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# HW1 - johnson(dej437 Homework 1 Sutclie(51060 This...

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johnson (dej437) – Homework 1 – Sutcliffe – (51060) 1 This print-out should have 16 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 Δ G rxn is positive, then the forward reaction is (spontaneous / nonspontaneous) and K is (less / greater) than one. 1. spontaneous, less 2. nonspontaneous; less correct 3. spontaneous, greater 4. nonspontaneous; greater 5. None of these; Δ G is not directly related to K . Explanation: A positive Δ G rxn (standard reaction free 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 of zero and one. 002 10.0 points Evaluate Δ G 0 at 724 C for a gas phase reac- tion for which K p = 7 . 91 × 10 4 at 724 C. Correct answer: - 93 . 4879 kJ / mol rxn. Explanation: K p = 7 . 91 × 10 20 T = 724 C + 273 = 997 K Δ G 0 = - RT ln K p = - (8 . 314 J / mol · K) (997 K) × ln(7 . 91 × 10 4 ) = - 93 . 4879 kJ / mol rxn 003 10.0 points Calculate the equilibrium constant at 307 K for the reaction HgO(s) Hg( ) + 1 2 O 2 (g) , given the data S Δ H f ( J K · mol ) ( kJ mol ) HgO(s) 70 . 29 - 90 . 83 Hg( ) 76 . 02 0 O 2 (g) 205 . 14 0 Correct answer: 1 . 59349 × 10 10 . Explanation: HgO(s) Hg( ) + 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 parenleftbigg 205 . 14 J K · mol parenrightbigg - 70 . 29 J K · mol = 108 . 3 J K · mol At 307 K, Δ G r(307 K) = 90 . 83 K - (307 K) × parenleftbigg 108 . 3 J K · mol parenrightbiggparenleftbigg 1 kJ 1000 J parenrightbigg = 57 . 5819 kJ / mol Δ G r(307 K) = - RT ln K ln K = - Δ G r(307 K) RT = - 57 . 5819 kJ (8 . 314 J / K)(307 K) × 1000 J 1 kJ = - 22 . 5599 K = e 22 . 5599 = 1 . 59349 × 10 10 .

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johnson (dej437) – Homework 1 – Sutcliffe – (51060) 2 004 10.0 points Which combination of ΔG and K is possible at standard conditions? 1. Δ G = - 42 . 8 J, K = 1 . 02 correct 2. Δ G = - 76 . 3 J, K = 8 . 11 × 10 13 3. Δ G = 69 . 6 kJ, K = 1 . 03 4. Δ G = - 77 kJ, K = 0 . 986 5. Δ G = 35 . 9 J, K = 2 . 38 × 10 13 Explanation: If ΔG is positive, K must be less than 1; if ΔG is negative, K must be greater than 1. Also, if ΔG is small (compared to RT), then K will be close to 1, and if Δ G is large, K will be much smaller or greater than 1.
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HW1 - johnson(dej437 Homework 1 Sutclie(51060 This...

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