Answer Key_Exam 2 - Version 098 – Exam 2 – rowland...

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Unformatted text preview: Version 098 – Exam 2 – rowland – (53210) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Remember that in the Kp to Kc conversion problems Quest refers to Kp as ’K’ and Kc as ’Kc’. 001 10.0 points At 700 K, K = 54 for the reaction H 2 (g) + I 2 (g) → 2 HI(g) . Calculate K c at 700 K for this reaction. 1. 3 . 2 × 10 − 4 2. 0.94 3. 0.45 4. 54 correct 5. 7 . 7 × 10 − 4 Explanation: 002 10.0 points A certain reaction has Δ H equal to 11.57 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 twice its value at 25 ◦ C? 1. 135 ◦ C 2. 31 ◦ C 3. 65 ◦ C 4. 50 ◦ C 5. 77 ◦ C correct 6. 94 ◦ C Explanation: T 1 = 25 ◦ C +273 = 298 K K 2 = 2 K 1 Δ H = 11570 J/mol ln K 2 K 1 = Δ H R parenleftbigg 1 T 1- 1 T 2 parenrightbigg R Δ H ln K 2 K 1 = 1 T 1- 1 T 2 T 2 = parenleftbigg 1 T 1- R Δ H ln K 2 K 1 parenrightbigg − 1 = parenleftbigg 1 298- 8 . 314 11570 ln 2 parenrightbigg − 1 = 349 . 94 K = 77 ◦ C 003 10.0 points Write the reaction quotient for 3 ClO − (aq) → 2 Cl − (aq) + ClO − 3 (aq) . 1. Q c = [Cl − ] 2 [ClO − 3 ] [ClO − ] 3 correct 2. Q c = [ClO − 3 ] [ClO − ] 3 [Cl − ] 3. Q c = [ClO − ] 3 [Cl − ] 2 [ClO − 3 ] 4. Q c = [ClO − 3 ] 2 [ClO − ] [Cl − ] 3 5. Q c = 2 [Cl − ] [ClO − 3 ] 3 [ClO − ] Explanation: Q c = activities of products activities of reactants , where the subscript c indicates the activities are given in terms of molarities: a J = [J] 004 10.0 points When the reaction CO 2 (g) + H 2 (g) ⇀ ↽ H 2 O(g) + CO(g) is at equilibrium at 1800 ◦ C, the equilib- rium concentrations are found to be [CO 2 ] = 0.24 M, [H 2 ] = 0.24 M, [H 2 O] = 0.48 M, and [CO] = 0.48 M. Then an additional 0 . 67 moles per liter of CO 2 and the same amount of H 2 are added. When the reaction comes to Version 098 – Exam 2 – rowland – (53210) 2 equilibrium again at the same temperature, what will be the molar concentration of CO? 1. . 653 M 2. . 927 M correct 3. . 733 M 4. . 867 M 5. . 813 M Explanation: [CO 2 ] = 0.24 M [H 2 ] = 0.24 M [H 2 O] = 0.48 M [CO] = 0.48 M K = [H 2 O] [CO] [CO 2 ] [H 2 ] = (0 . 48 M) (0 . 48 M) (0 . 24 M)(0 . 24 M) = 4 After the addition of the CO 2 and H 2 , [CO 2 ] ini = 0.24 M + 0 . 67 M = 0 . 91 M [H 2 ] ini = 0.24 M + 0 . 67 M = 0 . 91 M CO 2 (g) + H 2 (g) ⇀ ↽ H 2 O(g) + CO(g) . 91 . 91 0.48 0.48- x- x + x + x . 91- x . 91- x . 48 + x . 48 + x (0 . 48 + x )(0 . 48 + x ) (0 . 91- x )(0 . 91- x ) = 4 . 48 + x . 91- x = √ 4 = 2 . 48 + x = 2(0 . 91- x ) = 1 . 82- 2 x 3 x = 1 . 34 x = 0 . 446667 [CO] = (0 . 48 + x ) M = 0 . 926667 mol / L 005 10.0 points Given the following equilibrium data at 973 K MgCl 2 (s) + 1 2 O 2 (g) ⇀ ↽ MgO(s) + Cl 2 (g) K p = 2 . 95 atm MgCl 2 (s) + H 2 O(g) ⇀ ↽ MgO(s) + 2 HCl(g) K p = 8 . 40 atm Calculate the equilibrium constant K p...
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This note was uploaded on 01/30/2010 for the course GOV 312L taught by Professor Madrid during the Fall '07 term at University of Texas.

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Answer Key_Exam 2 - Version 098 – Exam 2 – rowland...

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