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Chapter 3

Chapter 3 - Instructor Supplemental Solutions to Problems...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 3 Acids and Bases. The Curved-Arrow Notation Solutions to In-Text Problems 3.1 (b) The nitrogen of ammonia donates an electron pair to electron-deficient boron of BF 3 . (The electron pairs on the fluorine are not shown because they have no direct part in the reaction.) 3.2 (b) (d) 3.3 (b) 3.4 (b) 3.6 Let H 2 O be the acid on the left. Then CH 3 OH is the acid on the right. (If you chose CH 3 OH as the acid on the left, then switch reactants and products in the reaction below.) 3.7 (a) The reverse of reaction 3.18a is a Lewis acid–base association reaction; there is no leaving group.

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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 3 2 (b) The reverse of reaction 3.18b is a Lewis acid–base dissociation reaction. There is no nucleophile. (c) Like the forward reaction, the reverse of reaction 3.18c is a Brønsted acid–base reaction. 3.8 (b) This is a Lewis acid–base association reaction. The reverse reaction is a Lewis acid–base dissociation reaction: 3.9 (b) The curved-arrow notation: The analogous Brønsted acid–base reaction replaces the ethyl group with H:
INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 3 3 3.12 Find the K a by changing the sign of the p K a and taking the antilog. (a) K a = 10 –4 (b) K a = 10 –7.8 = 1.58 ± 10 –8 (c) K a = 10 2 = 100 3.13 (b) The strongest acid in Problem 3.12 is the one with the largest K a —that is, (c). 3.14 (b) The acid–base reaction is The acid on the right side of the equation is H—F (p K a = 3.2), and the acid on the left side is H—CN (p K a = 9.4). Using the procedure in part (a), we find that log K eq = 3.2–9.4 = –6.2, and K eq = 10 –6.2 = 6.3 ± 10 –7 . 3.16 (b) For this calculation use Eq. 3.30 on text p. 106. ² G ° = –2.30 RT log K eq = –5.71 log(305) kJ mol –1 = –5.71(2.48) kJ mol –1 = –14.2 kJ mol –1 In these solutions and in the Study Guide and Solutions Manual, kJ mol –1 is used as the unit of energy. If you wish to convert kJ mol –1 to kcal mol –1 , simply divide by 4.184 kJ kcal –1 . Also, it is helpful to remember that 2.30 RT at 298 K (25 °C) is 5.71 kJ mol –1 . 3.17 (b) If the ² G ° is positive, the equilibrium constant is less than unity and the reaction is less favorable; consequently, there should be much less C and more A and B at equilibrium. An actual calculation shows that [ C ] = 0.0061. Then [ A ] = 0.1 – x = 0.0939 M , and [ B ] = 0.2 – x = 0.1939. The concentration of C is 10% of that in part (a), a result consistent with the statement in part (b).

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Chapter 3 - Instructor Supplemental Solutions to Problems...

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