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exam3a - Thomas Phillip Exam 3 Due Dec 5 2007 midnight Inst...

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Thomas, Phillip – Exam 3 – Due: Dec 5 2007, midnight – Inst: Brodbelt 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Brodbelt CH301 Exam 3 Fall 2007 Use of programmable calculators or cell phones is considered cheating and thus is pro- hibited. You may not leave the room for any reason. You may not start the exam after 7;15 pm. GOOD LUCK! 001 (part 1 of 1) 10 points Chloromethane (CH 3 Cl) forms a molecular solid. What type of forces hold it in a solid configuration? I) London forces II) dipole-dipole forces III) hydrogen bonding 1. I and II only correct 2. II only 3. I, II, and III 4. III only 5. I only 6. II and III only Explanation: Chloromethane is held in solid form by both London forces and dipole-dipole forces. 002 (part 1 of 1) 10 points Determine the enthalpy change of reaction at 25 C for 4 HNO 3 ( ) + 5 N 2 H 4 ( ) 7 N 2 (g) + 12 H 2 O(g) Δ H f for HNO 3 is - 174 . 1 kJ/mol; Δ H f for H 2 O is - 241 . 8 kJ/mol; Δ H f for N 2 H 4 is +50 . 63 kJ/mol. 1. 1952.1 kJ/mol rxn 2. - 2233.6 kJ/mol rxn 3. - 3851.9 kJ/mol rxn 4. - 2458.4 kJ/mol rxn correct 5. 3344.9 kJ/mol rxn 6. - 3344.9 kJ/mol rxn Explanation: Reactants: Δ H f HNO 3 ( ) = - 174 . 1 kJ/mol Δ H f N 2 H 4 (g) = 50 . 63 kJ/mol Products: Δ H f H 2 O(g) = - 241 . 8 kJ/mol Δ H 0 f N 2 = 0 Δ H 0 rxn = X n Δ H 0 f prod - X n Δ H 0 f rct = 12 - 241 . 8 kJ mol ¶‚ - 4 - 174 . 1 kJ mol + 5 50 . 63 kJ mol ¶‚ = - 2458 . 4 kJ mol rxn 003 (part 1 of 1) 10 points Which one of the following statements does NOT describe the general properties of solids accurately? 1. A solid sample has a characteristic volume that does not change greatly with changes in pressure. 2. Solids diffuse only very slowly when com- pared to liquids and gases. 3. A solid sample has a characteristic volume that does not change greatly with changes in temperature. 4. Solids are not fluid. 5. Most solids have high vapor pressures at room temperature. correct
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Thomas, Phillip – Exam 3 – Due: Dec 5 2007, midnight – Inst: Brodbelt 2 Explanation: Most solids do not vaporize easily at room temperature. 004 (part 1 of 1) 10 points Consider the reaction CH 4 (g) + I 2 CH 3 I(g) + HI(g) . Bond energy tables give the following values: C H : 411 kJ/mol I I : 149 kJ/mol H I : 295 kJ/mol C I : 213 kJ/mol The change in enthalpy for this reaction is: 1. +463 kJ/mol 2. - 52 kJ/mol 3. - 97 kJ/mol 4. +52 kJ/mol correct 5. There is no way to tell. Explanation: H C H H H + I I ----→ H C H I H + H I Δ H = BE reactants - BE products = 4 (C H) + (I I) - 3 (C H) - (C I) - (H I) = (C H) + (I I) - (C I) - (H I) = 411 kJ / mol + 149 kJ / mol - 213 kJ / mol - 295 kJ / mol = 52 kJ / mol 005 (part 1 of 1) 10 points The process of steam condensing to form liq- uid water is 1. an endothermic phase change. 2. an endothermic chemical reaction. 3. neither endothermic nor exothermic. 4. an exothermic chemical reaction. 5. an exothermic phase change. correct Explanation: Solid liquid gas is endothermic be- cause each consecutive phase has more en- ergy/heat. Thus gas liquid solid is exothermic because energy is conserved and is a state function. Phase changes are physical changes, not chemical changes.
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