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# G a b c d e rxn progress based on the figure the

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G A B C D E rxn progress Based on the figure, the standard reaction is 1. spontaneous. correct 2. nonspontaneous. Explanation: Δ G is negative (point E is lower free en- ergy than point A), so the standard reaction is spontaneous. 018 4.5points 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. 7 . 01 × 10 9 3. 1 . 02 × 10 10 4. 0.657 5. 0.145 6. None of these 7. 6.88 correct 8. 9 . 72 × 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 Δ G 0 = - RT ln K K = e Δ G 0 / ( RT ) = exp bracketleftbigg - - 4780 J / mol (8 . 3145 J / mol · K)(298 K) bracketrightbigg = 6 . 88395 019 4.5points The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at 298 K is 124.5 kJ/mol. What is the stan- dard molar Gibbs free energy of formation of gaseous benzene at 298 K? (Remember that vapor pressure is an equilibrium constant.) 1. 5.2 kJ/mol 2. None of the other answers is correct. 3. 124.7 kJ/mol

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casey (rmc2555) – Homework 4 – holcombe – (51395) 6 4. 119.3 kJ/mol 5. 129.7 kJ/mol correct Explanation: P C 6 H 6 = 94 . 4 mm of Hg Δ G 0 f , C 6 H 6 ( ) = 124 . 5 kJ / mol C 6 H 6 ( ) C 6 H 6 (g) K p = P C 6 H 6 = 94 . 4 760 = 0 . 12421 Δ G 0 = - RT ln K = - (8 . 314)(298) ln 0 . 12421 = 5 . 1677 kJ / mol C 6 H 6 (g) C 6 H 6 ( ) Δ G 0 rxn = - 5 . 1677 kJ / mol Δ G 0 rxn = Δ G 0 f , C 6 H 6 ( ) - Δ G 0 f , C 6 H 6 (g) Δ G 0 f , C 6 H 6 (g) = Δ G 0 f , C 6 H 6 ( ) - Δ G 0 rxn = 124 . 5 kJ / mol - ( - 5 . 1677 kJ / mol) = 129 . 7 kJ / mol 020 4.5points A gas-phase reaction has K p = 5 . 84 × 10 20 at 25 C. Calculate Δ G 0 for this reaction. Correct answer: - 118 . 493 kJ / mol rxn. Explanation: K p = 5 . 84 × 10 20 T = 25 C + 273 = 298 K Δ G 0 = - RT ln K p = - (8 . 314 J / mol · K) (298 K) × ln ( 5 . 84 × 10 20 ) = - 118 . 493 kJ / mol rxn 021 4.5points What is Δ G for the reaction N 2 O 4 (g) + 1 2 O 2 (g) N 2 O 5 (g) if each gas is present at a partial pressure of 0 . 1 atm and 25 C? 1. 17 . 31 kJ / mol 2. 18 . 04 kJ / mol 3. 17 . 81 kJ / mol 4. 16 . 68 kJ / mol 5. 20 . 03 kJ / mol correct 6. 18 . 32 kJ / mol 7. None of these Explanation: P N 2 O 5 = P N 2 O 4 = P O 2 = 0 . 1 atm T = 25 C = 298 K Q = P N 2 O 5 ( P N 2 O 4 )( P O 2 ) 1 / 2 = 0 . 1 0 . 1 0 . 1 = 3 . 16228 Δ G 0 = Δ G 0 f (N 2 O 5 (g)) - Δ G 0 f (N 2 O 4 (g)) - 1 2 Δ G 0 f (O 2 ) = 115 kJ / mol - 97 . 82 kJ / mol - 0 = 17 . 18 kJ / mol Δ G = Δ G 0 + RT ln K = 17 . 18 kJ / mol + parenleftbigg 8 . 314 J mol · K parenrightbigg (298 . 15 K) ln(3 . 16228) · 1 kJ 1000 J = 20 . 0338 kJ / mol 022 4.5points Consider the reaction: C graphite ( s ) + O 2 ( g ) CO 2 ( g ) ΔG = -400 kJ · mol 1 · K 1
casey (rmc2555) – Homework 4 – holcombe – (51395) 7
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