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Unformatted text preview: Version 337 – Exam 2 – McCord – (53215) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. McCord CH302 This exam is only for McCord’s CH302 class. PLEASE Bubble your EID and Ver sion Number Correctly! 001 10.0 points Which pH represents a solution with 1000 times higher [OH − ] than a solution with pH of 5? 1. pH = 0.005 2. pH = 3 3. pH = 4 4. pH = 8 correct 5. pH = 6 6. pH = 1 7. pH = 5000 8. pH = 2 9. pH = 7 Explanation: pH = 5 pOH = 14 pH = 14 5 = 9 [OH − ] = 10 − pOH = 10 − 9 M [OH − ] x = 1000 [OH − ] = (10 3 )(10 − 9 M) = 10 − 6 M pOH x = log(OH x ) = 6 pH x = 14 pOH x = 14 6 = 8 002 10.0 points The equilibrium constant K c for the reaction 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) is 11.7 at 1100 K. A mixture of SO 2 , O 2 , and SO 3 , each with a concentration of 0.015 M, was introduced into a container at 1100 K. Which of the following is true? 1. SO 2 (g) and O 2 (g) will be formed until equilibrium is reached. correct 2. [SO 3 ] = 0.045 M at equilibrium. 3. [SO 3 ] = 0.015 M at equilibrium. 4. [SO 3 ] = [SO 2 ] = [O 2 ] at equilibrium. 5. SO 3 (g) will be formed until equilibrium is reached. Explanation: 003 10.0 points What is the concentration of OH − ions in a 0.40 M solution of KCN? The ionization constant of HCN is 4 . × 10 − 10 . 1. 1 . 26 × 10 − 5 2. 3 . 16 × 10 − 3 correct 3. 6 . 25 × 10 − 5 4. 4 . 08 × 10 − 12 5. 1 . 60 × 10 − 10 Explanation: 004 10.0 points The standard molar Gibbs free energy of for mation of NO 2 (g) at 298 K is 51.30 kJ · mol − 1 and that of N 2 O 4 (g) is 97.82 kJ · mol − 1 . What is the equilibrium constant at 25 ◦ C for the reaction 2 NO 2 (g) ⇀ ↽ N 2 O 4 (g) ? 1. 7 . 01 × 10 − 9 Version 337 – Exam 2 – McCord – (53215) 2 2. 0.657 3. 9 . 72 × 10 9 4. None of these 5. 6.88 correct 6. 1.00 7. 1 . 02 × 10 − 10 8. 0.145 Explanation: Δ G products = 97 . 82 kJ · mol − 1 Δ G reactants = 51 . 30 kJ · mol − 1 Δ G rxn = summationdisplay n Δ G products summationdisplay n Δ G reactants = 97 . 82 (2)(51 . 30) = ( 4 . 78 kJ / mol) parenleftbigg 1000 J kJ parenrightbigg = 4780 J / mol Δ G = RT ln K K = e − Δ G / ( R T ) = exp bracketleftbigg 4780 J / mol (8 . 3145 J / mol · K)(298 K) bracketrightbigg = 6 . 88395 005 10.0 points A certain reaction has Δ H equal to 10.1 kJ/mol. This reaction is normally run at room temperature (25 ◦ C). At what new tem perature should the reaction be run so that K is 3 times its value at 25 ◦ C? 1. 77 ◦ C 2. 31 ◦ C 3. 151 ◦ C 4. 135 ◦ C correct 5. 65 ◦ C 6. 50 ◦ C 7. 94 ◦ C Explanation: K 2 = 3 K 1 Use the van’t Hoff equation. For the K ’s just put in 3 for K 2 and 1 for K 1 , and use the temperatures as appropriate ( T 1 is room tem perature, T 2 is the unknown temperature)....
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This note was uploaded on 10/21/2009 for the course CH 53395 taught by Professor Lyon during the Spring '09 term at University of Texas.
 Spring '09
 Lyon

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