CH302 Homework 1-solutions

# CH302 Homework 1-solutions - Bodet(ngb299 Homework 1...

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Unformatted text preview: Bodet (ngb299) 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 10.0 points If G rxn is positive, then the forward reaction is (spontaneous / nonspontaneous) and K is (less / greater) than one. 1. nonspontaneous; greater 2. nonspontaneous; less correct 3. None of these; G is not directly related to K . 4. spontaneous, less 5. spontaneous, greater 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. II only 3. IV only 4. I, II and III have equal highest values of K . 5. III only correct 6. I and II have equal highest values of K . 7. All four have equal values of K . Explanation: 003 10.0 points 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: G is negative (point E is lower free en- ergy than point A), so the standard reaction is spontaneous. 004 10.0 points A gas-phase reaction has K p = 5 . 36 10 20 at 25 C. Calculate G for this reaction. Correct answer:- 118 . 281 kJ / mol rxn. Explanation: K p = 5 . 36 10 20 T = 25 C + 273 = 298 K G =- RT ln K p Bodet (ngb299) Homework 1 sutcliffe (51630) 2 =- (8 . 314 J / mol K) (298 K) ln ( 5 . 36 10 20 ) =- 118 . 281 kJ / mol rxn 005 10.0 points A: The reaction free energy at equilibrium is zero. B: The standard reaction free energy at equi- librium is zero. C: A reaction stops when the equilibrium is reached....
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## This note was uploaded on 10/06/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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CH302 Homework 1-solutions - Bodet(ngb299 Homework 1...

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