Final MWF-solutions

Final MWF-solutions - Version 092 – Final MWF –...

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Unformatted text preview: Version 092 – Final MWF – sutcliffe – (50965) 1 This print-out should have 46 questions. Multiple-choice 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.

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Final MWF-solutions - Version 092 – Final MWF –...

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