EXAM4 solution_pdf - Version 250 Exam 4 mccord (50970) 1...

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Unformatted text preview: Version 250 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 Consider the equation NH 4 Br(s) NH 3 (g) + HBr(g) carefully, and think about the sign of S for the reaction it describes. H = +188.3 kJ. Which response describes the thermodynamic spontaneity of the reaction? 1. The reaction is spontaneous only at rela- tively low temperatures. 2. The reaction is spontaneous only at rela- tively high temperatures. correct 3. The reaction is not spontaneous at any temperatures. 4. All responses are correct. 5. The reaction is spontaneous at all tem- peratures. Explanation: Entropy ( S ) is high for systems with high degrees of freedom, disorder or randomness and low for systems with low degrees of free- dom, disorder or randomness. S(g) > S ( ) > S (s) . A reaction is spontaneous only when G is negative. H is positive for this reaction and S is positive. G = H- T S = (+)- T (+) = (+)- T G will be negative (spontaneous reactions) only at high values of T . 002 10.0 points Which of the following has the smallest molar entropy at 298 K? 1. Ne(g) 2. F 2 (g) 3. N 2 (g) 4. Cl 2 (g) 5. He(g) correct Explanation: To find the entropy we can use Boltzmanns formula S = k ln W where W is the number of microstates. At zero K, there is no mo- tion and W is given by the number of possible discernible orientations of the molecule in a crystal raised to the power of the number of molecules (1 mole for each in this case). If the substance forms a perfect crystal there should be zero residual entropy. We would expect this for these materials since they are either homonuclear diatomics or monoatomic species; in the solid, there is only one dis- cernible arrangement. We now consider the effect of being above 0 K. The greater the mass of the species, the closer together are the vibrational energy levels, so at a given temperature, more vibrational energy levels are occupied, thereby increasing the number of microstates. Thus the substance with the lowest standard molar entropy would be He, the lightest monoatomic species. 003 10.0 points A gas is isothermally expanded. Which plot best shows this type of expansion? 1. V P correct 2. V P 3. V P Version 250 Exam 4 mccord (50970) 2 4. V P 5. V P Explanation: The proper graph shows that P is inversely proportional to V . 004 10.0 points Consider the following data at 25.0 C and 1 atm. Species H f S G f kJ/mol J/mol K kJ/mol N 2 (g) . 191 . 5 . O 2 (g) . 205 . . N 2 O 5 (g) 11 . ? 115 . Calculate the absolute entropy of N 2 O 5 at 25.0 and 1 atm....
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This note was uploaded on 11/17/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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EXAM4 solution_pdf - Version 250 Exam 4 mccord (50970) 1...

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