solution4_pdf - elizondo(jie268 Homework 4...

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elizondo (jie268) – Homework 4 – fakhreddine – (51550) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 1.0points I: A reaction stops when the equilibrium is reached. II: An equilibrium reaction is not affected by increasing the concentrations of prod- ucts. III: If one starts with a higher pressure of reactant, the equilibrium constant will be larger. Which of these statements is/are true? 1. Only I and III are true. 2. Only II and III are true. 3. I, II, and III are true. 4. Only I is true. 5. None is true. correct 6. Only I and II are true. 7. Only II is true. 8. Only III is true. Explanation: I: 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. II: false; Equilibrium reactions are affected by the presence of both products and reac- tants. III: false; The value of the equilibrium con- stant is not affected by the amounts of reac- tants or products added as long as the tem- perature is constant. 002 1.0points Write the equilibrium constant for 2 NaBr(aq) + Pb(ClO 4 ) 2 (aq) PbBr 2 (s) + 2 NaClO 4 (aq) . 1. K = [PbBr 2 ] [Pb 2+ ][Br ] 2 2. K = [Pb 2+ ][Br ] 2 3. K = 1 [Pb(ClO 4 ) 2 ][NaBr] 2 4. K = [NaClO 4 ] 2 [NaBr] 2 [Pb(ClO 4 ) 2 ] 5. K = 1 [Pb 2+ ][Br ] 2 correct Explanation: 003 1.0points 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. 1.00 2. None of these 3. 0.145 4. 6.88 correct 5. 9 . 72 × 10 9 6. 1 . 02 × 10 10 7. 0.657 8. 7 . 01 × 10 9 Explanation: Δ G 0 products = 97 . 82 kJ · mol 1 Δ G 0 reactants = 51 . 30 kJ · mol 1 Δ G 0 rxn = summationdisplay n Δ G 0 products - summationdisplay n Δ G 0 reactants = 97 . 82 - (2)(51 . 30) = ( - 4 . 78 kJ / mol) parenleftbigg 1000 J kJ parenrightbigg = - 4780 J / mol
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elizondo (jie268) – Homework 4 – fakhreddine – (51550) 2 Δ G 0 = - RT ln K K = e Δ G 0 / ( RT ) = exp bracketleftbigg - - 4780 J / mol
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