Homework 1-solutions-2

Homework 1-solutions-2 - hayes (snh457) Homework 1 Sutclie...

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hayes (snh457) – Homework 1 – Sutclife – (51060) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. Note there are a number o± numerical (±ree 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 i± you can predict a qualitative version o± the problem to estimate the sign and possible size o± your answer. 001 10.0 points I± Δ G rxn is positive, then the ±orward reaction is (spontaneous / nonspontaneous) and K is (less / greater) than one. 1. spontaneous, greater 2. spontaneous, less 3. None o± these; Δ G is not directly related to K . 4. nonspontaneous; greater 5. nonspontaneous; less correct Explanation: A positive Δ G rxn (standard reaction ±ree 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 o± zero and one. 002 10.0 points Evaluate Δ G 0 at 775 C ±or a gas phase reac- tion ±or which K p =5 . 51 × 10 4 at 775 C. Correct answer: - 95 . 1198 kJ / mol rxn. Explanation: K p . 51 × 10 20 T = 775 C + 273 = 1048 K Δ G 0 = - ln K p = - (8 . 314 J / mol · K) (1048 K) × ln(5 . 51 × 10 4 ) = - 95 . 1198 kJ / mol rxn 003 10.0 points Calculate the equilibrium constant at 310 K ±or the reaction HgO(s) ± ² Hg( ³ )+ 1 2 O 2 (g) , given the data S Δ H f ( J K · mol )( kJ mol ) HgO(s) 70 . 29 - 90 . 83 Hg( ³ ) 76 . 02 0 O 2 (g) 205 . 14 0 Correct answer: 2 . 24861 × 10 - 10 . Explanation: HgO(s) ± ² Hg( ³ 1 2 O 2 (g) , Δ H r = - H f , HgO(s) ] = - ( - 90 . 83) = 90 . 83 kJ / mol Δ S r = S Hg( ± ) + 1 2 S O 2 (g) - S HgO(s) = 76 . 02 J K · mol + 1 2 ± 205 . 14 J K · mol ² - 70 . 29 J K · mol = 108 . 3 J K · mol At 310 K, Δ G r(310 K) = 90 . 83 K - (310 K) × ± 108 . 3 J K · mol ²± 1 kJ 1000 J ² = 57 . 257 kJ / mol Δ G r(310 K) = - ln K ln K = - Δ G r(310 K) = - 57 . 257 kJ (8 . 314 J / K)(310 K) × 1000 J 1 kJ = - 22 . 2155 K = e - 22 . 2155 =2 . 24861 × 10 - 10 .
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hayes (snh457) – Homework 1 – Sutclife – (51060) 2 004 10.0 points Which combination oF ΔG and K is possible at standard conditions? 1. Δ G = - 56 . 8 J, K =3 . 29 × 10 - 12 2. Δ G = - 60 . 9 kJ, K =0 . 982 3. Δ G = 45 . 3 J, K . 8 × 10 10 4. Δ G = 47 . 8 kJ, K =4 . 97 × 10 - 9 cor- rect 5. Δ G = 66 . 1 kJ, K =1 . 02 Explanation: IF ΔG is positive, K must be less than 1; iF ΔG is negative, K must be greater than 1.
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Homework 1-solutions-2 - hayes (snh457) Homework 1 Sutclie...

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