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Unformatted text preview: parvaresh (sep782) Homework 1 Sutcliffe (52440) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points If the reaction quotient Q for some hypothet- ical reaction is 100, how does G compare to G ? 1. G > G correct 2. G < G 3. There is insufficient information given to determine the answer. 4. G = G Explanation: G = G + RT ln Q If Q = 100, then the natural log term will be positive. This means that the non standard value of G will be bigger (more positive) than the value of G . 002 10.0 points If G = 27 . 1 kJ at 25 C for the reaction CH 3 COOH(aq) + H 2 O( ) CH 3 COO (aq) + H 3 O + (aq) , calculate K a for this reaction at 298 K. 1. 1 . 15 10 11 2. 1.01 3. 5 . 63 10 4 4. 9 . 89 10 1 5. 1 . 78 10 5 correct Explanation: 003 10.0 points Calculate the equilibrium constant at 305 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 1 2 O 2 (g) 205 . 14 Correct answer: 1 . 26185 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 305 K, G r(305 K) = 90 . 83 K- (305 K) parenleftbigg 108 . 3 J K mol parenrightbiggparenleftbigg 1 kJ 1000 J parenrightbigg = 57 . 7985 kJ / mol G r(305 K) =- RT ln K ln K =- G r(305 K) RT =- 57 . 7985 kJ (8 . 314 J / K)(305 K) 1000 J 1 kJ =- 22 . 7933 K = e 22 . 7933 = 1 . 26185 10 10 . 004 10.0 points The standard state for phosphorus is white phosphorus. (Theres a red form too, but thats not the standard). parvaresh (sep782) Homework 1 Sutcliffe (52440) 2 What is the numerical value of the equi- librium constant K p at 25 C for the de- composition of phosgine (PH 3 ) gas into the elements white phosphorus and hydrogen? G f = 13 . 4 kJ / mol for PH 3 (g). 1. None of these 2. 1 . 00 10 1 3. 9 . 95 10 1 4. 2 . 23 10 2 correct 5. 4 . 48 10 3 Explanation: G =- 13 . 4 kJ / mol =- 13400 J / mol T = 25 C = 298 K R = 8 . 314 J mol K K p = e G/ ( R T ) = exp bracketleftBigg 13400 J / mol (8 . 314 J mol...
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This note was uploaded on 02/06/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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solution_pdf - parvaresh (sep782) Homework 1 Sutcliffe...

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