Chem 161-2010 exam II + solutions

Chem 161-2010 exam II + solutions - Chemistry 161 Exam II...

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Chemistry 161 Exam II October 20, 2010 CHEM 161-2010 CHAPTER 4B - CHEMICAL REACTIONS IN AQUEOUS SOLUTIONS Oxidation-Reduction Rxns 1. 2 H 2 O + 4 MnO 4 - + 3 ClO 2 - 4 MnO 2 + 3 ClO 4 - + 4 OH - Which species acts as an oxidizing agent in the reaction represented above? A. H 2 O B. ClO 4 - C. ClO 2 - D. MnO 2 E . MnO 4 - +1 -2 +7 -2 +3 -2 +4 -2 +7 -2 -2 +1 2 H 2 O + 4 Mn O 4 - + 3 Cl O 2 - 4 Mn O 2 + 3 Cl O 4 - + 4 O H - Mn is reduced from +7 to +4. Cl is oxidized from +3 to +7. Therefore, MnO 4 - is the oxidizing agent. CHEM 161-2010 CHAPTER 6 - THERMOCHEMISTRY Heat and work 2. In which of the following processes is the sign of “w” negative when used to calculate ∆U? X. A gas is compressed under 2.0 atm pressure Y. H 2 (g) + Cl 2 (g) 2HCl(g) Z. H 2 O( ) H 2 O(g) A. X only B. X and Y only C . Z only D. Y only E. X and Z only ET note: The problem could have been better worded if it said, “In which of the following processes is the sign of “w” negative, for the system, when used to calculate ∆U?” “w” would be negative if the system does work on the surroundings, which means that the surroundings is compressed. 1
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X. False. If a gas is compressed, at any pressure, “w” for the system would be positive. Y. False. Going from two moles of gas to two moles of gas results in no change in volume; hence, the “w” is zero. Z. True. Going from a liquid to a gas expands the system, and compresses the surroundings. Hence, “w” for the system would be negative. CHEM 161-2010 CHAPTER 6 - THERMOCHEMISTRY Enthalpy 3. N 2 (g) + 3F 2 (g) 2NF 3 (g) H = -264 kJ How much heat is released when 0.256 mol of NF 3 (g) is formed in this reaction? A. 264 kJ B. 67.6 kJ C . 33.8 kJ D. 135 kJ E. -264 kJ N 2 (g) + 3F 2 (g) 2NF 3 (g) H = -264 kJ 264 kJ released when 2 mol NF 3 is formed. (264 kJ/2 mol) x 0.256 mol = 33.8 kJ CHEM 161-2010 CHAPTER 5 - GASES Partial pressure (mixtures of gases) 4. All of the following reactions occur in a fixed-volume container. Assume the temperature after the reaction is the same as the temperature before the reaction. Which reaction will result in a decrease in pressure inside the container? A. H 2 (g) + Cl 2 (g) 2HCl(g) B . 2NO(g) + O 2 (g) 2NO 2 (g) C. 2NH 3 (g) N 2 (g) + 3H 2 (g) D. 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) E. NH 4 Cl(s) NH 3 (g) + HCl(g) P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 V 1 = V 2 T 1 = T 2 P 1 /n 1 = P 2 /n 2 Hence, n 1 /n 2 = P 1 /P 2 . That is, the ratio of the moles is directly related to the ratio of the pressures. 2
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A. 2 moles of gas → 2 moles of gas Therefore, no change in pressure. B. 3 moles of gas → 2 moles of gas Therefore, decrease in pressure. C. 2 moles of gas → 4 moles of gas Therefore, increase in pressure. D.
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This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

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Chem 161-2010 exam II + solutions - Chemistry 161 Exam II...

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