Study Guides (238,524)
United States (119,825)
Chemistry (85)
CHM 2046 (54)
Christou (32)
Midterm

# Practice Exam 3 2009.doc

6 Pages
138 Views

School
University of Florida
Department
Chemistry
Course
CHM 2046
Professor
Christou
Semester
Spring

Description
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 O3 (g) + 2 N (g) 2 4 NH (g) + 3 3O (g) if you are given the information below: N (g) + 3 H (g) → 2 NH (g) ΔGº = -32.0 kJ 2 2 3 CH O3 (g) → 2 H (g) +2CO (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. a. -292 kJ b. -278 kJ c. -124 kJ d. -280 kJ e. none of these 3. What is the equilibrium constant of the reaction below at 298 K Xe (g) + F (2) ⇌ XeF (g) 2 given that ΔHº = -292 kJ, and ΔSº = -40.3 J/K? a. 1.21 x 10 49 -50 b. 8.29 x 10 c. 1.12 d. 0.893 e. none of these 4. A hot, molten mixture of NaCl (l), PbBr (l) a2d CuF (l) is 2laced in an electrolysis cell and electrolyzed. The products are: a. Cu (l) and Br (g) 2 b. Na (l) and Cl 2g) c. Pb (l) and Br 2g) d. Na (l) and F 2g) e. Cu(l) and F (g) 2 4. Consider the following three half-reactions: 2+ - Co (aq) + 2 e-→ Co (s- Eº = -0.28 V I2(s) + 2 e → 2 I (aq) Eº = 0.53 V Sn (aq) + 2 e → Sn (aq) 2+ Eº = 0.13 V the order of oxidizing agent strength is: 4+ 2+ a. I2(s) > Sn (aq) > Co (aq) b. I (aq) > Sn (aq) > Co (s) 2+ - c. Co (s) > Sn (aq) > I (aq) d. Co (aq) > Sn (aq) > I (s) 2 e. cannot be determined 6. Consider the following three half-reactions 4+ - 3+ Ce (aq) + e → Ce (aq) Eº = +1.61 V Sn (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 (aq)3+ 2+ + 4+ b. Sn (aq) > Ag (aq) > Ce (aq) c. Ce (aq) > Ag (s) > Sn (s) 4+ + 2+ d. Ce (aq) > Ag (aq) > Sn (aq) e. cannot be determined 7. In the redox reaction 3 Fe (s) + 2 NO (aq)3+ 8 H (aq) → 3 Fe (aq) + 2 NO (g) + 4 H O (l) 2 a. the Fe (s) becomes oxidized to Fe (aq) by reducing NO (aq) to NO (g) 3- + b. the Fe (s) is the reducing agent that reduces H (aq) to H O (l) 2 c. the NO (a3) is the oxidizing agent that is oxidized to NO (g) by the Fe (s) - + d. the Fe (s) and NO (aq)3are both oxidized by H (aq), which is reduced to H O (l) 2 e. the NO (a3) becomes oxidized to NO (g) by reducing Fe (s) to Fe (aq) 2+ 8. What is the cell reaction for the voltaic cell shown? 2+ + Mg (s) | Mg (aq) ║ H O (aq),2H 2aq) | graphite a. H 2 (2q) + 2 H (aq) + Mg (s) → Mg (aq) + 2 H O (l) 2 2+ + b. Mg (aq) + 2 H (aq) → Mg (s) + 2 H O (l) 2 2 c. Mg (aq) + H O (aq2 +22 H → Mg (s) + 2 H O (l) 2 + 2+ d. Mg (s) + H O (2q)2→ 2 H (aq) + H O (l) + Mg (2q) e. Mg (s) + 2 H (aq) + H O (l) 2 Mg (aq) + 2 H O (l) 2 2 9. What is the cell notation for the voltaic cell in which the following reaction occurs? H 2 (2q) + Hg 22+(aq) + 2 H (aq) → 2 Hg (aq) + 2 H O (l) 2 2+ 2+ + a. graphite | Hg 2 (aq), Hg (aq) ║ H O (aq)2 H2(aq) | graphite b. graphite | Hg (aq), Hg 2+(aq) ║ H O (aq), H (aq) | graphite 2+ 22+ + 2 c. graphite | Hg (aq), Hg 2 (aq) ║ H (aq), H O (aq2 |2graphite d. graphite | H O2(a2), H (aq) ║ Hg 22+ (aq), Hg (aq) | graphite + 2+ 2+ e. graphite | H O2(a2), H (aq) ║ Hg (aq), Hg 2 (aq) | graphite 10. A voltaic cell prepared using aluminumand nickel has the following cell notation: Al (s) | Al (aq) ║ Ni (aq) | Ni (s) Which of the following reactions occurs at the anode? 3+ - a. Al (s) → Al (aq) + 3 e b. Al (aq) + 3 e → Al (s) c. none of these 2+ - d. Ni (aq) + 2 e → Ni (s) e. Ni (s) → Ni (aq) + 2 e - 2+ 2+ 11. For the voltaic (galvanic) cell made up of the Fe (s)/Fe (aq, 1.0 M) and Pb (s)/Pb (aq, 1.0 M) half-cells, given that Fe (aq) + 2 e → Fe (s) Eº = -0.44 V 2+ - Pb (aq) + 2e → Pb (s) Eº = -0.13 V Which of the following statements is true? a. The lead (Pb) electrode is the cathode and electrons flow to it from the iron (Fe) electrode through an external circuit. b. Since both Eº half-potentials are negative, no spontaneous reactions are possible and no current willflow. 2+ c. The concentration of Fe (aq) decreases during operation of the cell. d. The iron (Fe) electrode is the cathode and electrons flow to it from the lead (Pb) electrode through the external circuit. 2+ 2+ e. Since [Fe ] = [Pb ], the cell is at equilibriumand no current willflow. 12. What is Eº cellor the voltaic cell with cell reaction 2 Al (s) + 3 Fe (aq) → 2 Al (aq) + 3 Fe (s)? a. 1.22 V b. 2.10 V c. -2.10 V d. -1.22 V e. 0.00 V 3+ 13. Which of the following can react spontaneously with gold (Au) metal to give Au (aq)? O 2g), Cl
More Less

Related notes for CHM 2046

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.