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Unformatted text preview: Version 092 – Final MWF – sutcliffe – (50965) 1 This printout should have 46 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What is the energy of a photon of light with a frequency of 6 . × 10 14 s − 1 ? 1. 5 . × 10 − 7 J 2. 1 . 3 × 10 − 27 J 3. 3 . 9 × 10 − 19 J correct 4. 1 . 2 × 10 − 18 J 5. 1 . 2 × 10 18 J Explanation: Energy is equal to Planck’s constant times frequency. Planck’s constant is 6 . 6262 × 10 − 34 J · s, and the frequency is given, so E = hν = ( 6 . 6262 × 10 − 34 J · s )( 6 . × 10 14 s − 1 ) = 3 . 9 × 10 − 19 J 002 10.0 points A friend states that water freezing is a vio lation of the second law of thermodynamics. Your best reply would be that 1. the second law does not apply at 0 ◦ C. 2. although the entropy of the water de creases the entropy of the universe increased. correct 3. the second law does not apply to water. 4. you must be thinking of the first law. Explanation: The Second Law of Thermodynamics says that in spontaneous changes, the universe tends toward a state of greater disorder. 003 10.0 points What volume of 1.5 M HCl solution do you need to use to make 500 mL of 0.25 M HCl solution by dilution? 1. 21 mL 2. 188 mL 3. 83 mL correct 4. 3.0 L Explanation: M 1 = 1.5 M M 2 = 0.25 M V 2 = 500 mL The formula to be used for calculating di lutions is V 1 M 1 = V 2 M 2 V 1 = V 2 M 2 M 1 = 500 mL × . 25 M 1 . 5 M = 83 mL 004 10.0 points Calculate the change in entropy when 2.00 moles of methane are compressed isother mally to one fifth of its original volume. Treat methane as an ideal gas. 1. 26.8 J · K − 1 correct 2. 1.39 J · K − 1 3. +13.4 J · K − 1 4. +26.8 J · K − 1 5. 13.4 J · K − 1 Explanation: If compressed to one fifth the original vol ume this means V 2 = V 1 5 or V 2 V 1 = 1 5 Δ S = nR ln parenleftbigg V 2 V 1 parenrightbigg = (2 . 00 mol)(8 . 314 J · mol − 1 · K − 1 ) × ln parenleftbigg 1 5 parenrightbigg = 26 . 7617 J · K − 1 Version 092 – Final MWF – sutcliffe – (50965) 2 We expect a negative answer since the volume decreased. 005 10.0 points A process CANNOT be spontaneous if 1. it is endothermic and there is a decrease in disorder. correct 2. it is endothermic and there is an increase in disorder. 3. it is exothermic and there a decrease in disorder. 4. it is exothermic and there is an increase in disorder. Explanation: Δ G = Δ H T Δ S For a process to be spontaneous, Δ G must be negative. Given Δ G = Δ H T Δ S , if Δ H is positive (endothermic) and Δ S is negative (decrease in disorder), Δ G has to be positive (nonspontaneous). T is in Kelvin so it is always positive....
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This note was uploaded on 10/20/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.
 Fall '07
 Fakhreddine/Lyon
 pH

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