chem hw4 - rose (kmr2427) Homework 4 fakhreddine (51615) 1...

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Unformatted text preview: rose (kmr2427) Homework 4 fakhreddine (51615) 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 1.0 points A: The standard reaction free energy at equi- librium is zero. B: A reaction stops when the equilibrium is reached. C: An equilibrium reaction is not affected by increasing the concentrations of products. Which of these statements is/are true? 1. Only B is true. 2. Only A and C are true. 3. Only B and C are true. 4. Only A and B are true. 5. A, B, and C are true. 6. None is true. correct 7. Only C is true. 8. Only A is true. Explanation: A: false; The standard reaction free energy G r is not 0 at equilibrium. The reaction free energy G r , which is dependent upon the concentrations of the products and reactants, is 0 at equilibrium. B: false; Equilibrium is dynamic. At equi- librium, the concentrations of reactants and products will not change, but the reaction will continue to proceed in both directions. C: false; Equilibrium reactions are affected by the presence of both products and reac- tants. 002 1.0 points Write the equilibrium constant for 2 NaBr(aq) + Pb(ClO 4 ) 2 (aq) PbBr 2 (s) + 2 NaClO 4 (aq) . 1. K = 1 [Pb(ClO 4 ) 2 ][NaBr] 2 2. K = 1 [Pb 2+ ][Br ] 2 correct 3. K = [NaClO 4 ] 2 [NaBr] 2 [Pb(ClO 4 ) 2 ] 4. K = [Pb 2+ ][Br ] 2 5. K = [PbBr 2 ] [Pb 2+ ][Br ] 2 Explanation: 003 1.0 points The standard molar Gibbs free energy of for- mation of NO 2 (g) at 298 K is 51.30 kJ mol 1 and that of N 2 O 4 (g) is 97.82 kJ mol 1 . What is the equilibrium constant at 25 C for the reaction 2 NO 2 (g) N 2 O 4 (g) ? 1. 1 . 02 10 10 2. 1.00 3. 7 . 01 10 9 4. 6.88 correct 5. 0.657 6. 9 . 72 10 9 7. None of these 8. 0.145 Explanation: G products = 97 . 82 kJ mol 1 G reactants = 51 . 30 kJ mol 1 G rxn = summationdisplay n G products- summationdisplay n G reactants = 97 . 82- (2)(51 . 30) = (- 4 . 78 kJ / mol) parenleftbigg 1000 J kJ parenrightbigg =- 4780 J / mol rose (kmr2427) Homework 4 fakhreddine (51615) 2 G =- RT ln K K = e G / ( R T ) = exp bracketleftbigg-- 4780 J / mol (8 . 3145 J / mol K)(298 K) bracketrightbigg = 6 . 88395 004 1.0 points At 25 C, K c = 1 . 58 10 8 for the reaction NH 4 (NH 2 CO 2 )(s) 2 NH 3 (g) + CO 2 (g) . Calculate K at 25 C for this reaction. 1. 9 . 45 10 5 2. 1 . 36 10 7 3. 3 . 87 10 7 4. 5 . 69 10 3 5. 2 . 31 10 4 correct Explanation: 005 1.0 points Consider the famous ammonia preparation 3 H 2 (g) + N 2 (g) 2 NH 3 (g) The equation K = [ x ] 2 [0 . 1- 3 x ] 3 [0 . 7- x ] is not a possible correct description of the equilibrium situation because 1. The equation is correct....
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This note was uploaded on 03/01/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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chem hw4 - rose (kmr2427) Homework 4 fakhreddine (51615) 1...

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