) and F 2 (g) e. Cu(l) and F 2 (g) 4. Consider the following three half-reactions: Co 2+ (aq) + 2 e - → Co (s) Eº = -0.28 V

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I 2 (s) + 2 e - → 2 I - (aq) Eº = 0.53 V Sn 4+ (aq) + 2 e - → Sn 2+ (aq) Eº = 0.13 V the order of oxidizing agent strength is: 6. Consider the following three half-reactions Ce 4+ (aq) + e - → Ce 3+ (aq) Eº = +1.61 V Sn 2+ (aq) + 2 e - → Sn (s) Eº = -0.14 V Ag + (aq) + e - → Ag (s) Eº = +0.80 V the order of reducing agent strength is: a. Sn (s) > Ag (s) > Ce 3+ (aq) b. Sn 2+ (aq) > Ag + (aq) > Ce 4+ (aq) c. Ce 3+ (aq) > Ag (s) > Sn (s) d. Ce 4+ (aq) > Ag + (aq) > Sn 2+ (aq) e. cannot be determined 7. In the redox reaction 3 Fe (s) + 2 NO 3 - (aq) + 8 H + (aq) → 3 Fe 2+ (aq) + 2 NO (g) + 4 H 2 O (l) a. the Fe (s) becomes oxidized to Fe 2+ (aq) by reducing NO 3 - (aq) to NO (g) b. the Fe (s) is the reducing agent that reduces H + (aq) to H 2 O (l) c. the NO 3 - (aq) is the oxidizing agent that is oxidized to NO (g) by the Fe (s) d. the Fe (s) and NO 3 - (aq) are both oxidized by H + (aq), which is reduced to H 2 O (l) e. the NO 3 -
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