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28 Chapter 2 STRUCTURE AND REACTIVITY Solutions to Problems 26. (a) Remember, (reaction) = (bonds broken) - (bonds formed). (i) To calculate the associated with breaking one of the two bonds in the carbon-carbon double bonds, use (C=C) as a bond-breaking contribution and (C-C) as a bond- forming contribution: = 146 + 46 - 83 - 2(68) = -27 kcal mol⁻¹ break break form form C=C Br-Br 2C-Br (ii) = 99 + 46 68 - 87 = 10 kcal mol⁻¹ break break form form C-H Br-Br C-Br H-Br (b) In reaction (i), two molecules combine to make one. This concentrates the energy content of the system into fewer particles, resulting in a large negative value for (-35 entropy units) for reaction (i). If you like, the system becomes more "ordered," which is saying much the same thing. In reaction (ii), two molecules react to make two different molecules. The dispersal of the energy content of the system undergoes no major change, resulting in a of approximately zero. (c) For (i) at 25°C, = = -27 298(-35 X 10⁻³) = -17 kcal mol⁻¹ For (i) at 600°C, = = -27 873(-35 10⁻³) = +4 kcal mol⁻¹ For (ii) at either 25°C or 600°C, ≈ = 10 kcal mol⁻¹ because ≈ 0. Both reactions have negative at 25°C, so both are thermodynamically favorable. At 600°C, the for reaction (i) has made its value positive: The reaction is therefore energetically unfavorable. Reaction (ii) is still just as good as it was at 25°C. 27. Don't identify the acids until you've looked to see which species give up protons; many of the species here can act both as acids and bases! The equilibrium lies to the side of the weaker acid/base pair, as indicated by the unequal lengths of the forward and reverse arrows. From the data in Table 2-2, you can identify the stronger acids as the species with the larger Kₐ or smaller (less positive or more negative) The equilibrium constant for each reaction is found by dividing Kₐ for the acid on the left by for the acid on the right. How did I know that? Here's how. For the following general reaction: we have = and = right? So, = = = (a) + HCN + CN⁻ = 1.3 X 10⁻¹¹ weaker base weaker acid stronger acid stronger base (b) + CH₃OH + NH₂⁻ = 3.1 X 10⁻²⁰ weaker base weaker acid stronger acid stronger base (c) HF + CH₃COO⁻ F⁻ + CH₃COOH = stronger acid stronger base weaker base weaker acid