Test 2 - Version 094 Exam 2 302 sutcliffe (51060) 1 This...

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Unformatted text preview: Version 094 Exam 2 302 sutcliffe (51060) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Assume the initial concentrations of N 2 O 4 and H 2 O are zero. Consider the reaction 4 NH 3 (g) + 7 O 2 (g) 2 N 2 O 4 (g) + 6 H 2 O(g) Initially, [NH 3 (g)] = [O 2 (g)] = 3.60 M; at equilibrium [N 2 O 4 (g)] = 0.60 M. Calculate the equilibrium concentration for NH 3 . 1. 3.20 M 2. 3.60 M 3. 2.40 M correct 4. 1.80 M 5. 2.80 M Explanation: 002 10.0 points Name the compound HBrO 2 . 1. hydrogen bromate 2. bromous acid correct 3. hydrogen bromite 4. hydrogen bromine oxide 5. hypobromic acid 6. bromic acid 7. perbromic acid 8. hypobromous acid Explanation: 003 10.0 points Arrange the bases I) ammonia (NH 3 ) K b = 1 . 8 10 5 ; II) hydrazine (NH 2 NH 2 ) K b = 1 . 7 10 6 ; III) deuterated ammonia (ND 3 ) K b = 1 . 1 10 5 ; IV) hydroxylamine (NH 2 OH) K b = 1 . 1 10 8 ; in increasing order of strengths. 1. II, IV, III, I 2. III, IV, II, I 3. I, III, II, IV 4. I, II, III, IV 5. I, III, IV, II 6. None of these 7. Cannot be determined 8. I, II, IV, III 9. IV, III, II, I 10. IV, II, III, I correct Explanation: The stronger the base, the higher the K b value and the lower the p K b value: K b = 10 p K b p K b =- log( K b ) I. For ammonia, p K b =- log(1 . 8 10 5 ) = 4 . 74473 II. For hydrazine, p K b =- log(1 . 7 10 6 ) = 5 . 76955 III. For deuterated ammonia, p K b =- log(1 . 1 10 5 ) = 4 . 95861 IV. For hydroxylamine, p K b =- log(1 . 1 10 8 ) = 7 . 95861 Version 094 Exam 2 302 sutcliffe (51060) 2 Thus NH 2 OH < NH 2 NH 2 < ND 3 < NH 3 004 10.0 points Consider the following reaction: PCl 5 (g) PCl 3 (g) + Cl 2 (g) . At equilibrium, [PCl 5 ] = 2.00 M and [PCl 3 ] = [Cl 2 ] = 1.00 M. If suddenly 1.00 M PCl 5 (g), PCl 3 (g), and Cl 2 (g) are each added, calculate the equilibrium concentration of Cl 2 (g). 1. 1.12 M 2. 2.32 M 3. 1.86 M 4. 1.35 M correct 5. 2.18 M 6. 3.35 M Explanation: K = (1)(1) 2 = 0 . 5 after the addition each concentration in- creases by 1 M and then the reaction must shift to the left to reestablish equilibrium. K = (2- x ) 2 (3 + x ) = 0 . 5 which converts to 0 = x 2- 4 . 5 x + 2 . 5 The quadratic formula will yield two an- swers for x . x = 0 . 6492 or 3 . 851 The 3.851 answer is not physically possible and therefore x = 0 . 6492. [Cl 2 ] = 2- x = 1 . 35 M 005 10.0 points The reaction Br 2 (g) + 3 F 2 (g) 2 BrF 3 (g) is exothermic in the forward direction. An increase in the partial pressure of BrF 3 in this reaction at equilibrium would be favored by a (higher, lower) total pressure and by a (higher, lower) temperature....
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Test 2 - Version 094 Exam 2 302 sutcliffe (51060) 1 This...

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