Exam 4-solutions

# Exam 4-solutions - Version 108 – Exam 4 – mccord...

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Unformatted text preview: Version 108 – Exam 4 – mccord – (50970) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In a system at equilibrium at constant tem- perature and pressure, Δ H is necessarily equal to 1. T Δ S. correct 2. 0. 3. P Δ V. 4. Δ U. 5. Δ G. Explanation: Δ G = 0 for a system at equilibrium, so Δ G = Δ H- T Δ S 0 = Δ H- T Δ S Δ H = T Δ S 002 10.0 points The standard molar enthalpy of formation of ammonium chloride, NH 4 Cl (s), is- 314 . 43 kJ/mol. What is the standard molar internal energy of formation of ammonium chloride? 1.- 314 . 43 kJ/mol 2.- 309 . 47 kJ/mol 3.- 306 . 99 kJ/mol correct 4.- 319 . 39 kJ/mol 5.- 321 . 87 kJ/mol 6. +311 . 95 kJ/mol 7.- 316 . 91 kJ/mol Explanation: Δ H =- 314 . 43 kJ/mol T = 298 . 15 K (standard temperature) 1 2 N 2 (g) + 2 H 2 (g) + 1 2 Cl 2 (g) → NH 4 Cl(s) n i = ( . 5 + 2 + . 5)mol = 3 mol n f = 0 mol Δ n = (0- 3) mol =- 3 mol Δ E = q + w, q = Δ H and w =- P Δ V =- Δ nRT =- (- 3 mol)(8 . 314 J / mol · K) × (298 . 15 K) parenleftbigg kJ 1000 J parenrightbigg = 7 . 43646 kJ Work is per 1 mole of N 2 H 4 formed, so Δ E = Δ H + w =- 314 . 43 kJ / mol + 7 . 43646 kJ / mol =- 306 . 994 kJ / mol 003 10.0 points Reactions with positive values of Δ S ◦ r always become spontaneous at low temperatures. 1. True 2. False correct Explanation: Δ G = Δ H- T Δ S is used to predict spon- taneity. (Δ G is negative for a spontaneous reaction.) T is always positive. When Δ S is positive and T is small, then in order for Δ G to be negative, Δ H must be negative, zero or have a very small positive value. But this is not always the case, so not ALL reactions with a positive Δ S are spontaneous at low temperatures. 004 10.0 points Calculate the standard entropy of condensa- tion of chloroform at its boiling point (335 K). The standard molar enthalpy of vaporiza- tion of chloroform at its boiling point is 31.4 kJ · mol − 1 . Version 108 – Exam 4 – mccord – (50970) 2 1. +506 J · K − 1 · mol − 1 2. +93 . 7 J · K − 1 · mol − 1 3. +31 . 4 kJ · K − 1 · mol − 1 4.- 93 . 7 J · K − 1 · mol − 1 correct 5.- 31 . 3 kJ · K − 1 · mol − 1 Explanation: Δ H vap =- 31 . 4 kJ · mol − 1 =- 31400 J · mol − 1 T BP = 335 K Δ S cond = q T = Δ H con T BP =- Δ H vap T BP =- 31400 J · mol − 1 335 K =- 93 . 7313 J · mol − 1 · K − 1 005 10.0 points A can of spray paint uses isobutane (C 4 H 10 ) as a propellant. The brand new can contains 15 g of propellant but after several uses the can now contains 13 . 5 g of propellant. What amount of work has been done by the propel- lant gas as it expands out of the can? Assume the can is stored and operated at 24 ◦ C....
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Exam 4-solutions - Version 108 – Exam 4 – mccord...

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