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Unformatted text preview: johnson (dej437) Homework 1 Sutcliffe (51060) 1 This printout should have 16 questions. Multiplechoice 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 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; G is not directly related to K . Explanation: A positive G rxn (standard reaction free energy) denotes an endothermic reation, which is nonspontaneous. Also, G rxn = RT ln K , so a positive G rxn would result in a K that is between the values of zero and one. 002 10.0 points Evaluate G 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 G = 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 H f ( J K mol ) ( kJ mol ) HgO(s) 70 . 29 90 . 83 Hg( ) 76 . 02 O 2 (g) 205 . 14 Correct answer: 1 . 59349 10 10 . Explanation: HgO(s) Hg( ) + 1 2 O 2 (g) , H r = [ H f , HgO(s) ] = ( 90 . 83) = 90 . 83 kJ / mol 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, 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 G r(307 K) = RT ln K ln K = 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 . johnson (dej437) Homework 1 Sutcliffe (51060) 2 004 10.0 points Which combination of G and K is possible at standard conditions? 1. G = 42 . 8 J, K = 1 . 02 correct 2. G = 76 . 3 J, K = 8 . 11 10 13 3. G = 69 . 6 kJ, K = 1 . 03 4. G = 77 kJ, K = 0 . 986 5. G = 35 . 9 J, K = 2 . 38 10 13 Explanation: If G is positive, K must be less than 1; if G is negative, K must be greater than 1.is negative, K must be greater than 1....
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This note was uploaded on 04/25/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.
 Spring '10
 McCord
 Chemistry

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