Ch 301 Exam 4 - Version Exam 4 Sutcliffe 1 This print-out...

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Unformatted text preview: Version Exam 4 Sutcliffe 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The heat of combustion of butane (C 4 H 10 ) is- 2878 kJ/mol. Assume a typical cigarette lighter holds about 1.75 g of butane. How much heat energy would be released if ALL of the butane in the lighter were combusted? 1. 95.59 kJ 2. 40.61 kJ 3. 86.63 kJ correct 4. 1.05 kJ Explanation: MW butane = 58.14 g/mol ? kJ =- 2878 kJ mol (1 . 75 g) 1 mol 58 . 14 g =- 86 . 6271 kJ 002 10.0 points A battery does 35 kJ of work driving an elec- tric motor. The battery also releases 7 kJ of heat. What is the change in internal energy of the system? Assume the motor is not part of the system. 1. +28 kJ 2.- 35 kJ 3.- 28 kJ 4. +42 kJ 5.- 42 kJ correct Explanation: U = q + w =- 7 kJ + (- 35) kJ =- 42 kJ . 003 10.0 points Remember when we say A is greater than B we mean A is more positive (less negative) than B. For an endothermic reaction in which more moles of gas are produced than are consumed, (at constant pressure) H is 1. unrelated to E . 2. equal to E . 3. less than E . 4. greater than E . correct Explanation: E = q + w q = H w =- P V =- ( n ) RT For the described situation we have + n . E = H- ( n ) RT Since H = E + ( n ) RT , H > E. 004 10.0 points The first law of thermodynamics states that 1. energy is the capacity to do work. 2. heat flows from a warmer body to a cooler body. 3. energy is neither created nor destroyed. correct 4. doing work is defined as causing move- ment against a resisting force. Explanation: Definition 005 10.0 points What is the entropy at T = 0 K for one mole Version Exam 4 Sutcliffe 2 of chloroform (CHCl 3 ), which can orient four ways? 1.- 11 . 5 J/K 2. 0 J/K 3. 1.38 J/K 4. 11.5 J/K correct 5. 1 . 9 10 23 J/K Explanation: 006 10.0 points Consider the following compounds and their thermodynamic data: Compound H f S G f ( kJ mol ) ( J mol K ) ( kJ mol ) CH 4- 75 186- 50 CH 2 O- 108 218- 102 C 6 H 5 NH 2 87 166 319 C 2 H 4 52 68 219 Using this data, which of the following an- swers includes the compounds that are ther- modynamically unstable? 1. All of the compounds are thermodynami- cally stable. 2. C 6 H 5 NH 2 , C 2 H 4 correct 3. CH 4 , C 2 H 4 4. CH 4 , CH 2 O, C 2 H 4 5. CH 2 O, C 6 H 5 NH 2 6. Cannot be determined from the data pro- vided. Explanation: 007 10.0 points What is the change in entropy when one mole of methane (CH 4 ) is frozen to a solid at its normal melting point of- 182 C? The heat of fusion of CH 4 is 0 . 92 kJ / mol. (Careful watch your signs.) 1. None of these 2.- 5 . 05 J K mol 3. 10 . 1 J K mol 4. 5 . 05 J K mol 5.- 10 . 1 J K mol correct Explanation: H = 0 . 92 kJ/mol T =- 182 C + 273 = 91 K . 92 kJ / mol 91 K = 0 . 0101099 kJ K mol Freezing increases the order of the system, so the entropy change is...
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Ch 301 Exam 4 - Version Exam 4 Sutcliffe 1 This print-out...

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