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Unformatted text preview: yeh (sy3475) Homework 1 sutcliffe (51630) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Note there are a number of numerical (free response) questions on this assignment. I rec ommend you use the accurate Kelvin  Cel sius conversion to help avoid rounding errors. Watch your signs and always see if you can predict a qualitative version of the problem to estimate the sign and possible size of your answer. 001 10.0 points If G rxn is positive, then the forward reaction is (spontaneous / nonspontaneous) and K is (less / greater) than one. 1. spontaneous, less 2. nonspontaneous; less correct 3. nonspontaneous; greater 4. spontaneous, greater 5. None of these; G is not directly related to K . Explanation: A positive G rxn (standard reaction free energy) denotes an endothermic reation, which is nonspontaneous. Also, G rxn = RT ln K , so a positive G rxn would result in a K that is between the values of zero and one. 002 10.0 points Knowing only the standard H rxn and S values of each of the different reactions I) H rxn = 10 J/mol, S = 1000 J/K/mol II) H rxn = 10 J/mol, S = 1000 J/K/mol III) H rxn = 20 J/mol, S = 1000 J/K/mol IV) H rxn = 40 J/mol, S = 2000 J/K/mol which will have the highest K eq value? As sume all reactions are run at 293 K. 1. I only 2. III only correct 3. I and II have equal highest values of K . 4. All four have equal values of K . 5. IV only 6. I, II and III have equal highest values of K . 7. II only Explanation: 003 10.0 points Note: The reaction progress axis goes from 100 The figure represents a reaction at 298 K. b b b b b G A B C D E rxn progress Based on the figure, the standard reaction is 1. nonspontaneous. 2. spontaneous. correct Explanation: G is negative (point E is lower free en ergy than point A), so the standard reaction is spontaneous. 004 10.0 points A gasphase reaction has K p = 6 . 49 10 20 at 25 C. Calculate G for this reaction. Correct answer: 118 . 755 kJ / mol rxn. Explanation: K p = 6 . 49 10 20 T = 25 C + 273 = 298 K G = RT ln K p yeh (sy3475) Homework 1 sutcliffe (51630) 2 = (8 . 314 J / mol K) (298 K) ln ( 6 . 49 10 20 ) = 118 . 755 kJ / mol rxn 005 10.0 points A: In an equilibrium reaction, the reverse re action will begin as soon any products are formed. B: If we make a reaction go faster, we can increase the amount of product at equilib rium. C: The reaction free energy at equilibrium is zero. Which of these statements is/are true?...
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This note was uploaded on 09/27/2011 for the course CH 302 taught by Professor Holcombe during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Holcombe

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