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Homework 1-solutions - yeh(sy3475 Homework 1 sutclie(51630...

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yeh (sy3475) – Homework 1 – sutcliffe – (51630) 1 This print-out should have 20 questions. Multiple-choice 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 10.0points If Δ 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; Δ G is not directly related to K . Explanation: A positive Δ G rxn (standard reaction free energy) denotes an endothermic reation, which is nonspontaneous. Also, Δ G rxn = - RT ln K , so a positive Δ G rxn would result in a K that is between the values of zero and one. 002 10.0points Knowing only the standard Δ H rxn and Δ S values of each of the different reactions I) Δ H rxn = - 10 J/mol, Δ S = 1000 J/K/mol II) Δ H rxn = 10 J/mol, Δ S = 1000 J/K/mol III) Δ H rxn = - 20 J/mol, Δ S = 1000 J/K/mol IV) Δ H rxn = 40 J/mol, Δ 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 10.0points Note: The ’reaction progress’ axis goes from 100 The figure represents a reaction at 298 K. G A B C D E rxn progress Based on the figure, the standard reaction is 1. nonspontaneous. 2. spontaneous. correct Explanation: Δ G is negative (point E is lower free en- ergy than point A), so the standard reaction is spontaneous. 004 10.0points A gas-phase reaction has K p = 6 . 49 × 10 20 at 25 C. Calculate Δ G 0 for this reaction. Correct answer: - 118 . 755 kJ / mol rxn. Explanation: K p = 6 . 49 × 10 20 T = 25 C + 273 = 298 K Δ G 0 = - RT ln K p
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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 10.0points 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? 1. Only A and B are true. 2. Only A is true. 3. Only B and C are true. 4. A, B, and C are true. 5. Only A and C are true. correct 6. Only C is true.
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