This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: parvaresh (sep782) Homework 1 Sutcliffe (52440) 1 This printout should have 18 questions. Multiplechoice 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...
View
Full
Document
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.
 Spring '07
 Holcombe
 Chemistry

Click to edit the document details